The August issue of Scientific American presented a cute mathematical problem which is very difficult unless you write a program to solve it. It does not look trivially easy with a computer, although I am sure that there are those at this forum who could solve it by writing a VB program.

If I tried to do it with pencil & paper or using a non-programmable calculator, I am sure that I would either lose interest or make a mistake before getting the correct answer.

I will describe the problem in this post, and provide the general method for solving the problem in my next post. I have not written a program to solve it, but have solved trivial variations by hand.

The idea is that you can bet on the flip of a coin at even money, and you have $100.00 available for betting. You can bet all that you have, nothing, or any amount in between on any given coin toss.

There is a predictor who is in fallible at predicting the next coin flip. After telling him how much you want to bet, he makes the bet for you. He will deliberately lose exactly once, so you can rely on winning all but one bet.

He will try to minimize your winnings. Until you lose one bet you dare not risk everything.

First some warmup problems to make sure you understand the game. The solutions to these trivial variations are below. Do not look if you want to try it on your own.
  • Decide how to bet if the game consists of only one bet on one coin flip.
  • Decide how to bet if the game consists of two bets on two coin flips.
  • Decide how to bet if the game consists of three bets on three coin flips.
The above are easy, with solutions at the end of this post if you want to look now without trying it first.

The real problem is to decide on how to bet if the game consists of ten bets on ten coin flips. Do not try to solve this without a computer. It is too tedious.

The following are solutions for the 3 easy games above.
  • If only allowed to make one bet, you must bet nothing. The Ba****rd predictor is out to hurt you and will lose the one and only bet.
  • If allowed two bets, bet $33.33 (as close as possible to 1/3) on the first flip. If you win, you know that the ba****rd will lose the next bet, so bet nothing on the second coin toss. If you lose the first bet, bet the remaining 2/3 of your money. In either case, you will gain 1/3 of your original bankroll.
  • If allowed three bets, bet $50.00 or half your money on the first toss. If you lose, bet all on the next two tosses, doubling your original bankroll. If you win the first toss, you now have $150.00 and should bet $50.00 (or 1/3 of your current bankroll) on the second toss. If you win, you have doubled your money and should bet nothing on the third toss (you would lose that final bet). If you lose the second toss, you are guaranteed to win the third, so bet your remaining $100.00, again doubling your original bankroll.
Now are you prepared to deal with ten bets, with nine wins and one loss guaranteed?