# If I have a 58% chance of winning a point, what's the chance of me winning a ping pong game to 21, win by 2?

I have a bet with a co-worker that out of 50 ping pong games (first to win 21 points, win by 2), I will win all 50. So far we've played 15 games and on average I win 58% of the points, plus I've won all the games so far. So we're wondering if I have a 58% chance of winning a point and he has a 42% chance of winning a point, what's the percent chance that I would win the game? Is there a formula that we can plug in difference % chances?

We've googled all over and even asked the data scientists at our company but couldn't find a straight answer.

Edit: Wow, I am blown away by the thoroughness of responses. Thank you all so much!!! In case people are curious, I have an update to how my bet is going: I've now won 18 out of 50 games, so I need to win 32 more games. I've won 58.7% of all points and my opponent has therefore won 41.3% of points. The standard deviation for my opponent is 3.52, his average score is 14.83, and his median score is 15.50. Below is a screenshot of the score of each game so far. I can keep updating as the bet goes on, if people are interested.

Edit #2: Unfortunately we've only been able to play a few more games, below are the results. I'm just going to keep replacing the picture so I don't have a bunch of screenshots of the score.

Final Update: I finally lost to my co-worker on game #28. He beat me 21-13. Thanks for all of your help!

• There is a formula: for $p=0.58,$ it's in the form $p^{21}/(1-2p+2p^2)$ times a degree-20 polynomial: 21 terms in all (with large coefficients, the largest exceeding $1.6\times 10^{16}$). If all the points are independent, you have only a $0.432\%$ chance of winning the next 35 games. – whuber Feb 19 '18 at 22:38
• I doubt that all points (and games) are independent of each other (for a variety of reasons). The non-independence could have a big impact on the answer. – Mark L. Stone Feb 20 '18 at 0:48
• Assuming it's the same game I played, I remember that the one serving has an advantage; so ignoring everything about "hot hand" it could be that you win 68% when serving and 48% when not - that will skew all probabilities; even if it evens out to 58% We thus don't have enough information.. – Hans Olsson Feb 20 '18 at 9:22
• Just a comment - 21 points? Table tennis switched to an 11 point format, best of 7 games, 2 serves per player at a time, back in 2001. – rcgldr Feb 23 '18 at 0:42
• I will continue posting updates on this bet every ~5 games or so. Unfortunately we only get to play a few games per week since we only play after work. – richard Feb 23 '18 at 17:23

The analysis is complicated by the prospect that the game goes into "overtime" in order to win by a margin of at least two points. (Otherwise it would be as simple as the solution shown at https://stats.stackexchange.com/a/327015/919.) I will show how to visualize the problem and use that to break it down into readily-computed contributions to the answer. The result, although a bit messy, is manageable. A simulation bears out its correctness.

Let $p$ be your probability of winning a point. Assume all points are independent. The chance that you win a game can be broken down into (nonoverlapping) events according to how many points your opponent has at the end assuming you don't go into overtime ($0,1,\ldots, 19$) or you go into overtime. In the latter case it is (or will become) obvious that at some stage the score was 20-20.

There is a nice visualization. Let scores during the game be plotted as points $(x,y)$ where $x$ is your score and $y$ is your opponent's score. As the game unfolds, the scores move along the integer lattice in the first quadrant beginning at $(0,0)$, creating a game path. It ends the first time one of you has scored at least $21$ and has a margin of at least $2$. Such winning points form two sets of points, the "absorbing boundary" of this process, whereat the game path must terminate.

This figure shows part of the absorbing boundary (it extends infinitely up and to the right) along with the path of a game that went into overtime (with a loss for you, alas).

Let's count. The number of ways the game can end with $y$ points for your opponent is the number of distinct paths in the integer lattice of $(x,y)$ scores beginning at the initial score $(0,0)$ and ending at the penultimate score $(20,y)$. Such paths are determined by which of the $20+y$ points in the game you won. They correspond therefore to the subsets of size $20$ of the numbers $1,2,\ldots, 20+y$, and there are $\binom{20+y}{20}$ of them. Since in each such path you won $21$ points (with independent probabilities $p$ each time, counting the final point) and your opponent won $y$ points (with independent probabilities $1-p$ each time), the paths associated with $y$ account for a total chance of

$$f(y) = \binom{20+y}{20}p^{21}(1-p)^y.$$

Similarly, there are $\binom{20+20}{20}$ ways to arrive at $(20,20)$ representing the 20-20 tie. In this situation you don't have a definite win. We may compute the chance of your win by adopting a common convention: forget how many points have been scored so far and start tracking the point differential. The game is at a differential of $0$ and will end when it first reaches $+2$ or $-2$, necessarily passing through $\pm 1$ along the way. Let $g(i)$ be the chance you win when the differential is $i\in\{-1,0,1\}$.

