Distribution of suits question

[AaronBrown]

Let's simplify things to test your logic. I take the four Aces and deal one to you and one to me, face down and for the moment neither one of us look. On average, there is one red card and one black card between us. Now I deal a card faceup from the two remaining Aces. It's red. So on average, there are two red Aces accounted for.

What is the probability that the remaining undealt card is red? Using your logic (which I admit I pushed to an extreme) it's zero since two red Aces are already accounted for.

There are two ways to get the correct answer, and many years ago Reverend Thomas Bayes proved they always give the same answer.

First is to ignore the two down cards in front of each of us and treat them as still in the deck. Then it's obvious that one of the three unknown cards is red, so the chance of the next face up card being red is 1/3.

Second is to consider the three possibilities. We can't both have red cards, because a red card came face up. I could have red and you could have black, in which case the next up card will be black. I could have black and you could have red, in which case the next up card will be black. Or I could have black and you could have black, in which case the next up card will be red. So the chance of another red upcard is 1/3.

With 20 cards in 10 hands you can do either of the same two things: treat all the unknown cards as still in the deck or consider every possible distribution of suits (not just the expected distribution).