Friday Puzzler -- another three curtains

Re: Friday Puzzler -- another three curtains

Larry Barrett

When A goes first, he has a probability of 1/3 finding $100 on the first try. If he fails (p=2/3), then B has a probability of 1/2 of finding the $100 his try, so overall probability for B is 2/3 x 1/2 = 1/3. And if B fails (p = 2/3 x 1/2 = 1/3) then A opens last box and finds the $100. So overall, when A goes first he finds the $100 with p = 1/3 + 1/3 = 2/3.

The opposite is the case when B goes first.

After the first three games, the expected winnings for A = 2/3 x $300= $200 and expected winnings for B = $100.

The opposite is the case for games 4, 5, and 6, so after game 6 the expected winnings for A will be $300 and the same for B.

In game 7, B goes first again and has p = 2/3 chance of winning the $100.

So overall, B has the best chance of winning $400, and thus gets the first chance of winning the car.

Right so far

Alex Y

Very good! Looking at all seven games and who is likely to win the most (who will of necessity be the first to reach $400) is the hard-to-see key to that part -- answering who will be most likely to win the car.

But how about the other question--who is likely to win the most money?

Re: Right so far

Larry Barrett

If B is the most likely to get to $400 first, it seems to me that B is the most likely to win the most money. I can't think of any other way for A to end up with more money than B.

So it would seem

Alex Y

But paradoxically, while B is more likely to get to $400 first and thus compete for the car, A has a higher expectation of money winnings.

At a high level, it is because A is much more likely to win by a large margin.

Looking at the 128 possible A/B combinations for all seven rounds, we see that for the 16 cases that $400 is earned in just 4 rounds, it is 4 times more more likely that A will have won (and B got nothing). There is a slight advantage to A in the 32 cases where $400 is reached in the 5th round. While the advantage shifts to B for the cases where $400 is reached in the 6th or 7th round, the advantage is not as great as A had in the other cases.

I can email you a spreadsheet (I had to convince myself) if you would like.

Re: So it would seem

Larry Barrett

I see that the probability that A has $400 at the end of 4 rounds vs the prob that B does is ((2/3)^3)(1/3) vs ((1/3)^3)(2/3) = 8/81 vs 2/81.

I suppose that the probability that the game is tied (each has $300) at the end of 6 rounds is greater than the probability that either A wins in 6 rounds or that B wins in 6 rounds.

Please send the spreadsheet.