Since your chance of winning in any situation is $p$, we have

\eqalign{ g(0) &= p g(1) + (1-p)g(-1), \\ g(1) &= p + (1-p)g(0),\\ g(-1) &= pg(0). }

The unique solution to this system of linear equations for the vector $(g(-1),g(0),g(1))$ implies

$$g(0) = \frac{p^2}{1-2p+2p^2}.$$

This, therefore, is your chance of winning once $(20,20)$ is reached (which occurs with a chance of $\binom{20+20}{20}p^{20}(1-p)^{20}$).

Consequently your chance of winning is the sum of all these disjoint possibilities, equal to

\eqalign{ &\sum_{y=0}^{19}f(y) + g(0)p^{20}(1-p)^{20} \binom{20+20}{20} \\ = &\sum_{y=0}^{19}\binom{20+y}{20}p^{21}(1-p)^y + \frac{p^2}{1-2p+2p^2}p^{20}(1-p)^{20} \binom{20+20}{20}\\ = &\frac{p^{21}}{1-2p+2p^2}\left(\sum_{y=0}^{19}\binom{20+y}{20}(1-2p+2p^2)(1-p)^y + \binom{20+20}{20}p(1-p)^{20} \right). }

The stuff inside the parentheses on the right is a polynomial in $p$. (It looks like its degree is $21$, but the leading terms all cancel: its degree is $20$.)

When $p=0.58$, the chance of a win is close to $0.855913992.$

You should have no trouble generalizing this analysis to games that terminate with any numbers of points. When the required margin is greater than $2$ the result gets more complicated but is just as straightforward.

Incidentally, with these chances of winning, you had a $(0.8559\ldots)^{15}\approx 9.7\%$ chance of winning the first $15$ games. That's not inconsistent with what you report, which might encourage us to continue supposing the outcomes of each point are independent. We would thereby project that you have a chance of

$$(0.8559\ldots)^{35}\approx 0.432\%$$

of winning all the remaining $35$ games, assuming they proceed according to all these assumptions. It doesn't sound like a good bet to make unless the payoff is large!

I like to check work like this with a quick simulation. Here is R code to generate tens of thousands of games in a second. It assumes the game will be over within 126 points (extremely few games need to continue that long, so this assumption has no material effect on the results).

n <- 21      # Points your opponent needs to win
m <- 21      # Points you need to win
margin <- 2  # Minimum winning margin
p <- .58     # Your chance of winning a point
n.sim <- 1e4 # Iterations in the simulation

sim <- replicate(n.sim, {
x <- sample(1:0, 3*(m+n), prob=c(p, 1-p), replace=TRUE)
points.1 <- cumsum(x)
points.0 <- cumsum(1-x)
win.1 <- points.1 >= m & points.0 <= points.1-margin
win.0 <- points.0 >= n & points.1 <= points.0-margin
which.max(c(win.1, TRUE)) < which.max(c(win.0, TRUE))
})
mean(sim)


When I ran this, you won in 8,570 cases out of the 10,000 iterations. A Z-score (with approximately a Normal distribution) can be computed to test such results:

Z <- (mean(sim) - 0.85591399165186659) / (sd(sim)/sqrt(n.sim))
message(round(Z, 3)) # Should be between -3 and 3, roughly.


The value of $0.31$ in this simulation is perfectly consistent with the foregoing theoretical computation.

### Appendix 1

In light of the update to the question, which lists the outcomes of the first 18 games, here are reconstructions of game paths consistent with these data. You can see that two or three of the games were perilously close to losses. (Any path ending on a light gray square is a loss for you.)

Potential uses of this figure include observing:

• The paths concentrate around a slope given by the ratio 267:380 of total scores, equal approximately to 58.7%.

