## Playing With an Unfair Dice

Suppose two gamblers are playing dice. Each throws the dice 5 times, and the winner is the one with most sixes.
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

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