Two Probability Exercises

Exercise 1

A and B are playing a game (e.g. table tennis), which comprises 5 sets; whoever firstly wins 3 sets is the winner. We make two assumptions: (a) The probability that A wins a set is $p$ ($ 0 < p < 1 $), and the probability that B wins a set is $ 1-p $; (b) the 5 sets are independent. What is the probability that A wins the game?

Solution. Let $X$ be the number of sets that A wins in a game. Then, $X$ has the binomial distribution $ B(5,\ p) $. The probability that A wins the game is

\begin{align} \hbox{Pr}(X\ge 3) & =\sum_{i = 3}^5 {5 \choose i}p^i(1-p)^{5-i} \\ & = 6p^5 -15p^4 + 10p^3. \end{align}

I asked myself this question: Is this winning probability $\hbox{Pr}(X\ge 3)$ greater or less than $p$? To find out the answer to this question, I plot the winning probability against $p$ and the 45-degree line:

n_out_of_2n_minus_1 <- function(n, p)
{m <-  2*n - 1
 s <- 0
 for(j in n:m) {
   s <- s + choose(m, j) * p^j * (1-p)^(m-j)
 }
 return(s)
}

x <- seq(0.001, 0.999, length.out = 1000)
y <- n_out_of_2n_minus_1(n = 3, p = x)

plot(x, y, type = 'l', xlab = 'p', ylab = 'Winning probability')
abline(a = 0, b = 1, col = 'red', lty = 2)
abline(v = 0.5, col = 'red', lty = 2)  

The above plot shows that: the winning probability is

• $ > p $, if $ p > 0.5 $,
• $ = p $, if $ p = 0.5 $,
• $ < p $, if $ p < 0.5 $.

This is interesting!

Exercise 2

Two gamblers, A and B, are playing a game of 13 rounds. The rule is: Whoever firstly wins 7 rounds will collect the entire prize. All the rounds are independent; the probability that A wins a round is $p$ ($ 0 < p < 1 $), and the probability that B wins a round is $ 1-p $. The following situation is of interest: After 9 rounds, A has won 5 rounds and B has won 4 rounds. However, for some reason, they must stop the game here. How do they divide the prize?

So, here I am talking about the famous problem of points (AKA division of the stakes).This wikipedia article provides a solution. The key is to figure out A’s winning probability if they carry on the game until round 13. Now, I want to find A’s winning probability with a Markov chain approach.

The state space is $ { 5, 6, 7+ } $ — each state is a count of how many rounds that A has won. The initial distribution is:

$$ \left( \begin{array}{lll} 5 & 6 & 7+\\ 1 & 0 & 0 \end{array} \right). $$

The transition matrix is

$$ {\boldsymbol P} = \left[ \begin{array}{ccc} 1-p & p & 0 \\ 0 & 1-p & p \\ 0 & 0 & 1 \end{array} \right] $$

Note that if they carry on the game A’s winning probability is given by

$$ [1,\ 0,\ 0]{\boldsymbol P}^4 {\boldsymbol e}_3, $$

where $ {\boldsymbol e}_3^{T} = [0,\ 0,\ 1] $. It’s easy to find

$$ {\boldsymbol P}^2 = \left[ \begin{array}{ccc} (1-p)^2 & 2p(1-p) & p^2\\ 0 & (1-p)^2 & 2p-p^2 \\ 0 & 0 & 1 \end{array} \right] $$

Therefore,

$$ [1,\ 0,\ 0]{\boldsymbol P}^4 {\boldsymbol e}_3 = [(1-p)^2,\ 2p(1-p),\ p^2] \left[ \begin{array}{c} p^2\\ 2p-p^2\\ 1 \end{array} \right], $$

which is equal to $ 3p^4 - 8p^3 + 6p^2 $.