• The scatter of the paths around that slope shows the variation expected when points are independent.

• If points are made in streaks, then individual paths would tend to have long vertical and horizontal stretches.

• In a longer set of similar games, expect to see paths that tend to stay within the colored range, but also expect a few to extend beyond it.

• The prospect of a game or two whose path lies generally above this spread indicates the possibility that your opponent will eventually win a game, probably sooner rather than later.

### Appendix 2

The code to create the figure was requested. Here it is (cleaned up to produce a slightly nicer graphic).

library(data.table)
library(ggplot2)

n <- 21      # Points your opponent needs to win
m <- 21      # Points you need to win
margin <- 2  # Minimum winning margin
p <- 0.58     # Your chance of winning a point
#
# Quick and dirty generation of a game that goes into overtime.
#
done <- FALSE
iter <- 0
iter.max <- 2000
while(!done & iter < iter.max) {
Y <- sample(1:0, 3*(m+n), prob=c(p, 1-p), replace=TRUE)
Y <- data.table(You=c(0,cumsum(Y)), Opponent=c(0,cumsum(1-Y)))
Y[, Complete := (You >= m & You-Opponent >= margin) |
(Opponent >= n & Opponent-You >= margin)]
Y <- Y[1:which.max(Complete)]
done <- nrow(Y[You==m-1 & Opponent==n-1 & !Complete]) > 0
iter <- iter+1
}
if (iter >= iter.max) warning("Unable to find a solution. Using last.")
i.max <- max(n+margin, m+margin, max(c(Y$You, Y$Opponent))) + 1
#
# Represent the relevant part of the lattice.
#
X <- as.data.table(expand.grid(You=0:i.max,
Opponent=0:i.max))
X[, Win := (You == m & You-Opponent >= margin) |
(You > m & You-Opponent == margin)]
X[, Loss := (Opponent == n & You-Opponent <= -margin) |
(Opponent > n & You-Opponent == -margin)]
#
# Represent the absorbing boundary.
#
A <- data.table(x=c(m, m, i.max, 0, n-margin, i.max-margin),
y=c(0, m-margin, i.max-margin, n, n, i.max),
Winner=rep(c("You", "Opponent"), each=3))
#
# Plotting.
#
ggplot(X[Win==TRUE | Loss==TRUE], aes(You, Opponent)) +
geom_path(aes(x, y, color=Winner, group=Winner), inherit.aes=FALSE,
data=A, size=1.5) +
geom_point(data=X, color="#c0c0c0") +
geom_point(aes(fill=Win), size=3, shape=22, show.legend=FALSE) +
geom_path(data=Y, size=1) +
coord_equal(xlim=c(-1/2, i.max-1/2), ylim=c(-1/2, i.max-1/2),
ratio=1, expand=FALSE) +
ggtitle("Example Game Path",
paste0("You need ", m, " points to win; opponent needs ", n,
"; and the margin is ", margin, "."))

• How are the $f(y)$ disjoint? Don't your repeat configurations? For example, when $y=0$ the binomial coefficient is $1$. When $y=1$ then $\binom{21}{20} = 21$. But one of the latter configurations is exactly the one found for $y=0$ (i.e. 21 points won for our player, 0 for opponent). Should we not subtract probabilities of intersections? This is what blocked me in the first place. – Easymode44 Feb 20 '18 at 9:09
• @whuber: Great, would be also the "nice visualization" part of the r code implementation disclosure? Many thanks. – Maximilian Feb 20 '18 at 20:31
• @Stefan My value was computed using exact rational arithmetic (in Mathematica) and rounded at the end. I suspect yours might have been computed using only double precision floating point, and therefore assume your last few digits are incorrect. As a rational number the value is $$\frac{24949298160611146419680580467045837441748491517750191635779953104861}{29149305191822350025177001953125000000000000000000000000000000000000}.$$ – whuber Feb 21 '18 at 15:59
• @Maximilian I posted code for the visualization. – whuber Feb 21 '18 at 16:00
• I think a simpler way to handle overtime would be to take points in pairs once 20-20 is reached. The only things that matter are either the first player wining both (probability 0.58²) or the second winning both (0.42²). If anything else happens, ignore it and keep playing until one fo the above occurs. The first player's win probability after 20-20 is thus 0.58²/(0.58²+0.42²) and the second player's is 0.42²/(0.58²+0.42²). – supercat Feb 23 '18 at 21:27

