[Guest]

Two tennis players, A and B, are playing a game. In that game, A has a 60% probability of winning any given point (with B obviously having a 40% probability). If the game gets to deuce, what is the probability that A will win the game?

I'm interested to see the reasoning behind your answer.

For anyone not familiar with tennis scoring, deuce works as follows: if the score is deuce, then whichever player wins the next point will get the advantage. If the player with the advantage wins the point after that they win the game, if they lose it the score goes back to deuce.

[Glycereine]

If 2 more points are played the game will either be back at deuce (starting point) or one of the two players will win.

Player A has a .6 X .6 probability of winning = .36

Player B has a .4 x .4 probability of winning = .16

There is a .6 x .4 probability that player A will win then lose resulting in deuce = .24

There is a .4 x .6 probability that player B will win then lose resulting in deuce = .24

since 48% of the time the conditions reset back to a deuce those probabilities can be ignored in-so-far as determining win probability and not the number of points it takes to win.

so .36/.52 is player A's probability of winning.

69.23%

edit: messed up spoiler

Edited May 26, 2010 by Glycereine

[Guest]

9/13 if i remember my math formula correctly

The logic is you start looking at the probabilities for each outcome a more likely to gain A point B more likely to gain a point. From those you can see the probabilities for A to get two points (let's call that x) and B to get two points. In those two cases it's clear which probability is in favor of A and which is against. The remaining % is a tie(let's call this y) => it can be split again in the x probability of y that A will win and a y probability of y that we're interested in the tie appearing again. We can keep doing this splitting forever until we reach the following equation

x + x*y + x*y² + ... + x*yⁿ

where x=.36 and y=.48

This is solvable since y < 1

From then on it's mostly manipulations of the equation.

[Guest]

From deuce the chances are:

A = 36% A wins

D = 48% return to Deuce

B = 16% B wins

A = 0.36 * E_n=0^infDⁿ

A = 0.36*S₀^inf 0.48ⁿ

A ~ 0.490 = 49.0% chance for A to win

[Glycereine]

From deuce the chances are:

A = 36% A wins

D = 48% return to Deuce

B = 16% B wins

A = 0.36 * E_n=0^infDⁿ

A = 0.36*S₀^inf 0.48ⁿ

A ~ 0.490 = 49.0% chance for A to win

I don't follow this at all.

[Guest]

I don't follow this at all.

Me either, but mainly for the same reason that I don't follow your answer...(OK, I don't follow the notation either, but...)

Both of you start your post with something like "A has an x% chance of winning."...then do 'some stuff' and end with something like "So, then, A has a y% chance of winning."

I think you are both overthinking this...

...A has a .6 probability of scoring any point,

and 2 consecutive points are needed to win,

then the probability of A getting two consecutive points (the only way to win) is

0.6 * 0.6 = 0.36 or 36%

Passing advantage back and forth doesn't affect the odds of getting two consecutive points...

Smoo

[superprismatic]

Me either, but mainly for the same reason that I don't follow your answer...(OK, I don't follow the notation either, but...)

Both of you start your post with something like "A has an x% chance of winning."...then do 'some stuff' and end with something like "So, then, A has a y% chance of winning."

I think you are both overthinking this...

...A has a .6 probability of scoring any point,

and 2 consecutive points are needed to win,

then the probability of A getting two consecutive points (the only way to win) is

0.6 * 0.6 = 0.36 or 36%

Passing advantage back and forth doesn't affect the odds of getting two consecutive points...

Smoo

Well, one of A or B will win. If, as you say, A's probability of winning is .36, then B's

probability of winning is .64 -- larger than that of A, the better player. How strange!

[Guest]

Me either, but mainly for the same reason that I don't follow your answer...(OK, I don't follow the notation either, but...)

Both of you start your post with something like "A has an x% chance of winning."...then do 'some stuff' and end with something like "So, then, A has a y% chance of winning."

I think you are both overthinking this...

...A has a .6 probability of scoring any point,

and 2 consecutive points are needed to win,

then the probability of A getting two consecutive points (the only way to win) is

0.6 * 0.6 = 0.36 or 36%

Passing advantage back and forth doesn't affect the odds of getting two consecutive points...

Smoo

The question asks what A's total chances of winning a game starting from deuce.

A's chance to win any deuce is 36% = A₀.

Their chance to return to deuce is 48%.

So the chance to reach the nth deuce is:

D₁ = 0.48

D₂ = 0.48 * 0.48 = 0.2304

.

.

.

D_n = 0.48ⁿ

so A's chance for winning the nth deuce is the chance to reach that deuce times A's chance to win:

A_n = A₀ * D_n

therefore A's total chance to win starting from the original deuce is:

A_t = A₀ + A₀ * D₁ + A₀ * D₂ + ... + A₀ * D_n

we can generalize D₀ = 0.48⁰ = 1, which makes sense since that's where we started anyway.

A_t = A₀ * Sum(D_n from n=0 to +inf) = A_i Sum(0.48ⁿ from n=0 to +inf) ~ 49.0%

[Guest]

The question asks what A's total chances of winning a game starting from deuce.

