+0  
 
+20
964
72
avatar+2353 
Hi!

Welcome to this probability puzzle thread.
Every once in a while I will upload some interesting puzzles related to probability.

Feel free to join in!

Here is the very first puzzle!

edit: Please be aware! This is the very first post of this thread! Therefore this puzzle has already been solved many times.
If you want the latest probability puzzle, go to the last page of this thread!


Easier one:

There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000
What are the odds that that persons birthday is a monday?

Hard one (quite famous actually):

You are on a tv-show where you can win a million dollars. The flamboyant host asks you to pick one of three boxes.
In two boxes you will find nothing, but one box will give you a million dollars.
You pick a box and the host then decides to play a trick with you.
Since he knows what is inside each box he opens one of the boxes that has nothing in it.
He then asks you the million dollar question.
Will you stick with the box you picked, or will you switch to the other box?

Should you stick with the bos you picked, should you switch or is it irrelevant what you do?
Explain why please
reinout-g  Feb 26, 2014

Best Answer 

 #52
avatar+63 
+14

Hard one (still unanswered): 

You and your best friend have been arrested for several white collar crimes. (Or something else which is more classy than stealing a candy bar) 
Your attorney has told you the police has only got enough evidence to put you and your friend behind bars for 3 years. 
However, the both of you are given the option to confess. If both of you confess you will both serve 10 years, 
if one of you confesses and the other does not, the one who did not confess will serve 20 years. 
Obviously, if you both remain silent you will both serve 3 years. 
You are not allowed to have any communication with your friend. 
Suppose your main priority is to keep yourself out of jail, and your secondary priority is to keep your partner out of jail. 
You have decided that you find it twice as important to keep yourself out of jail as you find it important to keep your partner out of jail. 
(So for example, in your opinion him serving 10 years is equal to you serving 5). 
You estimate the chance of your partner confessing p. 
For which values of p will you choose to confess and for which values will you keep silent? 

HINT: make a formula for the amount of suffering you expect, also taking the suffering for your friends jailtime into account.

-----------------------------------------------------------------------------------------------------------

2 Points for every year you're in jail, 1 Point for every year your partner is in jail.

Therefore...

 

Both don't confess- if 50% chance friend will confess

3 years for you (6 points), 3 years for your partner (3 points)

[9] points total

 

Both confess- if 50% chance friend will confess

10 years for you (20 points), 10 years for your partner (10 points)

[30] points total

 

You confess and your friend doesn't- if 50% chance friend will confess

0 years for you (0 points), 20 years for your partner (20 points)

[20] points total

 

Your friend confesses and you don't- 50% chance friend will confess

20 years for you (40 points), 0 years for your partner (0 points)

[40] points total

 

Therefore...

 

If you confess you will have an average of 25 "points"

(30+20)/2=25

 

If you don't you have an average of 24.5 "points"

(40+9)/2=24.5

 

Therefore, in most cases you shouldn't confess :D

CreeperCosmo  May 20, 2014
 #1
avatar+92781 
+11
Thanks reinout-g,

I am bound to be wrong but I think that it is irrelevant.
Whether you picked the right one or the wrong one to start with there is still a box with nothing in it that can be opened so what difference does it make.
I am obviously wrong since Reinout Says that this is a hard problem.
Anyone got any other (better) ideas
Melody  Feb 26, 2014
 #2
avatar+2353 
+8
Melody:

Thanks reinout-g,

I am bound to be wrong but I think that it is irrelevant.
Whether you picked the right one or the wrong one to start with there is still a box with nothing in it that can be opened so what difference does it make.
I am obviously wrong since Reinout Says that this is a hard problem.
Anyone got any other (better) ideas



You might want to reconcider, after all there is a million dollars at stake here

A small hint: the host knows which box is empty and which box isn't
reinout-g  Feb 26, 2014
 #3
avatar+92781 
+6
No I don't get it !
Melody  Feb 26, 2014
 #4
avatar
+11
Use Bayesian statistics
Overall probability of wining increases to 66.6% by switching.
This is not intuitive with 3 options, but is very obvious if you started with 10 boxes.
Choose 1 of 10 boxes. After choosing, the MC exposes 8 of the 9 remaining boxes.
Now do you think you chose the correct box the first time? You could have but it's unlikely.
Guest Feb 26, 2014
 #5
avatar+92781 
+11
Hard one (quite famous actually):

You are on a tv-show where you can win a million dollars. The flamboyant host asks you to pick one of three boxes.
In two boxes you will find nothing, but one box will give you a million dollars.
You pick a box and the host then decides to play a trick with you.
Since he knows what is inside each box he opens one of the boxes that has nothing in it.
He then asks you the million dollar question.
Will you stick with the box you picked, or will you switch to the other box?

Should you stick with the bos you picked, should you switch or is it irrelevant what you do?
Explain why please
---------------------------------------------------------------------------------------------------------------------------------------------------------
Okay Reiner-g, so you are saying that the probability that you chose correctly in the first place is 1/3
But the probability that the other cup iscorrect is 1/2
therefore if you swap cups your chance of being correct increase from 1/3 to 1/2 (an increased chance of 1/6 = 16.6% ? )
No I am still not right. Where did the 66.6% come from
Melody  Feb 27, 2014
 #6
avatar+2353 
+11
Melody:

Hard one (quite famous actually):

You are on a tv-show where you can win a million dollars. The flamboyant host asks you to pick one of three boxes.
In two boxes you will find nothing, but one box will give you a million dollars.
You pick a box and the host then decides to play a trick with you.
Since he knows what is inside each box he opens one of the boxes that has nothing in it.
He then asks you the million dollar question.
Will you stick with the box you picked, or will you switch to the other box?

Should you stick with the bos you picked, should you switch or is it irrelevant what you do?
Explain why please
---------------------------------------------------------------------------------------------------------------------------------------------------------
Okay Reiner-g, so you are saying that the probability that you chose correctly in the first place is 1/3
But the probability that the other cup iscorrect is 1/2
therefore if you swap cups your chance of being correct increase from 1/3 to 1/2 (an increased chance of 1/6 = 16.6% ? )
No I am still not right. Where did the 66.6% come from



vestri amicus is correct,

Think of all the possibilities with their probability

box 1 - you have the million dollar box, 33%
box 2 - you have an empty box 33%
box 3 - you have an empty box 33%

Now there are three scenarios given that you switch

1) you have the million dollar box, the host removes box (2 or 3) and you switch to box 3 which gives you nothing

2) you have empty box 2, the host removes box 3 you switch and win a million dollars,

3) you have empty box 3, the host removes box 2 you switch and win a million dollars.

Note that all these scenarios have equal probability since you can have either box 1,2,3.

I also like the other explaination of vestri amicus thinking of the same problem with more boxes.

