You and your friend are playing a game. You both start with a score of $0$. Also, you both start with $\$1$. At each step, you're allowed to bet a fraction of your $\$1$, and whoever bets more money wins that "round". However, whoever wins the round (bets more money) loses whatever they bet, and whoever loses that round (bets less money) keeps whatever they bet. If you win, your score increases by $1$; if you lose, your score decreases by $1$. The game terminates when a player gets a score of $-3$ (they lose) or $+3$ (they win). If you bet the same amount of money as your opponent, then your opponent wins.

What's the optimal amount of money you should bet in the first round?

I was asked this question for a quantitative research position, but I couldn't solve the problem. They seemed to suggest that the answer was irrational, but I still can't figure it out. Does anyone have any ideas?

Some extreme strategy like betting everything in one round isn't feasible since you lose the money if you win.

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    $\begingroup$ "However, whoever wins the round loses whatever they bet, and whoever wins that round keeps whatever they bet. " Which player keeps his money and which loses it? $\endgroup$
    – Moko19
    Commented Sep 1, 2020 at 21:21
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    $\begingroup$ Yep, one of those "whoever wins" should be "whoever loses", but which? $\endgroup$ Commented Sep 1, 2020 at 21:32
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    $\begingroup$ Also, what happens if both bet the same amount? $\endgroup$ Commented Sep 1, 2020 at 21:35
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    $\begingroup$ @saulspatz The OP should clarify that theit mistype is interpreted in that way (and still - what happens in case of a tie) $\endgroup$ Commented Sep 1, 2020 at 21:39
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    $\begingroup$ This sounds more like an auction for the score point than a bet $\endgroup$ Commented Sep 1, 2020 at 21:40

2 Answers 2


I would argue that the optimal initial betting is $\frac{2 - \sqrt{2}}{2} \sim 0.29$.

To do so, we need a few arguments:

  1. Whatever the initial bet is that you use to go from $0$ to $+1$, the second bet will be the same. Suppose not. Let $x$ be the bet you made to go from $0$ to $+1$, and let $y$ be the bet you make to go from $+1$ to $+2$, with $y = x + \epsilon$ where $\epsilon > 0$. Now, your opponent initially will bet $x + \epsilon/ 4$, and then bet $x + \epsilon/ 2$. Now, either you:

a) Let your opponent win the first round, and then win the second (in which case, now, the money in game is $(1 - (x + \epsilon/2), 1 - (x + \epsilon /4))$, so your opponent gained a net advantage. You are doomed to lose if you repeat this strategy.

b) Your opponent lets you win both rounds, so you are now at $+2$ with $(1 - (x + y), 1)$

c) You let your opponent win both rounds, so your opponent is now at $+2$ with $(1, 1-(2x + 3 \epsilon / 4))$.

Since you want to make sure you are doing as well as your opponent, you need to pick $y$ arbitrarily close to $x$.

Once you reach $+2$ with $(1 - 2x, 1)$, your opponent has to counter this by paying $(1 - 2x)$, so now, we are back at $+1$ with $(1 - 2x, 2x)$.

Note that the game is scale invariant, and if you ended up strictly better up to scaling, your opponent could have copied you, and if you ended up strictly worse, then you will just have to repeat the same game, and will eventually lose, so we assume that the proportions of money is equal i.e. $(1 - x, 1) \propto (1 -2x, 2x)$ so we let $1 - x = \frac{1 - 2x}{2x}$ and solve the quadratic for an $x$ value less than 1, which gives us the answer mentioned before.

