Difficult Math Problem #87 - Probability

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800guy
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Topic: Difficult Math Problem #87 - Probability
PostWed Jan 17, 2007 11:33 am
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As a part of a game, four people each much secretly chose an integer between 1 and 4 inclusive. What is the approximate likelihood that all four people will chose different numbers?

this question comes from the difficult math doc on this website, OA coming after some people answer.
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maxim730
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PostWed Jan 17, 2007 3:14 pm
4!?
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sumitkrishna
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PostThu Jan 18, 2007 9:08 pm
1/24 ???
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sarika.k
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PostThu Jan 18, 2007 9:36 pm
3/32 ?
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aim-wsc
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PostFri Jan 19, 2007 1:56 am
c'mon guys come up with explanation.

why do you think so?
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thankont
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PostFri Jan 19, 2007 3:45 am
Choosing the same number is (1/4)^4 since say our chosen # is 2 then prob. of first person choosing 2 is 1/4 etc.
Now here we want different numbers so let first person choose 1 with prob. (1/4), second chooses number 2 with p = 1/4 so I think that probability is the same with the first case and p=(1/4)^4
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800guy
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PostFri Jan 19, 2007 9:31 am
i second aim-wsc's comments..the purpose of sharing these questions is to get more discussion around explanations, sharing of knowledge.

here's the OA:

The probability that the first person will pick unique number is 1 (obviously) then the probability for the second is 3/4 since one number is already picked by the first, then similarly the probabilities for the 3rd and 4th are 1/2 and 1/4 respectively. Their product 3/4*1/2*1/4 = 3/32
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