The problem is known as the monty hall problem ( http://en.wikipedia.org/wiki/Monty_Hall_problem )

The answer to the problem is very counterintuïve, which is also how it became so famous.
reinout-g  Feb 27, 2014
 #7
avatar+26750 
+11
This is known as the Monty Hall problem. There is an interesting Youtube video at http://www.youtube.com/watch?v=o_djTy3G0pg where mathematics professor Marcus Du Sautoy takes comedian Alan Davies through the problem.
Alan  Feb 27, 2014
 #8
avatar+92781 
+6
Thanks you vestri amicus, reinout and Alan. I will definitely check out that web site. But it will have to wait for morning .
Melody  Feb 27, 2014
 #9
avatar+2353 
+11
And here is a new puzzle!

Easier one:

There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000
What are the odds that that persons birthday is a monday?

Hard one (little easier than last time):

My slick host, which I've decided to name Steve, has again chosen to play a game with you.

He has given you two urns and 100 marbles to divide among them. 50 of these marbles are black and the other 50 are white.

After You've divided them among the urns, he is going to blindfold you and mix the urns.

You then choose one of the urns and pick one marble from that urn.

If the marble is black, you win a bus ticket home and if it is white you win a million dollars.

How should you divide the marbles between the urns assuming you prefer the million dollars?


Explain why please
reinout-g  Mar 2, 2014
 #10
avatar+92781 
+11
He has given you two urns and 100 marbles to divide among them. 50 of these marbles are black and the other 50 are white.
After You've divided them among the urns, he is going to blindfold you and mix the urns.
You then choose one of the urns and pick one marble from that urn.
If the marble is black, you win a bus ticket home and if it is white you win a million dollars.
How should you divide the marbles between the urns assuming you prefer the million dollars?

[size=150]Hi Reinout-g. Thanks for the puzzle[/size]

Once again I will state from the outset that I am bound to be wrong BUT
I would put just 1 white marble in one urn and the rest of the marbles in the other urn.
So my chance of winning the money will be

1/2 * 1 + 1/2 * 49/99 = 1/2 + 49/198 = 148/198 = 74/99

That sounds like pretty good odds to me.
Now you can tell me why i am wrong
Melody  Mar 2, 2014
 #11
avatar+2353 
+8
Melody:

He has given you two urns and 100 marbles to divide among them. 50 of these marbles are black and the other 50 are white.
After You've divided them among the urns, he is going to blindfold you and mix the urns.
You then choose one of the urns and pick one marble from that urn.
If the marble is black, you win a bus ticket home and if it is white you win a million dollars.
How should you divide the marbles between the urns assuming you prefer the million dollars?

[size=150]Hi Reinout-g. Thanks for the puzzle[/size]

Once again I will state from the outset that I am bound to be wrong BUT
I would put just 1 white marble in one urn and the rest of the marbles in the other urn.
So my chance of winning the money will be

1/2 * 1 + 1/2 * 49/99 = 1/2 + 49/198 = 148/198 = 74/99

That sounds like pretty good odds to me.
Now you can tell me why i am wrong



Haha, no actually you are right. Thanks for the detailed answer.
I'm trying to find the level where you still need to work your brain a little but won't find it impossible to solve the puzzle.
I'll post another one in a couple of minutes
reinout-g  Mar 2, 2014
 #12
avatar+2353 
+11
And here is the new puzzle!
If you have trouble solving it, feel free to give the easier one a try.

Easier one:

There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000
What are the odds that that persons birthday is a monday?

Hard one (I'm curious to see whether anyone can find a different strategy than I found.) :

Steve again wants to play a game with you but you need $2 for each time you want to play.
Now, each time you play the game, Steve has the computer print 100 blank cards with 100 random different integers.
After that you can flip the cards for as long as you like and if your last card flipped is the highest card in the deck you will win $10.

Think of a strategy that will win you money in this game.


Hint: The cards you previously flipped can give you some kind of information.

[/quote]
reinout-g  Mar 2, 2014
 #13
avatar+2353 
+11
reinout-g:

And here is the new puzzle!
If you have trouble solving it, feel free to give the easier one a try.

Easier one:

There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000
What are the odds that that persons birthday is a monday?

Hard one (I'm curious to see whether anyone can find a different strategy than I found.) :

Steve again wants to play a game with you but you need $2 for each time you want to play.
Now, each time you play the game, Steve has the computer print 100 blank cards with 100 random different integers.
After that you can flip the cards for as long as you like until you choose to stop flipping. If your last card flipped is the highest card in the deck you will win $10.

Think of a strategy where you expect to win money in this game.


Hint: The cards you previously flipped can give you some kind of information.

[/quote]
reinout-g  Mar 2, 2014
 #14
avatar+92781 
+11
Hi reinout-g,
Arn't there any constraints on the integers? If they can be absolutely anything I don't see how this can be done.
Melody  Mar 3, 2014
 #15
avatar+2353 
+11
Melody:

Hi reinout-g,
Arn't there any constraints on the integers? If they can be absolutely anything I don't see how this can be done.



Nope, I can give you another hint though...
Think of the problem as if it were two decks
reinout-g  Mar 3, 2014
 #16
avatar+92781 
+11
Hard one (I'm curious to see whether anyone can find a different strategy than I found.) :

Steve again wants to play a game with you but you need $2 for each time you want to play.
Now, each time you play the game, Steve has the computer print 100 blank cards with 100 random different integers.
After that you can flip the cards for as long as you like until you choose to stop flipping. If your last card flipped is the highest card in the deck you will win $10.

Think of a strategy where you expect to win money in this game.
Hint: The cards you previously flipped can give you some kind of information.
-----------------------------------------------------------------------------------------------------------------------------------------------------------
Alright then. I am assuming that I can only go through the 100 cards once.
I'll split the deck into 2deck of 50 cards each.
I'll go through the first 50 cards and note the highest integer.
Then I will stop at the 1st integer higher than this one in the second 50 cards.
If there are non higher then, I would have lost already.

The chance of success is not particularly high but it is better than no strategy at all.
Have you got a better idea reinout-g?
Melody  Mar 4, 2014
 #17
avatar+2353 
+11
Melody:

Hard one (I'm curious to see whether anyone can find a different strategy than I found.) :

Steve again wants to play a game with you but you need $2 for each time you want to play.
Now, each time you play the game, Steve has the computer print 100 blank cards with 100 random different integers.
After that you can flip the cards for as long as you like until you choose to stop flipping. If your last card flipped is the highest card in the deck you will win $10.

Think of a strategy where you expect to win money in this game.
Hint: The cards you previously flipped can give you some kind of information.
-----------------------------------------------------------------------------------------------------------------------------------------------------------
Alright then. I am assuming that I can only go through the 100 cards once.
I'll split the deck into 2deck of 50 cards each.
I'll go through the first 50 cards and note the highest integer.
Then I will stop at the 1st integer higher than this one in the second 50 cards.
If there are non higher then, I would have lost already.

The chance of success is not particularly high but it is better than no strategy at all.
Have you got a better idea reinout-g?



Actually your strategy is the same as I had, and the odds will win you money.

