Alf and Ben play a game. They take turns to select a card at random from a pack of 52 shuffled cards. If it is an ace they win. If not they replace it in the pack. Alf draws the first card:
a) what is the probability that Alf wins on his 3rd go?
b) Show that the prob. that Alf wins prior to his (n+1)th go is = 13/25(1-(12/13)^2n)

Question

Alf and Ben play a game. They take turns to select a card at random from a pack of 52 shuffled cards. If it is an ace they win. If not they replace it in the pack. Alf draws the first card: 
a) what is the probability that Alf wins on his 3rd go?
b) Show that the prob. that Alf wins prior to his (n+1)th go is = 13/25(1-(12/13)^2n)

dumbcow · Answer

this is a geometric distribution http://en.wikipedia.org/wiki/Geometric_distribution in this case prob of picking an ace is 1/13 $$P(X=k) = (\frac{12}{13})^{k-1} (\frac{1}{13})$$ where k is num of turns since we only want Alfs turns, these are all the odd numbered turns .... 1,3,5,7... k = 2n - 1 $$P(X=n) = (\frac{12}{13})^{2n-2} (\frac{1}{13}) $$ to find prob of Alf winning on his 3rd go, plug in n=3