# game theory - coin flipping game

Lets say $2$ players $A$ and $B$ make a bet, who can have more money at the end after playing the following game:

a coin is flipped:

with $51\%$ probability it lands tails, with $49\%$ probability it lands heads

you win if it lands heads, where you get back your bet $\times 2$.

e.g. you bet $\$1$and it lands heads, then you get back$\$2$

e.g. you bet $\$2$and it lands tails, then you get back$\$0$

here are the rules to the bet between A and B (the winner of the bet wins $\$100000$): • you both start with$\$100$ (given to you for free, you're not allowed to cash this out nor the money you make from the coin game)
• each player may play the game as many times as they want and bet as much as they want for each time they play the game
• player A must go first (player A plays the casino games as many times as he wants then decides to stop, after that, A can't play the game anymore)

obviously the optimal strategy for player $B$ involves playing until $B$ either goes bankrupt or has more money than $A$ (although it's not obvious what bet sizes to use).

what would be the optimal strategy for $A$?

• B's strategy isn't all that obvious either. Yes he keeps playing until he either has more money than A or he goes bankrupt. But how much to bet at a time? If B has less than 100 A need not bet at all. What if B has 200? Does A bet all 100 in one go? Or bet small amounts? This question seems pretty computationally intense...a good way to answer it might be to code up some Monte Carlo simulations and see what results different strategies yield. Jan 22, 2013 at 7:31
• @Greg, betting 2 dollars gives 49% chance of going up $2 by the problem statement Jan 22, 2013 at 9:24 • To see that, as @Russ pointed out, even just$B$'s strategy isn't obvious, you might want to take a look at this question and my answer to it; the situation there is basically the one$B$faces after$A$finishes playing. Jan 22, 2013 at 9:54 • @OP correct me if I'm wrong to be sure about the structure: (1) A plays as much as she wants (namely, goes broken or stops at a certain point); (2) she quits the game with a certain amount$x$(from 0 to whatever); (3) B's turn to play; (4) B either stops or goes broken; (5) end of the game. Right? PS: Some points are redundant,but it is to have some sort of picture of the game. Jul 5, 2013 at 10:45 • @soakley Right. If A doesn't bet at all then B has a good strategy it seems of betting$1+\epsilon$, then doubling up until she wins or is bankrupt. She wins with probability greater than$1-0.51^6 \approx 0.98$I think. However if A bets everything once she wins with probability$0.2499$it seems. Jul 5, 2013 at 16:14 ## 3 Answers Once$A$has finished playing, ending with an amount$a$, the strategy for$B$is simple and well-known: Use bold play. That is, aim for a target sum of$a+\epsilon$and bet what is needed to reach this goal exactly or bet all, whatever is less. As seen for example here, the probability of$B$reaching this target is maximized by this strategy and depends only on the initial proportion$\alpha:=\frac{100}{a+\epsilon}\in(0,1)$. (Of course,$B$wins immediately if$a<100$). While the function$p(\alpha)$that returns$B$'s winning probability is fractal and depends on the dyadic expansion of the number$\alpha$, we can for simplicyity (or a first approximate analysis) assume that$p(\alpha)=\alpha$: If the coin were fair, we would indeed have$p(\alpha)=\alpha$, and the coin is quite close to being fair. Also, we drop the$\epsilon$as$B$may chose it arbitrarily small. (This is the same as saying that$B$wins in case of a tie). In view of this, what should$A$do? If$A$does not play at all,$B$wins with probability$\approx 1$. If$A$decides to bet$x$once and then stop,$B$wins if either$A$loses ($p=0.51$) and$B$wins immediately or if$A$wins$p=0.49$and then$B$wins (as seen above) with$p(\frac{100}{100+x})\approx \frac{100}{100+x}$. So if$A$decides beforehand to play only once, she better bet all she has and thus wins the grand prize with probaility$\approx 0.49\cdot(1-p(\frac12))\approx \frac14$. Assume$A$wins the first round and has$200$. What is the best decision to do now? Betting$x<100$will result in a winning probability of approximately $$0.49\cdot(1-\frac{100}{200+x})+0.51\cdot(1- \frac{100}{200-x})$$ It looks like the best to do is stop playing (with winning probability$\approx\frac12$now). Alernatively, let us assume instead that$A$employs bold play as well with a target sum$T>100$. Then the probability of reaching the target is$\approx \frac{100}{T}$, so the total probability of$A$winning is approximately $$\frac{100}T\cdot(1-\frac{100}T)$$ and this is maximized precisely when$T=200$. This repeats what we suspect from above: The optimal strategy for$A$is to play once and try to double, resulting in a winning probability$\approx \frac14$. Admittedly, the optimality of this strategy for$A$is not rigorously shown and especially there may be some gains from exploiting the detailed shape of$B$'s winnign probability function, but I am pretty sure this is a not-too-bad approximation. • I agree completely it is likely to be not too bad an approximation. The mystery is if it is exactly the best answer. According to mathoverflow.net/questions/119537/…$A$can improve their probability of winning from$0.2499$to$0.249999385$by being a little cleverer. Jul 10, 2013 at 12:39 • Do you have any idea how one might even hope to show a strategy is optimal for this kind of game? Jul 10, 2013 at 12:43 • @marshall That's a smart answer there ("Bold play to target that can be reached with probability$\frac12$") and even lets the deviation from linearity cancel itself out! And you may have misread the answer: With that value,$A$wins with$\frac14$precisely (the$0.249999385$were for bold play with target$196$instead of$195.67803788$). Jul 10, 2013 at 13:16 • Oh you are right! My mistake. Jul 10, 2013 at 13:22 • this answer seems incorrect - the setup of the game says "each player starts with$100" - if player A wins the first round, player B is bankrupt, correct?
– Jim
Aug 26, 2020 at 5:48

The best answer provided by marshall / Hagen von Eitzen on the comments of his answer:

That's a smart answer there ("Bold play to target that can be reached with probability $$\frac12$$") and even lets the deviation from linearity cancel itself out! And you may have misread the answer: With that value, $$A$$ wins with $$\frac 14$$ precisely (the $$0.249999385$$ were for bold play with target $$196$$ instead of $$195.67803788$$). – Hagen von Eitzen Jul 10 '13 at 13:16

So, $$A$$ should not just play all in once, but use a bold play strategy to a goal that is reachable by $$p=50\%$$. If he loses, he lost, if he wins, the best strategy for B is to also bold play up to that goal, which he can reach with $$p=50\%$$ (since A and B start from the same money). So, A will win with exactly p=25%, and may play many times.

Note here $$A$$ may be forced to play fractional dollars.

what would be the optimal strategy for A?

Since the game has negative expectancy (-0.02 of the bet), the optimal strategy for A is not to play.

• That's not right. If $A$ doesn't play $B$ almost certainly wins. Jul 10, 2013 at 12:40
• @marshall: even if you are right (which is not obvious to me ATM), it does not mean that A should play. If he plays, his probability to win decreases.
– sds
Jul 10, 2013 at 13:28
• See the answer above. A can make her chances of winning be precisely $1/4$. Jul 10, 2013 at 13:30
• So that it is clear: if A does not play, B bets e, a very small quantity. if he wins he is done, otherwise he bets 3e. if he wins he is done, otherwise he bets 5e. if he wins at step i he is done, otherwise he bets e+lost_so_far. Even if the casino allows e to be no less than 1, he will most probably win.
– ntg
Aug 23, 2016 at 16:22