Frank was worried about his school team. The start quarterback, who had led them to a win in last year's superbowl, was hurt. The coach said there was a 75% chance he was going to play this weekend, but Frank wasn't happy about those odds.

If the star QB played, Frank figured his team had a 75% chance of winning. Without him, Frank gave the cross town rivals a 50% shot of upsetting his team.

Unfortunately, Frank had to go away for the weekend and was unable to watch the game. On Monday, he found out his team had won.

What are the odds the star QB played? Assume Frank was right about the odds he gave.

## Friday, January 16, 2009

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I think I've got a thought:

ReplyDeleteThere was a 75% chance of the quarterback playing and a 25% chance of him not playing before we had any more information.

Of the 75% of times we'd expect him to play in repeated runs of the scenario, the team would win 75% of the time (or .75*.75=56.25% of the total) and lose 25% of the time (or .75*.25=18.75% of the total).

Of the 25% of times we'd expect him not to play in repeated runs of the scenario, the team would win 50% of the time (or .25*.5=12.5% of the total) and lose 50% of the time (or 12.5% of the total).

So in total, you would expect the team (before getting any more information) to win 68.75% of the time and lose 31.25% of the time.

Now, that 68.75% expected winning percentage is composed of 56.25% of scenarios in which the QB played and 12.5% of scenarios in which he did not. So, the percentage of winning scenarios involving his play is 56.25/68.75 = 81.82%.

So I believe that that is our answer: Given the information that the team won, the QB is 81.82% likely to have played.

Let P denote the event that the star player played, and W the event that the team wins. Now, we have p(P) = 3/4, p(W|P) = 3/4 and p(W|not-P) = 1/2.

ReplyDeleteUsing Bayes' theorem, we know that since the team won, we're looking at p(P|W) which is

= p(W|P) * p(P)

-------------

p(W)

= p(W|P) * p(P)

-------------

p(P|W) + p(P|W-not)

= 3/4 * 3/4

---------

3/4 + 1/2

= 9/20

Therefore, there is a 45% chance that the star player played.

I agree with everything you said Sudipta, except for the P(W). That seems high to me. I believe P(W) = 9/16 + 1/8 = 11/16.

ReplyDeleteSo, the probability the star player played was 9/11.

I could be wrong, but 45% seems very low.

You guys are thinking too much.

ReplyDeleteTheres a 75% chance he played :)