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# Dice anyone?

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What is the total, between 12 and 72,  of rolling a dozen dice that will give the greatest probability of occurrence? Secondly, what is that probability?
Thank you very much for help.

Jul 21, 2018

### 1+0 Answers

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The first part is easy enough. Will simply take the average of the 12 dice as follows:[1 + 2 + 3 + 4 + 5 + 6=21/6 =3.5 x 12 =42 - this is the most likely total for rolling 12 dice, or the expected total.]

Now, the second part, however, is much more difficult to calculate. There is a rather complicated formula that will calculate the total, and it comes from this page of "Mathword": http://mathworld.wolfram.com/Dice.html {Equations 7, 8, 9, 10}, where s = number of sides on each die = 6, n = number of dice = 12, p = total of dice rolled = 42, k=number of summations =[ (42 - 12)/6] =0 to 5.
∑(-1)^k*bin(12, k)*bin[(42 - 6k - 1), (12 -1)], k=0 to [(42 - 12)/6]
I entered the above into my computer and it summed them up pretty quickly: = 144,840,476 / 6^12 =0.0665387960.......etc., which is the probability of a total of 42.

Note: The above number of 144,840,476 can also be obtained, as the coefficient of x^42 in the expansion of this polynomial: [x + x^2 + x^3 + x^4 + x^5 + x^6]^12.

Jul 21, 2018
edited by Guest  Jul 21, 2018