September 18th, 2020 at 6:04:01 PM
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So, here are the Beatles per/game bowling averages and per/game SD and variance:

John - 200 avg. 15 SD. (Var. 225)

Paul - 190 avg. 17 SD. (Var. 289)

George - 180 avg. 18 SD. (Var. 324)

Ringo - 170 avg. 20 SD. (Var. 400)

They bowl a one-game match. What I want to know is:

1) What percentage of the time will each one win

2) What would be the fair betting odds for this match

2) Is there a way to solve this problem numerically.

First, I realize that this problem can be "solved" through a Monte Carlo simulation....and that's fine. But, I'm curious as to the numeric solution. There IS a method for solving a two-person match using the NORMDIST function in excel. For example, the probability of John vs. Paul = 1-NORMDIST(0,20, sqrt(514) ,TRUE). (Where 20 is the mean difference in scores and sqrt(514) is the combined SD of both players. I don't know how to do this for more than 2 players.

Second, could this system work --> I compute the two-person percentage for John winning the match against each opponent = (John beating Paul *John beating George*John beating Ringo). Repeat for each player. Sum the total percentages of each player winning and divide by each player's winning percentage.... So, the fair betting odds for John = (John % beating all three/ (John % beating all three+Paul % beating all three+ George % beating all three +Ringo % beating all three))

To put this second method another way...... if you had 3 guys flipping coins in a round-robin, one-flip match.... the odds of each player winning is .5*.5 = .25. So, the fair betting odds would be .25/.75 = .33 = +200....which intuitively makes sense. (Notice that the total wins don't sum to 1.00 that's because in this system there are matches in which A beats B, B beats C, but C beats A. I conveniently ignore these possibilities, because this type of result can't happen in my bowling match.....I'm only interested in the odds of each one winning outright.) This seems like it should work.

Comment if you'd like. If not... Let It Be.

- railer

John - 200 avg. 15 SD. (Var. 225)

Paul - 190 avg. 17 SD. (Var. 289)

George - 180 avg. 18 SD. (Var. 324)

Ringo - 170 avg. 20 SD. (Var. 400)

They bowl a one-game match. What I want to know is:

1) What percentage of the time will each one win

2) What would be the fair betting odds for this match

2) Is there a way to solve this problem numerically.

First, I realize that this problem can be "solved" through a Monte Carlo simulation....and that's fine. But, I'm curious as to the numeric solution. There IS a method for solving a two-person match using the NORMDIST function in excel. For example, the probability of John vs. Paul = 1-NORMDIST(0,20, sqrt(514) ,TRUE). (Where 20 is the mean difference in scores and sqrt(514) is the combined SD of both players. I don't know how to do this for more than 2 players.

Second, could this system work --> I compute the two-person percentage for John winning the match against each opponent = (John beating Paul *John beating George*John beating Ringo). Repeat for each player. Sum the total percentages of each player winning and divide by each player's winning percentage.... So, the fair betting odds for John = (John % beating all three/ (John % beating all three+Paul % beating all three+ George % beating all three +Ringo % beating all three))

To put this second method another way...... if you had 3 guys flipping coins in a round-robin, one-flip match.... the odds of each player winning is .5*.5 = .25. So, the fair betting odds would be .25/.75 = .33 = +200....which intuitively makes sense. (Notice that the total wins don't sum to 1.00 that's because in this system there are matches in which A beats B, B beats C, but C beats A. I conveniently ignore these possibilities, because this type of result can't happen in my bowling match.....I'm only interested in the odds of each one winning outright.) This seems like it should work.

Comment if you'd like. If not... Let It Be.

- railer

September 19th, 2020 at 8:42:01 AM
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I think you need a function that maps the interval (0,1) to a normal distribution corresponding to one of the bowlers.

For example, assume Paul is playing Ringo.

Let r be a uniform random variable, 0 < x < 1.

Find Paul's score on the normal curve used for Paul's score (the one with mean score 190 and SD 17) such that the probability that the score < that value = r.

Now, use the normal curve for Ringo's score to determine the probability that Ringo's score > Paul's score.

I think the overall probability =the integral from 0 to 1 of (the integral from P'(x) to positive infinity of R(y) dy) dx, where R(x) is Ringo's score and P'(x) is the mapping of the interval 0 < x < 1 to the Paul score graph.

The problems are, the second function does not have a closed form solution, but requires approximation, and I am assuming the first function does as well.

Also, this doesn't take into account that scores are discrete values. The most accurate way is to calculate the probabilities of Paul getting each possible score from 0 to 300, doing the same for Ringo, and summing the 45,150 pairs of scores where Ringo beats Paul.

For example, assume Paul is playing Ringo.

Let r be a uniform random variable, 0 < x < 1.

Find Paul's score on the normal curve used for Paul's score (the one with mean score 190 and SD 17) such that the probability that the score < that value = r.

Now, use the normal curve for Ringo's score to determine the probability that Ringo's score > Paul's score.

I think the overall probability =the integral from 0 to 1 of (the integral from P'(x) to positive infinity of R(y) dy) dx, where R(x) is Ringo's score and P'(x) is the mapping of the interval 0 < x < 1 to the Paul score graph.

The problems are, the second function does not have a closed form solution, but requires approximation, and I am assuming the first function does as well.

Also, this doesn't take into account that scores are discrete values. The most accurate way is to calculate the probabilities of Paul getting each possible score from 0 to 300, doing the same for Ringo, and summing the 45,150 pairs of scores where Ringo beats Paul.

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