The strategy will pay off when the second-highest card of the 100 is in the first 50 cards and the highest card is in the last 50 cards.

Now there are four (almost) equally alike situations
Both the second highest card and the highest card are in the last deck - lose (at least, most of the times)
Both the second highest card and the highest card are in the first deck - lose (you won't find a value higher than the highest card)
The second highest card is in the last deck and the highest is in the first deck - lose (you won't find a value higher than the highest card)
The second highest card is in the first deck and the highest is in the second deck - win

So your chances of winning are a little over 25% (given that you might also stumble upon the highest card when you for example have the third highest card set as your value)
Now the game costs $2 to play and the winning are $10. So the expected value per game is at least E(X) = -2*0.75 + 10*0.25 = -1.50 + 2.50 = $1.
reinout-g  Mar 4, 2014
 #18
avatar+2353 
+11
Since I do not have time for a new puzzle, I'll post something which people can comment on.

Steve flips a coin,
If tail appears in the first flip, you receive $2, if tail appears in the second flip, you receive $4, if tail appears in the third flip, you receive $8, if tail appears in the fourth flip, you receive $16, etc. The money will keep on doubling until you hit a tail.

How much would you be willing to pay to play this game?
reinout-g  Mar 4, 2014
 #19
avatar+92781 
+11
reinout-g:

Since I do not have time for a new puzzle, I'll post something which people can comment on.

Steve flips a coin,
If tail appears in the first flip, you receive $2, if tail appears in the second flip, you receive $4, if tail appears in the third flip, you receive $8, if tail appears in the fourth flip, you receive $16, etc. The money will keep on doubling until you hit a tail.

How much would you be willing to pay to play this game?



There is no talk of losing any money here. Does it cost something to play? If it doesn't cost anything and the coin is 'fair' (I can't think of the proper word) Then I would play until i got too too bored, or until I won enough money for that trip to The Netherlands that i have always desired. .
I didn't read it properly - what a shame.

If I played $1 each game, eventually I would have to win.
If I paid $2 per flip then I would break even if tails came up on the first toss and win otherwise. So long as i keep putting my money down.
If i paid $4 per flip then I would only lose money if tails came up on the 1st toss, I would break even if tails came second and after that I would win.
So, it doesn't matter how much I pay (assuming each toss costs the same). so long as I get the chance to play again after each win, I cannot lose (eventually)
NOW you can tell me that I misinterpreted the question again (I always have a problem following the dots) ?
Melody  Mar 5, 2014
 #20
avatar
+6
just switch to the other box so you get a 66 percent chance instead of a 33 percent chance.
Guest Mar 5, 2014
 #21
avatar+92781 
+11
superextremebro:

just switch to the other box so you get a 66 percent chance instead of a 33 percent chance.



What other box. What question are you answering?

When a thread is long, it is always a good idea to include the question with the answer.

reinout-g will probably know what you are talking about. This thread is his baby ( I hope that sounds positive ).
Melody  Mar 5, 2014
 #22
avatar+2353 
+11
Melody:
reinout-g:

Since I do not have time for a new puzzle, I'll post something which people can comment on.

Steve flips a coin,
If tail appears in the first flip, you receive $2, if tail appears in the second flip, you receive $4, if tail appears in the third flip, you receive $8, if tail appears in the fourth flip, you receive $16, etc. The money will keep on doubling until you hit a tail.

How much would you be willing to pay to play this game?



There is no talk of losing any money here. Does it cost something to play? If it doesn't cost anything and the coin is 'fair' (I can't think of the proper word) Then I would play until i got too too bored, or until I won enough money for that trip to The Netherlands that i have always desired. .
I didn't read it properly - what a shame.

If I played $1 each game, eventually I would have to win.
If I paid $2 per flip then I would break even if tails came up on the first toss and win otherwise. So long as i keep putting my money down.
If i paid $4 per flip then I would only lose money if tails came up on the 1st toss, I would break even if tails came second and after that I would win.
So, it doesn't matter how much I pay (assuming each toss costs the same). so long as I get the chance to play again after each win, I cannot lose (eventually)
NOW you can tell me that I misinterpreted the question again (I always have a problem following the dots) ?




I posted this one a little differently because it is more of a philosophical probability question than a riddle. It points out the flaw in expected value.

I like your idea of missing a number of times one can play the game, however people also buy lottery tickets which is pretty much the same for the less probable outcomes.

Actually, you could rephrase the question to a single lottery ticket where the odds of winning are equally distributed (so E(x) = (1/2) * $2 + (1/4) * $4 + (1/8) * $8.... => infinity)

I'll post a riddle (no paradox this time) somewhat later today...
reinout-g  Mar 5, 2014
 #23
avatar+2353 
+8
And here is the new puzzle!
If you have trouble solving it, feel free to give the easier one a try.

Easier one: (still to be solved)

There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000
What are the odds that that persons birthday is a monday?

Hard one :

Three kids named Brooks, Clint and Riley have a new game using water balloons.
They each take turns trying to throw one balloon at one of the other two. If they hit, the person is out, otherwise the next person may take his turn.

Brooks hits 1/3 of the time, Clint hits 2/3 of the time and Riley never misses.

Brooks is allowed to start, then Clint (given that he is still in the game), then Riley (given that he is still in the game), etc.

Where should brooks aim during his first turn?
reinout-g  Mar 5, 2014
 #24
avatar+92781 
+11
*
Hard one :
Three kids named Brooks, Clint and Riley have a new game using water balloons.
They each take turns trying to throw one balloon at one of the other two. If they hit, the person is out, otherwise the next person may take his turn.

Brooks hits 1/3 of the time, Clint hits 2/3 of the time and Riley never misses.
Brooks is allowed to start, then Clint (given that he is still in the game), then Riley (given that he is still in the game), etc.
Where should brooks aim during his first turn?

Well if Brooks misses both targets then Clint will definitely try to hit Riley. Because if Clint hits Brooks, Riley will get him next anyway.
Therefore there is no point in Brooks going for Clint.
Now if Brook hits Riley then Brook has a 2/3 chance of being hit next
If Brook misses altogether then he has no chance of being hit next, If Clint hits Riley then it is Brooks turn next anyway (he'll get another shot at clint). If Brook misses Riley then there is probably a less than 2/3 chance that Riley will target Brook next.

So it seems to me that Brook's best strategy is to purposefully miss both targets.

IS that right ??
--------------------------------------------------------------------------------------------------------------------------------------

Easier one: (still to be solved)
There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000
What are the odds that that persons birthday is a monday?

Er let me see... There are seven days in the week. Everyone has to be born on one of them.
So the chance of anywone being born on a monday is 1/7
-----------------------------------------------------------------------------------------------------------------------------------------
Melody  Mar 10, 2014
 #25
avatar+2353 
+8
Melody:

*
Hard one :
Three kids named Brooks, Clint and Riley have a new game using water balloons.
They each take turns trying to throw one balloon at one of the other two. If they hit, the person is out, otherwise the next person may take his turn.

