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Probability of dice sum just greater than 100

A die is rolled several times and the number appearing is summed. We stop when this sum becomes greater than or equal to 100. What value of sum in the end is the most probable (out of 100, 101, 102, 103, 104, 105)?

105 can be generated in only 1 way : 99 + 6

104 can be generated in only 2 ways: 99 + 5, 98 + 6

and so on.

Assuming that the dice rolling will reach either of numbers in range [94,99] is equally likely, 100 is the most likely. Am I correct?

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marked as duplicate by joriki, Mike Spivey, draks ..., Cameron Buie, Arkamis Nov 28 '12 at 20:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Related : math.stackexchange.com/questions/12433/… –  Jean-Sébastien Nov 28 '12 at 19:04
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"Assuming that the dice rolling will reach either of numbers in range [94,99] is equally likely, 100 is the most likely." That's a big assumption. –  Thomas Andrews Nov 28 '12 at 19:08

1 Answer 1

up vote 6 down vote accepted

Let an be the number of ways to get a sum of n:

a100 = a99 + a98 + a97 + a96 + a95 + a94

a101 = a99 + a98 + a97 + a96 + a95

a102 = a99 + a98 + a97 + a96

a103 = a99 + a98 + a97

a104 = a99 + a98

a105 = a99

So you can see just from this that a100 is the most likely since there are more ways to get to a sum of 100.

In plain language...

In other words, you might have the opportunity to end a game at 100 by rolling a 1, 2, 3, 4, 5, or 6, but you can only end a game at 105 by rolling a 6 on your last roll.

Sum at the penultimate roll

I disagree that sums of [94,99] are all equally likely for your penultimate roll, but it doesn't matter if they are equal or not. It's the last roll that determines the total score, which is the focus here. There are still a98 + a97 + a96 + a95 + a94 more ways to get a sum of 100 than there are to get a sum of 105, for example.

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This suppose that that getting $94,95,...99$ is equally likely, but this is not true –  Jean-Sébastien Nov 28 '12 at 19:27
    
@Jean-Sébastien: It doesn't suppose that. We have $a_{100}\gt a_{101}$ independent of the values of $a_{94}$ through $a_{99}$, simply because the difference is $a_{94}$ and hence positive. –  joriki Nov 28 '12 at 19:32
    
@Jean-Sébastien They don't have to be equally likely. There are still $a_{98}$ + $a_{97}$ + $a_{96}$ + $a_{95}$ + $a_{94}$ more ways to get 100 than there are to get 105. –  woz Nov 28 '12 at 19:33
    
Right, sorry, don't know where i got this off –  Jean-Sébastien Nov 28 '12 at 19:39

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