Puzzle: The Monty Hall problem
Posted: Mon Jun 26, 2006 11:33 am
Linus Van Pelt wrote:1. This is best of 400, not first to 400, right?
Yes. You could call it a "race to 201."
2. If, after the 400th game, each player has won 200, for the purposes of your side bet, is that a draw (no one wins), or is a tiebreaker played?
The series is to 400 games, the bet is on my losing the series, not anybody winning. There may not be a winner of the series, in which case, I lose my bet because I didn't lose the series.
3. If, after the 400th game, there is a set still uncompleted, is that set completed or are you done?
The series is 400 games. There may be a set undecided after 400 games, but that doesn't matter. We may play another 100 sets after the series, that doesn't matter. During the series of 400 games, we will be playing sets of race-to-10.
4. If, after the side bet has been finished (after the first 400 or 401 games, depending on the answer to #2), and more games are played, they make no difference to the side bet, right?
exactly. The series is 400 games. What happens after that is of no interest to my bookie.
Assuming the answers are 1.yes, 2.draw, 3.the set is completed, and 4.yes, here is my answer:
Win 19 sets 10-9. Lose 1 set 10-9. Lose 2 sets 10-0.
You get:
From sets: $1900 - $300 = $1600.
From games: $1990 - $2010 = -$20.
From the side bet: $10,000
Total: $11,580.
There's one solution. Any others?