Quote:
Originally Posted by Brocktoon
Wow, super tough spots tonight.
Final totals are: 17,400; 9400; and 12400.
I wouldn't really blame anyone for betting anything here but I will basically guarantee that P1 bets $7401 or more to cover P3's double, and this is really the only assumption I was ever making.
BTW, it is after 730 but I haven't seen the wagers yet.
My thoughts on this scenario.
P1 - 17,400
P2 - 9400
P3 - 12,400
Let's just say for argument's sake P1 makes exactly the shutout bet of $7401, and shows this bet to P2 and P3 before they wager.
Now P2 should say to himself, "Even if P1 gets it wrong, he will have 9999, which is more than I currently have. Therefore I must get this question right to win the game. Therefore I bet all 9400 to maximize the amount of money I get in the cases where I win the game."
Now P3 is caught in the middle. Given the leader (P1's) wager, P1 has to get it wrong for P3 to win. If P2 did not exist, P3 could bet such that he would be guaranteed to win every time P1 got it wrong. Unfortunately for P3, P2 does exist and is close enough that P2 cannot guarantee this.
Holding the fact that P1 gets it wrong constant, there are 4 cases:
P2 gets it right / P3 gets it right - P3 should bet it all
P2 gets it right / P3 gets it wrong - this case doesn't matter for winning the game because P3 is screwed because P2 went all in
P2 gets it wrong / P3 gets it right - this case doesn't matter for winning the game, P3 will win no matter what he wagered, to maximize winnings in this case you should bet it all
P2 gets it wrong / P3 gets it wrong - P3 should bet $0, he will remain above P1 if he does so
Basically P3 is caught between the other two. He cannot bet small enough to guarantee victory against P1 in the case of a P1 miss and at the same time large enough to shut out P2. P3 needs to decide which scenario is more likely:
Triple Stumper (aka all three miss) - he should bet $0 if this is more likely
P1 is the only one to miss - he should bet all $12,400 if this is more likely