Abrazolica

Samuel Pepys who was a president of the Royal Society, and apparently a gambling man, once asked Isaac Newton which of the following events had the highest probability:

1. Throwing 6 dice and getting at least one six.
2. Throwing 12 dice and getting at least two sixes.
3. Throwing 18 dice and getting at least three sixes.

Pepys thought C had the highest probability but Newton set him straight and told him A had the highest probability. Probability has had over 300 years to mature since Newton looked at this problem so today it's not a hard problem to solve. What makes it interesting is that you can solve it in two seemingly different ways. One way is more combinatorial where you find the number of favorable outcomes and divide by the number of possible outcomes to get the probability. The other way is purely probabilistic and involves turning the problem into an equivalent biased coin problem.

First the combinatorial approach. Each of the dice have 6 faces so if you throw \(N\) dice there are \(6^N\) possible outcomes. The number of outcomes with no six on any of the dice is \(5^N\). The number of outcomes with one of the dice showing a six is \(N\cdot 5^{N-1}\). In general it's not hard to see that the number of outcomes with \(k\) of the dice showing a six is \(\binom{N}{k}5^{N-k}\). Using this result the probabilities for A, B, and C can be written as a ratios of favorable to total number of outcomes.

\[ P(A)=\frac{6^6-5^6}{6^6}=\frac{31031}{46656}=0.665102 \]

\[ P(B)=\frac{6^{12}-5^{12}-12\cdot 5^{11}}{6^{12}}=\frac{1346704211}{2176782336}=0.618667 \]

\[ P(C)=\frac{6^{18}-5^{18}-18\cdot 5^{17}-153\cdot 5^{16}}{6^{18}}=\frac{15166600495229}{25389989167104}=0.597346 \]

Now for the probabilistic approach. First of all note that tossing \(N\) dice is equivalent to randomly putting \(N\) balls into 6 boxes labeled 1 to 6. You just let each ball represent one of the dice and the box it goes into is given by the value that comes up on the die. The balls have an equal probability of 1/6 for ending up in any of the boxes. Every ball has a probability of 1/6 for ending up in the box labeled 6 and a probability of 5/6 for ending up in one of the other boxes. So randomly putting a ball into the boxes is like flipping a biased coin with a probability for heads (box 6) of 1/6 and a probability for tails (boxes 1,2,3,4,5) of 5/6. If \(p=1/6\) is the probability of getting heads in a single toss then the probability of getting \(k\) heads in \(N\) tosses is given by the Binomial distribution

\[ b(N,k)=\binom{N}{k}p^k(1-p)^{N-k} \]

The probabilities for A, B, and C can now be expressed in terms of the binomial distribution

\[ P(A) = 1 - b(6,0) \]

\[ P(B) = 1 - b(12,0) - b(12,1) \]

\[ P(C) = 1 - b(18,0) - b(18,1) - b(18,2) \]

These equations are equivalent to the previous equations but they were found using purely probabilistic arguments. An advantage of this is that the problem can easily be generalized to the case of loaded dice. For example if the dice are loaded so that 6 has a higher or lower probability than 1/6 of coming up then you only need to change the value of \(p\) in the Binomial distribution. With loaded dice you cannot use the combinatorial approach since it implicitly assumes that all 6 sides of the dice are equally likely.