twenty mathematicians

I like to think that it would average at 10 (50%) but that's probably too obvious.

The question is, or should have said/read: How many can they GUARANTEE will be saved?

btw, the average saved is higher than 50%

19 - it's a parity thing. 20th guy has a 50% chance of survival.

markr, please explain further...

The guy at the back will see 99 hats, so there will be an odd number of one color and an even number of the other. They agree ahead of time that he will call out the color that appears an odd (or even) number of times. Each person keeps track of the parity (odd/even) of the number of hats of the originally called color that is called out behind him. That information combined with what he sees in front of him allows each person to call out the correct color.

For instance, if the first guy calls out black (indicating an odd number of black hats in front of him), and an odd number of black calls have been made since the first call, then the next guy will call black if he sees an odd number of black hats in front of him (odd+odd=even, therefore his must be black to make the total odd) and white if he sees an even number of black hats (odd+even=odd, therefore his must be white to keep the total odd).

The first guy has a 50/50 chance of getting his correct.

First there's only twenty.....

Second the question doesn't say there are 10 black and 10 white.....

Oops, I goofed the total. Replace 99 with 19.

What I described doesn't assume a 10/10 split. In fact, it will work if all of the hats are the same color. However, this made me realize that they have to agree that the first guy will call out the odd color. There won't be an even color if they are all the same color.

Not only is that not permitted (at least in other instances of the problem I've seen), it is not necessary. The solution I provided will ensure that 19 are saved.

Here's an example with ten people.

From last to first (the order they will be calling out) their hats are:
B B B B W B W W B W

The last (tenth/leftmost) guy, seeing an odd number of black hats, calls out black.
Here's the thought process for the ninth through first persons:

9: There are an odd number of black hats (the color of the tenth guy's hat doesn't count or matter). I see an even number (4), so mine must be black. He calls black.
8: Black is odd. I can account for an even number (1 behind, 3 in front), so mine must be black. He calls black.
7: Black is odd (I'm dropping this on the rest). 2 behind, 2 in front (4 is even), so mine is black.
6: 3 behind, 2 in front (5 is odd), so mine is white.
5: 3 behind, 1 in front (4 is even), so mine is black.
4: 4 behind, 1 in front (5 is odd), so mine is white.
3: 4 behind, 1 in front (5 is odd), so mine is white.
2: 4 behind, 0 in front (4 is even), so mine is black.
1: 5 behind, 0 in front (5 is odd), so mine is white.

Super