A's chance to win any deuce is 36% = A₀.

Their chance to return to deuce is 48%.

So the chance to reach the nth deuce is:

D₁ = 0.48

D₂ = 0.48 * 0.48 = 0.2304

.

.

.

D_n = 0.48ⁿ

so A's chance for winning the nth deuce is the chance to reach that deuce times A's chance to win:

A_n = A₀ * D_n

therefore A's total chance to win starting from the original deuce is:

A_t = A₀ + A₀ * D₁ + A₀ * D₂ + ... + A₀ * D_n

we can generalize D₀ = 0.48⁰ = 1, which makes sense since that's where we started anyway.

A_t = A₀ * Sum(D_n from n=0 to +inf) = A_i Sum(0.48ⁿ from n=0 to +inf) ~ 49.0%

B's total chance to win the game is:

B_t ~ 21.8%

with a 29.2% chance of getting stuck infinitely playing the same game of Tennis

[Guest]

Here's another way to look at it:

We are looking for the probability that Player A wins given that the game is currently at deuce

Let A represent the case where Player A wins

Let D represent the case where the game is currently at deuce

Now there are two scenarios that could take place from deuce: the next two points could result in a winner or the next two points could result in deuce.

Let W represent the case where the next two points result in a winner

Let T represent the case where the next two points result in deuce

The probability that A wins given that the game is currently at deuce =

The probablity that A wins given that the next two points result in a winner given that the game is currently at deuce *

The probability that the next two points result in a winner given that the game is currently at deuce +

The probablity that A wins given the next two points result in deuce given that the game is currently at deuce *

The probability that the next two points result in deuce given that the game is currently at deuce

We can write this as P(A|D) = P((A|W)|D) * P(W|D) + P((A|T)|D) * P(T|D)

But the probablity that A wins given the next two points result in deuce given that the game is currently at deuce

is the same as the probability that A wins given that the game is currently at deuce, i.e. P(A|D) = P((A|T)|D)

So we now have P(A|D) = P((A|W)|D) * P(W|D) + P(A|D) * P(T|D)

We can rewrite this as P(A|D) * [1-P(T|D)] = P((A|W)|D) * P(W|D)

But 1-P(T|D) = P(W|D)

So we now have P(A|D) * P(W|D) = P((A|W)|D) * P(W|D)

Which means that P(A|D) = P((A|W)|D)

So the Probability that A wins given that the game is currently at deuce =

The probablity that A wins given that the next two points result in a winner given that the game is currently at deuce

P((A|W)|D) = (3/5)(3/5) / [(3/5)(3/5) + (2/5)(2/5)] = (9/25) / (13/25) = 9/13

So Player A would win about 69% of the time.

[Glycereine]

B's total chance to win the game is:

B_t ~ 21.8%

with a 29.2% chance of getting stuck infinitely playing the same game of Tennis

Unfortunately I don't understand the formulas enough to tell you where you went wrong, but I can logically explain why your answer is incorrect.

From deuce there is a 36% chance A will win the game a 16% chance B will win and a 48 percent chance the game will return to deuce.

Once the game is back at deuces there is again a 36% chance A will win, 16% chance B will win and a 48% chance the game will return to deuce.

Therefore each time two games are played there is a .48^n chance that the game is back to deuce. Since this number is less than 1, as n apporaches infinity (the case we're talking about here because we want the game to end) the probability of being at deuce approaches 0.

That said, every time the game was at deuce the same probability existed for A or B to win, so the overall chance for them to win is proportional to the 36 and 16%. 36% is to X as 52% (16 and 36) is to 100. X is roughly 69%, or exactly 9/13.

[Guest]

Me either, but mainly for the same reason that I don't follow your answer...(OK, I don't follow the notation either, but...)

Both of you start your post with something like "A has an x% chance of winning."...then do 'some stuff' and end with something like "So, then, A has a y% chance of winning."

I think you are both overthinking this...

...A has a .6 probability of scoring any point,

and 2 consecutive points are needed to win,

then the probability of A getting two consecutive points (the only way to win) is

0.6 * 0.6 = 0.36 or 36%

Passing advantage back and forth doesn't affect the odds of getting two consecutive points...

Smoo

The problem with this line of thinking is the assumption that winning 2 point in a row guarantees a win which is not the case. If B scores the next point, A scoring 2 consecutive points does not give a win but only advantage.

[Guest]

Well Glycereine got the answer first. I'd worked out the same answer using similar reasoning, but I don't think either of us had a really good justification for it. Exagorazo came up with a really good justification, though, using conditional probability to prove the intuitive argument Glycereine used. Chui approached the problem from a different angle, using infinite series to arrive at the same answer. I'd be really interested to see the manipulations that derived the answer from the series.

