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Scientific American

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Math reveals the one game of chance you should always accept
2026-05-13 · via Scientific American

This article is from Proof Positive, our friendly newsletter that explores the joys and peculiarities of math. Sign up today for a weekly math essay and puzzle in your email inbox.


I challenge you to a game. We roll a fair die, and if it lands on 1 or 2, you get $10. If it lands on 3, you get $20. Otherwise you go home empty-handed. Because you can't lose anything this way, I ask for a stake of $10 for each roll. Do you accept?


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You could decide based on your gut. But there is also a systematic way to determine whether the risk is worth it. For example, you can consider probability theory. A typical die has six sides, so there are six outcomes. In just two of the six possibilities (that is, 1/3 of the time), you win $10. Furthermore, the chance of winning $20 (that is, of the die landing on 3) is 1/6. If you multiply these probabilities by the money amounts and add them up, you get 40/6 = 20/3. That means, statistically speaking, you win an average of $6.66 per game.

But I demanded a $10 stake per roll. That means I will win, on average, $3.33 per game. All things considered, you should decline.

Back in 1713 mathematicians Nicolaus I Bernoulli and Pierre de Montmort exchanged ideas about a somewhat more complex scenario. It involved tossing a coin until it landed on heads for the first time. In this idea, the more tosses you perform, the more you will win: the proceeds are always doubled. So the first time you get tails, you get $1, and on subsequent times, you get $2, then $4, and so on. Imagine that I offer to play this game with you—and I ask for an extremely high stake of $2,000. Will you accept?

Playing at Any Stake

Probably every sensible person will answer with a clear “no.” But what would be a reasonable stake? For this, you can consult mathematics and look at the expected value: with a probability of 1/2, the coin lands on tails on the first toss, getting tails twice in a row corresponds to a probability of 1/4, doing so three times corresponds to a probability of 1/8, and so on.

At the same time, the win doubles in each case. The expected value is thus the infinite sum:

An equation calculates the expected value of winning in a game where you receive more money every time a coin flip lands heads

That means, mathematically speaking, no stake is too high—you should always play the game.

Because Bernoulli and Montmort chose a casino in Saint Petersburg as the setting for their thought experiment, they henceforth called the counterintuitive result the Saint Petersburg paradox. But this does not denote a paradox in the strict sense; the only paradox is that people would probably never follow the recommended course of action.

These results are counterintuitive in part because they involve infinity. The expected value results from an addition of an infinite number of summands, so the profit increases rapidly. If you have six successful tosses, you get $32; if you have six more, your winnings become $2,048. If you have a lucky streak and fall on tails six more times, you get $131,072.

The game, in short, is extremely unrealistic: it only works if the challenger has infinite resources. I personally do not. Even casinos with bulging coffers have an upper limit. Assuming finite capital, the game cannot continue indefinitely.

Setting Limits, and Engaging in a Billion-Dollar Battle

Let's say I have $1,050 in my account and am willing to bet anything to challenge you with the coin toss. I can’t ask you to bet $2,000 when the most you’ll get from me is a little more than a grand. Therefore, I offer you a friendly $6 stake to play. Do you accept?

Because I only have $1,050 at my disposal, the expected value of the game changes. If you roll 11 numbers in a row, I already owe you $1,024, so I might not be able to finance a 12th roll. Therefore the changed expected value now corresponds to 1 × 1/2 + 2 × 1/4 + 4 × 1/8 + ... + 1,024 × 1/2,048 = 1/2 × 11 = 5.5.

Given my limited fortune, the situation has changed completely. Instead of an infinite expected value, you now get the result $5.5. And that means, at $6 dollars for the stake, you should refuse the game. On the other hand, if you bargain me down to $5 (taking advantage of the fact that I'm not particularly gifted at mental arithmetic), then your chances of making some profit are good.

What if a billionaire challenged you to play? To calculate the stake you’re be willing to play for, you again have to determine the maximum number of rounds this fat cat can last before they go broke. For a sense of scale, after just 38 rounds, your challenger would already owe you more than $137 billion. (That figure would make Warren Buffett uncomfortable about tossing the coin again and would bankrupt Bill Gates.) Assuming that’s as much as your opponent can offer, the expected value (the maximum stake you should agree to) corresponds to just $19.

This article originally appeared in Spektrum der Wissenschaft and was reproduced with permission. It was translated from the original German version with the assistance of artificial intelligence and reviewed by our editors.