Q. 3.28

Question

A total of $2 n$ 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?

Step-by-Step Solution

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Answer

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

When $n \to \infty$ aces are independent $lim_{n \to \infty} P_{n} = \frac{1}{3}$

1Step 1: 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 $(\begin{matrix} 2 n \\ n \end{matrix})$ cards.

For in an occurrence $A$ that comprises $k (A)$ 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:

$P (A) = \frac{k (A)}{(\begin{matrix} 2 n \\ n \end{matrix})}$

$P (A_{0}) = \frac{(\begin{matrix} 2 n - 2 \\ n \end{matrix})}{(\begin{matrix} 2 n \\ n \end{matrix})} = \frac{n (n - 1)}{2 n (2 n - 1)} = \frac{n - 1}{2 (2 n - 1)}$

$P (A_{2}) = \frac{(\begin{matrix} 2 n - 2 \\ n - 2 \end{matrix})}{(\begin{matrix} 2 n \\ n \end{matrix})} = \frac{(\begin{matrix} 2 n - 2 \\ n \end{matrix})}{(\begin{matrix} 2 n \\ n \end{matrix})} = \frac{n - 1}{2 (2 n - 1)}$

$P (A_{1}) = 1 - P (A_{0}) - P (A_{2}) = 1 - 2 \frac{n - 1}{2 (2 n - 1)} = \frac{2 n - 1 - (n - 1)}{2 n - 1} = \frac{n}{2 n - 1}$

$P (A_{0} \cup A_{1} \cup A_{2}) = 1, A_{i} \cap A_{j} = \emptyset$

2Step 2: Probabilities Stated

$I)$ Computed $P (A_{2} ∣ A_{0}^{c})$

$A_{2} \Leftrightarrow$ The second player seems to own no aces.

$A_{0}^{c} \Leftrightarrow$ The primary player possesses a minimum one ace.

Likelihood function is formulated as having:

$P (A_{2} ∣ A_{0}^{c}) = \frac{P (A_{2} \cap A_{0}^{c})}{P (A_{0}^{c})}$

$\begin{matrix} A_{2} \cap A_{0}^{c} = A_{2} \\ & \\ P (A_{0}^{c}) = 1 - P (A_{0}) \end{matrix}$

$P (A_{2} ∣ A_{0}^{c}) = \frac{P (A_{2})}{1 - P (A_{0})} = \frac{\frac{n - 1}{2 (2 n - 1)}}{1 - \frac{n - 1}{2 (2 n - 1)}} = \frac{n - 1}{3 n - 1}$

3Step 3: Independent Aces

$I I) a) n = 2$

$P_{2} (A_{2} ∣ A_{0}^{c}) = \frac{2 - 1}{3 \times 2 - 1} = \frac{1}{5}$

$b) n = 10$

$P_{2} (A_{2} ∣ A_{0}^{c}) = \frac{10 - 1}{3 \times 10 - 1} = \frac{9}{29}$

$c) n = 100$

$P_{2} (A_{2} ∣ A_{0}^{c}) = \frac{100 - 1}{3 \times 100 - 1} = \frac{99}{299}$

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.

$lim_{n \to \infty} P_{n} (A_{2} ∣ A_{0}^{c}) = lim_{n \to \infty} \frac{n - 1}{3 n - 1} = \frac{1}{3}$

Q. 3.29

Q. 3.27

Other exercises in this chapter

Q. 3.30

Show that for any events Eand F, P(E∣E∪F)≥P(E∣F)Hint: Compute P(E∣E∪F) by conditioning on whether F occurs.

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Q. 3.29

There are n distinct types of coupons, and each coupon obtained is, independently of prior types collected, of type i with probability pi,∑i=1n&

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Q. 3.27

An urn initially contains 1 red and 1 blue ball. At each stage, a ball is randomly withdrawn and replaced by two other balls of the same color. (For i

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Q. 3.26

show that P(A∣B)=1, then PBc∣Ac=1. P(A∣B)=1, then PBc∣Ac=1

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