Brooks hits 1/3 of the time, Clint hits 2/3 of the time and Riley never misses.
Brooks is allowed to start, then Clint (given that he is still in the game), then Riley (given that he is still in the game), etc.
Where should brooks aim during his first turn?

Well if Brooks misses both targets then Clint will definitely try to hit Riley. Because if Clint hits Brooks, Riley will get him next anyway.
Therefore there is no point in Brooks going for Clint.
Now if Brook hits Riley then Brook has a 2/3 chance of being hit next
If Brook misses altogether then he has no chance of being hit next, If Clint hits Riley then it is Brooks turn next anyway (he'll get another shot at clint). If Brook misses Riley then there is probably a less than 2/3 chance that Riley will target Brook next.

So it seems to me that Brook's best strategy is to purposefully miss both targets.

IS that right ??
--------------------------------------------------------------------------------------------------------------------------------------

Easier one: (still to be solved)
There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000
What are the odds that that persons birthday is a monday?

Er let me see... There are seven days in the week. Everyone has to be born on one of them.
So the chance of anywone being born on a monday is 1/7
-----------------------------------------------------------------------------------------------------------------------------------------




Well done melody!

I like how you explained your reasoning.

I will see whether I can post some new ones a little later today
reinout-g  Mar 12, 2014
 #26
avatar+2353 
+11
Here are the new probability puzzles!
--------------------------------------------------------------------------------------------------------------------------------------
Easier one:
A special machine contains 100 red, 100 green and 100 blue marbles.
The machine performs the following four operations.

First, the machine randomly picks 50 marbles from the entire batch.

Secondly the machine divides the marbles into one group of 20 and one group of 30

Furthermore, the machine picks and shuffles 5 marbles of the group of 20 and 10 marbles of the group of 30.

Finally, the machine randomly chooses one of the final 15 marbles it picked and shows it to you.

Assuming you cannot see anything which happens inside the machine.

What are the odds the machine will show you a red marble?

--------------------------------------------------------------------------------------------------------------------------------------
Hard one :

Steve has two routes (A and B) to travel to his girlfriend which he is indifferent about (see my amazing illustrational skills )

knipsel11.png

Steve has decided to play a game which makes him decide which route to pick.
He always waits where the road splits between route A and B until he either sees a train going from A to B or from B to A.
If the first train Steve sees goes from A to B he always takes route A and if the first train Steve sees is a train going from B to A he always takes route B.
There are equally as many trains going from A to B as there are going from B to A.
Nevertheless Steve has noticed that for every five times he visits his girlfriend he takes route A four times and route B only once.

How can this be?

--------------------------------------------------------------------------------------------------------------------------------------

Reinout-g

edit: I updated a flaw in the illustration, 'work' was meant to be 'girlfriend'
reinout-g  Mar 22, 2014
 #27
avatar
+6
This is so because the majority of the B trains must pass while he is at work, and then the A trains start to roll in when he leaves work for his girlfriend.
Guest Mar 22, 2014
 #28
avatar+13 
+11
reinout-g:

Here are the new probability puzzles!
--------------------------------------------------------------------------------------------------------------------------------------
Easier one:
A special machine contains 100 red, 100 green and 100 blue marbles.
The machine performs the following four operations.

First, the machine randomly picks 50 marbles from the entire batch.

Secondly the machine divides the marbles into one group of 20 and one group of 30

Furthermore, the machine picks and shuffles 5 marbles of the group of 20 and 10 marbles of the group of 30.

Finally, the machine randomly chooses one of the final 15 marbles it picked and shows it to you.

Assuming you cannot see anything which happens inside the machine.

What are the odds the machine will show you a red marble?

--------------------------------------------------------------------------------------------------------------------------------------



Alright, I'm gonna \try this one. ^_^
Since there are all 100 marbles per color, and the machine chooses randomly, then there would be an equal chance that you would get any color.
So there would be a 33.33...% chance of red, blue, or green. Right? ^_^

And I'd agree with @stuff on the other one.
Nobody Special  Mar 23, 2014
 #29
avatar+2353 
+11
stuff:

This is so because the majority of the B trains must pass while he is at work, and then the A trains start to roll in when he leaves work for his girlfriend.



Good thinking!

Nevertheless, there is also another possibility.
Would you be able to think of an answer when both trains are scheduled to go every 10 minutes?
reinout-g  Mar 23, 2014
 #30
avatar+889 
+11
reinout-g:

Here are the new probability puzzles!
--------------------------------------------------------------------------------------------------------------------------------------
Easier one:
A special machine contains 100 red, 100 green and 100 blue marbles.
The machine performs the following four operations.

First, the machine randomly picks 50 marbles from the entire batch.

Secondly the machine divides the marbles into one group of 20 and one group of 30

Furthermore, the machine picks and shuffles 5 marbles of the group of 20 and 10 marbles of the group of 30.

Finally, the machine randomly chooses one of the final 15 marbles it picked and shows it to you.

Assuming you cannot see anything which happens inside the machine.

What are the odds the machine will show you a red marble?

--------------------------------------------------------------------------------------------------------------------------------------
Hard one :

Steve has two routes (A and B) to travel to his girlfriend which he is indifferent about (see my amazing illustrational skills )

knipsel11.png

Steve has decided to play a game which makes him decide which route to pick.
He always waits where the road splits between route A and B until he either sees a train going from A to B or from B to A.
If the first train Steve sees goes from A to B he always takes route A and if the first train Steve sees is a train going from B to A he always takes route B.
There are equally as many trains going from A to B as there are going from B to A.
Nevertheless Steve has noticed that for every five times he visits his girlfriend he takes route A four times and route B only once.

How can this be?

--------------------------------------------------------------------------------------------------------------------------------------

Reinout-g

edit: I updated a flaw in the illustration, 'work' was meant to be 'girlfriend'



Easier one.
Ignore the flim-flam, via a circuitous route a single marble has been taken at random from the original 300. It's 2 to 1 against it being a red marble.

Harder one.
In principle there can be any number of trains travelling between A and B, (and B and A), but start by thinking in terms of one train in either direction.

Suppose that the train from A to B passes on the hour and the train from B to A at twelve minutes past the hour. If 'our man' arrives at the split in the road at some random time, there will be a 48 minute window within which the next train will be the one from A to B, (12 min past the hour through to 60min past the hour, or 12.01 and 59.99 min past the hour if you're going to be pedantic) and a 12 minute window when the next train will be the one from B to A, (0 min to 12 min). The time interval for A is four times as long as the interval for B so it's four times more likely that the next train will be the A to B train.

Now suppose that there are two trains in either direction, the A to B ones on the hour and half hour, the B to A ones at 6 minutes and 36 minutes past the hour. Combining the windows, there will be a total of 48 minutes, (24+24), during which the next train will be from A to B, and 12 minutes, (6+6), when the next train will be from B to A. Again, that's 4 to 1 in favour of A.