[Guest]

.6 x .6 = 36 %

Edited May 27, 2010 by johnflabriola

[Guest]

p(A Wins) = 0.6932

A simpler method would be to consider the simple cases after 'deuce' for A to win the match which gives rise to an infinite GP:

P(A wins)= 2^0[{(0.4)(0.6)}^0*(0.6)(0.6)] + 2^1[{(0.4)(0.6)}^1*(0.6)(0.6)] + 2^2[{(0.4)(0.6)}^2*(0.6)(0.6)] + 2^3[{(0.4)(0.6)}^3*(0.6)(0.6)] + ...................

Summing up gives p(A Wins) = 0.6932

[Guest]

Each summation in the above solution gives 1 case:

1st: when A wins in 2 points

2nd: when A wins in 4 points

3rd: when A wins in 6 points

and so on.....

The cases can be easily evaluated

[witzar]

Game cannot end after odd number of points (since we start from a deuce).

Let's call a two consecutive points a turn (game can end after some number of turns).

Let:

a = .6*.6 = .36 = probability for player "A" to win in 1 turn from deuce

b = .4*.4 = .16 = probability for player "B" to win in 1 turn from deuce

d = 1 - (.36+.16) = .48 = probability of deuce after 1 turn from deuce.

Let A, and B be the probabilities for players "A" and "B" respectively to win the game.

Obviously:

A + B = 1

since "infinite deuce" probability is zero.

Now player "A" can win in 1 turn, 2 turns, and so on.

Probability for player "A" to win in exactly n turns equals d^(n-1)*a (since it only can happen with (n-1) consecutive d-turns followed by a-turn).

Therefore we have:

A = a + da + dda + ddda + ... = a(1 + d + d^2 + ...)

Let

X = (1 + d + d^2 + ...)

Now we have:

A = aX

We can apply the same for the player "B", so:

B = bX

We know that A + B = 1, therefore

aX + bX = 1

From above equation we can find X:

X = (a+b)^-1 = (.36 + .16)^-1 = .52^-1

Now:

A = aX = .36 * .52^-1 which is approximately 0.6923 and this is the correct answer.

[Guest]

Can we not look at it this way? THe question stated that "at any given point in the game A has a 60% chance of winning and B has a 40% chance". At duece, that is a given point in the game, hence A has a 60% chance an d heB has a 40 % chance of winning.

[Guest]

Can we not look at it this way? THe question stated that "at any given point in the game A has a 60% chance of winning and B has a 40% chance". At duece, that is a given point in the game, hence A has a 60% chance an d heB has a 40 % chance of winning.

You're looking for the probability of each player winning two consecutive points. I go with 60/40 which is the same as each player winning a particular point.

My logic is that we are only looking at the possibility of A winning two consecutive points or B winning two conecutive points. We can ignore all other combinations. If we were tossing a coin and the question was which side is most likley to turn up on two consecutive throws, you would have no problem saying they have equal an equal probability. Likewise here, they have a 60/40 probability.

[Guest]

Can we not look at it this way? THe question stated that "at any given point in the game A has a 60% chance of winning and B has a 40% chance". At duece, that is a given point in the game, hence A has a 60% chance an d heB has a 40 % chance of winning.

It actually says "In that game, A has a 60% probability of winning any given point"

You're looking for the probability of each player winning two consecutive points. I go with 60/40 which is the same as each player winning a particular point.

My logic is that we are only looking at the possibility of A winning two consecutive points or B winning two conecutive points. We can ignore all other combinations. If we were tossing a coin and the question was which side is most likley to turn up on two consecutive throws, you would have no problem saying they have equal an equal probability. Likewise here, they have a 60/40 probability.

The difference here is that once Player A has the advantage he is more likely to win than go back to Deuce whereas when B has the advantage, he's more likely to go back to Deuce than win.

Edited May 27, 2010 by Tuckleton

[Guest]

It actually says "In that game, A has a 60% probability of winning any given point"

The difference here is that once Player A has the advantage he is more likely to win than go back to Deuce whereas when B has the advantage, he's more likely to go back to Deuce than win.

I think from a "duece" situation you are still left with 60/40. If we were to say that A had 100% chance of winning the next point, then you would say that they will have 100% chance of being the first to win two consecutive points. If (as with the coins) A has 50% chance of winning the next point then he will have 50% chance of being the first to win two consecutive points. Likewise if A has a 0% chance of winning the next point then he has 0% chance of being the first to win two consecutive points. Why does this logic not prevail for all other combinations of probabilities?

[Guest]

Well, one of A or B will win. If, as you say, A's probability of winning is .36, then B's

probability of winning is .64 -- larger than that of A, the better player. How strange!

Point taken... (no pun intended, really)

It appears that I am under-thinking it...

Smoo

[Guest]

I there is 40% and 60% probability how did they get to duce in the first place? The game itself shows they both at that point have a 50% 50%.

[Guest]

I there is 40% and 60% probability how did they get to duce in the first place? The game itself shows they both at that point have a 50% 50%.

Random variance.

[Quantum.Mechanic]

Just for my education, is there a closed form solution for the infinite sum (1 + y + y^2 + ...) ???

[superprismatic]

Just for my education, is there a closed form solution for the infinite sum (1 + y + y^2 + ...) ???

If y<1, then there is and it is 1/(1-y).