Does Deuce Help?

Probability Level 1

In a game of tennis, does having "deuce" (as compared to a simple "first to 4 points" system) help, hurt, or not affect the better player?

Details and Assumptions:

The "better player" wins each point with fixed probability $p > 0.5.$
In "deuce" rules, if the game reaches 3-3, the winner is the first person to subsequently lead by 2 points (colloquially, "win by 2").

Help Hurt Not affect

3 solutions

Eli Ross Staff
Aug 17, 2017

Intuitively, anything that reduces the "randomness" (more technically, variance) will help the better player.

Mathematically, we are replacing the "win 4-3" case with "win after a deuce". Let's compare those probabilities: in both cases, we need to first get to 3-3, so we're just comparing "win 1 additional point" and "win a deuce".

Winning 1 point has probability $p.$ For the deuce case, you have two ways in which case you can win deuce:

Win two points
Win and lose in the first two points (in either order), and then win deuce (you're back to being tied)

This gives the probability of winning a deuce, $D,$ as $D = p^2 + 2p(1-p)\cdot D \quad \Rightarrow \quad D = \frac{p^2}{1-2p(1-p)}.$

Some algebra shows that $D > p$ precisely when $p>\frac{1}{2}.$

Can we apply such intuition to winning of better player in first case giving some formula?

Bijay Shah - 3 years, 9 months ago

Who is serving....?

Greg Litcher - 3 years, 9 months ago

Please can someone do the "Some algebra" as that's the bit I can't work out?

Justin Roughley - 3 years, 9 months ago

Have you tried experimenting by substituting values less than $\frac{1}{2}$ and more than $\frac{1}{2}$ to check how it works out?

Agnishom Chattopadhyay - 3 years, 8 months ago

$\frac{p^2}{1-2p(1-p)}>p\Rightarrow \frac{p}{1-2p(1-p)}>1\Rightarrow p>1-2p(1-p)\Rightarrow p>1-2p+2p^2$ $\Rightarrow 0>2p^2-3p+1 \Rightarrow 0>(2p-1)(p-1)\Rightarrow 0>2p-1$ and $0<p-1$ or $0<2p-1$ and $0>p-1\Rightarrow 1/2>p$ and $1<p$ or $1/2<p$ and $1>p$ . The first pair of inequalities is impossible so it must be the second set, i.e., $1/2<p<1$ .

James Wilson - 3 years, 7 months ago

Stephen Mellor
Aug 28, 2017

Relevant wiki: Probability - Problem Solving

In a game, the helpfulness of having or not having a deuce rule only comes into effect at 3-3.

Without a deuce rule, despite the better player having more of a chance than the other player, the better player could still lose.

With a deuce rule, the chance of the better player losing two points in a row is much smaller as p > $\frac{1}{2}$

Therefore, the deuce rule helps the better player

Good intuition, but with an important missing point: there is a third possibility, namely that the players go back to equal scores, after two plays.

Now, the original intuition can be applied to all such situations with an equal score. This shows that the odds of the better player winning are precisely $p^2$:$(1-p)^2$.

Eric Lebigot - 3 years, 9 months ago

Oliver Piattella
Aug 28, 2017

Relevant wiki: Game Theory

The probabilities of the better player winning 4-0, 4-1, 4-2, are the same in the two typologies of game. The only difference is in the probability of winning 4-3 which we are going to analyze in a moment. The total probability of winning is the sum

$P(\mbox{Win}) = P(4-0) + P(4-1) + P(4-2) + P(4-3)\;.$

since the four events are mutually exclusive.

For both the typologies of games we have:

$P(4-0) = p^4\;, \qquad P(4-1) = 4p^4(1 - p)\;, \qquad P(4-2) = 10p^4(1 - p)^2\;, \qquad P(3-3) = 20p^3(1 - p)^3\;.$

Now, for the game "first to 4" we have:

$P(4-3) = pP(3-3)\;.$

Whereas for the "Deuce" game we have:

$P(4-3) = P(D)P(3-3)\;,$

where $P(D)$ is the probability of the better player winning from a deuce. So, in order to understand which kind of game is better for the better player we need to compare $p$ with $P(D)$ . Therefore, let's calculate the latter. You can build the following system:

$P(D) = pP(Adv) + (1-p)P(Dis)\\ P(Adv) = p + (1 - p)P(D)\\ P(Dis) = pP(D)\;,$

where $P(Adv)$ is the probability of the better player to win from the advantage (i.e. he scores a point after the deuce) and $P(Dis)$ is the probability of the better player to win from the disadvantage (i.e. his opponents scores a point after the deuce).

Solving the above system, one finds:

$P(D) = \frac{p^2}{(1 - p)^2 + p^2}\;.$

This is larger than $p$ , for $p > 1/2$ and thus having deuce helps the better player.

Why is P(D)>p for p>1/2 ?

Justin Roughley - 3 years, 9 months ago

Suppose P(D) < p. Then p^2 < p[(1 - p)^2 + p^2]. Since p > 0 you can simplify one p and get the second-order inequality

2p^2 + 1 - 3p > 0

The roots of the polynomial are p = 1/2 and p = 1. The discriminant is 1 > 0. Hence the inequality is satisfied if p < 1/2 or p > 1. The first case is not possible by hypothesis and the second is impossible because p is a probability. That is, we must have that 1/2 < p < 1 and hence P(D) > p.

Oliver Piattella - 3 years, 9 months ago

Maybe a clearer way is to write the polynomial p - P(D) as

2(p - 1)(p - 1/2)

and then to study its sign. It is negative, hence P(D) > p, when the two factors have opposite signs, i.e.

p > 1 and p < 1/2 (which is impossible)

1/2 < p < 1 (which is our initial hypothesis on p)

Oliver Piattella - 3 years, 9 months ago

Does Deuce Help?

3 solutions

0 pending reports