We can do this for higher numbers of trains, though it does become somewhat impracticable after a while !
For three trains in either direction the timings for the A to B trains could be** on the hour, 20 minutes past and 40 minutes past, while the timings for the B to A trains could be at 4 minutes, 24 minutes and 44 minutes past the hour.

** The timings need not be exactly as stated, and the intervals need not be equal. So long as the 'windows' for the A to B trains add to 48 minutes it will always be 4 to 1 in favour of A to B.
Bertie  Mar 23, 2014
 #31
avatar+2353 
+11
Bertie:
reinout-g:

Here are the new probability puzzles!
--------------------------------------------------------------------------------------------------------------------------------------
Easier one:
A special machine contains 100 red, 100 green and 100 blue marbles.
The machine performs the following four operations.

First, the machine randomly picks 50 marbles from the entire batch.

Secondly the machine divides the marbles into one group of 20 and one group of 30

Furthermore, the machine picks and shuffles 5 marbles of the group of 20 and 10 marbles of the group of 30.

Finally, the machine randomly chooses one of the final 15 marbles it picked and shows it to you.

Assuming you cannot see anything which happens inside the machine.

What are the odds the machine will show you a red marble?

--------------------------------------------------------------------------------------------------------------------------------------
Hard one :

Steve has two routes (A and B) to travel to his girlfriend which he is indifferent about (see my amazing illustrational skills )

knipsel11.png

Steve has decided to play a game which makes him decide which route to pick.
He always waits where the road splits between route A and B until he either sees a train going from A to B or from B to A.
If the first train Steve sees goes from A to B he always takes route A and if the first train Steve sees is a train going from B to A he always takes route B.
There are equally as many trains going from A to B as there are going from B to A.
Nevertheless Steve has noticed that for every five times he visits his girlfriend he takes route A four times and route B only once.

How can this be?

--------------------------------------------------------------------------------------------------------------------------------------

Reinout-g

edit: I updated a flaw in the illustration, 'work' was meant to be 'girlfriend'



Easier one.
Ignore the flim-flam, via a circuitous route a single marble has been taken at random from the original 300. It's 2 to 1 against it being a red marble.

Harder one.
In principle there can be any number of trains travelling between A and B, (and B and A), but start by thinking in terms of one train in either direction.

Suppose that the train from A to B passes on the hour and the train from B to A at twelve minutes past the hour. If 'our man' arrives at the split in the road at some random time, there will be a 48 minute window within which the next train will be the one from A to B, (12 min past the hour through to 60min past the hour, or 12.01 and 59.99 min past the hour if you're going to be pedantic) and a 12 minute window when the next train will be the one from B to A, (0 min to 12 min). The time interval for A is four times as long as the interval for B so it's four times more likely that the next train will be the A to B train.

Now suppose that there are two trains in either direction, the A to B ones on the hour and half hour, the B to A ones at 6 minutes and 36 minutes past the hour. Combining the windows, there will be a total of 48 minutes, (24+24), during which the next train will be from A to B, and 12 minutes, (6+6), when the next train will be from B to A. Again, that's 4 to 1 in favour of A.

We can do this for higher numbers of trains, though it does become somewhat impracticable after a while !
For three trains in either direction the timings for the A to B trains could be** on the hour, 20 minutes past and 40 minutes past, while the timings for the B to A trains could be at 4 minutes, 24 minutes and 44 minutes past the hour.

** The timings need not be exactly as stated, and the intervals need not be equal. So long as the 'windows' for the A to B trains add to 48 minutes it will always be 4 to 1 in favour of A to B.




Excellent answers Bertie!

Thank you for your descriptive explanation with examples of trains leaving.

In fact, I think I don't even have anything else to add to your answer

I will upload new puzzles shortly
reinout-g  Mar 24, 2014
 #32
avatar+2353 
+11
Nobody Special:
reinout-g:

Here are the new probability puzzles!
--------------------------------------------------------------------------------------------------------------------------------------
Easier one:
A special machine contains 100 red, 100 green and 100 blue marbles.
The machine performs the following four operations.

First, the machine randomly picks 50 marbles from the entire batch.

Secondly the machine divides the marbles into one group of 20 and one group of 30

Furthermore, the machine picks and shuffles 5 marbles of the group of 20 and 10 marbles of the group of 30.

Finally, the machine randomly chooses one of the final 15 marbles it picked and shows it to you.

Assuming you cannot see anything which happens inside the machine.

What are the odds the machine will show you a red marble?

--------------------------------------------------------------------------------------------------------------------------------------



Alright, I'm gonna \try this one. ^_^
Since there are all 100 marbles per color, and the machine chooses randomly, then there would be an equal chance that you would get any color.
So there would be a 33.33...% chance of red, blue, or green. Right? ^_^

And I'd agree with @stuff on the other one.



You're absolutely right.
Everything the machine does is irrelevant since you cannot see any of the in-between steps.
I'll try and make the easier puzzle a little more difficult when I upload the new ones.
For the answer with the trains, see whether you also understand Bertie's answer.


Reinout
reinout-g  Mar 24, 2014
 #33
avatar+2353 
+11
Here are the new probability puzzles!
--------------------------------------------------------------------------------------------------------------------------------------
Easier one:

Ben the hiker wants to spend the night on top of a mountain. He hikes all day long towards the top and camps there overnight. The following day he is taking the same path down towards the bottom when he suddenly looks at his watch en notices that he is at the very same spot at the very same time of the day as the day before on his way up. 'What a coïncidence!' he exclaims!
Give an estimate of the probabilty that a hiker will be at exactly the same spot on the mountain at the same time of the day on his return trip as he was the previous day's hike up the mountain and explain why.

--------------------------------------------------------------------------------------------------------------------------------------
Hard one :

You and your best friend have been arrested for several white collar crimes. (Or something else which is more classy than stealing a candy bar)
Your attorney has told you the police has only got enough evidence to put you and your friend behind bars for 3 years.
However, the both of you are given the option to confess. If both of you confess you will both serve 10 years,
if one of you confesses and the other does not, the one who did not confess will serve 20 years.
Obviously, if you both remain silent you will both serve 3 years.
You are not allowed to have any communication with your friend.
Suppose your main priority is to keep yourself out of jail, and your secondary priority is to keep your partner out of jail.
You have decided that you find it twice as important to keep yourself out of jail as you find it important to keep your partner out of jail.
(So for example, in your opinion him serving 10 years is equal to you serving 5).
You estimate the chance of your partner confessing p.
For which values of p will you choose to confess and for which values will you keep silent?

