Puzzle: The Monty Hall problem

131
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?
steve albini
Electrical Audio
sa at electrical dot com
Quicumque quattuor feles possidet insanus est.

Puzzle: The Monty Hall problem

132
Here is my solution:

I win 18 sets by a score of 10-9
I win one set by a score of 10-8
I lose one set by a score of 9-10
I lose 2 sets by a score of 0-10.

At this point, I have won 199 games, my opponent has won 200 games. I begin another set by letting him win one more, giving him 201. This loses the 400 game match for me, winning my sauce bet. I then go on to win the current set by a score of 10-1.

Final win (after finishing the set): $11780

Redistributing the losses allows me to begin another set while the series is still underway. As long as I lose the first game of this new set, I win the sauce bet.

Is there any way to get more?
steve albini
Electrical Audio
sa at electrical dot com
Quicumque quattuor feles possidet insanus est.

Puzzle: The Monty Hall problem

133
steve wrote:Here is my solution:

I win 18 sets by a score of 10-9
I win one set by a score of 10-8
I lose one set by a score of 9-10
I lose 2 sets by a score of 0-10.

At this point, I have won 199 games, my opponent has won 200 games. I begin another set by letting him win one more, giving him 201. This loses the 400 game match for me, winning my sauce bet. I then go on to win the current set by a score of 10-1.

Final win (after finishing the set): $11780

Redistributing the losses allows me to begin another set while the series is still underway. As long as I lose the first game of this new set, I win the sauce bet.

Is there any way to get more?


I was trying to get something like that, but couldn't get it. So the rule is, any set left uncompleted after the 400 gets finished, and gets counted in your winnings, but no games or sets after that set, right?

I think my assumptions above are still fundamentally sound, except that you can only win 19 sets before the 400 is completed. Of course, you can only win 1 set afterwards, since the only set after the 400 that may count is a set that is incomplete at the 400th game, and there can only be one incomplete set at any given time. So: if 19 sets before is a maximum, and 1 set after is a maximum, then 20 sets total should be a maximum. Next step, minimizing sets lost: the most games you can let your opponent lose in your 19 winning sets is 171. As we see, this is not optimal, as it doesn't let you count that extra set in your winnings. So, if we cut that down to 170 (your solution) or even as low as 162 (win 18 10-9, win 1 10-0, lose 1 10-9, lose 2 10-0, lose the first 9 of the next set, then go on to win it 10-9, which gives the same result), we can add the extra set. But if you bring it to 161 (or fewer), that means your opponent has to get 40 more wins (or more) while you only get 9, which means your opponent needs (at least) 4 more sets. As long as your opponent doesn't win xx1 games in the first 19 sets, you can still get that extra set at the end, but if your opponent loses fewer than 162 games in the first 19 sets, he'll have to win more than 3 sets after that. So now I think your solution (or the variant I gave above) is optimal.

So what am I missing now?
Why do you make it so scary to post here.

Puzzle: The Monty Hall problem

135
Linus Van Pelt wrote: So the rule is, any set left uncompleted after the 400 gets finished, and gets counted in your winnings, but no games or sets after that set, right?

Every game I play against the sucker "counts," in that I get to keep his money if I win. The problem is in distributing the losses throughout a 400-game series so that I get the most money. You can assume he and I will continue playing indefinitely.
steve albini
Electrical Audio
sa at electrical dot com
Quicumque quattuor feles possidet insanus est.

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