KCC's Quizzes AQQ287 about throwing 3 dice

With 3 dice, the range of sums will be from 3 to 18 and the most common sums will be 10 and 11 with equal probability of 27 times out of 216 - P(10) = P(11) = 27/216 = 0.125.

With n dice, the probability of rolling a particular number on any die is the same and is 1/s where s is the number of sides so 1/6 in this case.  For two dice or more, the number of permutations is given by s^n = 6^2 = 36 for two dice rolls and 6^3 = 216 for three dice rolls.  The sum of the rolls follow a gaussian distribution as there are 6 ways of getting 7 with a 2 rolls but only one way of getting 2 or 12.  The range of values will be integers from n to 6n for a 6-sided dice and the most probable sum will be median value (or two values if the number of dice is odd).

1. Mean value of a dice is (1+2+3+4+5+6)/6 = 3.5

To know the value S where probability P(S) is the maximum, it must be in the mean value.

Here for 3 dice, mean value is 3.5*3 = 10.5.

So the value S where probability P(S) is the maximum are 10 and 11 (same probability).

2. For n number of dice --> S for which P(S) is maximum = 3.5*n.

It took me a while to figure out the solution, because I initially (and mistakenly) thought the average value of a die was 6 divided by 2. Maybe my background in electronics is to blame... As we always start counting from 0! 

Thanks for this nice puzzle KC !

Great answers and reasoning MTF_Walker ! Thus for n dice, the median value for the sum is n+((6n-2)/2 or 7n/2. If n s even it is just 7n/2 and 7/n/2 +/-0.5 if n is odd. But what is the probability to get that sum? I have to confess I don( have the answer...

Congratulations FlorentDENIS ! The most frequent sum is indeed 3.5 n with n the number of dice. 3.5 n if n is even and 3.5 n +/-0.5 if n is odd. You are in fact nearly there! The only difficult point is what is the probability? I confess I on't have the answer; and count on a miraculous participant...

1) Enumeration of the results of each throw gives:
    1 way  for 3
    3 ways for 4
    6 ways for 5
    10 ways for 6
    15 ways for 7
    21 ways for 8
    25 ways for 9
    27 ways for 10
    27 ways for 11
    25 ways for 12
    21 ways for 13
    15 ways for 14
    10 ways for 15
    6 ways for 16
    3 ways for 17
    1 ways for 18
Two numbers, 10 and 11, have the highest count of 27 so they have the highest chance of occurring. P(S) => 27/216=0.125

2) For n dice the number(s) with the highest count is(are) at the mid range of S. This gives the most range of selection to equal S.
   Min count = n, Max count = 6n
   Mid range => (6n-n)/2 + n = 3.5n
   S = floor(3.5n) and ceil(3.5n); Both numbers are equal when n is an even integer.
   Example: S=7 for 2 dice and S=10 and 11 for 3 dice.

Excellent retiredEE  and with a solid and clear development, congratulations!

1. When you throw 3 dice, max probability is for S=10, 11. You can see this by calculating all the occurrences or by using excel or (better) noting that the average value (expectation value) is 3x (avg outcome 1 dice) = 3x 3.5=10.5 (in the limit of high number of rolls). Since you cannot get a fractional number, max probability will be for the closest integers 10 and 11. The avg outcome of 1 dice is (1+2+3+4+5+6)/6=3.5.

2. Similarly, for n dice you expect the average outcome= 3.5*n being the most probable outcome.

Yes, I agree with your inputs! The relation 3.5*n with n the number of dice can be easily verified for n up to 3. The only pending question is what is the probability (thus the number of times the 3.5 * n result will appear. As I mentioned earlier: I don't have that answer...

You have a a discrete probability distribution so you need to calculate the "positive" number of cases and divide it by the total number of cases. For n dice, total number of cases is 6ⁿ. Difficult part is to calculate the number of positive cases. Be G(x)= x¹ + x² + x³ +x⁴ +x⁵ +x⁶ the polynomial function associated with a single dice. If you have n dice, the total number of ways you can get 3.5*n is the coefficient of x^3.5*n  when you, or most probably an algorithm or a tool like wolfram alpha, expand [G(x)]ⁿ (let's assume 3.5*n is an integer, otherwise we follow same process for the closest integers).

For example, suppose we want to know the probability we get S=35 by rolling 10 dice. By letting wolfram alpha to expand [G(x)]10 , you get 4395256 as a coefficient for x³⁵. So the probability to get S=35 by rolling 10 dice is P(S=35)= 4395256 / 6¹⁰ = 0.0727 or something like 7.3%. 

The max probability will depend on n. I calculated the probability to get S=3.5*n for n=20 and the result is about 5.2%. On the other hand, the probability to get 3.5*n with n=30 is about 4.2%. So the higher the value of n, the lower the max probability, as you might expect by considering that  when n grows, you can get much more different sum results.   

Wow, I am impressed gpiarino , I was totally ignorant about the wolfram alpha technique. Will dig on it when during my summer vacation... Anywau, many thanks!