--------------------------------------------------------------------------------------------------------------------------------------

Reinout-g

edit; I changed 'at what value of p' to 'for which values of p'
reinout-g  Apr 4, 2014
 #34
avatar+92781 
+11
Easier one:

Quote:

Ben the hiker wants to spend the night on top of a mountain. He hikes all day long towards the top and camps there overnight. The following day he is taking the same path down towards the bottom when he suddenly looks at his watch en notices that he is at the very same spot at the very same time of the day as the day before on his way up. 'What a coïncidence!' he exclaims!
Give an estimate of the probabilty that a hiker will be at exactly the same spot on the mountain at the same time of the day on his return trip as he was the previous day's hike up the mountain and explain why.


I think that the probability is 1. That means it has to happen.
This is why.
Ben's friend Sven is doing the same hike at exactly the same pace today, from the bottom to the top of the mountain that Ben did yesterday. Ben is on his way down. Somewhere Ben and Sven must pass each other. Obviously it will be the same time for each of them.
Although, Ben might have forgotten to set his watch back to mark the end daylight saving time. (Daylight savings finished on Sunday night in Sydney.) This could cause a problem.
No really, I think that the probablility is one - unless one of them falls and brakes his legs and has to get air lifted out. - then their paths may not cross.
Melody  Apr 8, 2014
 #35
avatar+2353 
+8
Melody:

Easier one:

Quote:

Ben the hiker wants to spend the night on top of a mountain. He hikes all day long towards the top and camps there overnight. The following day he is taking the same path down towards the bottom when he suddenly looks at his watch en notices that he is at the very same spot at the very same time of the day as the day before on his way up. 'What a coïncidence!' he exclaims!
Give an estimate of the probabilty that a hiker will be at exactly the same spot on the mountain at the same time of the day on his return trip as he was the previous day's hike up the mountain and explain why.


I think that the probability is 1. That means it has to happen.
This is why.
Ben's friend Sven is doing the same hike at exactly the same pace today, from the bottom to the top of the mountain that Ben did yesterday. Ben is on his way down. Somewhere Ben and Sven must pass each other. Obviously it will be the same time for each of them.
Although, Ben might have forgotten to set his watch back to mark the end daylight saving time. (Daylight savings finished on Sunday night in Sydney.) This could cause a problem.
No really, I think that the probablility is one - unless one of them falls and brakes his legs and has to get air lifted out. - then their paths may not cross.



You're right, assuming that all events like a dinosaur eating Ben are very close to 0, the probability of this happening is very close to 1. Did you think about the hard one yet?
reinout-g  Apr 8, 2014
 #36
avatar+92781 
+6
Quote:

Hard one :

You and your best friend have been arrested for several white collar crimes. (Or something else which is more classy than stealing a candy bar)
Your attorney has told you the police has only got enough evidence to put you and your friend behind bars for 3 years.
However, the both of you are given the option to confess. If both of you confess you will both serve 10 years,
if one of you confesses and the other does not, the one who did not confess will serve 20 years.
Obviously, if you both remain silent you will both serve 3 years.
You are not allowed to have any communication with your friend.
Suppose your main priority is to keep yourself out of jail, and your secondary priority is to keep your partner out of jail.
You have decided that you find it twice as important to keep yourself out of jail as you find it important to keep your partner out of jail.
(So for example, in your opinion him serving 10 years is equal to you serving 5).
You estimate the chance of your partner confessing p.
At what value of p will you choose to confess and at what value will you keep silent?



Maybe it doesn't make any difference to the question BUT If one confesses and the other doesn't then the non confessor gets 20yrs. HOW many years will the confessor get?
Melody  Apr 9, 2014
 #37
avatar+2353 
+3
Melody:
Quote:

Hard one :

You and your best friend have been arrested for several white collar crimes. (Or something else which is more classy than stealing a candy bar)
Your attorney has told you the police has only got enough evidence to put you and your friend behind bars for 3 years.
However, the both of you are given the option to confess. If both of you confess you will both serve 10 years,
if one of you confesses and the other does not, the one who did not confess will serve 20 years.
Obviously, if you both remain silent you will both serve 3 years.
You are not allowed to have any communication with your friend.
Suppose your main priority is to keep yourself out of jail, and your secondary priority is to keep your partner out of jail.
You have decided that you find it twice as important to keep yourself out of jail as you find it important to keep your partner out of jail.
(So for example, in your opinion him serving 10 years is equal to you serving 5).
You estimate the chance of your partner confessing p.
At what value of p will you choose to confess and at what value will you keep silent?



Maybe it doesn't make any difference to the question BUT If one confesses and the other doesn't then the non confessor gets 20yrs. HOW many years will the confessor get?



The confessor will get 0 years, but remember you do care about your friend. You want to keep yourself out of jail twice as much as you want to keep your partner out of jail. Therefore, you'd rather serve 4 years than have your friend serve 10 years (since 2*4 = 8 < 10), but you'd be indifferent between you serving 4 years and him serving 8. (Yes this puzzle does require some calculation)



edit: I just made an edit to the original question, I changed 'at which value of p' to 'for which values of p'. Sorry if I might have put you on the wrong foot
reinout-g  Apr 9, 2014
 #38
avatar+92781 
+6
I don't know reinout,

If p =0 then I won't confess and we both get 3 years
If p = 1 then I do confess and we both get 10 years.
If p is something else then I think I would have to leave it to the gods because I DO NOT KNOW
Melody  Apr 9, 2014
 #39
avatar
+6
I helps to explain it with visuals

1st Pick Reveal Switch?
Nothing Box 1 Yes-Million
Nothing Box 1 No-Nothing
Nothing Box 2 Yes-Million
Nothing Box 2 No-Nothing
Million Box 1-or-2 Yes-Nothing
Million Box 1-or-2 No-Million
As you can see NOT switching ONLY benefits you when you first chose the million (1/3 Chance) and Switching benefits you when ever you choose nothing (2/3). It now comes down to: What is more likely 1/3 or 2/3?
Guest Apr 10, 2014
 #40
avatar+2353 
+3
Melody:

I don't know reinout,

If p =0 then I won't confess and we both get 3 years
If p = 1 then I do confess and we both get 10 years.
If p is something else then I think I would have to leave it to the gods because I DO NOT KNOW




Here's a hint: Try to find some formula for the amount of suffering you'll have to deal with.
1 year of jail for your friend amounts to 1 unit of suffering and 1 year of jail for you amounts to 2 units of suffering.
reinout-g  Apr 10, 2014
 #41
avatar+2353 
+3
Guest:

I helps to explain it with visuals

1st Pick Reveal Switch?
Nothing Box 1 Yes-Million
Nothing Box 1 No-Nothing
Nothing Box 2 Yes-Million
Nothing Box 2 No-Nothing
Million Box 1-or-2 Yes-Nothing
Million Box 1-or-2 No-Million
As you can see NOT switching ONLY benefits you when you first chose the million (1/3 Chance) and Switching benefits you when ever you choose nothing (2/3). It now comes down to: What is more likely 1/3 or 2/3?



