Amusing math problem

[Justin7]

I was reminiscing on some of the stuff I've dabble in, and I felt minorly clever for figuring this out.

In a best of 3 sets tennis match, player "A" is a -200 favorite (no juice) to win the match against "B". If they played the best of 5 sets, what would be the fair price on "A"? Assume the players are robots, and don't change mentally or physically.

[dwaechte]

I tried going about this in an absolutely ridiculous manner and ended up having to solve a cubic, at which point I gave up.

I'm guessing there's a damn easy way to go about this that I'm missing.

[Justin7]

Actually, going from best of 3 to best of 5 is easy - just quadratic. Going from 5 to 3 is much harder.

[Cabo]

Where's Ganch when you need him?

[pico]

what is the odds of him winning 1st set. once you figure that out, it would be pretty easy to extroplate.

[Cappy]

I don't know, but if these guys are robots, then someone who has a small skill advantage (at -200 they should have an incredible advantage) ends up with an enormous advantage in 5 sets, so I would imagine you'll end up with a line (this is a guess) somewhere areound -600

[Cappy]

It's very strange because, tennis is more mental and includes a lot of mental and physical exhaustion

[Ganchrow]

I was reminiscing on some of the stuff I've dabble in, and I felt minorly clever for figuring this out.

In a best of 3 sets tennis match, player "A" is a -200 favorite (no juice) to win the match against "B". If they played the best of 5 sets, what would be the fair price on "A"? Assume the players are robots, and don't change mentally or physically.

Let P_A3 = probability of player A winning a 3-set match = 2 3
Let P_A5 = probability of player A winning a 5-set match
Let P_A = probability of player A winning any given set

Hence:

P_A3 = P_A² * ( combin(2,2) + (1 - P_A) * combin(2,1) ) = 2 3
P_A3 = P_A² * ( 1 + 2*(1 - P_A) ) = 2 3

Solving the above cubic equation for P_A yields:

P_A ≈ 61.3037%

So to solve for P_A5 we could using Excel's BINOMDIST() function as follows:

P_A5 = 1 - BINOMDIST(3-1, 5, P_A, 1)
P_A5 ≈ 1 - BINOMDIST(2, 5, 61.3037%, 1)
P_A5 ≈ 70.483%

This implies that the fair odds on Player A winning the 5-set match would be roughly -238.8.

Easy peasy lemon squeezy.

[AgainstAllOdds]

Ganch...did you know that was the formula that you had to use right when you saw the problem or did you have to think about and/or look it up first?

[Bet Shooter]

Ganch...did you know that was the formula that you had to use right when you saw the problem or did you have to think about and/or look it up first?

Does it matter? The rest of us couldn't do it with three weeks time and Sagan as our neighbor!

[LT Profits]

I wish I were smart.

[Cappy]

I wish I were smart.

I wish you were too

[Cappy]

Ganch...did you know that was the formula that you had to use right when you saw the problem or did you have to think about and/or look it up first?

eff that, he sat on the keyboard

[Ganchrow]

Ganch...did you know that was the formula that you had to use right when you saw the problem or did you have to think about and/or look it up first?

It's really just a straightforward problem of probabilistic combinatorial logic.

Think about it and come up with a solution.

[dwaechte]

Well what do you know, I actually did have it if I had kept going with it. Unlike Ganch, I had to screw around for a while to get the formula but I ended up with the same thing, and would've done the last step the same way as well.

How did you manage with just a quadratic Justin?

[Justin7]

How did you manage with just a quadratic Justin?

Forget that. It wasn't quadratic.

Now, if you want something harder...

Best of 5 match - favorite is -150.

What is fair price for best of 3?

[Ganchrow]

Forget that. It wasn't quadratic.

Now, if you want something harder...

Best of 5 match - favorite is -150.

What is fair price for best of 3?

This is obviously fundamentally no different than the problem as posed above.

Let P_x = probability of player winning a x-set match (for all odd integers x > 0)
Let P = P₁ = probability of player A winning a single set match

Then in general:

[nbtable][tr][td]P_x = P ^{(x+1) 2} * [/td][td][/td][td]{ (1-P) ^{(2i-x-1) 2} * combin(i-1, (x-1) 2 ) }[/td][/tr][/nbtable]

So given x = 5 and P_x = 150 250 = 60%:

P₅ = P * ( 1 + (1 - P) * 3 + (1 - P)² * 6) = 60%

Solving the quintic equation for P yields a single real root of P ≈ 55.375%.

Plugging back into the summation above:

P₃ = P² * ( 1 + (1 - P) * 2)
P₃ ≈ 55.375%² * ( 1 + (1 - 55.375%) * 2)
P₃ ≈ 58.031%

For a fair money line of roughly -138.3.

Note that we can also obtain the same probability via the Binomial distribution:

P₃ = 1 - BINOMDIST(2-1, 3, 55.375%, 1)
P₃ ≈ 58.031%

The very simple attached spreadsheet will determine answers to problems such as these using the Binomial Distribution+Solver.

