[gumpzilla]

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There are 4 aces or tens left. we know 5 of the cards, so the chance he is holding an A OR a 10 = 4/47 + 4/46. This is about 8/48 or 1/6.

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From nitpick to quasi-nitpick to not really nitpick at all:

1) We know 7 cards, not 5, since we can also see the two in our hand.

2) Suppose there are 3 unseen cards (A,B,C), you have two random cards, and I want to know if you have A. The chance of this is 2/3 - you have either (A,B), (A,C), or (B,C). Your method tells me that it is 1/3 + 1/2 = 5/6. There are a couple of ways you could get at the probability: 1 - 41 choose 2 / 45 choose 2, the number of ways of pulling two cards that aren't an A or T from the 45 remaining cards divided by the total number of ways of pulling two cards, is one way. Alternatively, say there is a 41 / 45 chance that he has one card that is neither A nor T, and then a 40 / 44 chance that he has another , multiply these and subtract from one. This is the same thing. Your method will work approximately in cases where things are unlikely, but I see people use it pretty frequently around here and I like to make sure that they realize that it's not really correct.

3) This probability calculation only makes sense if we allow that our opponent has two random cards. But, I think to assume that our opponent has two random cards at the end here is not right. Consider a hypothetical opponent who will only bet with an A or T here. The probability following a bet by him that he holds an A or T is 100%, not ~ 1/6.

Now, I'm not saying that our opponents are only going to bet on the end with an A or T, but they are substantially more likely to bet on the end with an A or T than with just about anything else. Consequently, they are substantially more likely than random chance would suggest to be holding one. When people mention Bayes's Theorem, this is the kind of thing they are getting at.