Three persons A, B and C, in order, cut a pack of cards replacing them after each cut on the condition that the first who cuts a spade shall win the prize. Find their respective chances.

Answer

`16//37,12//37,9//37`

Explanation

Let p be the chance of cutting a spade and q be the chance for not cutting a spade from the pack of 52 cards.
Then `p=13//52=1//4and q=1-1//4=3//4.`
Now, A will win a proze if the cuts spade at `1^(st),4^(th),7^(th),10^(th)` turns etc. Note that A will get a second chance if A , B and C all fail to cut spade once ane then A cuts a spade at the `4^(th)` turn.
Similarly he will cut a spade at the `7^(th)` turn when each of A, B, C fail to cut spade twice etc. Hence A's chance of winning the prize
`=(p)/(1-q^(3))=((1)/(4))/(1-((3)/(4))^(3))=(16)/(37)`
Similarly, B's chance
`=(qp+q^(4)p+q^(7)p+...)`
`=q(p+q^(3)p+q^(6)p)=12//37`
`and C's"chance"=1-(16//37)-(12//37)=9//37`