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) $n=100?$To what does the probability converge as n goes to infinity? Why?

Short answer

Use the probability on the equally likely set of events $-P_n=\frac{n-1}{3n-1}$

When $n\to \mathrm{\infty }$aces are independent$\underset{n\to \mathrm{\infty }}{lim} P_n=\frac{1}{3}$

Step-by-step solution

Events and Probabilities

$A_0$-The very first player has precisely$0$trumps.

$A_1$-The very first player has precisely $1$trumps.

$A_2$- The very first player has precisely $2$trumps.

Each participant receives haphazardly from a deck of $\left(\begin{array}{c}2n\\ n\end{array}\right)$cards.

For in an occurrence $A$that comprises $k\left(A\right)$distinct first field of play, there's many feasible selections for the trumps of its first player, and because all of them would be equally probable:

localid="1649669641130" $P\left(A\right)=\frac{k\left(A\right)}{\left(\begin{array}{c}2n\\ n\end{array}\right)}$

localid="1649669645656" $P\left(A_0\right)=\frac{\left(\begin{array}{c}2n-2\\ n\end{array}\right)}{\left(\begin{array}{c}2n\\ n\end{array}\right)}=\frac{n(n-1)}{2n(2n-1)}=\frac{n-1}{2(2n-1)}$

localid="1649669649939" $P\left(A_2\right)=\frac{\left(\begin{array}{c}2n-2\\ n-2\end{array}\right)}{\left(\begin{array}{c}2n\\ n\end{array}\right)}=\frac{\left(\begin{array}{c}2n-2\\ n\end{array}\right)}{\left(\begin{array}{c}2n\\ n\end{array}\right)}=\frac{n-1}{2(2n-1)}$

localid="1649669660127" $P\left(A_1\right)=1-P\left(A_0\right)-P\left(A_2\right)=1-2\frac{n-1}{2(2n-1)}=\frac{2n-1-(n-1)}{2n-1}=\frac{n}{2n-1}$

localid="1649669670663" $P\left(A_0\cup A_1\cup A_2\right)=1,A_i\cap A_j=\varnothing$

Probabilities Stated

$I)$Computed $P\left(A_2\mid A_0^c\right)$

$A_2\iff$The second player seems to own no aces.

$A_0^c\iff$The primary player possesses a minimum one ace.

Likelihood function is formulated as having:

$P\left(A_2\mid A_0^c\right)=\frac{P\left(A_2\cap A_0^c\right)}{P\left(A_0^c\right)}$

$\begin{array}{c}{A}_2\cap A_0^c=A_2\\ \&\\ P\left(A_0^c\right)=1-P\left(A_0\right)\end{array}$

$$\begin{gathered} P\left(A_2\mid A_0^c\right)=\frac{P\left(A_2\right)}{1-P\left(A_0\right)} \\ =\frac{\frac{n-1}{2(2n-1)}}{1-\frac{n-1}{2(2n-1)}} \\ =\frac{n-1}{3n-1} \end{gathered}$$

Independent Aces

$II)a)n=2$

$P_2\left(A_2\mid A_0^c\right)=\frac{2-1}{3\times 2-1}=\frac{1}{5}$

$b)n=10$

$P_2\left(A_2\mid A_0^c\right)=\frac{10-1}{3\times 10-1}=\frac{9}{29}$

$c)n=100$

$P_2\left(A_2\mid A_0^c\right)=\frac{100-1}{3\times 100-1}=\frac{99}{299}$

If $n$gets larger, dual aces will unilaterally handed either as first or second player $-4$fairly probable possibilities, in $3$in that the first player does have an ace, within the one of these aces.

$\underset{n\to \mathrm{\infty }}{lim} P_n\left(A_2\mid A_0^c\right)=\underset{n\to \mathrm{\infty }}{lim} \frac{n-1}{3n-1}=\frac{1}{3}$