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Marbles and Bags

Two bags contain different numbers of red and blue marbles. A marble is removed from one of the bags. The marble is blue. What is the probability that it was removed from bag A?

Coin Tossing Games

You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?

Win or Lose?

A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has he more money than he started with?

Gambling at Monte Carlo

Age 14 to 16
Challenge Level

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
&= 1 - \left(\frac{5}{6}\right)^2 \\
&= 1 - \frac{25}{36} \\
&= \frac{11}{36} \\
&= 0.306\quad\text{(3.d.p.)}

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
&= 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.)}

A similar argument for the original problem follows, where $\mathbb{P}(\text{something})$ means 'the probability of something occurring'.

\mathbb{P}(\text{at least 1 six from 6 dice}) &= 1 - \left(\frac{5}{6}\right)^6 \\
&= 0.665\quad\text{(3.d.p.)}

\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.)}

\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.)}

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.