# Tennis probabilities

1. Aug 28, 2009

### musicgold

Hi,

There are two tennis players. Player A wins a point 50% of the time (against any rival), while Player B wins a point 60% of the time against any rival. That is P(A) = 0.50 and P(B) = 0.60.

If Player A plays with Player B, what is the probability that Player A beats Player B, i.e. P(A, B) and the probability that Player B beats Player A, P(B, A) ?

I am not sure how I should simulate this problem.

Thanks,

MG.

2. Aug 28, 2009

### mXSCNT

Intuitively, since A is an "average" player and B beats the "average" player 60% of the time, perhaps B should win 60% of the time against A.

But it doesn't necessarily work like that. For example, perhaps A and B are equally skilled when playing against people at a given experience level, except that B is better at taking advantage of the weaknesses of poor players than A is. That would give B a higher win percentage, without giving him any advantage against A.

Conclusion: you just don't have enough information.

3. Aug 30, 2009

### winterfors

Player B has a P(A)*P(B) / (P(A)*P(B) + (1-P(A))*(1-P(B))) = 60% chance of winning a point against player A.

The probability of then winning a MATCH against A will then be significantly higher (close to 100%) depending on if they play best of three or five sets.

4. Aug 30, 2009

### mXSCNT

Justify your claim. What assumptions are you making?

Last edited: Aug 30, 2009
5. Aug 31, 2009

### musicgold

I am reading a book: Calculated Bets by Steven Skiena. The book explains mathematical modeling to beat the JAI ALAI betting system.
The author used the following function get the answer to the above problem.

$$P (A , B ) = \frac { 1 + [ P (A) - P(B) ]^ \alpha} {2}$$ if P(A) >= P(B)

$$P (A , B ) = \frac { 1 - [ P (B) - P(A) ]^ \alpha} {2}$$ if P(A) <= P(B)

The constant $$\alpha >= 0$$ is used as a fudge factor to tune the results of this function to the observed stats.

6. Sep 1, 2009

### winterfors

I was using a Bayesian approach.

Let's denote the probability that player $$X_1$$ wins against player $$X_2$$ :
$$P(W | X_1, X_2)$$

We know the following:

$$P(W | A) = \sum\limits_{X}P(W | A, X)P(X) = 0.5$$ (called P(A) in posts above)
$$P(W | B) = \sum\limits_{X}P(W | B, X)P(X) = 0.6$$ (called P(B) in posts above)

Furthermore, chance of anyone beating anyone is 50%
$$P(W) = \sum\limits_{X_1}\sum\limits_{X_2}P(W, X_1, X_2)= 0.5$$
due to symmetry.

We want to calculate $$P(W | A, B)$$. This can be done from assuming that
$$P(W, X_1, X_2) = P(X_1 | W)P(X_2 | W)P(W)$$.

Then,
$$P(W| A, B) = \frac{P(W, A, B)}{P(A, B)}$$
$$= \frac{P(A | W)P(B | W)P(W)}{P(A, B)}$$
$$= \frac{P(W|A)P(A) P(W|B)P(B) /P(W)}{(P(W|A)P(A) P(W|B)P(B) /P(W)) + P(L|A)P(A) P(L|B)P(B) /P(L)}$$
$$= \frac{P(W|A) P(W|B) } {(P(W|A) P(W|B)) + P(L|A)P(L|B) }$$
$$= \frac{0.5 * 0.6 } {0.5 * 0.6 + 0.5 * 0.4 }$$
$$= 0.6$$

where $$P(L|A)=1-P(W|A)$$ and $$P(L|B)=1-P(W|B)$$ is the probability of losing.

It all got a bit technical, sorry about that, but I can't find any more simple derivation...

Last edited: Sep 1, 2009
7. Sep 1, 2009

### musicgold

winterfors,

Thanks for the detail explanation. However, I am stuck at the following equation.
I would appreciate it if you could explain this to me.

I don't understand how we can sum individual probabilities to get a collective probability of A winning against anybody, as the sum may be bigger than 1.

Let us say, there are 4 players, A, B, X1 and X2, and P(W | A, X1) = 0.4, P(W | A, X2) = 0.7.

Then P(W | A, X1) + P(W | A, X2) > 1

IMHO, we should take an average (or weighted average) of individual winning probabilities to get a collective probability.

8. Sep 1, 2009

### winterfors

You're perfectly right it dosn't make sense. I meant the expectation over $$X$$ rather than the sum.
$$P(W | A) = \sum\limits_{X}P(W | A, X)P(X) = 0.5$$

(I have corrected my original post)

9. Sep 1, 2009

### musicgold

winterfors,

Thanks. I am still digesting your explanation. I am a bit confused with your notations. Could you explain what the following notations stand for?

$$P(X)$$

$$P(W, X_1, X_2)$$ versus $$P(W | X_1, X_2)$$

$$P(X_1 | W)$$

$$P(W)$$

10. Sep 1, 2009

### winterfors

$$P(X)$$ Probability that any random player will have the value $X$. If for instance $X$ is a number describing player's strength, there might be 10% of players with $X=1$, 40% with $X=2$ and 50% with $X=3$.

