## The effect of playing more points

Suppose you have two players of differing abilities playing a match with an odd number $$n$$ of points to be played and the winner of the match is the player who wins more points. If the better player wins each point with probability greater than $$\frac12$$, and each point is independent of the others, then it seems intuitively obvious that a larger $$n$$ increases the chance that the better player will win the match.

What seems intuitively obvious may in fact not always be true. Three examples are given below:

• the first shows an intuitive case where the probability that the better player wins each point stays constant and independent of the score so far, so the skill of the better player increasingly overwhelms the uncertainty of each individual point
• the second shows a case where the probability that the better player wins each point decreases as more points are played to such an extent the probability that the better player wins the match also reduces for longer matches
• the third shows a case where the marginal probability that the better player wins each point stays constant but the conditional probability of winning a point depends on the score so far, with the effect of making the likely outcomes either an overwhelming win for the better player or a close loss for the better player, resulting in the better player being more likely than not to lose the match if it is long enough and the better player is not too good

This note was stimulated by a question on math.stackexchange https://math.stackexchange.com/q/2793166/6460 posed by antkam. It is composed using R Markdown (see http://rmarkdown.rstudio.com) and includes the R code used to do calculations and produce the figures, allowing them to be reproduced or developed.

# Playing more points helping the better player

The probability $$m_n$$ of the better player winning a match of an odd number $$n$$ points is $m_n=\sum\limits_{i=(n+1)/2}^n {n \choose i} p^i (1-p)^{n-i} \approx \Phi\left(\sqrt{n}\frac{p-\frac12}{\sqrt{p(1-p)}}\right)$ where $$\Phi(x)$$ is the cumulative distribution function of a standard normal random variable. The actual probability is increasing with odd $$n$$, while the normal approximation is an increasing function of an increasing function of $$n$$. The same probabilities would apply if the match was “best of $$n$$” or “first to $$(n+1)/2$$”.

As an illustration, suppose $$p=0.6$$. This graph shows the probability of the better player winning the match for given $$n$$: the black circles are the actual probabilities of the better player winning the match $$(0.6$$ when $$n=1$$ and about $$0.978$$ when $$n=99)$$ and the red line the normal approximation, while the blue crosses show the constant probability of the better player winning a particular point.

p    <- 0.6 # probability better player wins a particular point
maxn <- 99  # maximum number of points in a match
n <- (1:maxn)[(1:maxn) %% 2 == 1]
mn <- rep(NA, maxn)
oddn <- (1:maxn)[(1:maxn) %% 2 == 1]
mn[oddn] <- 1 - pbinom((oddn-1) / 2, oddn, p)
approx_mn <- function(n){ pnorm(sqrt(n) * (p - 1/2) / sqrt(p*(1-p))) }
plot(1:maxn, rep(p, maxn), pch=3, col="blue", ylim=c(p,1),
ylab="probability for better player", xlab="points", main="Figure 1. Usually the better player is more likely to win \n a match which requires more points to be played")
points(oddn, mn[oddn], pch=1, col="black")
legend("right", legend=c("prob. better player wins point", "prob. better player wins match", "approximate match prob."), col=c("blue", "black", "red"), lty=c(0,0,1), pch=c(3,1,NA))

This illustrates that the better player is more likely to win if the number of points played increases: more points being played gives skill a greater opportunity to dominate luck.

This need not be the case in more complicated circumstances, for example where the probability of the better player winning a point changes through the match, or where the marginal probability for the better player winning particular points stays constant but the conditional probabilities change.

# Staying better but with reduced chances

If the aim is show a case where the better player have a lower probability of winning the match the longer the match is, a possibility would to reduce the probability that the better player wins later points, while still staying over $$\frac12$$. Perhaps the worse player improves over time, or the better player suffers from fatigue. Let’s define

• $$p_n$$ as the probability that Player A wins point $$n$$
• $$q(n,j)$$ as the probability that after $$n$$ points, Player A has won exactly $$j$$ of them
• $$m_{n}=\sum\limits_{j=(n+1)/2}^{n} q(n,j)$$ is the probability that Player A has won the majority of an odd number $$n$$ points

so we have the recurrence $q(n,j)= p_n \,q(n-1,j-1) +(1-p_n)\,q(n-1,j)$ starting at $$q(0,0)=1$$ and $$q(0,j)=0$$ for $$j \not =0$$, and given $$p_1,p_2,\ldots,p_n$$ we can calculate all $$q(n,j)$$. We want