Using the binomial distribution and assuming every point is independent:

• The probability the $58\%$ player gets to $21$ in the first $40$ points (taking account of the fact the last point must be won) is $\sum_{n=21}^{40} {n-1 \choose 20} 0.58^{21}0.42^{n-21}$ $=\sum_{k=21}^{40} {40 \choose k} 0.58^{k}0.42^{40-k}$ $\approx 0.80695$

• The probability $58\%$ player gets $20$ from $40$ points played is the binomial ${40 \choose 20} 0.58^{20}0.42^{20} \approx 0.074635$. Conditioned on that, the probability the $58\%$ player then wins with the two point margin is $\frac{0.58^2}{0.58^2+0.42^2}\approx 0.656006$

So the overall probability the $58\%$ player wins is about $0.80695+0.074635\times 0.656006$ $\approx 0.8559$

The probability of the $58\%$ player winning the first $15$ games is then about $0.85559^{15} \approx 0.0969$ which is fairly unlikely. The probability of the $58\%$ player winning the final $35$ games is about $0.85559^{35} \approx 0.0043$ which is very unlikely.

• The part "the probability the $58\%$ player then wins with the two point margin is $0.58^2/(0.58^2+0.42^2)\approx 0.656006$" could use some explanation, as it is probably the most difficult part in this problem. – JiK Feb 23 '18 at 8:31
• @JiK: Once at $20-20$ or later equality, the probability of a decisive couple of points is $0.58^2+0.42^2$ and so the probability that the better player gets two ahead rather than the worse player doing so is $\frac{0.58^2}{0.58^2+0.42^2}$ - otherwise they return to the same position – Henry Feb 23 '18 at 11:00
• A much more concise answer than the top answer, but I guess since it didn't have pictures and was posted 12 hours later, it gets 80 less votes? =| – Attackfarm Feb 25 '18 at 16:57

I went with a computational answer. Here is an R function that simulates a ping-pong game where the winner has to win by 2. The only argument is the probability that you win a point. It will return the final score of that game:

## data simulation function ----------------------------------------------------
sim_game <- function(pt_chance) {
them <- 0
you <- 0
while (sum((them < 21 & you < 21), abs(them - you) < 2) > 0) {
if (rbinom(1, 1, pt_chance) == 1) {
you <- you + 1
them <- them + 0
} else {
you <- you + 0
them <- them + 1
}
}
return(list(them = them, you = you))
}


Let's first make sure it works by simulating 10,000 games where you have a 50% chance of winning each point. We should observe that your win percentage is about 50%:

## testing 10,000 games --------------------------------------------------------
set.seed(1839)
results <- lapply(1:10000, function(x) sim_game(.5))
results <- as.data.frame(do.call(rbind, results))
results$you_win <- unlist(results$you) > unlist(results$them) mean(results$you_win)


This returns .4955, about what we would expect. So let's plug in your 58%:

## simulate 10,000 games -------------------------------------------------------
set.seed(1839)
results <- lapply(1:10000, function(x) sim_game(.58))
results <- as.data.frame(do.call(rbind, results))
results$you_win <- unlist(results$you) > unlist(results$them) mean(results$you_win)


This returns .8606. So you have about an 86.06% chance of winning one game.

We can now simulate across 35 game batches and see how many times you would win all 35:

## how often do you win all 35? ------------------------------------------------
set.seed(1839)
won_all_35 <- c()
for (i in 1:10000) {
results <- lapply(1:35, function(x) sim_game(.58))
results <- as.data.frame(do.call(rbind, results))
results$you_win <- unlist(results$you) > unlist(results$them) won_all_35[i] <- mean(results$you_win) == 1
}
mean(won_all_35)


This returns .0037, which means you have about a 0.37% chance of winning the next 35 games. This assumes that all games and all points are independent of one another. You could program that explicitly into the function above, if you wanted to.

Note: I'm doing this on the fly. I'm sure there is a more computationally efficient way of programming this.

