Gambling at Monte Carlo
Monte Carlo is famous not only for motor racing, but also for its large number of casinos.
A man went to Monte Carlo to try and make his fortune. Whilst he was there he had an opportunity to bet on the outcome of rolling dice. He was offered the same odds for each of the following outcomes:
- At least 1 six with 6 dice.
- At least 2 sixes with 12 dice.
- At least 3 sixes with 18 dice.
The man decided that these were all the same so he could choose one at random and bet on that. Was he right?
At first you may think that all the probabilities are the same, but consider a similar problem about tossing coins. The probability of one head with one coin is 1/2, the probability of two heads with two coins is 1/4 and the probability of three heads with three coins is 1/8, definitely not the same!
The problem given was difficult and it might be best understood by considering a similar, but simpler, problem.
Which is more likely, at least 1 six from rolling two dice or at least 2 sixes from rolling 4 dice?
To find the solution we use the fact that the probabilities must sum to 1:
The probability of at least 1 six from 2 dice = 1 - probability of no sixes from 2 dice
\begin{align*}
&= 1 - \left(\frac{5}{6}\right)^2 \\
&= 1 - \frac{25}{36} \\
&= \frac{11}{36} \\
&= 0.306\quad\text{(3.d.p.)}
\end{align*}
The probability of at least 2 sixes from 4 dice = 1 - probability of no sixes from 4 dice - probability of 1 six from 4 dice
\begin{align*}
&= 1 - \left(\frac{5}{6}\right)^{4} - 12\times\frac{1}{6}\times\left(\frac{5}{6}\right)^{3} \\
&= 1 - \frac{625}{1296} - \frac{500}{1296} \\
& = \frac{171}{1296} \\
&= 0.132\quad\text{(3.d.p.)}
\end{align*}
A similar argument for the original problem follows, where $\mathbb{P}(\text{something})$ means 'the probability of something occurring'.
\begin{align*}
\mathbb{P}(\text{at least 1 six from 6 dice}) &= 1 - \left(\frac{5}{6}\right)^6 \\
&= 0.665\quad\text{(3.d.p.)}
\end{align*}
\begin{align*}
\mathbb{P}(\text{at least 2 sixes from 12 dice}) &= 1 - \left(\frac{5}{6}\right)^{12} - 12\times\frac{1}{6}\times\left(\frac{5}{6}\right)^{11} \\
&= 0.619\quad\text{(3.d.p.)}
\end{align*}
\begin{align*}
\mathbb{P}(\text{at least 3 sixes from 18 dice}) &= 1 - \left(\frac{5}{6}\right)^{18} - 18\times\frac{1}{6}\times\left(\frac{5}{6}\right)^{17} - 153\times\left(\frac{1}{6}\right)^2\times\left(\frac{5}{6}\right)^{16}\\
&= 0.597\quad\text{(3.d.p.)}
\end{align*}
The person gambling at Monte Carlo would have gained the best advantage from choosing to gamble on "at least 1 six from six dice". This had the highest probability.