• $$p_n \gt \frac12$$ implying $$r_{n} \gt \frac12$$ for all positive integer $$n$$ since Player A is the better player
• $$m_{n-2} \gt m_{n}$$ for all odd positive integer $$n$$ to suggest Player A’s chances get worse with a longer match

Since playing two extra points only makes a difference to the match if the score was already almost equal, we have $m_{n-2} - m_{n} = (1-p_{n-1})(1-p_{n})q\left(\frac{n-1}2,n-2\right) - p_{n-1} p_{n} q\left(\frac{n-3}2,n-2\right)$ and so for the longer match to be worse for the better player we want this to be positive and so $\dfrac{(1-p_{n-1})(1-p_{n})}{p_{n-1} p_{n} } > \dfrac{q\left(\frac{n-3}2,n-2\right)}{q\left(\frac{n-1}2,n-2\right)}.$

Essentially, the fall in the probability that the better player wins the match when you extend the match by two points is the probability that the better player won the shorter match by exactly one point and then lost an additional two points minus the probability the better player lost the shorter match by exactly one point and then won an additional two points.

A simple function of $$p_1$$ which works is to have the probability that Player A wins point $$n$$ being $$p_n = \frac12 +\frac{p_1-\frac12}{n}$$ for all $$n$$. An approximation to the probability of the better player winning a match of odd $$n$$ points is $$m_n \approx \Phi \left(\frac{\log_e(n)+\gamma+\frac{1}{2n}}{\sqrt{\frac{n}{\left(2p_1-1\right)^2}-\frac{\pi^2}{6}}} \right)$$ for large $$n$$.

The effect is illustrated in the following chart with $$p_1=0.6$$. It shows the probability $$p_n$$ of the better player winning the $$n$$th point with blue crosses $$(p_1=0.6$$ and $$p_{99}\approx 0.50101)$$ and the probability $$m_n$$ of the better player winning a match of odd $$n$$ points with the black circles $$(r_1=0.6$$ and $$r_{99}\approx 0.54156)$$ and the red line as an approximation to this; all the curves converge to $$\frac12$$. Again, the same probabilities would apply if the match was “best of $$n$$” or “first to $$(n+1)/2$$”.

p    <- 0.6 # probability better player wins first point
maxn <- 99  # maximum number of points in a match
qmat <- matrix(numeric((1+maxn)^2), ncol=1+maxn)
qmat[1,1] <- 1 # remember counting from 0,0 in this matrix
mn <- rep(NA, maxn)
pn <- 1/2 + (p-1/2)/(1:maxn)
for (n in 1:maxn){
qmat[n+1, 1]       <- qmat[n, 1]*(1-pn[n])
qmat[n+1, 2:(n+1)] <- qmat[n, 1:n]*pn[n] + qmat[n,2:(n+1)]*(1-pn[n])
if (n %% 2 == 1){ mn[n] <- sum(qmat[n+1, ((n+3)/2):(n+1)])     }
}
approx_mn <- function(n){ pnorm((log(n)+0.5772156649+1/(2*n))/sqrt(n/(2*p-1)^2-pi^2/6)) }
oddn <- (1:maxn)[(1:maxn) %% 2 == 1]
plot(1:maxn, pn, pch=3, col="blue", ylim=c(0.5,p),
ylab="probability for better player", xlab="points",
main="Figure 2. A better but weakening player can see a reduced chance \n of winning a match which has more points to be played")
points(oddn, mn[oddn], pch=1, col="black")
legend("topright", legend=c("probability better player wins point", "probability better player wins match", "approximate match probability"), col=c("blue", "black", "red"), lty=c(0,0,1), pch=c(3,1,NA))

This illustrates that it is possible for the better player to have a lower chance of winning the match when the match is longer, though this happens with a reducing probability of the better player winning later points, and the better player still has a probability of over $$\frac12$$ of winning the match no matter how many points are involved.

Here the probability of winning each point was independent of the results of the previous points; by removing that condition, it is possible for the the better player to have a constant marginal probability of winning each point but have a probability below $$\frac12$$ of winning the match.