• Try pbetterwins <- pbinom(19,40,0.42) + dbinom(20,40,0.42) * 0.58^2/(0.58^2+0.42^2); pbetterwins; pbetterwins^35 for a calculation using the binomial distribution. Close enough to your simulation – Henry Feb 25 '18 at 18:24

# Should we assume that the 58% chance of winning is fixed and that points are independent?

I believe that Whuber's answer is a good one, and beautifully written and explained, when the consideration is that every point is independent from the next one. However I believe that, in practice it is only an interesting starting point (theoretic/idealized). I imagine that in reality the points are not independent from each other, and this might make it more or less likely that your co-worker opponent gets to a win at least once out of 50.

At first I imagined that the dependence of the points would be a random process, ie not controlled by the players (e.g. when one is winning or loosing playing differently), and this should create a greater dispersion of the results benefiting the lesser player to get this one point out of fifty.

A second thought however might suggest the opposite: The fact that you already "achieved" something with a 9.7% of chance may give some (but only slight) benefit, from a Bayesian point of view, to ideas about favouring mechanisms that get you to win more than 85% probability to win a game (or at least make it less likely that your opponent has a much higher probability than 15% as argued in the previous two paragraphs). For instance, it could be that you score better when your position is less good (it is not strange for people scoring much more different on match points, in favor or against, than on regular points). You can improve estimates of the 85% by taking these dynamics into account and possibly you have more than 85% probability to win a game.

Anyway, it might be very wrong to use this simple points statistic to provide an answer. Yes you can do it, but it won't be right since the premises (independency of points) are not necessarily correct and highly influence the answer. The 42/58 statistic is more information but we do not know very well how to use it (the correctness of the model) and using the information might provide answers with high precision that it actually does not have.

# Example

Example: an equally reasonable model with a completely different result

So the hypothetical question (assuming independent points and known, theoretical, probabilities for these points) is in itself interesting and can be answered, But just to be annoying and skeptical/cynical; an answer to the hypothetical case does not relate that much to your underlying/original problem, and might be why the statisticians/data-scientists at your company are reluctant to provide a straight answer.

Just to give an alternative example (not neccesarily better) that provides a confusing (counter-) statement 'Q: what is the probability to win all of the total of 50 games if I already won 15?' If we do not start to think that 'the point scores 42/58 are relevant or give us better predictions' then we would start to make predictions of your probability to win the game and predictions to win another 35 games solely based on your previously won 15 games:

• with a Bayesian technique for your probability to win a game this would mean: $p(\text{win another 35 | after already 15}) = \frac{\int_0^1 f(p) p^{50}}{\int_0^1 f(p) p^{15}}$ which is roughly 31% for a uniform prior f(x) = 1, although that might be a bit too optimistic. But still if you consider a beta distribution with $\beta=\alpha$ between 1 and 5 then you get to:

which means that I would not be so pessimistic as the straightforward 0.432% prediction The fact that you already won 15 games should elevate the probability that you win the next 35 games.

# Note based on the new data

Based on your data for the 18 games I tried fitting a beta-binomial model. Varying $\alpha=\mu\nu$ and $\beta=(1-\mu)\nu$ and calculating the probabilities to get to a score i,21 (via i,20) or a score 20,20 and then sum their logs to a log-likelihood score.

It shows that a very high $\nu$ parameter (little dispersion in the underlying beta distribution) has a higher likelihood and thus there is probably little over-dispersion. That means that the data does not suggest that it is better to use a variable parameter for your probability of winning a point, instead of your fixed 58% chance of winning. This new data is providing extra support for Whuber's analysis, which assumes scores based on a binomial distribution. But of course, this still assumes that the model is static and also that you and your co-worker behave according to a random model (in which every game and point are independent).

Maximum likelihood estimation for parameters of beta distribution in place of fixed 58% winning chance:

Q: how do I read the "LogLikelihood for parameters mu and nu" graph?

