# Probability of each of the three Christmas puddings having exactly 2 coins

A cook makes plum puddings for Christmas. He stirs 6 coins into the pudding mixture thoroughly before dividing it into three equal portions. What is the probability there are 2 coins in each pudding?

The answer is 10/81, but I cannot work out how to arrive at this answer.

What I worked out so far:

A = pudding A has exactly 2 coins

$P(A) = P(B) = P(C) = {6\choose 2}(1/3)^2(2/3)^4$

$P(A~\text{or}~B~\text{or}~C) = 3P(A) - 2P(A~\text{and}~B~\text{and}~C)$ [because $P(A~\text{and}~B) = P(A~\text{and}~B~\text{and}~C)$]

so

$P(A~\text{and}~B~\text{and}~C) = 3P(A)/2P(A~\text{or}~B~\text{or}~C)$

so I guess I just need to figure out what $P(A~\text{or}~B~\text{or}~C)$ is.

• What is the distribution of coins in pudding? Does the centrifugal force of mixing send them flying out to the extremum of the pud, or does their greater density cause them to concentrate to the bottom? I confess my physics skillz are a bit rusty, but so may be the coins. Dec 20, 2016 at 14:53

Call the number of pieces in each section $A$, $B$, and $C$. Because $A+B+C=6$, you are interested in $Pr(A=2, B=2) = Pr(B=2|A=2)Pr(A=2)$.

$Pr(A=2)$ is a simple binomial calculation: $A\sim Binom(6, 1/3)$, so $Pr(A=2) = {6\choose 2}(1/3)^2(2/3)^4 = 80/243$.

Conditioned on $A$ having two pieces, $B\sim Binom(4, 1/2)$, so $Pr(B=2|A=2) = {4\choose 2}(1/2)^4 = 3/8$.

Multiplying these together, we conclude that $Pr(A=2, B=2) = 10/81$.

• Thanks for this, because technically the question I wanted to answer didn't require us to know the multinomial distribution. Dec 19, 2016 at 2:29
• There is an implicit assumption that the mixing places the coins uniformly in the pudding before division. Dec 19, 2016 at 3:22

You shouldn't use the binomial distribution here as it is a multinominal distribution problem (a generalization of the binomial).

So let's gather what we have:

n = 6 (total number of events)
n1 = 2 in part 1 (pudding #1)
n2 = 2 in part 2 (pudding #2)
n3 = 2 in part 3 (pudding #3)
p1 = 2/6 (probability to get 2 from n1)
p2 = 2/6 (probability to get 2 from n2)
p3 = 2/6 (probability to get 2 from n3)


the formula goes like this:

so let's put the numbers in motion:

$p = \frac{6!}{(2!*2!*2!)} *(\frac{2}{6})^2 * (\frac{2}{6})^2 * (\frac{2}{6})^2 = 0.12345$

and we get

10/81