# The better player being unlikely to win the match

It is possible to have the better Player A more likely than not to lose the match. Consider for example a $$3$$ point match where the possible ordered outcomes for the points have the probabilities $$\mathbb P(AAA) = 0.4$$, $$\mathbb P(ABB) = 0.2$$, $$\mathbb P(BAB) = 0.2$$ and $$\mathbb P(BBA) = 0.2$$. Then the marginal probability that player A wins a particular point is a constant $$0.6$$, and the expected number of points Player A will win is $$1.8$$ which is more than half of the $$3$$ points played, but the probability player A wins the match is only $$0.4$$.

There is a question of how badly the chances can move against Player A with a long enough match when that player’s marginal probability of winning a point is a constant $$p\gt \frac12$$. It turns out that there is an approachable but unachievable lower bound of $$2p-1$$, so for example with $$p=0.6$$ this would be $$0.2$$, while with $$p=0.9$$ it would be $$0.8$$. To have the better Player A more likely than not to lose the match, i.e. to have this bound below $$\frac12$$, requires $$\frac12 \lt p \lt \frac34$$, so to get this counter-intuitive result the better player should not be too good.

To show this bound is not achievable, consider the expected number of points Player A will have won after an odd number $$n$$ of points; this is obviously $$np$$. Using the earlier notation, we will have $np = \sum\limits_{j=0}^n j\, q(n,j) = \sum\limits_{j=(n+1)/2}^n j\, q(n,j) + \sum\limits_{j=0}^{(n-1)/2} j\, q(n,j) \le n\,m_n + \frac{n-1}{2}(1-m_n) = \frac{n m_n+ m_n +n-1}{2}$ which when $$\frac12 \lt p \lt 1$$, i.e. $$0 \lt 2p-1 \lt 1$$, implies $m_n \ge \frac{2np-n +1}{n+1} \gt \frac{2np-n +2p-1}{n+1} = 2p-1.$

To show the bound is approachable, consider the following possible pattern for $$p_{n,j}$$, the probability of the better player winning the $$n$$th point conditional on that player having won $$j$$ of the previous $$n-1$$ points: $p_{n,j} =\left\{\begin{array}{ll} 1-\frac{1-p}{(n-1)(2p-1)+1}, & \text{for } j=n-1 \text{ and } n \text{ a power of } 2 \\ 1, & \text{for } j=n-1 \text{ and } n \text{ not a power of } 2 \\ 1, & \text{for } j=\frac{n-3}{2} \text{ and } n \text{ odd} \\ \frac12, & \text{for } j=\frac{n-2}{2} \text{ and } n \text{ even} \\ 0, & \text{otherwise} \end{array}\right.$

which gives the following probabilities $$q(n,j)$$ of the better player having won a total of $$j$$ points out of $$n$$ using $$q(n,j) = p_{n,j-1} q(n-1,j-1) + (1-p_{n,j-1}) q(n-1,j)$$ and starting from $$q(0,0)=1$$ :

$q(n,j) =\left\{\begin{array}{ll} 1, & \text{for } j=0 \text{ and } n=0 \\ 2p-1 +\frac{1-p}{j}, & \text{for } n=j \text{ and } j \text{ a power of } 2 \\ 2p-1 +\frac{1-p}{2^{\lfloor \log_2(j)\rfloor}}, & \text{for } n=j \\ \frac{1-p}{j+1}, & \text{for } n \lt 2j \lt 2n \text{ and } j+1 \text{ a power of } 2 \\ 1-p, & \text{for } n = 2j \text{ and } j+1 \text{ a power of } 2 \\ (1-p)\left(1 - \frac1{2j}\right), & \text{for } n = 2j \text{ and } j \text{ a power of } 2 \\ (1-p)\left(1 - \frac1{2^{1+\lfloor \log_2(j)\rfloor}}\right), & \text{for } n = 2j \text{ and } j+1 \text{ not a power of } 2 \\ (1-p)\left(2-\frac1{j+1}\right), & \text{for } n = 2j+1 \text{ and } j+1 \text{ a power of } 2 \\ (1-p)\left(2-\frac1{2^{\lfloor \log_2(j+1)\rfloor}}\right), & \text{for } n = 2j+1 \\ (1-p)\left(1-\frac1{2(j+1)}\right), & \text{for } n = 2j+2 \text{ and } j+1 \text{ a power of } 2 \\ (1-p)\left(1-\frac1{2^{1+\lfloor \log_2(j+1)\rfloor}}\right), & \text{for } n = 2j+2 \\ 0, & \text{otherwise} \end{array}\right.$

with the properties we expect of $$\sum\limits_{j=0}^n q(n,j) = 1$$ and $$\sum\limits_{j=0}^n j \, q(n,j) = np$$ and $$\sum\limits_{j=0}^{n-1} p_{n,j} \, q(n-1,j) = p$$.

