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reinout-g

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 #2
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PabloDons:

Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo



There are two ways in which I can interpret your question, You're either asking what the odds are of throwing one six in six throws, or you're asking the odds of throwing at least one six in six throws.

The odds of throwing one six in six throws can be calculated in the following way;

We know that there is a (1/6)th chance of throwing a six with one dice and a chance of (5/6) of not throwing a six with one dice.

So the odds of throwing for example [6] [no 6] [no 6] [no 6] [no 6] [no 6] = (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6)

However this is only the chance of throwing a six in the first throw and no six in the next 5 throws.

Another possibility would be throwing [no 6] [6] [no 6] [no 6] [no 6] [no 6] or [no 6] [no 6] [no 6] [no 6] [no 6] [6]

All these possibilities have the same probability as the one I just calculated ( (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) )

There are 6 possibilities for which in six throws you throw 1 six (in the first, second, third, fourth, fifth, or sixth throw)

So the odds of throwing 1 six in six throws is 6 * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) = 0.4019 ( = 40.19%)

These odds take into account the cases where there is only one six, and it leaves out the cases where you throw six twice, three times, or even more times

To calculate the odds where you want at least one six you need to make use of the following;

The chance of something happening = 1 - the chance of that thing NOT happening

So for example if I have 0.30 (=30%) chance of winning $1000, I have 0.70 (=70%) chance of NOT winning $1000.

Similarly, to calculate the probability of throwing at least 1 six, It is more convenient to calculate the probability of throwing less than 1 six.
Or more clearly, the probability of throwing no sixes at all.

The probability of throwing no sixes means that each dice hits the (5/6)th chance of NOT throwing a six.
Consequently the probability of NOT throwing a six = (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6) = 0.3349 (or 33.49%)

Now to find the odds of throwing at least one six, we simply need to calculate 1-0.3349 = 0.6651

So there is a 66.51% chance of throwing at least one six.

As you can see, the formulation of the question is very important in calculating the right probability
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...
Mar 5, 2014
 #8
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bump
Mar 5, 2014