Probability -Difficult

[gmatrant]

A and B pick up a card at random from a well shuffled pack of cards, one after the other, replacing it every time till one of them gets a heart. A begins the game , then the probability that the game ends with B is

1) 3/7 2)4/7 3)3/4 4)1/4

Ans : [spoiler]1)[/spoiler]

Solution: I couldnt get to the answer , but this is what I tried. Where could I be going wrong?

p(A does not pick heart) = 3/4
p(B picks heart) = 1/4

p(of B winning is) = (3/4)* (1/4)= 3/16.
But thats no where near the answer.

Please correct me where I am going wrong.

Thanks

[agni_mba]

The trick is in understanding that this game may never end. The probability of B winning will be summation of probs. associated winning in the first run, second run, third run and so and so forth.

P(B)1st run = (3/4)*1/4 = 3/16
P(B)2nd run = [No result in 1st run]* (3/16) = [9/16]*3/16
P(B)3rd run = [no result in 1st AND 2nd run]*3/16 = [9/16*9/16]*3/16
and so and so forth...

when you sum this infinite GP up you will get the answer.

cheers

[gmatrant]

The trick is in understanding that this game may never end. The probability of B winning will be summation of probs. associated winning in the first run, second run, third run and so and so forth.

P(B)1st run = (3/4)*1/4 = 3/16
P(B)2nd run = [No result in 1st run]* (3/16) = [9/16]*3/16
P(B)3rd run = [no result in 1st AND 2nd run]*3/16 = [9/16*9/16]*3/16
and so and so forth...

when you sum this infinite GP up you will get the answer.

cheers

Oh yes your are right.. great .thanks for the solution

[jangojess]

simply superb man.....where did u get this pblm from???