In particular, when $$n$$ is one less than a power of $$2$$ we have $$q(n,n)= 2p-1 +\frac{2(1-p)}{n+1}$$ and $$q\left(\frac{n-1}{2},n\right)=\frac{2n(1-p)}{n+1}$$ adding to $$1$$, and so making the probability of the better player winning a match $$m_n = 2p-1 +\frac{2(1-p)}{n+1}$$, which reduces towards $$2p-1$$ as $$n$$ take increasing values one less than a power of $$2$$. This probability does not change for other odd $$n$$ not one less than a power of $$2$$, with $$q(n,j)$$ being positive for only three $$j$$s, so more generally $$m_n = 2p-1 +\frac{1-p}{2^{\lfloor \log_2(n+1)\rfloor-1}}$$.

p    <- 0.6 # probability better player wins a particular point
maxn <- 99  # maximum number of points in a match
power2below <- function(x){ 2^floor(log(x,2)) }
mn <- rep(NA, maxn)
qmat <- matrix(numeric((1+maxn)^2), ncol=1+maxn)
qmat[1, 1] <- 1 # remember counting from 0,0 in this matrix
pmat <- matrix(numeric((1+maxn)^2), ncol=1+maxn) # also from 0,0
for (n in 1:maxn){
pmat[n+1, n-1+1] <- 1
if(n == power2below(n)){ pmat[n+1, n-1+1] <- 1 - (1-p)/((n-1)*(2*p-1)+1) }
if(n %% 2 == 1){ pmat[n+1, (n-3)/2+1] <- 1 }
if(n %% 2 == 0){ pmat[n+1, (n-2)/2+1] <- 1/2 }
qmat[n+1, 1]       <- qmat[n, 1]*(1-pmat[n+1,1])
qmat[n+1, 2:(n+1)] <- qmat[n, 1:n]*pmat[n+1, 1:n] + qmat[n, 2:(n+1)]*(1-pmat[n+1, 2:(n+1)])
if (n %% 2 == 1){ mn[n] <- sum(qmat[n+1,((n+3)/2):(n+1)])     }
}
rownames(qmat) <- 0:maxn
colnames(qmat) <- 0:maxn
rownames(pmat) <- 0:maxn
colnames(pmat) <- 0:maxn
approx_mn <- function(n){ 2*p-1 + 2*(1-p)/(n + 1) }
# 1 - rowSums(qmat)                                 # should be 0 to rounding
# (0:maxn)*p - colSums((0:maxn)*t(qmat))            # should be 0 to rounding
# p - rowSums(pmat[-1,] * qmat[-(maxn+1),])         # should be 0 to rounding
# 2*p-1 + 2*(1-p)/power2below(oddn + 1) - mn[oddn]  # should be 0 to rounding

If for example we have constant marginal probability of the better player winning a point being $$p=0.6$$, we get the following values for the probability $$p_{n,j}$$of the better player winning the $$n$$th point conditional on that player having won $$j$$ of the previous $$n-1$$ points. Each row corresponds to a different $$n$$ and each column to a different $$j$$.

print(pmat[2:11, 1:10])
##      0         1   2    3   4 5 6         7 8 9
## 1  0.6 0.0000000 0.0 0.00 0.0 0 0 0.0000000 0 0
## 2  0.5 0.6666667 0.0 0.00 0.0 0 0 0.0000000 0 0
## 3  1.0 0.0000000 1.0 0.00 0.0 0 0 0.0000000 0 0
## 4  0.0 0.5000000 0.0 0.75 0.0 0 0 0.0000000 0 0
## 5  0.0 1.0000000 0.0 0.00 1.0 0 0 0.0000000 0 0
## 6  0.0 0.0000000 0.5 0.00 0.0 1 0 0.0000000 0 0
## 7  0.0 0.0000000 1.0 0.00 0.0 0 1 0.0000000 0 0
## 8  0.0 0.0000000 0.0 0.50 0.0 0 0 0.8333333 0 0
## 9  0.0 0.0000000 0.0 1.00 0.0 0 0 0.0000000 1 0
## 10 0.0 0.0000000 0.0 0.00 0.5 0 0 0.0000000 0 1