Note that this probability is not actually used to calculate $P(W | A, B)$, all terms $P(X)$ cancel out along the way.

$$P(W | X_1, X_2)$$ Probability of player $X_1$ winning over player $X_2$ (vertical bar indicated conditional probability of W (=winning), given players $X_1$ and $X_2$.

$$P(W, X_1, X_2)$$ Probability of player $X_1$ winning over player $X_2$ times relative abundances of players with $X=X_1$ and $X=X_2$, respectively.

$$P(W)$$ Probability that any random player will win over another random player

$$P(X_1 | W)$$ The probability of a random player $X$ will have the value $X = X_1$, given that he has just won a match.

11. Sep 1, 2009

### musicgold

winterfors,

Thanks for bearing with me. I get all the notations except $$P(W, X_1, X_2)$$.
Could you please explain that concept with an example?

12. Sep 1, 2009

### mXSCNT

You have a notational problem. The events separated by commas after the conditional bar should commute, but in your notation P(W|A,B) is apparently not the same as P(W|B,A). You need to keep track of which player the W refers to. For example, you might say P(W(A) | A, B), where W(A) means "A wins," A means "A plays," and B means "B plays." It looks like the fact that you don't do this leads to some confusion later on.

13. Sep 2, 2009

### winterfors

As mXSCNT points out, I have messed up the notation in my previous post (definitions of $P(W |X_1)$ and $P(W |X_2)$ are ambiguous). I have tried to make a more formal derivation below.

PRELIMINARIES AND NOTATION

Let's denote the probability that player $A$ wins against player $B$ :
(1) $$P(W | A_1, B_2)$$
where the indices 1 and 2 are only to mark that it is player 1 winning over player 2 (and not the other way around)

This implies that $P(W | A_1, B_2)$ is equal to the probability that player $B$ loses to player $A$
(2) $$P(W | A_1, B_2) = P(L | B_1, A_2)$$

One can define joint probability distributions over the space of all (pairwise) combinations of players and all outcomes (Win / Loss) by multiplying with $P(A)P(B)$
(3) $$P(W, A_1, B_2) = P(W | A_1, B_2) P(A) P(B)$$

One can also define marginal probability distributions by summing over all $X_1$ or $X_2$
(4) $$P(W , A_1) = \sum\limits_{X_2}P(W, A_1, X_2)$$

(5) $$P(W , B_2) = \sum\limits_{X_1}P(W, X_1, B_2)$$

(6) $$P(W) = \sum\limits_{X_1}\sum\limits_{Y_2}P(W, X_1, Y_2)$$

From equation (2) we can deduce also that $P(W, A_1) =P(L, A_2)$ and $P(W , B_1) =P(L , B_2)$.

Finally, we can define the conditional probabilites:
(7) $$P(W | A_1) P(A) = P(A_1|W) P(W) = P(W, A_1)$$

(8) $$P(W | B_2) P(B) = P(B_2|W) P(W) = P(W, B_2)$$

WHAT WE KNOW

(9) $$P(W | B_1) = 0.6$$

(10) $$P(W | A_2) = 0.5$$

(11) $$P(W) = P(L)= 0.5$$
The probability of a random player winning (or losing) against another random player is exactly 50%

(12) $$P(W, B_1, A_2) = P(B_1 | W) P(A_2 | W) P(W)$$

This assumption can be justified by the maximum entropy principle, since it minimizes the Shannon information of $P(O, B_1, A_2)$ (O= W or L) given that we know $P(B_1 | W)$, $P(A_2 | W)$ and $P(W)$.

DEDUCTIONS FROM ABOVE

We want to calculate
$$P(W| B_1, A_2) = \frac{P(W, B_1, A_2)}{P(B) P(A)}$$

Inserting (12) gives
$$P(W| B_1, A_2) = \frac{P(B_1 | W) P(A_2 | W) P(W)}{P(B) P(A)}$$

Using (7) and (8) we can finally write
$$P(W| B_1, A_2) = \frac{P(W| B_1) P(W| A_2 ) }{P(W)}$$

Inserting (9) (10) and (11) gives
$$P(W| B_1, A_2) = \frac{0.6 * 0.5 }{0.5} = 0.6$$

That is, the probability of player B beating player A is 60%

Last edited: Sep 2, 2009
14. Sep 3, 2009

### musicgold

winterfors,

Thanks a lot. Though your solution is not fully clear to me, I appreciate your efforts.
I think I need to do more reading on the Bayesian probability theory.

MG.

15. Sep 4, 2009

### Kittel Knight

Then, considering long time of observation (or "after a large amount of points played" ), for each 10 points played, Player A wins 5 points, and Player B wins 6 points...

:yuck:

Last edited: Sep 4, 2009
16. Sep 5, 2009

### Kittel Knight

It seems the OP, the way it is, leads to a strange conclusion!

Who is wrong: me, or the assumptions in the OP?