Thank you guest, would you also like to try the new one?
Why don't you sign up so we can identify your posts?
reinout-g  Apr 10, 2014
 #42
avatar+2353 
+3

BUMP!!! 

reinout-g  May 6, 2014
 #43
avatar+2353 
+8

Here is an update on the probability puzzles!
-------------------------------------------------------------------------------------------------------------------------------------- 
Easier one: 

The fabulous game host Steve offer you a bag with 5001 pearls of which 2501 are white and 2500 are black. He asks you to draw as many pearls from the bag as you desire without looking. Steve then tells you how many black and white pearls you drew. If there are equally as much white as black pearls in the draw you gain €10 for each pearl you drew. How many pearls should you draw?

-------------------------------------------------------------------------------------------------------------------------------------- 
Hard one (still unanswered): 

You and your best friend have been arrested for several white collar crimes. (Or something else which is more classy than stealing a candy bar) 
Your attorney has told you the police has only got enough evidence to put you and your friend behind bars for 3 years. 
However, the both of you are given the option to confess. If both of you confess you will both serve 10 years, 
if one of you confesses and the other does not, the one who did not confess will serve 20 years. 
Obviously, if you both remain silent you will both serve 3 years. 
You are not allowed to have any communication with your friend. 
Suppose your main priority is to keep yourself out of jail, and your secondary priority is to keep your partner out of jail. 
You have decided that you find it twice as important to keep yourself out of jail as you find it important to keep your partner out of jail. 
(So for example, in your opinion him serving 10 years is equal to you serving 5). 
You estimate the chance of your partner confessing p. 
For which values of p will you choose to confess and for which values will you keep silent? 

HINT: make a formula for the amount of suffering you expect, also taking the suffering for your friends jailtime into account.


-------------------------------------------------------------------------------------------------------------------------------------- 

Reinout-g 

reinout-g  May 7, 2014
 #44
avatar+4150 
+6

i dont know but i suffered alot reading this question just take me to jail!!!!

zegroes  May 15, 2014
 #45
avatar+2353 
+3

bump bump bump

reinout-g  May 15, 2014
 #46
avatar+4150 
+6

well actually going off of logic both persons will stay silent because if you both stay silent its only 3 years but if you do the other choices you will get a longer sentence but im probably wrong since it probably takes alot of overly complicated math to solve this but all well i tried

zegroes  May 15, 2014
 #47
avatar+2353 
+8

Okay let me give you a hint;

Let p be the probability that your partner confesses

Choosing to confess gives a probability p of two confessions and a probability (1-p) that only you implicate your partner. If you don't confess, there is a probability p that he implicates you and a probability (1-p) that neither of you confess.

Furthermore, 1 year in jail gives you 2 suffering points and 1 year of your partner in jail gives you 1 suffering point. 

reinout-g  May 15, 2014
 #48
avatar+4150 
+6

.........you know reinout i.....need to save my brain power for questions on the forum ill come back later.....maybe

zegroes  May 15, 2014
 #49
avatar+2592 
+11

Harder one: just confess because if you confess you will have 1/2 the time he has if he doesnt confess( you=10, him=20). And if he confesses then you both get 10 years therefore you will have your partner out as the same time as you 

SpawnofAngel  May 15, 2014
 #50
avatar+2353 
+8

Hey SpawnAngel,

 

I think you misread the question,

if I confess and my friend does not I will get 20 years and he will get 0.

The other way around I get 0 and he gets 20.

 

Also it's not a matter of confessing or not confessing.

The question is how big the odds of your friend confessing should be in order for you to choose to not confess or confess.

 

Have another look at it. It is not as difficult as it may seem, however it does require a little calculation 

reinout-g  May 18, 2014
 #51
avatar+92781 
+6

Points given for joining in - I want everyone to get involved! 

Not silly involvement - either proper maths or proper humour or both!

Melody  May 19, 2014
 #52
avatar+63 
+14
Best Answer

Hard one (still unanswered): 

You and your best friend have been arrested for several white collar crimes. (Or something else which is more classy than stealing a candy bar) 
Your attorney has told you the police has only got enough evidence to put you and your friend behind bars for 3 years. 
However, the both of you are given the option to confess. If both of you confess you will both serve 10 years, 
if one of you confesses and the other does not, the one who did not confess will serve 20 years. 
Obviously, if you both remain silent you will both serve 3 years. 
You are not allowed to have any communication with your friend. 
Suppose your main priority is to keep yourself out of jail, and your secondary priority is to keep your partner out of jail. 
You have decided that you find it twice as important to keep yourself out of jail as you find it important to keep your partner out of jail. 
(So for example, in your opinion him serving 10 years is equal to you serving 5). 
You estimate the chance of your partner confessing p. 
For which values of p will you choose to confess and for which values will you keep silent? 

HINT: make a formula for the amount of suffering you expect, also taking the suffering for your friends jailtime into account.

-----------------------------------------------------------------------------------------------------------

2 Points for every year you're in jail, 1 Point for every year your partner is in jail.

Therefore...

 

Both don't confess- if 50% chance friend will confess

3 years for you (6 points), 3 years for your partner (3 points)

[9] points total

 

Both confess- if 50% chance friend will confess

10 years for you (20 points), 10 years for your partner (10 points)

[30] points total

 

You confess and your friend doesn't- if 50% chance friend will confess

0 years for you (0 points), 20 years for your partner (20 points)

[20] points total

 

Your friend confesses and you don't- 50% chance friend will confess

20 years for you (40 points), 0 years for your partner (0 points)

[40] points total

 

Therefore...

 

If you confess you will have an average of 25 "points"

(30+20)/2=25

 

If you don't you have an average of 24.5 "points"

(40+9)/2=24.5

 

Therefore, in most cases you shouldn't confess :D

CreeperCosmo  May 20, 2014
 #53
avatar+4150 
+6

its not possible to answer

zegroes  May 20, 2014
 #54
avatar+2353 
+8

Okay, so I think Creepercosmo was so close to the answer that I'm gonna give him 5 points...

 

Here's how it can be done.

 

Let's say there are indeed 4 options.

 

1. I confess, my friend does too and we both spend 10 years in jail

2. I confess, my friend does not and he spends 20 years on his own

3. I do not confess, my friend does and I spend 20 years on my own

4. I do not confess, my friend doesn't either and we both spend 3 years

 

Let's see how many suffering points each situation gives me;

 

1. s = 2*10 + 10 = 30

2. s = 2*0 + 20 = 20

3. s = 2*20 + 0 = 40

4. s = 2*3 + 3 = 9

 

Let p be the estimated probability that your partner confesses.

Then the expected amount of suffering is

If I confess;

E(S|C) = 30p+20(1-p) = 10p+20

If I do not confess;

E(S|NC) = 40p+9(1-p) = 31p+9

 

We want to find a value for p where you're indifferent between confessing or not confessing.

Therefore

$$10p+20 = 31p+9 \Rightarrow 21p = 11 => p = \frac{11}{21} = 0.5238$$

Now the slope of 31p+9 is obviously higher than that of 10p+20 (and we want to minimize the amount of suffering)

Therefore if $$p>0.5238$$ You'd choose to confess and

if $$p<0.5238$$ You'd choose not to confess.