The next table has the corresponding values of $$q(n,j)$$ the probability of the better player having won a total of $$j$$ points out of a possible $$n$$. Again each row corresponds to a different $$n$$ (so each row sums to $$1$$) and each column corresponds to a different $$j$$.

print(qmat[1:10, 1:10])
##     0   1   2    3    4   5   6    7    8    9
## 0 1.0 0.0 0.0 0.00 0.00 0.0 0.0 0.00 0.00 0.00
## 1 0.4 0.6 0.0 0.00 0.00 0.0 0.0 0.00 0.00 0.00
## 2 0.2 0.4 0.4 0.00 0.00 0.0 0.0 0.00 0.00 0.00
## 3 0.0 0.6 0.0 0.40 0.00 0.0 0.0 0.00 0.00 0.00
## 4 0.0 0.3 0.3 0.10 0.30 0.0 0.0 0.00 0.00 0.00
## 5 0.0 0.0 0.6 0.10 0.00 0.3 0.0 0.00 0.00 0.00
## 6 0.0 0.0 0.3 0.40 0.00 0.0 0.3 0.00 0.00 0.00
## 7 0.0 0.0 0.0 0.70 0.00 0.0 0.0 0.30 0.00 0.00
## 8 0.0 0.0 0.0 0.35 0.35 0.0 0.0 0.05 0.25 0.00
## 9 0.0 0.0 0.0 0.00 0.70 0.0 0.0 0.05 0.00 0.25

So we could say with $$n=1$$ we had $$P(A)=0.6$$ and $$P(B)=0.4$$ and splitting these with $$n=3$$ a possibility was $$\mathbb P(AAA) = 0.4$$ and $$\mathbb P(ABB) =$$ $$\mathbb P(BAB) =$$ $$\mathbb P(BBA) = 0.2$$, then we could extend this to $$n=7$$ by splitting each of those with something like $$\mathbb P(AAAAAAA) = 0.3$$ and $$\mathbb P(AAABBBB) =$$ $$\mathbb P(ABBABAB) =$$ $$\mathbb P(ABBBABA) =$$ $$\mathbb P(BABABAB) =$$ $$\mathbb P(BABBABA) =$$ $$\mathbb P(BBAABAB) =$$ $$\mathbb P(BBABABA) = 0.1$$, and similarly extending these to larger $$n$$s also one less than a power of $$2$$; the short sequences and probabilities can then be seen as the inital parts of the longer sequences. There are similar patterns for $$n=5$$ and other odd $$n$$ which also have the probability of the better player winning being exactly $$2p-1 +\frac{2(1-p)}{n+1}$$ but these are not extensions or the initial parts of the patterns for neighbouring cases one less than a power of $$2$$.

The overall effect is illustrated in the following chart with constant marginal probability of the better player winning a point $$p=0.6$$ with blue crosses. It shows the probability $$m_n$$ of the better player winning a match of odd $$n$$ points with the black circles and the red line as an approximation to this (exact when $$n$$ is one less than a power of $$2$$); both converge towards $$2p-1$$ which is $$0.2$$ here. Unlike the earlier examples, the marginal probabilities might not be considered to be the same if the match was “best of $$n$$” or “first to $$(n+1)/2$$” as some later points may not need to be played, affecting the calculations.

plot(1:maxn, rowSums(pmat[-1,] * qmat[-(maxn+1),]), pch=3, col="blue", ylim=c(2*p-1,p),
ylab="probability for better player", xlab="points",
main="Figure 3. An overall better player can be unlikely \n to win a match which requires more points to be played")
points(oddn, mn[oddn], pch=1, col="black")
legend("right", legend=c("marginal probability better player wins point", "probability better player wins match", "approximate match probability"), col=c("blue", "black", "red"), lty=c(0,0,1), pch=c(3,1,NA))