Three Doors with one prize

I got .1 for each of the first two callers Gus, 1 in 10 for the first and 9/10 by 1/9 for the second. The tenth comes in at a healthy .63 probability.[/quote]

1st caller 1/10 to win, trivial.

2nd - to win, 1 to miss, and then to pick from the 9 remaining (9/10 x 1/9) = 1/10.

3rd to win - 1 to guess wrong (9/10) times 2 to guess wrong (8/9) and then picks correctly (1/ = 1/10.

4th: first 3 guess wrong (9/10 x 8/9 x 7/ and picks correctly @ 1/7 = 1/10.

And so on. So at the start all contestants have an equal 1/10 chance of winning regardless of where they are in the queue.

I think you're right Gus, mind you I had to run a simulation to be sure. I was using only the probability that the previous caller had guessed wrong e.g. Caller 3 was 9/10 ( the probability that caller 2 got it wrong) * 1/8. Anyway this appears to be wrong. Well done to you.