Probability of winning a game in tennis?

Solution 1:

Here is my attempt, I would appreciate feedback:

Let $p$ be the probability that player A wins a single point.

(1) Player A can win after 40-0, with probability $p^4$

(2) Player A can win after 40-15 with probability ${4\choose 1}\times p^4(1-p)$

(3) Player A can win after 40-30 with probability ${5\choose 2}\times p^4(1-p)^2$

(4) Player A can reach deuce, with probability ${6\choose 3}\times p^3(1-p)^3$, and once deuce is reached, player A will win with a probability of $\frac{p^2}{1-2p(1-p)}$ (http://www.austinrochford.com/posts/2013-04-25-probability-and-deuces-in-tennis.html)

Therefore the probability player A will win the game is given by,

$$P(Win) = p^4 + {4\choose 1}\cdot p^4(1-p) + {5\choose2}\cdot p^4(1-p)^2 + {6\choose 3}\cdot \frac{p^5(1-p)^3}{1-2p(1-p)}$$

Solution 2:

Numerically, this problem is easy using Markov chains. It is a little bit tricky to create the transition matrix. I will paste some R code I created just for fun.

transition_matrix <- function(p_win) {

    half_states <- c(0, 15, 30, 40)

    state <- function(own, opp) paste0(half_states[own], '/', half_states[opp])

    states <- c(state(expand.grid(1:4, 1:4)[, 1], expand.grid(1:4, 1:4)[, 2]), 'won', 'lost')

    tm <- array(0, dim = c(length(states), length(states)))

    colnames(tm) <- states
    rownames(tm) <- states

    rex <- '^([0-9]+)/([0-9]+)$'

    for (s in states) {
        if (grepl(rex, s)) {
            own <- which(half_states == gsub(rex, '\\1', s))
            opp <- which(half_states == gsub(rex, '\\2', s))

            stopifnot(length(own) == 1, length(opp) == 1)

            if (own < 4) {
                j_win <- which(states == state(own + 1, opp))
            } else {
                if (opp < 4) {
                    j_win <- which(states == 'won')
                } else {
                    j_win <- which(states == '40/30')
                }
            }
            if (opp < 4) {
                j_lost <- which(states == state(own, opp + 1))
            } else {
                if (own < 4) {
                    j_lost <- which(states == 'lost')
                } else {
                    j_lost <- which(states == '30/40')
                }
            }

            stopifnot(length(j_win) == 1, length(j_lost) == 1)

            i  <- which(rownames(tm) == s)

            tm[i, j_win]  <- p_win
            tm[i, j_lost] <- 1 - p_win

        } else {
            if (s == 'won') tm['won', 'won'] <- 1 else tm['lost', 'lost'] <- 1
        }
    }

    tm
}

For example, transition_matrix(0.3) is:

      0/0 15/0 30/0 40/0 0/15 15/15 30/15 40/15 0/30 15/30 30/30 40/30 0/40 15/40 30/40 40/40 won lost
0/0     0  0.3  0.0  0.0  0.7   0.0   0.0   0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0 0.0  0.0
15/0    0  0.0  0.3  0.0  0.0   0.7   0.0   0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0 0.0  0.0
30/0    0  0.0  0.0  0.3  0.0   0.0   0.7   0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0 0.0  0.0
40/0    0  0.0  0.0  0.0  0.0   0.0   0.0   0.7  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0 0.3  0.0
0/15    0  0.0  0.0  0.0  0.0   0.3   0.0   0.0  0.7   0.0   0.0   0.0  0.0   0.0   0.0   0.0 0.0  0.0
15/15   0  0.0  0.0  0.0  0.0   0.0   0.3   0.0  0.0   0.7   0.0   0.0  0.0   0.0   0.0   0.0 0.0  0.0
30/15   0  0.0  0.0  0.0  0.0   0.0   0.0   0.3  0.0   0.0   0.7   0.0  0.0   0.0   0.0   0.0 0.0  0.0
40/15   0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.7  0.0   0.0   0.0   0.0 0.3  0.0
0/30    0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.3   0.0   0.0  0.7   0.0   0.0   0.0 0.0  0.0
15/30   0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.3   0.0  0.0   0.7   0.0   0.0 0.0  0.0
30/30   0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.3  0.0   0.0   0.7   0.0 0.0  0.0
40/30   0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.7 0.3  0.0
0/40    0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0  0.0   0.3   0.0   0.0 0.0  0.7
15/40   0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.3   0.0 0.0  0.7
30/40   0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.3 0.0  0.7
40/40   0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.3  0.0   0.0   0.7   0.0 0.0  0.0
won     0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0 1.0  0.0
lost    0  0.0  0.0  0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0  0.0   0.0   0.0   0.0 0.0  1.0
>

Now, what you are looking for is the result after playing infinite points. For practical purposes, multiplying the matrix by itself 20 times is like playing 2^20 == 1,048,576 points. You can expect all paths to be either 'won' or 'lost' after that many points.

for (time in 1:20) tm <- tm %*% tm

And what you get is the probability of winning for each possible initial state, not only 0/0.

> tm[1:16, c('won', 'lost')]
              won      lost
0/0   0.099211034 0.9007890
15/0  0.210981724 0.7890183
30/0  0.412168966 0.5878310
40/0  0.710224138 0.2897759
0/15  0.051309310 0.9486907
15/15 0.124758621 0.8752414
30/15 0.284431034 0.7155690
40/15 0.586034483 0.4139655
0/30  0.019831034 0.9801690
15/30 0.056327586 0.9436724
30/30 0.155172414 0.8448276
40/30 0.408620690 0.5913793
0/40  0.004189655 0.9958103
15/40 0.013965517 0.9860345
30/40 0.046551724 0.9534483
40/40 0.155172414 0.8448276
>

Additionally, you can solve the inverse problem using optimize()

sq_error <- function (p_win, p_win_game) {
    tm <- transition_matrix(p_win)

    for (time in 1:20) tm <- tm %*% tm

    (tm['0/0', 'won'] - p_win_game)^2
}

p_win <- optimize(function(p) sq_error(p, p_win_game = 0.3), c(0, 1))$minimum

> p_win
[1] 0.4166342

Probability of winning a game in tennis?

Solution 1:

Solution 2:

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