  • $\begingroup$ @RossMillikan But I will bet $x$, so if they bet less than $x$, I would win? (Since if I am willing to lose to a bet y < x, then, we would be back at the origin with me having less money than my opponent) With that said, it is quite possible (likely, even) that there are mistakes in this answer, so thank you for reading through it! $\endgroup$
    – E-A
    Commented Sep 6, 2020 at 3:30
  • $\begingroup$ If I know you are playing this, I bet $0.3$ the first round, win the point, and bet $0.35$ every round thereafter. I claim I will win. $\endgroup$ Commented Sep 6, 2020 at 4:06
  • $\begingroup$ @RossMillikan Ok, so we get to +2 with (0.35, 1), since I will for sure let you win. Now, I have to pay 0.35+\epsilon to take us back to +1 so that is (0.35, 0.65 - \epsilon). If you pay your remaining 0.35 to jump back up to +2, you are out of money, and I win. Going through an example helps by the way; thanks! $\endgroup$
    – E-A
    Commented Sep 6, 2020 at 4:36
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    $\begingroup$ You have made me realize that my claim of equivalence to the winner getting one point and the loser getting zero is wrong. You need to win by three, not to get three. I will think some more. $\endgroup$ Commented Sep 6, 2020 at 4:40

Denote by $(a, b, \sigma)$ the game state where you have $\$a$, your opponent has $\$b$, and your score is $\sigma$. Clearly the game is scale-independent, so the ratio of your money to your opponent's is all that matters to determine the winner. Also, clearly it can only help you to have more money at any point; so there must be exactly one transition (as the ratio is increased) from losing to winning for each score. Let $R_\sigma$ be the critical ratio when your score is $\sigma$; that is, if the ratio is larger, you win, and if the ratio is smaller, you lose. By symmetry, $R_0=1$. (The winner at the critical ratio for each score could still go either way; certainly it depends on the tie-breaker rules, while the rest of this analysis does not.)

First consider $\sigma=+2$. You win games $(r,1,+2)$ with $r > 1$; for $r \le 1$, your opponent can (must) match your bet, which should therefore be as large as possible, leading to the game $(r,1-r,+1)$, which goes from losing to winning at $r/(1-r)=R_1$. So $R_2/(1-R_2)=R_1$, or $R_2=R_1/(1+R_1)$.

Now consider $\sigma=+1$; you again clearly win for $r > 1$, but for $r \le 1$, your opponent can choose whether to match your bet or not. Let $s(r) \le r$ be your bet. Then your opponent can choose whether the next state is $(r-s(r), 1, +2)$ or $(r, 1-s(r), 0)$. The latter is a loser for you if $r < 1-s(r)$, so you must have $s(r) \ge 1-r$. The former is a loser for you if $r-s(r)< R_2$, so you must also have $s(r) \le r-R_2$. These conditions become incompatible (that is, your opponent can always win) if $1-r>r-R_2$, or $r<(1+R_2)/2$. We conclude that $R_1=(1+R_2)/2$. Combining this with the previous relation between $R_1$ and $R_2$, we find that $R_2=-1+\sqrt{2}$, and that $R_1=\frac{1}{2}\sqrt{2}$.

Finally, let's determine your optimal bet from $(1,1,0)$. If you bet $s$, your opponent can choose whether the outcome is $(1-s,1,+1)$ or $(1,1-s,-1)$. One of these is a clear loser unless $1-s=R_1$, or $$s=1-R_1=1-\frac{1}{2}\sqrt{2}.$$ This is your optimal initial play.

Note that because of the unfair tiebreaker rules, your opponent will win this game... if you bet exactly the critical amount each round, then your opponent will exactly match your bets and win in three rounds. However, there is a silver lining: you can draw out your defeat, making your opponent spend an arbitrarily large number of rounds reaching her inevitable victory. This can be done by betting your opponent's wealth plus a sufficiently small $\varepsilon$ when the score reaches $-2$.

  • $\begingroup$ You can upgrade your arbitrarily large to infinitely long if you add \epsilon/2, \epsilon/4, \epsilon/8, ... instead. $\endgroup$
    – E-A
    Commented Sep 14, 2020 at 4:00
  • $\begingroup$ @E-A: I don't believe that's so (though it might be an interesting follow-up question to fill in the details). I think after you win your first point with a bid that is too large by $\varepsilon$, you are doomed to lose in $O(1/\varepsilon)$ additional rounds even if the tiebreaker rules were switched to be in your favor. $\endgroup$
    – mjqxxxx
    Commented Sep 14, 2020 at 20:10

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