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I have a game where each game starts with \$200 and you have to bet on the outcome of a fair coin flip. You can bet any positive integer amount as long as you have the money, and the payout ratio is 1:1 (If you win, you get double the bet, otherwise you lose all your bet). You continue to bet money on coin flips, until you either lose all your money or you won a specific target amount of money, for example \$1000, and the next game begins. What is the best betting strategy to minimize the number of games you have to play before winning the target amount?

My first instinct is that you want to minimize the number of bets you make, because the probability of winning decreases exponentially the more coin flips you play (assuming you bet everything you have every time). Thus, the sequence to a successful game is:

  1. bet 200 (win), bet 400 (win), bet 800 (win) -> win a total of 1600

The probability of this case happening is $(\frac{1}{2})^3=0.125$

However, notice that we actually surpass the goal, and if we can bet just the difference between the goal and what we have we can reach the same target with the same probability, but without risking the game ending immediately. If we lose in this case, we can continue to bet and have a chance at reaching the target. One such strategy has the following sequences of events:

  1. bet 200 (win), bet 400 (win), bet 200 (win) -> win a total of 1000
  2. bet 200 (win), bet 400 (win), bet 200 (lose), bet 400 (win) -> win a total of 1000
  3. bet 200 (win), bet 400 (win), bet 200 (lose), bet 400 (lose), (200 remain, start over...)

I'm not quite sure how to calculate probability for this, since case 3 essentially resets the game, but if we consider just the first two cases then the probability for reaching the target is at least $(\frac{1}{2})^3+(\frac{1}{2})^4=0.1875$

So I believe that this shows the second strategy is better than the first. But this can be extended to the first couple of coin tosses as well, so that you don't bet your entire pot on any of the coin flips (except the last before you run out of money). My question is how do you analyze this to find what amount to bet at each coin toss, given a target amount \$X, for a best strategy that minimizes the number of games played?

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I think the answer is that all the strategies are equally good.

Suppose that you always bet \$1. Then the problem is equivalent to the "drunk sailor's random walk" problem. (By the way, the probability you win is $\$200 / (\$200 + \$1000) = 1/6$.)

The possibility to bet not 1, but several dollars would not change the outcome (providing the bet is small enough so that both in case of win or lose your amount stays within specified limits). "I bet \$10" is equivalent to "I will bet \$1 as many times as necessary until I win or loose \$10" - in both cases sooner or later you will get into situation when you either got or lost \$10, and probability of any of these situations is 1/2.

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