A:

• 1) Maximum likelihood estimate (MLE) is a way to fit a model. Likelihood means the probability of the data given the parameters of the model and then we look for the model that maximizes this. There is a lot of philosophy and mathematics behind it.
• 2) The plot is a lazy computational method to get to the optimum MLE. I just compute all possible values on a grid and see what the valeu is. If you need to be faster you can either use a computational iterative method/algorithm that seeks the optimum, or possibly there might be a direct analytical solution.
• 3) The parameters $\mu$ and $\nu$ relate to the beta distribution https://en.wikipedia.org/wiki/Beta_distribution which is used as a model for the p=0.58 (to make it not fixed but instead vary from time to time). This modeled 'beta-p' is than combined with a binomial model to get to predictions of probabilities to reach certain scores. It is almost the same as the beta-binomial distribution. You can see that the optimum is around $\mu \simeq 0.6$ which is not surprising. The $\nu$ value is high (meaning low dispersion). I had imagined/expected at least some over-dispersion.

code/computation for graph 1

posterior <- sapply(seq(1,5,0.1), function(x) {
integrate(function(p) dbeta(p,x,x)*p^50,0,1)[1]$value/ integrate(function(p) dbeta(p,x,x)*p^15,0,1)[1]$value
}
)

prior <- sapply(seq(1,5,0.1), function(x) {
integrate(function(p) dbeta(p,x,x)*p^35,0,1)[1]$value } ) layout(t(c(1,2))) plot( seq(1,5,0.1), posterior, ylim = c(0,0.32), xlab = expression(paste(alpha, " and ", beta ," values for prior beta-distribution")), ylab = "P(win another 35| after already 15)" ) title("posterior probability assuming beta-distribution") plot( seq(1,5,0.1), prior, ylim = c(0,0.32), xlab = expression(paste(alpha, " and ", beta ," values for prior beta-distribution")), ylab = "P(win 35)" ) title("prior probability assuming beta-distribution")  code/computation for graph 2 library("shape") # probability that you win and opponent has kl points Pwl <- function(a,b,kl,kw=21) { kt <- kl+kw-1 Pwl <- choose(kt,kw-1) * beta(kw+a,kl+b)/beta(a,b) Pwl } # probability to end in the 20-20 score Pww <- function(a,b,kl=20,kw=20) { kt <- kl+kw Pww <- choose(kt,kw) * beta(kw+a,kl+b)/beta(a,b) Pww } # probability that you lin with kw points Plw <- function(a,b,kl=21,kw) { kt <- kl+kw-1 Plw <- choose(kt,kw) * beta(kw+a,kl+b)/beta(a,b) Plw } # calculation of log likelihood for data consisting of 17 opponent scores and 1 tie-position # parametezation change from mu (mean) and nu to a and b loglike <- function(mu,nu) { a <- mu*nu b <- (1-mu)*nu scores <- c(18, 17, 11, 13, 15, 15, 16, 9, 17, 17, 13, 8, 17, 11, 17, 13, 19) ps <- sapply(scores, function(x) log(Pwl(a,b,x))) loglike <- sum(ps,log(Pww(a,b))) loglike } #vectors and matrices for plotting contour mu <- c(1:199)/200 nu <- 2^(c(0:400)/40) z <- matrix(rep(0,length(nu)*length(mu)),length(mu)) for (i in 1:length(mu)) { for(j in 1:length(nu)) { z[i,j] <- loglike(mu[i],nu[j]) } } #plotting levs <- c(-900,-800,-700,-600,-500,-400,-300,-200,-100,-90,-80,-70,-60,-55,-52.5,-50,-47.5) # contour plot filled.contour(mu,log(nu),z, xlab="mu",ylab="log(nu)", #levels=c(-500,-400,-300,-200,-100,-10:-1), color.palette=function(n) {hsv(c(seq(0.15,0.7,length.out=n),0), c(seq(0.7,0.2,length.out=n),0), c(seq(1,0.7,length.out=n),0.9))}, levels=levs, plot.axes= c({ contour(mu,log(nu),z,add=1, levels=levs) title("loglikelihood for parameters mu and nu") axis(1) axis(2) },""), xlim=range(mu)+c(-0.05,0.05), ylim=range(log(nu))+c(-0.05,0.05) )  • +1 I appreciate the new perspective. But I would challenge the assertion that dependence among points makes it more likely the opponent will win in the next 35 games. In fact, it could go either way. A plausible mechanism for the opposite conclusion is that you are far stronger than the 58-42 edge in points would suggest, and that when called on, you can always rally to win any game even if far behind. The real problem in not assuming independence concerns how to model the non-independence. – whuber Feb 21 '18 at 15:38 • @whuber, you are right. I also argue for either ways. 1) My first thoughts went into one direction the dependency would be random, e.g. people have uncontrolled ups and downs good moments and bad moments, and this I imagine will create a larger dispersion of the results pushing up the probability of the lesser player. 2) However, then I was thinking of Bayesian principles and how the 15 won games may influence the analysis (at least the question in the post is a different situation from the question in the title), and there could be possible mechanisms that benefit the stronger player. – Martijn Weterings Feb 21 '18 at 15:45 • In the second half of my post I give just one example which argues that the probability to win should be larger than 86%. But while all this mathematics sounds very precise, in reality we are not really that sure since our models are bad (with lot's of additional, accuracy decreasing, subjective information) given this little amount of information. – Martijn Weterings Feb 21 '18 at 15:52 • @whuber I have edited my answer. That was a good comment, and I hope it is more clear now in the answer. – Martijn Weterings Feb 21 '18 at 15:59 • 2) The plot is a lazy computational method to get to the optimum MLE. I just compute all possible values on a grid and see what the valeu is. If you need to be faster you can either use a computational iterative method/algorithm that seeks the optimum, or possibly there might be a direct analytical solution. – Martijn Weterings Feb 22 '18 at 17:10 Much effort could be spent on a perfect model. But sometimes a bad model is better. And nothing says bad model like the central limit theorem -- everything is a normal curve. We'll ignore "overtime". We'll model the sum of individual points as a normal curve. We'll model playing 38 rounds and whomever has the most points win, instead of first to 20. This is quite similar game wise! And, blindly, I'll claim we get close to the right answer. Let$X$be the distribution of a point.$X$has value 1 when you get a point, and 0 when you don't. So$E(X)$=~$0.58$and$Var(X)$=$E(X)*(1-E(X))$=~$0.24$. If$X_i$are independent points, then$\sum_{i=1}^{38}{X_i}$is the points you get after playing 38 rounds.$E(\sum_{i=1}^{38}{X_i})$=$38*E(X)$=~$22.04Var(\sum_{i=1}^{38}{X_i})$= 38*Var($X$) =~$9.12$and$SD(\sum_{i=1}^{38}{X_i})$=$\sqrt{38*Var(X))}$=~$3.02$In our crude model, we lose if$\sum_{i=1}^{38}{X_i} < 19$and win if$\sum_{i=1}^{38}{X_i} > 19$.$\frac{22.04-19}{3.02}$is$1.01$standard deviations away from the mean, which works out to a$15.62\%$chance of failure after consulting a z score chart. If we compare to the more rigorous answers, this is about$1\%$off the correct value. You'd generally be better off examining the reliability of the$58\%$victory chance rather than a more rigorous model that assumes$58\%\$ chance and models it perfectly.

