Question

A total of 2n cards, of which 2 are aces, are to be randomly divided among two players, with each player receiving n cards. Each player is then to declare, in sequence, whether he or she has received any aces. What is the conditional probability that the second player has no aces, given that the first player declares in the affirmative, when (a) n = 2? (b) n = 10? (c) n = 100?

Answer 1

`P(X_s^c|X_F) =0.2`

`P(X_s^c|X_F) =0.31`

`P(X_s^c|X_F) =0.331`

Explanation:

From the given information:

Let represent
`X_F` as the first player getting an ace

Let
`X_S` to be the second player getting an ace and

`\sim X_S` as the second player not getting an ace.

So;

The probabiility of the second player not getting an ace and the first player getting an ace can be computed as;

`P(\sim X_S| X_F) = 1 - P(X_S|X_F)`

`P(X_S|X_F) = (P(X_SX_F))/(P(X_F))`

Let's determine the probability of getting an ace in the first player

i.e

`P(X_F) = 1 - P(X_F^c)`

`= 1 -((^(2n-2)_n))/((^(2n)_n))}`

`= 1 - (n-1)/(2(2n-1))`

`= (3n-1)/(4n-2) --- (1)`

To determine the probability of the second player getting an ace and the first player getting an ace.

`P(X_sX_F) = \text{ (distribute aces to both ) and (select the left over n-1 cards from 2n-2 cards}`
`P(X_sX_F) = (2(^(2n-2)C_(n-1)))/(^(2n)C_n)`

`P(X_sX_F) = (n)/(2n -1)---(2)`

`P(X_s|X_F) = (2)/(1)`

`P(X_s|X_F) = (2n)/(3n -1)`

Thus, the conditional probability that the second player has no aces, provided that the first player declares affirmative is:

`P(X_s^c|X_F) = 1- (2n)/(3n -1)`

`P(X_s^c|X_F) = (n-1)/(3n -1)`

Therefore;

for n= 2

`P(X_s^c|X_F) = (2-1)/(3(2) -1)`

`P(X_s^c|X_F) = (1)/(6 -1)`

`P(X_s^c|X_F) = (1)/(5)`

`P(X_s^c|X_F) =0.2`

for n= 10

`P(X_s^c|X_F) = (10-1)/(3(10) -1)`

`P(X_s^c|X_F) = (9)/(30 -1)`

`P(X_s^c|X_F) = (9)/(29)`

`P(X_s^c|X_F) =0.31`

for n = 100

`P(X_s^c|X_F) = (100-1)/(3(100) -1)`

`P(X_s^c|X_F) = (99)/(300 -1)`

`P(X_s^c|X_F) = (99)/(299)`

`P(X_s^c|X_F) =0.331`