## The boxer, the wrestler, and the coin flip, again

Mike Grosskopf writes:

I came across your blog the other day and noticed your paper about “The Boxer, the Wrestler, and the Coin Flip” . . . I do not understand the objection to the robust Bayesian inference for conditioning on X=Y in the problem as you describe in the paper. The paper talks about how using Robust Bayes when conditioning on X=Y “degrades our inference about the coin flip” and “has led us to the claim that we can say nothing at all about the coin ﬂip”. Does that have to be the case however, because while conditioning on X=Y does mean that p({X=1}|{X=Y}I) = p({Y=1}|{X=Y}I), I don’t see why it has to mean that both have the same π-distribution where Pr(Y = 1) = π.

Which type of inference is being done about Y in the problem?

If you are trying to make an inference on the results of the fight between the boxer and the wrestler that has already happened, in which your friend tells you that either the boxer won and he flipped heads with a coin or the boxer lost and he flipped tails, then it makes sense that the two situations would be given the same inferential status. The distribution of π’ defined as (Pr(Y=1|{X=Y},I) = Pr(X=1|{X=Y},I) ~ a delta function at 0.5. I am not sure why this is objectionable.

π’ ≠ p(π|{X=Y},I) however because p(π|{X=Y},I) = p(p(Y=1|I)|{X=Y},I), which is basically saying how does conditioning on X=Y effect the distribution of possible probabilities for the prior of Y.

If you are trying to better understand what chance the boxer had going into the fight after conditioning on the information that X=Y, then p(π|{X=Y}I) is the relevant inference instead of π’. This is where you would expect no change in uncertainty in π when conditioning on the coin flip. As I laid out in the first email (though in a particularly messy and illegible format), this inference is not changed by conditioning on the coin flip in Bayesian analysis. We are still just as uncertain about what the boxer’s chances were (π). I would think that Bayesian analysis gives the correct analysis in both cases.

A better way to clarify what I was thinking is considering where, instead of conditioning on the result of the coin flip (Y=X), condition on the result of something essentially certain, like the sun will rise tomorrow (Y=A). If someone presented you with the information that “the boxer won just as sure as the sun will rise tomorrow” you would give the same inferential status as certain (p()=1 for both) and the distribution of π” = Pr(Y=1|{Y=A},I) would be basically a delta function at 1. However, if you were doing inference on what the boxer’s chances were going into the fight, p(π|{Y=A},I), you would not be certain if the boxer was unlikely to win and pulled an upset or if the boxer was a heavy favorite and easily followed through. Your distribution for π would be updated to be p(π|{Y=A},I) = 2*π (from the first email). All you would really be certain of is that the boxer had some chance, and you would feel that it was now more likely that he had a good chance to win instead of being the underdog. This again seems to work out fine using Bayesian with an uninformative prior.

My reply: I’m too tired to think about this, but I’ll post it and then maybe others will have some thoughts. The one thing I can tell you is that it’s an old example–I came up with it in discussions with Augustine Kong back around 1988.

1. Paul says:

Sounds like you want a probability distribution of probability distributions. If you take a coin and flip it a million times it'll probably end up 50/50 heads vs. tails. But it's possible there's some minor manufacturing defect, such that heads is very slightly preferred and you end up with 51/49. Thus rather than stating we expect a normal curve centered at 50%, we expect a normal curve where the mean itself is drawn from a tight normal distribution with a small s.d.

If we set a boxer and wrestler fighting over and over again (tricky experiment to design since you phrased it as to the death) we may find that the boxer wins 10% of the time, or 50% or 99%. Thus we expect a normal curve where the mean is drawn from a uniform distribution from [0,1]. In both cases the most likely result remains a 50/50 split, but in the boxer case we're now acknowledging that repeated experiments are unlikely to actually result in half wins for the boxer, but rather they'll collapse to some other win distribution.

2. aslak says: