Post
by Prasutagus » Sat Jun 17, 2017 7:26 pm
If the casino is so sketchy that there's only a 0.95 probability you'll be dealt an honest game, and you think one form of cheating might be repeatedly dealing blackjacks to the house, and then the first two hands are dealer blackjacks, you can say it's now likely you are being cheated. With the new information, your estimated probability of being dealt an honest game has dropped from 0.95 to something much lower.
(But if you're at a reputable casino, the prior probability of being dealt an honest game is so high, and moreover if they do cheat it's likely to be much subtler, that two dealer blackjacks does not cause concern about the integrity of the game.)
As for Jeopardy auditions, you have misinterpreted my argument. I stated three times that "A" in my example is the event that invitations to Jeopardy auditions continue to be made without regard to sex, from a population whose ratio is approximately 70:30 male to female. Therefore a significant change in the 70:30 ratio, for any reason, falls under "not A". I've shown that if P(A) was initially 95%, it's now around 5% to 15%. I never suggested that this decrease is because "women must be treated better than men". One way in which "A" might no longer be true is a significant change in the sex ratio of those who pass the online test (which might be caused either by a relative increase in the number of women who take the test, or by a relative increase in the average score of women who take it).
You pointed out that in my calculation of P(B|~A)=0.3459 I assumed that under "~A", the ratio would be 50:50. This was a simplifying assumption and does not dramatically change the results. If you think that my estimate of P(B|~A) is too high, you can replace the 0.3459 by, for example, 0.25, which results in a value of P(B)=0.015 instead of 0.02, and then P(A|B) is 14.6% rather than 10.9%. Or if you think that my estimate of P(B|~A) is too low, replace the 0.3459 by, say, 0.6, resulting in P(B)=0.032 and P(A|B)=6.8%. In any case, we get the result that P(A|B) is much less than P(A)=95%.
Over the last several years (up until a few weeks ago) it's been observed that men outnumber women at in-person auditions, but there are roughly the same number of male and female contestants appearing on the show. In other words, women who audition get the Call at a higher rate than men who audition. That doesn't necessarily mean there is an affirmative-action effect. Another explanation might be this: if men are more likely to cheat on the online test, then women would be more likely than men to pass the in-person written test. Another possibility is that women at the auditions have had on average more energy, better enunciation, or some other quality the contestant coordinators are looking for. The underlying reasons, we do not know. We do know that (over the past few years) women at in-person auditions are selected for the show at a higher rate than men. Therefore, it's plausible that the contestant coordinators recently decided to invite more women to auditions, since they expect to select an equal number of men and women to appear on the show. That's one possibility. Another possibility is that significantly more women are taking the test, resulting in a change of the ratio from 70:30 to something else.
The point is, with Bayes' theorem we can infer that these two recent examples are probably indicative of a new pattern, rather than just being outliers.