[Justin7]

Ganch, you lost me buddy. Your proofs may or may not be true, but I'd love to see more explanation behind them, instead of just the answer. In particular, your "In general" for Px skips a few steps of explanation.

Simplify it for us pions, please! If you can gear your explanations so that anyone who stopped taking math classes after Calc I can follow, you'll have a fan club so big, you'll have to have security to keep out the teeny-bop sex freaks.

[Ganchrow]

Ganch, you lost me buddy. Your proofs may or may not be true, but I'd love to see more explanation behind them, instead of just the answer. In particular, your "In general" for Px skips a few steps of explanation.

The same question was posed twice, simply with several parameters modified. Rather than once again offer a one-off solution to the problem as I did the first time, the second time around I simply presented the general solution to what's really a fairly simple problem of probabilistic combinatorics. If one understood the answer to problem as first posed (probability of winning a 3-set match known, probability of winning a 5-set match unknown), I'd think the the general reasoning would be readily apparent via induction.

In this way we'd be able to avoid having to once again manually spell out the logic for a problem such as:

Best of 189 match - favorite is -185.

What is fair price for best of 101?

The real question, however, is anyone (other than myself) really going to care enough to read this post in its entirety? Over/under at 0.5, perhaps?

That said, here's a description of the terms involved.

[nbtable][tr][td]P_x = P ^{(x+1) 2} * [/td][td][/td][td]{ (1-P) ^{(2i-x-1) 2} * combin(i-1, (x-1) 2 ) }[/td][/tr][/nbtable]

Term #1: x -- This refers to the match length in sets. This number must be odd.

Term #2: (x+1)/2 -- This refers to the number of sets either player would need to win in order to win the match. For example, if the match length were 3, then (3+1)/2 = 2 sets would need to be won in order to win the match. if the match length were 5, then (5+1)/2 = 3 sets would need to be won in order to win the match.

Term 3: P ^{(x+1) 2} -- This refers to the probability of winning precisely the required number of sets in order win the match without a single loss. For example, in a 5-set match, this would be probability of winning the fisrt 3 sets.

Term 4: [nbtable][tr][td][/td][td]  [/td][td]This summation runs from the minimum number of possible sets to the maximum number of possible sets. So for example given a hypothetical 11-set match, this summation would run from 6 sets to 11 sets.[/td][/tr][/nbtable]

Term 5: (2i-x-1) 2 -- This term refers to the number of sets the player in question would need to lose in order for the match to go exactly i sets but still win the match. For example, in a 5-set match, were i=4, then this term would be 1, meaning that the player would lose exactly 1 set were he to win a 5-set match in exactly 4 sets.

Term 6: (1-P) ^{(2i-x-1) 2} -- This term refers to the probability of the player losing exactly (2i-x-1) 2 sets.

Term 7: combin(i-1, (x-1) 2 ) -- This term refers to the number of ways a player could win (x-1) 2 = (x+1) 2 - 1 out of i-1 games. We subtract 1 from each term because we know that the winning player must win the final game. So for example, in a 5-set match, there would be combin(4-1, (5-1) 2 ) = combin(3, 2) = 3 ways for the match to terminate in exactly 3 sets. Similarly, in a 3-set match, there would be combin(3-1, (3-1) 2 ) = combin(2, 1) = 3 ways for the match to terminate in exactly 3 sets.

So in plain English:

Probability of winning a 5-set match =
 prob(winning 3 of 3) * [# of ways to win exactly 2 games out of 2] +
 prob(winning 3 of 3) * prob(losing 1 of 1) * [# of ways to win exactly 2 of 3] +
 prob(winning 3 of 3) * prob(losing 2 of 2) * [# of ways to win exactly 2 of 4]

And of course:

prob(winning 3 of 3) = prob(single set win)³
prob(losing 1 of 1) = 1-prob(single set win)
prob(losing 2 of 2) = (1-prob(single set win))²
# of ways to win exactly 2 games out of 2 = combin(2,2)
# of ways to win exactly 2 games out of 3 = combin(3,2)
# of ways to win exactly 2 games out of 4 = combin(4,2)

Oh and in case anyone was actually curious:

Best of 189 match - favorite is -185.

What is fair price for best of 101?

Answer: ~ -156.7 (or about 61.036%)

[dwaechte]

Yeah the method for the next one is obviously the same but I probably would've made a stupid mistake on the math itself if I had done it.

And although I read through and understood the series being used, there's no way I would've come up with it on my own. But Ganch, I believe I just pushed it over the 0.5 line you set

[LT Profits]

1.5 so far. Cash those Over tickets.

[Dark Horse]

[nbtable][tr][td]P_x = P ^{(x+1) 2} * [/td][td][/td][td]{ (1-P) ^{(2i-x-1) 2} * combin(i-1, (x-1) 2 ) }[/td][/tr][/nbtable]

Yeah, but can you get the girl?

Oh, that's right, you can.

[B1GER1C828]

i got infinity...damnit

[St.Aquinas]

Nice work!

Ganch smart me dumb.