 

Now obviously, I highly doubt someone would say his friend has a 52.39% chance of confessing, so well done Creepercosmo! 

 

I'm going to think of a new one 

reinout-g  May 21, 2014
 #55
avatar+2353 
+11

Here is an update on the probability puzzles!
-------------------------------------------------------------------------------------------------------------------------------------- 
Easier one (hasn't been answered yet): 

The fabulous game host Steve offer you a bag with 5001 pearls of which 2501 are white and 2500 are black. He asks you to draw as many pearls from the bag as you desire without looking. Steve then tells you how many black and white pearls you drew. If there are equally as much white as black pearls in the draw you gain €10 for each pearl you drew. How many pearls should you draw?

-------------------------------------------------------------------------------------------------------------------------------------- 
Hard one: 

All right, so here's a classic one.

Jason and Tyrell are in a classroom of 23 students.

At some point Tyrell finds out that some of his classmates have the same birthday.

In surprise he exclaims; 'what a surprise!'.

Jason however, says it is more likely for two or more people to have the same birthday in their class than it is for noone to have the same birthday.

 

Assume any birthday is equally likely to another and pretend leap years don't exist. (it makes it a lot easier)

What are the odds of two or more people sharing their birthday?

 


Tip: You might find that the web2.0calc.com calculator has trouble calculating this. If you prefer not to do it by hand I suggest using www.wolframalpha.com 

 

Reinout 

reinout-g  May 21, 2014
 #56
avatar
+6

Only in 6th grade!!  HEHEH

Guest May 22, 2014
 #57
avatar+2353 
+3

Try the easier one then,

It doesn't really require any calculation, just some thinking. 

reinout-g  May 22, 2014
 #58
avatar
+3

I'd just do eeny meeny miny moe and get it over with

Guest May 22, 2014
 #59
avatar
+3

I'm in 6th grade 2

Guest May 22, 2014
 #60
avatar
+11

I'd say your best bet is to draw 5,000 pearls. That way you have a 50/50 chance of having all pairs or not. If a white is left, that means you have the same amount of each. If a black is left, then too bad. In that case, snatch the entire bag of pearls and start running.

Guest May 24, 2014
 #61
avatar+87301 
+8

I'll take a run at the pearl problem......although I kind of like the idea that Anonymous had...snatching the bag and running!!!

I also believe that drawing all the pearls out of the bag, except one, is the correct strategy. Since, after any number of "even" draws, we never know how many white or black ones we have, we can never be sure that their totals are equal...if we could, "fabulous" Steve would soon be "broke" Steve.

Notice that after just 2 draws, there are 3 possibilities.....0W 2B, 1W 1B, 2W 0B  ........Thus only one thing out of 3 wins for us.

Notice that after just 4 draws, there are 5 possibilities.....0W 4B, 1W 3B , 2W 2B, 3W 1B, 4W 0B........Thus only one thing out of 5 wins for us

And after 100 draws, again, only one thing out of 101 wins for us - 50W 50B, and 100 other combinations lose for us.

But, consider the scenario in which two pearls remain in the bag......If we draw either a white or black one on the next draw and a white one remains, we win. Thus, (draw white)(white remains) or (draw black)(white remains) are two winning scenarios.

But, if we draw either a black or white one and a black remains, we lose no matter what. Thus, out of the four remaining scenarios, 2 win for us and 2 lose, so we have a 50% chance of winning. And drawing all but one pearl out of the bag results in a greater chance of winning than stopping at any other point of even draws. Also note, that when we draw 2 pearls out ot the bag at first, we have a 33% chance of winning, but when two pearls remain, we are even money to win !!!

One final note....you could turn the tables on "fabulous" Steve and tell him that you will take one pearl out of the bag, and whatever remains in the bag is your "draw." Thus, if you draw a white one out, you win......otherwise, no.

That's my story, and I'm sticking to it!! (But, it might be incorrect, too......)

CPhill  May 24, 2014
 #62
avatar+2353 
+11

What a wonderful explanation CPhill 

I have nothing to add

 

reinout-g  May 24, 2014
 #63
avatar+126 
+6

well for the pearl one to get the most you cna you draw 2500 of each one with one white pearl left over.

does that count?

ThAnswerGuy  Sep 9, 2014
 #64
avatar+3 
+3

i think the answer to the pearl one would be 5000, 2500 of white, and 2500 of black. Since you said they must be equal in draw, this game is impossible to win entirely because the amount of pearls begins uneven, but the way to win maximum money is to draw 5000. (hopefully you don't have to draw them one at a time, otherwise this wouldn't work. :D)

NuclearGearhead  Sep 26, 2014
 #65
avatar+24 
0

well with the game host question then for starters i belive they are rigged of all technicallities then they wouldnt keep the money in the box plus you said decided to play a trck so theres no money in the box you picked so thats out of the question so then if youve watched the show alot go for the mllion if not still go for the million if you have nothing to lose so yeah thers my answer

thelonewolf  Oct 30, 2014
 #66
avatar+107 
0

Actually, if you switch, your probability of getting the million dollars, pounds, etc. increases

Dragontmaermaths  Nov 14, 2014
 #67
avatar+271 
+5

Since I do not have time for a new puzzle, I'll post something which people can comment on. 

Steve flips a coin, 
If tail appears in the first flip, you receive $2, if tail appears in the second flip, you receive $4, if tail appears in the third flip, you receive $8, if tail appears in the fourth flip, you receive $16, etc. The money will keep on doubling until you hit a tail. 

How much would you be willing to pay to play this game?

I am not sure what is less and what is more. I would had paid less than normal because there is a tiny bit of chance on landing on its side.

Pyramid  Nov 18, 2014
 #68
avatar+92781 
0

Thanks for this Pyramid:)

how much are you gambling each time?

If you get $4 does that mean that you get $4 plus the amount that you gambled back again?

I hate to nit pick but that is what mathematicians are good at       :)

Melody  Nov 18, 2014
 #69
avatar+7188 
0

Less.......you have a 50-50 on landing on tails.......

happy7  Nov 18, 2014
 #70
avatar+271 
0

Melody:

Thanks for this Pyramid:)

how much are you gambling each time?

If you get $4 does that mean that you get $4 plus the amount that you gambled back again?

I hate to nit pick but that is what mathematicians are good at       :)

People say I have strange ideas. So, I'd rather not know how the coin will land. If the chance of landing on its side is so small that I have to make a difference less than a cent, I'll round to the nearest cent. I had a think. I guess the normal price is 50c, because if you win, you'll get more money and if you lose, you won't waste much money.

Pyramid  Nov 19, 2014
 #71
avatar+271 
0

Anonymous:

I'm in 6th grade 2

Guess what? I'm in year 2 (I wish if I can be in year 7 because as soon as I do year 2 maths, I get bored).

Pyramid  Nov 19, 2014

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