# How to play a betting game

I have been interviewing in a few trading firms recently. I came up with the following question myself, but it is similar to some of the questions they ask and ways of thinking they expect.

Suppose you have some capital to invest (for example \$100). You can play a game where you bet \$x of your money and with probability $$\frac{2}{3}$$ your bet is doubled (so now you have \$(100 + x)) and with probability $$\frac{1}{3}$$ you lose your bet (so now you have \$(100 - x)). How much should you bet on this game.

I can see two ways of thinking about this problem. Firstly, there is the expected value maximisation approach. It can be easily seen that your expected gain in this game is $$\frac{x}{3}$$. So in order to maximise EV you should bet all of your money instantly. And if you were to play this game a million times, you should bet all of your money each time.

Of course this approach has the obvious flaw that when you play a few times, you will almost certainly go bankrupt. So we decide not to maximise EV and instead first make sure that we never go bankrupt. We do it by deciding to, at each point of the game, always bet exactly the same proportion of our money, say $$p$$. Then after $$n=n_1+n_2$$ games, where our bet was doubled $$n_1$$ times and we lost $$n_2$$ times, we will have $$M \cdot (1+p)^{n_1} \cdot (1-p)^{n_2}$$ money, where $$M$$ was our initial amount. Differentiating the log of this with respect to $$p$$ we can see that this function has its maximum for $$p=\frac{n_1-n_2}{n} \rightarrow \frac{2}{3}-\frac{1}{3} = \frac{1}{3}$$ as $$n \rightarrow \infty$$. So if we bet just a third of our money every time, we are (almost) guaranteed not to go bankrupt and, out of the strategies that bet a constant proportion every time, this one maximises our gain in the most likely outcome.

So here is my question - does this second strategy make sense to you? If you were to play this game with your own money would you use it? Does it make any sense to use a different strategy if you only play once, and not many times? I personally would be tempted to bet more than a third if I only got one chance, because it would increase my EV even if it is potentially bad in the long term. Does this sentiment make any sense?

Also, I just described two ways of thinking about the game above. Do you know of any other ways to think about it? Other strategies?

Your approach is remarkably on point. This issue is generally discussed in terms of portfolio construction in finance and depending on risk tolerance we define different functions to optimize. You choose to maximize the log of the expected payoff after n games, which is the same as Kelly criteria. For further and more detailed discussion of it you can check https://en.wikipedia.org/wiki/Kelly_criterion

• Correct me if I am wrong here: I don't think I am maximising the log of the expected payoff - I think I am maximising the log of the most likely payoff. The expected payoff is still the largest if I bet everything I have every time, right? Also, taking the log shouldn't change anything in terms of maximisation, because log is strictly increasing. So I am really maximising the most probable payoff, no? – user132290 Dec 11 '18 at 10:25
• Well, in effect you are basically maximizing the log of expected payoff. So you first wrote the payoff equation given $p, n_1, n_2$. While finding the most probable payoff, you plug $\frac 2 3$ for $\frac {n_1} n$, which is in practice taking the expected value of $n_1$ and $n_2$ with respect to $n$, hence taking the expected value of payoff. – Ofya Dec 11 '18 at 15:19

Your intuition is generally captured in a "utility function" which is supposed to describe how happy you are to have a certain amount of money. You then maximize the expected value of this function. Your thought to maximize the log of the amount of money you have is one utility function and not an unreasonable one. There is good psychological evidence that more money makes people happier, but much more slowly once they have some. The log function is in this vein, but there are many others as well. Once you define the function, the maximization process is the one you have used.

I would suggest that no simply described function can capture utility properly. If you were allowed to play the game fifty times but had to bet one dollar each time, you would probably play. You would probably win about $$\25$$ and be happy about it, but it wouldn't really change your life. As the bet rises the impact on your life does too. At the start it is only good because the chance you lose is almost zero so more is better. Eventually it may get to the point that you become risk averse. If your income is large compared to your cash assets it may make sense to bet everything you have because you can replace it easily. If you are living on assets you may become risk averse at a small fraction of your assets. This is all supposed to be captured in the utility function, which indicates why a simple answer like log is not appropriate for real life.

• Correct me if I am wrong please, but you seem to be treating the log as if it was meaningful in my example (because it is the utility function of my wealth - so the second million I make will make me less happy than the first million). But I am just taking the log here for simpler differentiation and to maximise the log is the same as to maximise its argument. So the log should not be relevant at all here, no? – user132290 Dec 11 '18 at 10:28
• Yes, I am treating the log as meaningful. If your utility is linear in money you should bet all your money each time. I answered this here. Yes, you will almost surely be broke, but the tiny chance of winning a huge amount of money overwhelms that. The utility function should apply to all the money you have, not just to what you win from the game. The log is a nice function because it is increasing but slower and slower as the argument gets larger, which we want utility todo – Ross Millikan Dec 11 '18 at 15:19
• Changing the log to a different function will shift the amount you should play with. – Ross Millikan Dec 11 '18 at 15:20

Straight out of my files:

For a P% profit level and D odds to 1:

RequiredPercentageCorrectBets = (P + 100) / (D + 1) .

This posting is simply viewpoint of pari-mutuel wagering where the probability is derived from the wagering.

In other words, in pari-mutuel wagering all the information that there is, is supposed to be represented by the wagering on the tote board. So in pari-mutuel wagering, odds of 1.00 represents a probability of 50%. Then the Kelly Criteria calls that situation a no-bet. However, some bettors do wager profitably and so I suppose that there is a personal probability of winning that can be applied to calculating the percentage of the stake to bet that maximizes profit. However, a personal probability of winning would still tend to go up with lower odds and tend to go down with higher odds.

In roulette or keno I suppose that the gambler can keep track of a number not coming up and then expect an increasing probability of the number coming up. That situation assumes a faith of an honest game instead of a discovery of a dishonest game.

I previously suggested a one-number Keno game that pays 3 to 1 for odds of 2 to 1. Twenty numbers are drawn out of 80 for a probability of 25%. But the probability of hitting the number in two draws is 50%, the probability of hitting the number in three draws is 75%, and the probability of hitting the number in four draws is 100%. So the idea is to decide how much to increase the wager each time the number doesn't come up and keep playing the number until it does come up.