Expected value about a tennis match's prize

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tomcps

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There is a tennis match which who first win 6 game would win the match. Winning a match would get a prize of $1280, it is known that A and B are equally good. A had already won 5 games, however the match stopped due to weather issue, how much should A get so that it is fair?



The answer is $1260
 
This is a classic type of problem! As this link says,

But the problem is not merely one of calculation; it also involves deciding what a "fair" division actually is.​

What thoughts do you have on that? (I should point out that the classic problem involves a game of chance, in which each person has the same chance of winning each round; that is not generally true in tennis, but is stipulated in the problem statement. It's important!)

You evidently have some reason to think the problem involves expected value; can you explain?
 
I actually found this question in an online expected value exercise posted by National Taiwan Normal University, and it only says "fair" on the question and that's it.

My thought on this is
"A and B are equally good" implies P(A wins a game)=0.5

P(A wins the match)
=(0.5)6+ (0.5)7 (7C6)+ (0.5)8 (8C6)+ (0.5)9 (9C6)+ (0.5)10 (10C6)+ (0.5)11 (11C6)
=0.774414062
Expected value of A winning: 1280*(0.774414062) =991.2499994
Which means winner getting $796.0976563 is "fair" or it is the ideal expected value

P(A wins the match with only 5 match)
=(0.5)5+ (0.5)6 (6C5)+ (0.5)7 (7C5)+ (0.5)8 (8C5)+ (0.5)9 (9C5)
=0.75390625

Set x be the money that A can get when A only wins 5 match
0.75390625x = 991.2499994
x=1314.818652

But this doesn't make sense as the chance that A wins seem a bit too high, and the probability of A winning 5 games should be higher than 6 games and the prize shouldn't be larger than the original prize.
 
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I am not sure why you converted 0.774414062 to 793/1024 but you did. You do have one typo as 1280 became 1080. Please change that.
 
Jomo said:
I am not sure why you converted 0.774414062 to 793/1024 but you did. You do have one typo as 1280 became 1080. Please change that.
Click to expand...
Sorry, I calculated 793/1024 at first and then converted it to 0.774414062 later on, and forgot to change all of them.
 
Suppose that at some point A has a 75% chance of winning and B has a 25% chance of winning where the total winning is $1280. if you (A or B) were offered to stop the game now and take a fair amount do you see that A should get 75% of the money and B should get 25% of the money?

We can even use exceptions to arrive at that. P(A wins) is .75 and P(A loses)=.25. Now if A wins then A gets $1280 and if A loses then A gets $0

E(A) = .75(1280) + .25(0) = .75(1280) = 960

Using this method can you get your answer?

Do you really think that it is possible for A to fairly receive 1314.818652 which is more than 1280? Even if A won the first 5 games there is a chance that A will lose the set! Also, would you ever turn down more than $1280 to end the game early. Of course not. Why, because it is not a fair amount and the unfairness is in your favor!!
 
Oh, I see that you did what I suggested. But then you changed your thinking and did something else.

You calculated the P(A wins after 5 matches) and got something other than 0. A (or B) can not win after 5 matches according to the rules that says to win the match you need to win 6 games. What you did there is all wrong.

Unless I am missing something what you did below is correct.
P(A wins the match)
=(0.5)6+ (0.5)7 (7C6)+ (0.5)8 (8C6)+ (0.5)9 (9C6)+ (0.5)10 (10C6)+ (0.5)11 (11C6)
=0.774414062
Expected value of A winning: 1280*(0.774414062) =991.2499994
Which means winner getting $796.0976563 is "fair" or it is the ideal expected value.

Now you said that winning $796.... is fair. So why did you change your mind?????
 
tomcps said:
Sorry, I calculated 793/1024 at first and then converted it to 0.774414062 later on, and forgot to change all of them.
Click to expand...
Had you read the link Prof Peterson posted you would have seen this is one of the oldest questions in modern probability: The Problem of Points.
But to solve one needs to know how many points each player has. As I read the post you say \(\bf{A}\) has five points out of six needed to win. Without knowing how many \(\bf{B}\) has there is no way to come up with a fair division. Consider \(\bf{B}'s\) having only \(\bf{1}\) point as oppose to \(\bf{B}'s\) having four points. Clearly those two possibilities would yield different fair divisions.
Here again is the link.
 
I was going to say the same thing: We aren't told how many games B has won. Since the only way it can be solved is to assume that means that B hasn't won any (that is, "A has already won 5 games" means "A has won all 5 games so far"), that's what I would assume; and that does give the claimed answer, fairly easily.

But I don't see any correct work yet.

If A has won 5 games, he needs only one more, while B needs 6. So what ways are there in which A will win? We could list them as A, BA, BBA, BBBA, BBBBA, BBBBBA. (If the winners were BBBBBBA, then B would have won.) What is the probability of each case? Then, what is the expected value for A?

And, of course, you need to explain why that is the fair amount to give A.
 
[MATH]\text{A has won 5 games and B has won an unknown $N$ games, $0\leq N\leq 5$}\\ P[\text{A wins 6 games|B has won $k$ games}] = \left(\dfrac 1 2\right)^k \cdot \dfrac 1 2\\ P[\text{A wins 6 games}] = \sum \limits_{k=0}^5 \left(\dfrac 1 2\right)^k \cdot \dfrac 1 2 = \dfrac{63}{64} \text{ (I'll let you work that out)}\\ \dfrac{63}{64} \cdot 1280 = 1260[/MATH]
 
P(A wins the match) =(0.5)6+ (0.5)7 (7C6)+ (0.5)8 (8C6)+ (0.5)9 (9C6)+ (0.5)10 (10C6)+ (0.5)11 (11C6)=0.774414062

I guess that what I agreed with above is wrong.
It is a bit late in NY now so I do not want to think about this right now. But I think about it soon as I was sure that it was correct. Oh well.

Ah, the last win must be by A. I will think about how to fix the above work, if possible.
 
Here is what I think p(A wins) should be.
P(A wins) = P( wins in 6 games) + P( wins in 7 games) + P( wins in 8 games)+ P( wins in 9 games)+ P( wins in 10 games)+ P( wins in 11 games) = (0.5)6+ (0.5)7 (6C5)+ (0.5)8 (7C5)+ (0.5)9 (8C5)+ (0.5)10 (9C5)+ (0.5)11 (10C5)
Now that equals 1/2 which is wrong. Where is not thinking going sour?
 
OK, I see it.
It is as if it is a new match. A wins if they win 1 game before B wins 6 games.

So P(A wins) = P(A) + P(BA) + P (BBA) + P(BBBA) + P(BBBBA) + P(BBBBBA) where for example BBA means B wins the 6th and 7th game and A wins the 8th game.
P(A wins 6 games) [MATH]= \sum \limits_{k=0}^5 \left(\dfrac 1 2\right)^k \cdot \dfrac 1 2[/MATH]
 
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