I watched Jeopardy last nite, and it was an interesting finish. All three players were in it, but one had no way of winning. So I'm going to focus on the other two.
Player 1 (P1) has 16,600, P2 has 11,400.
P1 bets 6201, which makes sense. If P2 bets it all, he ends up with 22,800. So P1 needs to have 22,801 to win, hence the 6201 bet.
P1 bet 1000. Hmm. That seems odd. Until they both got the question wrong.
So, P1 ends up with (16,600-6201) = 10,399.
P2 ends up with 10,400, and the win. He bet that both players would get the answer wrong, and won. I don't recall ever seeing that before, but I don't watch all the time, so maybe it happens a lot.
Anyway, I knew there had to some math to explain this, and I just had to figure it out. So, here's the math:
Pn = Player n amount left before final jeopardy
Pnb = Player n bet
Pnf = Player n final amount
PnG = Player n goal amount
P1b = (2P2 - P1) + 1
in this case: 2*11400 - 16,600 + 1 = 6201
If P1 wins, P1f = P1 + P1b = P1 + (2P2 - P1) + 1 = 2P2 + 1
If P1 loses, P1f = P1 - P1b = P1 - ((2P2 - P1) + 1) = 2(P1 - P2) - 1
in this case: 2(16,600-11,400) - 1 = 10,399
P2g = P1f + 1 (i.e. P2 needs one dollar more than what P1 ends up with)
from above, P1f = 2(P1 - P2) - 1, so
P2g = 2(P1-P2)
P2b = P2 - (2(P1 - P2)) = 3P2 - 2P1
in this case: 3(11,400) - 2(16,600) = 34,200 - 33,200 = 1000
This only works, however, if 3P2 >= 2P1. And, of course you have to assume P1 is going to bet the minimum to win, and both answers have to be wrong.
Now, maybe this is some well known Jeopardy thing, but I thought it took serious balls for that guy to bet they'd both get it wrong. And, he knew how to bet properly for it. And he was able to the the math - I would have forgotten to carry a 3 or something.
This is what I was doing until 11:30 last night (DVR'd the show, watched it at 10:30).
For the record, the "answer" was something about this country allows US hurricane hunters into it's airspace. The three answers were: North Korea, China and Iran.
questionanswer (which I got): What is Cuba.