We've got two teams that are playing a bo5 match.

Bookies odds are:

Team 1: 1.45
Team 2: 2.70

So the win probability should be more than 69% and 37% if we want to bet and be profitable. The question is, how to figure out the (implied) odds for each game?

Is there a fast way to figure out an accurate estimate?

Thank you!

Theres an equation that will let you figure this out. I forget what it is named but I know it exists, I used it before. Theres one equation that works if the outcome of the first event changes the odds of the outcome of the second event, and theres a different equation if the outcomes dont change based on previous events.

try googling for probability equations.

It is probably best to move this post to the Probability Forum.

There does not appear to be sufficient information to answer your question. I guess you are assuming that the prob of each team winning each game remains the same, but this need not be true.

In baseball, the starting pitchers can greatly affect the prob of individual games. For example, think back to World Series between San Francisco and Kansas City. The Giants were favored in games that Madison Bumgarner started but underdogs in games that he did not.

Anyway, in the case in which the probabilities do not change from game to game, the overall probabilities are given by the Binomial Distribution (in fact the cumulative binomial distribution).

Given each game's probs, it is straightforward to calculate the overall best-of-five series prob that each team wins. You want to reverse the direction so that you can derive the individual game probs given the overall best-of-five series probabilities.

The formula for the cumulative binomial probabilities are available in many places on the internet and available on many calculators and computer programs (such as Excel).

Let us know if you have any other questions.

From:

P(x) = C(2, 2)·x³ + C(3, 2)·x³·(1 - x) + C(4, 2)·x³·(1 - x)²

simplifies to:

P(x) = ((6·x - 15)·x + 10)·x³

Setting P(x) = .69 and entering this into WolframAlpha to solve for x gives: x ≈ 60.4%.
Setting P(x) = .37 and entering this into WolframAlpha to solve for x gives: x ≈ 43.0%.

There may be a nifty way to solve for x without resorting to WolframAlpha, but I'm not very good at figuring those out. Maybe someone out there wants to give it a shot?

I used the following Python program as a sanity check and it is in agreement with the above derived formula:

An interesting approximation for P(x) limited to the range 0 ≤ x ≤ .50 is:

P(x) ≈ (12·x - 1)·x/5

Note: To compute P(x) = .69, solve P(x') = 1 - .69 for x', then the answer will be x = 1 - x'.


Another one limited to the range 0.30 < x ≤ .50 is:

P(x) ≈ 1.74·x - .37

This one is useful for matches that are not so lopsided and is suitable for paper and pencil computation FWIW.

The following recursive program is more elegant:
Testing as follows verifies the program:
Which produces the familiar 69% and 37% figures. The .604321 and .429748 numbers came from the WolframAlpha runs. If you wanted to generate these without using WolframAlpha, you would have to do a binary search to find them or use something like the secant method.

BTW, the recursive solution can be applied to Best-Of-Three, Best-Of-Seven, etc. It can also be used for partially completed matches as well e.g. if you are a 60% favorite to win an individual game and are leading 2-0 in a best of seven, you would compute BestWL(.60, 2, 4). The 2 is how many wins you need and the 4 is how many the opponent needs.

This is very nicely done and presented.

And, yes, closed form solutions exist for the general cubic equation.

It's a 5th degree polynomial.

Ha ha. Nevermind. I am a doofus.