## Playing With an Unfair Dice

One of the gamblers is not playing fair however. They are playing with a dice designed to throw a sixth one quarter of the time. What is the probability of the cheating gambler winning the game?

The number of sixes thrown by each player may be modelled by a binomial distribution.

The cheating gamblers throws sixes modelled by

\[X \sim B(5,1/4)\]

.The other gambler scores sixes modelled by a biomia distribution

\[Y \sim B(51/6)\]

.The cheat will win if he throws six and the other player throws no sixes:

\[P(X=1) \times P(Y=0)= 0.6328 \times 0.4019=0.2543\]

The cheat will win if he throws two sixes and the other player throws at most one six:

\[P(X=2) \times P(Y \le 1)= 0.2637 \times 0.8038=0.2119\]

The cheat will win if he throws three sixes and the other player throws at most two sixes:

\[P(X=3) \times P(Y \le 2)= 0.0879 \times 0.9645=0.0848\]

The cheat will win if he throws four sixes and the other player throws at most three sixes:

\[P(X=4) \times P(Y \le 3)= 0.0146 \times 0.9967=0.0146\]

The cheat will win if he throws 5 sixes and the other player throws at most four sixes:

\[P(X=2) \times P(Y \le 1)= 0.001 \times 0.9999=0.001\]

Adding these gives 0.5666