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Probability of an odd number of 6’s when tossing a bunch of dice

January 16, 2018 By Gary Ernest Davis

Alexander Bogomolny (@CutTheKnotMath) has a post on his website Cut The Knot dealing with the probability of an odd (or of an even) number of 6’s when tossing a bunch of fair dice:

The answer, at least to this bear with a small brain, was initially confusing: there is, at least for a small number of dice – say 7 or so – a somewhat greater chance of getting an even number of 6’s than an odd number.

My confusion lifted when I plotted the result and remembered that, for the purposes of this problem, 0 is a perfectly good even number.

Here is the solution given on Alexander’s post prior to my tweeting him a plot (now shown on his web page):

Here’s the plot for P (in blue) and Q (in red):

So throwing no die does give us an even number of 6’s, namely 0 6’s, with probability 1.


We might, in a somewhat devilish mood, ask: what is the probability of getting a number of 6’s that leaves a remainder of 0 (or 1 or 2) mod 3 when we toss n dice?

When k = 3p is a multiple of 3 the probability of k 6s when tossing n dice is

P_0(n) =\sum_{k\equiv 0(\textrm{mod }3)}{n\choose k}(\frac{5}{6})^{n-k}(\frac{1}{6})^k

= sum of every 3^{rd} term in the binomial expansion of (a+b)^n where a=\frac{5}{6}, b=\frac{1}{6}

= \sum_{p=0}^{\lfloor n/3\rfloor}{n\choose 3p}(\frac{5}{6})^{n-3p}(\frac{1}{6})^{3p}

We can compute these probabilities in Mathematica as follows:

Prob[0,n_]:=Total[Table[Binomial[n,3*p]*(5/6)^(n-3*p)*(1/6)^(3*p),{p,0,Floor[n/3]}]]

with the result (for 0\leq n \leq 40:

Unlike the odd/even cases, these probabilities are not a monotone function of n, and decrease until n =12 in which case they then converge, as we expect, to 1/3 (but not monotonically).

The value of Prob(0,500), accurate to 60 decimal places, is

0.333333333333333333333333333333333333333333333333333333333347

Here are more details:

We can calculate the probability that the number of 6s is of the form 3p+1 or 3p+2 similarly, and here is a plot of those probabilities for n ranging from 0 though 30:

The probabilities converge, as we expect, on 1/3 as n increases.

http://www.crikeymath.com/wp-content/uploads/2017/09/crikey.wav

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