Mobius Strip
John P. Robertson

Many of you will recall that a peculiar property of the Mobius strip is that it has only one side. One method of creating a Mobius strip is to take a long rectangular piece of paper, and join the short ends with a half-twist.

Now for the puzzlement: you have a book with a front and back cover, and only one leaf. Equivalently, you have three   sheets of paper joined along a single edge. Using just a scissors, cut a Mobius strip out of the book. No taping or gluing is allowed, just cutting.

## Interrupted Betting

In Tom Struppeck's puzzle, you had bet \$100 on a best-of-seven series between two evenly matched teams, A and B, and you had to leave after your team, A, had lost the first game. You gave your agent some money, and some instructions on how to bet on the remaining games. The goal was for the agent to settle up on your behalf when the series ended, and have the agent come out exactly even at that point (although she could be out-of-pocket at intermediate points).

Alex Kozmin's solution is to give the agent \$31.25, with the instructions to place bets on team B, following amounts:

1. \$31.25 immediately

2. \$12.50 if current score is 0:3

3. \$25 if current score is 1:3
(either way) or 0:2

4. \$37.50 if current score is 1:2
or 1:1

5. \$50 if current score is 2:3 or 2:2

6. \$100 if current score is 3:3

7. No gambling otherwise!

She will need up to \$100 of her money (\$37.50 for the third game, another \$37.50 for fourth and additional \$25 for game five).

Alex also observes that \$31.25 is the expected loss given the result of the first game, the above strategy involves an average bet of \$31.25 across all possible games, and that the puzzle has roots in SOA course 6 (multi-period immunization, Financial Economics).

Malcolm Handte notes that the amount bet at each point is \$100 times the conditional probability that the series will go to seven games.

Kirk Bitu, Robert Lapson, David Uhland, David Westerberg, and Shawn Wright also solved this puzzle.