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Our team is playing another team that wins 70% of its games. Our team is pretty good but wins only 60% of its games. If these teams have never played before and they've played equally hard opponents, what is the chance our team will be victorius? I think I solved it by comparing the game to flipping coins, but was looking for an answer a little more mathematical or rigorous.

My approach is based on flipping coins. The other team's "coin" comes up heads 70% of the time (and 30% tails) and our team's "coin" comes up heads 60% of the time (and 40% tails). If I flip the coins many times and disregard the outcomes when both coins are heads or both coins are tails, I would expect 18% of the flips our team's coin would have a head and their team's coin would be a tail (0.6*0.3); and 28% of the flips the other team would get a head and our team a tail (0.7*0.4). Thus, our team would be expected to win 39% of the games (0.18/(0.18+0.28) and the other team 71%.

I thought I'd ask if this makes sense and for a more mathematical approach. Thank you

StubbornAtom
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tedretiree
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    There is no rigid answer to this, just about any result is possible. Of course you can add assumptions and come up with an answer that way. here is a related question with some discussion. – lulu Mar 25 '20 at 17:01
  • Note: I'd say that the model you propose is too lopsided. In practice, I (for one) would not expect such an overwhelming advantage to the stronger team. You could, of course, vary matters by saying that the teams split the $HH$ and $TT$ cases event, or something like that. As I say, it is not difficult to come up with possible models. – lulu Mar 25 '20 at 17:04
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    $39%+71%=110%$ – saulspatz Mar 25 '20 at 17:09
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    To win, ya gotta give 110% – David P Mar 25 '20 at 19:17

1 Answers1

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There is no correct answer, but the theory underlying Elo ratings in chess implicitly uses odds-ratios. This is similar to what you have done with your flipping argument. One way of applying it here might look like:

  • If the first team has a probability of winning against a standard team of $70\%$ and losing of $30\%$ then the odds of it winning are $\frac73$

  • If the second team has a probability of winning against a standard team of $60\%$ and losing of $40\%$ then the odds of it winning are $\frac64=\frac32$

  • If you divide one by the other, you get an odds-ratio of $\frac{14}{9}$ in favour of the first team which might suggest its probability of beating the second team could be around $\frac{14}{23} \approx 60.87\%$

  • Similarly you get an odds-ratio of $\frac{9}{14}$ for the second team which might suggest its probability of beating the first team could be around $\frac{9}{23} \approx 39.13\%$

Those are the same results as your method should give. It has a degree of transitivity when applied to more than two teams, which is an attractive property though could only really be justified or not empirically.

Henry
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