# Which sum shows up most often when you roll 10 dice?

Which sum shows up most often when you roll 10 dice?

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Please turn off caps lock when you write the title. What have you tried-do you know the answer for 2 dice? –  Ross Millikan Feb 10 '11 at 15:06
please note that this is at least the fourth time in the recent days another member of the site has requested that you not to use all caps in the question title. Consider this your moderator warning. Any further abuse will result in graduated suspension from the site. –  Willie Wong Feb 10 '11 at 15:19
How many sides do your dice have? –  user489 Feb 10 '11 at 18:47

The sum that shows up most often when you roll 10 dice is 35.

Are you also interested to see a reason why this is so?

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I am interested –  ajbeaven Jun 22 '11 at 22:26
@ajbeaven: I think Yuval Filmus suggestion for a proof is very good. See also this question. –  Rasmus Jun 23 '11 at 20:52

The key to this question is the concept of unimodality (q.v.).

Try to prove by induction that if you take the sum of $n \geq 2$ cubes then the distribution has the following form:

• It is supported (has non-zero probability) on the numbers $n, \ldots, 6n$.
• It is symmetric around $3.5n$.
• It is unimodal: the sequence of probabilities increases monotonically until it reaches its mode (the most probable value), then decreases monotonically.
• If $n$ is even then the mode is unique (and so $3.5n$).
• If $n$ is odd then there are two modes ($\lfloor 3.5n \rfloor$ and $\lceil 3.5n \rceil$).

If you find the proof difficult, try again with two-sided dice.

Of course, as $n \rightarrow \infty$ the distribution will approach (in some sense) a Gaussian, but I'm not sure this is enough to prove anything specific about the mode (other than its asymptotics).

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For each individual die, its 'expected value' is 7/2. This is because it can be either 1, 2, 3, 4, 5 or 6, each with a probability of 1/6 (assuming it's a fair die). If you want to know what value will come up most often for 1 die, this is the sum of each of it's possible values times the probability of that value. Which means 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 1/6(1 + 2 + 3 + 4 + 5 + 6) = 7/2. Obviously, 3.5 will never show up on a die, and this makes sense because if each side has equal probability, you shouldn't expect to see one side more than the others.

To find the expected value of multiple die, you can simply add up each individual expected value. For all 10 dice, this value is 7/2, so you have 7/2 * 10 = 35

Thus, the expected value of rolling 10 dice is 35.

Wikipedia has more information on probability and expected values, if it interests you:

http://en.wikipedia.org/wiki/Expected_value

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Expected value != Most likely value. –  Aryabhata Mar 25 '11 at 17:16
hmm, would you mind explaining how to go about solving this problem then? I'm curious now! –  mozi Mar 25 '11 at 17:57
Perhaps this will help: math.stackexchange.com/questions/4632/… . Computer software can easily compute the the required coefficients. There might even be a proof that 35 has the highest coefficient... –  Aryabhata Mar 25 '11 at 19:19