# What number would you bet on to maximize your chances of winning?

You have an asymmetrical die with $$20$$ faces: it has $$30\%$$ chance for $$20$$ to be rolled, while the other faces ($$1,2,\dots,19)$$ are equiprobable.

You and your friend each choose a number, and the die is rolled once. Whoever's number is closer to the result will win. What number do you choose?

Do you just compute the expected value of the roll and round to the closest integer? My intuition says "yes" but I'm not sure.

• It depends on what you think your friend will choose. Suppose you know your friend will bet on $20$. Then you should bet $19$, because you will win whenever the die rolls $19$ or below, which happens $70\%$ of the time. If you bet $18$ or below, you will win less often than $70\%$. (I assume you tie and split the stakes when you bet $18$, your friend bets $20$, and the die rolls $19$.) – Theoretical Economist Nov 15 '19 at 5:11
• @TheoreticalEconomist good point, but what if I had to bet first? – tamamshud Nov 15 '19 at 5:32
• Well, it still depends on what you think your friend will choose. (In the language of game theory, it depends on the strategy they're using.) Suppose that you knew they were going to bet $20$ whatever you choose. Then, again, you should still choose $19$. I think what you need to do is be more specific about what you know about how your friend will bet. – Theoretical Economist Nov 15 '19 at 5:45
• If you want this question to work like I think you do (where what you should bet is independent of what your friend bets), you may want to make the problem "You chose a number and roll the special asymmetrical die, calculating how far off you were. Your friend then chooses their own number and rolls the die themselves, calculating how far off they were. Whoever calculated the lower number wins." – scohe001 Nov 15 '19 at 21:53
• What happens if you both choose the same number? – alephzero Nov 16 '19 at 1:00

As mentioned in my comment, this depends on what you think your friend will bet.

If your friend bets $$14$$, then you should bet $$14$$. (Assuming ties count as half a win.)

If your friend bets $$n > 14$$, then you should bet $$n-1$$. Similarly, if your friend bets $$n<14$$, then you should bet $$n+1$$.

I'll let you verify that this strategy in fact maximises your probability of winning. One way to help you see this is to fix an $$n$$ (e.g. $$n=20$$), and then check that betting anything other than the bet prescribed above (e.g. when $$n=20$$, you should bet $$n-1=19$$) will have you winning less often.

• +1 for 'this depends on what you think your friend will bet.' I'm not sure if the OP realised this before the answers came in. – Toby Mak Nov 15 '19 at 5:35
• If you have to bet first and your friend knows your bet before deciding their own, I suppose you should bet 14 to prevent this strategy from giving your friend the advantage. But if the bets are blind (e.g., simultaneous), I guess that also means you should assume the friend will bet 14… – Arkku Nov 15 '19 at 14:13
• @Arkku If the friend has the same payoff function and is acting rationally, then what you say is correct. However, the most we can say without making said additional assumptions (which aren’t outlined in the question) is probably just my answer (or something similar). – Theoretical Economist Nov 15 '19 at 14:16
• @Arkku But if your friend bets n>15, then n-1 > 14. So if you bet 14, then can't your friend just bet something > 15? – Michael Nov 15 '19 at 23:42
• @Michael Not sure I follow: the best you can do against a bet of 14 is to bet the same for a tie. And if they can't or won't bet the same, their best choice is 15. (With the bets 14 and 15, 14 wins about 51.6% of the time, or 70% / 19 × 14. Betting above 15 only makes their odds worse, as does betting 13 or below.) – Arkku Nov 16 '19 at 0:50

The weighted sum that appears in the expected value doesn't really help with "closest to the value" two-person games. For that, you really want the median.

Consider for instance you choose 13 (the expected value) and I know that somehow. I choose 14 (the median). What is the probability you win?

What if i choose 14 first and you know that? What is the best win probability your guess can give you?

In other words, the maximin strategy in such games is to bet the median.

Considered as a two-person game, this is incompletely specified, both because there is some residual ambiguity in the specification of your own objective (do you consider a tie to be just as bad as a loss, for instance, or not so bad as a loss but not as good as a win), and no specification whatever of your opponent's objective.

Even if you ignore your opponent's objective entirely, the fact remains that your probability of winning still depends partially on his or her choice of number, so nothing you do can maximise your probability of winning tout court, because you have no control over one of the inputs on which that probability depends. There are, however, a couple assumptions you might make to resolve this difficulty:

• Assume your opponent's goal is the same as yours—i.e. to maximise his or her probability of winning. This turns the problem into a two-person game, for which you can try to find a Nash equilibrium. The game is not constant-sum however, because if both you and your opponent choose the same number, the probabilities of each of you winning sum to zero instead of $$1$$.
• Ignore your opponent's objective completely, and try instead to maximise what your minimum probability would be if your opponent's choice is the worst possible for whatever strategy you adopt. In this case, you must use a mixed strategy if the worst case is to be one where you have a positive probability of winning).

As it happens, the solution in both cases turns out to be the same: choose $$13$$ with probability $$\ \frac{99}{190}\$$ and $$\ 14\$$ with probability $$\ \frac{91}{190}\$$ (and keep your choice concealed from your opponent until after he or she has irrevocably chosen his or hers—assuming this is possible). This is your unique Nash equilibrium strategy under the first assumption, and it guarantees that you win with probability $$\ \frac{9009}{36100}\approx0.25\$$, which is also the maximum winning probability you can be assured of, because if your opponent chooses the same strategy, then your probability of winning is at most $$\ \frac{9009}{36100}\$$, no matter what you do.

You could also turn the game into a $$2$$-person constant sum game by assuming that both you and your opponent consider a tie to be half as valuable as a win, while still regarding a loss as worthless. In that case your optimal strategy is to choose $$14$$, which guarantees you an outcome at least as good as a tie. If your opponent chooses any other strategy your expected payoff will be greater than $$\ \frac{1}{2}\$$—that is,$$\ \frac{1}{2}\text{Prob}(\text{tie})+\text{Prob}(\text{you win})>\frac{1}{2}\ .$$

• This should plainly be the top answer – David Diaz Dec 8 '19 at 1:08

If you choose 14, you win over half the time when your opponent chooses any other number.

• This is true, but betting 14 doesn't always maximise the OP's chances of winning. – Theoretical Economist Nov 15 '19 at 5:26
• I don't know what to assume about the opponent's behaviour, since the OP said nothing about how they behave. If they bet $20$, for example, then you should clearly bet $19$. – Theoretical Economist Nov 15 '19 at 5:32
• If you don't know what to assume about your opponent's behaviour, how are you going to guess at their choice? If you can't guess at their choice, why wouldn't you choose 14? – David Diaz Nov 15 '19 at 5:40
• Personally, I would assume they would be playing an equilibrium strategy (e.g. the one I specify in my answer), but the OP hasn't mentioned anything about equilibrium. But now you're really asking me how I would play this game, when what I'm really trying to do is answer the question as laid out by the OP. – Theoretical Economist Nov 15 '19 at 5:45
• This is more of a comment than an answer... – Please stop being evil Nov 15 '19 at 22:24