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My friend and I were playing a Minecraft "hunger games" server.

At the start of the game, 24 people are randomly placed on 24 different pedestals.

My friend happened to be on the pedestal right next to mine, and I was wondering what the probability of this happening was.

I'm not in probability and statistics yet, but I was hoping for a more-basic explanation of how to solve this and maybe a generalization of how to solve problems similar to this.

I'll be sure to give you a guys a thumbs up and 5 points for a good explanation!

NinjaDevo Jun 27, 2014

#11**+10 **

Assuming that your initial location doesn't matter and that you just want to know the chance of your friend being placed on a pedestal right next to you, we can do the following:

Place you in a random pedestal.

Place your friend in another of the 23 remaining pedestals.

Your friend can be put in 23 different pedestals, but only two of those pedestals will be next to you.

We have 23 events of which 2 are positive events. The chance of your friend being next to you is $${\frac{{\mathtt{2}}}{{\mathtt{23}}}}$$, which is aproximately equal to 8,7%. Basically that will happen once in every 12 rounds.

Guest Jun 27, 2014

#1**+10 **

Hi Ninja,

Probability is not mystrong suit but I'll give it a shot.

I am assuming that the pedestals are all in a line. If they are in a circle the answer would be different.

Firstly, how many ways can 24 people be put on 24 pedestals.

24 possiblilities for the 1st one

23 possibilities for the 2nd one

22 people left for the third one etc

so this is 24! ways.

Now lets think of you and your freind joined together and 2 of the pedestals jointed together. How many ways can you all be put on the pedestals now?

Well this would be 23! ways except you could be either on the left or on the right of your friend so that would be 23!*2

So I think the probability would be 2*23!/24! = 2/24=1/12

That really does not sound right to me. Like I said I am not very good at probability.

Melody Jun 27, 2014

#2**+10 **

I kinda' suck at probability questions, ND....but I'll take the plunge, here. (I'm sure someone on here will correct me, if I'm wrong!!!........LOL!!)

Let's take a simple example and then apply it to your situation. Instead of 24 people, let's just suppose that we only have four. And let's let each person be represented by a different letter (person 1 =A , person 2 =B, etc.)......Now, we can think of these people as occupying "slots."

Note that there are four letters, A B C D, and there are 24 possible arrangements of these. To see this, note that there are 4 letters possible for slot 1 and three letters possible for slot 2 and two letters possible for slot 3 and one choice for slot 4. So.....4 x 3 x 2 x 1 = 24.

And let's let you and your friend be "A" and "B"......Note, that there are 2 possible arrangements for both of you.....either AB or BA. And both of you could occupy three different positions...(positions 1,2 .....positions 2,3....or positions 3,4). And note that, the other two people "C" and "D," can be arranged in 2 possible ways for each arrangement.

So, for instance....let's see what it looks like when you and your friend occupy positions 1,2. We have.....

ABCD or BACD or ABDC or BADC......so, you and your friend are arranged in two possible ways and the other two people are arranged in two possible ways. And since you and your friend could occupy any of three positions, the total arrangements possible where you and your friend are next to each other is given by....

(2 ways for AB) x (2 ways for CD) x (3 possible positions occupied by you and your friend) = 12.

So the probability that you and your friend are next to each other when four people are involved =

(12 arrangements) / (the total arrangements) = 1/2

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

So applying the same reasoning to 24 people...we have 2 arrangements possible for you and your friend in each position x 23 possible positions that you both could occupy x 22! arrangements for the other people in each arrangement. And the total arrangements possible are 24!

So we have [(2) x 23 x 22!] / 24! = 1/12

Same as Melody, I believe......however, don't bet your bank account on this!!....we've been known to "b**w a fuse" before .....!!!

CPhill Jun 27, 2014

#4**+5 **

Yeah....but remember...we thought that about the red/white card problem, too...(as I recall)....!!!

LOL!!!!

CPhill Jun 27, 2014

#5**0 **

Yes i know- My comment was tongue in cheek.

It is most likely a lot more common for 2 stupid people to think alike.

I wonder what the probablilities are ?

Melody Jun 27, 2014

#6**0 **

Well....I wouldn't term either of us as "stupid." Maybe probability just isn't one of our "stronger" points.....

Doesn't really matter.......if we're correct, someone will let us know (high "probability" of this)...and same "probabilty" if we're not......!!!

CPhill Jun 27, 2014

#7**0 **

I wasn't talking personally Chris.

There are more people in the world than just us you know!

Melody Jun 27, 2014

#8**+10 **

Here's my reasoning:

If you are on pedestal 1 the probability of your friend being next to you is 1/23

Same if you are on pedestal 24

If you are on any of the other 22 pedestals the probability is 2/23

Hence the overall weighted probability is (2/24)*(1/23) + (22/24)*(2/23)

$$\left({\frac{{\mathtt{2}}}{{\mathtt{24}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{23}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{22}}}{{\mathtt{24}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{23}}}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{12}}}} = {\mathtt{0.083\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

.Alan Jun 27, 2014

#10**+8 **

Melody if you are saying " WE all are GENIUSES" then this means all of us so i am also one of those "all of us" so are u calling me also a genius or just you,CPhll and Alan are GENIUSES!if u r calling me a genius (too) then thank you very much!

rosala Jun 27, 2014

#11**+10 **

Best Answer

Assuming that your initial location doesn't matter and that you just want to know the chance of your friend being placed on a pedestal right next to you, we can do the following:

Place you in a random pedestal.

Place your friend in another of the 23 remaining pedestals.

Your friend can be put in 23 different pedestals, but only two of those pedestals will be next to you.

We have 23 events of which 2 are positive events. The chance of your friend being next to you is $${\frac{{\mathtt{2}}}{{\mathtt{23}}}}$$, which is aproximately equal to 8,7%. Basically that will happen once in every 12 rounds.

Guest Jun 27, 2014

#12**+5 **

OK.....I guess all of us can put ourselves on pedestals .....!!!!

After all....maybe we can't ALL be smart eggs.....but at least we're "colorful"......!!!!!!

CPhill Jun 27, 2014

#13**0 **

Thanks for your imput guys.

I forgot to meantion, this was a circle of pedestals, I think some of you might have assumed that this was a line. (sorry about that)

I didn't really need a completely accurate answer either, just was a bit curious.

Points are awarded where they're due. :)

NinjaDevo Jun 27, 2014

#14**0 **

Actually...I'm starting to think that anonymous's answer hit it right on the dot.

I believe he assumed this was a circle, which would give you a little more of a chance of being next to your friend.

That's for your imput anonymous. You should create an account here, you seem to have some nice answers in your brain. :)

NinjaDevo Jun 27, 2014

#15**0 **

I think if it is a circle you drop everything down by one because there is no specific endpoint.

So the original answer was 2*23!/24! = 2/24=1/12

and this becomes 2*22!/23! = 2/23

so yes I think anonymous is right also and

Of course Rosala is correct. she is a genious too!!!!

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Melody Jun 28, 2014