• @Yakk, where the heck did the 38 come from?? Also I'm pretty sure var(38*x) = 38^2 * var(X), not 38*var(X). How does your "very nice back of the envelope calculation" hold up after you correct that error? – use_norm_approx Feb 26 '18 at 2:12
• @use_ I am using a sloppy 38*X as "sum of 38 independent Xs", not "one X times 38". 38 comes from "whomever gets more than 19 wins first wins the game". I could have used 39 games and first one > 19.5 instead; result would be similar. – Yakk Feb 26 '18 at 12:13

Based on simulation, it looks like the probability of winning any given game is about 85.5%.

The probability of winning by exactly 2 (which is how I read the title, but doesn't seem to be what you're asking) is about 10.1%.

Run the code below.

set.seed(328409)
sim.game <- function(p)
{
x1 = 0
x2 = 0
while( (max(c(x1,x2)) < 21) | abs(x1-x2)<2  )
{
if(runif(1) < p) x1 = x1 + 1 else x2 = x2 + 1
}
return( c(x1,x2) )
}

S <- matrix(0, 1e5, 2)
for(k in 1:1e5) S[k,] <- sim.game(0.58)

mean( (S[,1]-S[,2]) == 2 ) #chance of winning by 2
mean(S[,1]>S[,2]) #chance of winning

• This gets very close to Whuber's analytical solution: dbinom(20,40,0.58)*0.58^2/(1-2*0.58+2*0.58^2)+dbinom(20,39,0.58)*0.58 giving 10.04 % – Martijn Weterings Feb 21 '18 at 17:28