# Picking a time to bet

You play a game with a friend: he deals cards face up one a time from a uniformly shuffled deck of 26 black cards and 26 red cards. At some point before all the cards have been dealt, you must call "bet". When you do, your friend flips the next card, and you win if it's black and lose if it's red.

What's the highest win probability you can achieve with any strategy?

• What happens when I don't say "bet" before the deck is empty? – Philipp Jan 15 '15 at 14:46
• @Philipp I would think that you lose. If not, you can play without losing by waiting until the last card (the color of which you will know) and either betting or passing. – KSmarts Jan 15 '15 at 15:31
• @Philipp: If you are required to yell "bet" before all cards are dealt, then if you haven't yelled "bet" by the time the 51st card was dealt you would be required to yell it then, even if such action was a guaranteed loss. – supercat Jan 15 '15 at 18:11
• @KSmarts: I would interpret the rules as saying that failure to bet on the 51st card simply isn't an option; failure to bet before the 50th is dealt would constitute an automatic bet on the 51st. – supercat Jan 15 '15 at 18:12
• @Philipp Yes, you are required to bet on the last card if you haven't already. – xnor Jan 15 '15 at 20:04

## 4 Answers

It is

$50-50$, i.e a fair game. In fact, any strategy will give you 50%

To show that

there are no strategy that maximizes gain, look at the game once we call stop. Suppose there are $k$ cards left. The probability that the next card is black is the same as the probability that the last card is black. But to bet on the next card is the same as betting on the last one whatever strategy I use, I have 50% chance of winning.

• I'm not convinced that's correct. If I call stop with 4 cards remaining and I've counted 25 red cards that means there's only one red card left. So isn't my change of winning 3/4? – Ash Burlaczenko Jan 15 '15 at 8:23
• @Ash, let me restate the second spoiler. Suppose there's 4 cards left and you're ready to stop. Your friend fans the last four cards out and says "Eh, take any one you want." Remember, the deck is randomly shuffled. So from your perspective, there is no difference between any of the four cards. So it doesn't matter which one you take. So you might as well take the bottom card. Do you agree? – Lopsy Jan 15 '15 at 9:17
• A great answer! Clear and simple. – Gamow Jan 15 '15 at 9:56
• @Lopsy, if I counted that of 4 remaining cards only one is red, my win chance is 75% regardless of which card I pick. A proper proof of the answer needs to show that advantageous and disadvantageous situations even out, regardless of strategy. – Cephalopod Jan 15 '15 at 20:43
• @Cephalopod That's right! The probability is 75% regardless of which one you pick, so you might as well take the bottom card. Of course, the exact numbers don't matter. If there are 19 cards left, 12 black, and you're ready to stop, the same argument applies. You might as well take the bottom card, because it's the same probability either way. In this game, at some point, you'll declare that you're ready to stop. Then, I'll fan out the cards, and you might as well take the bottom one. Hey, I'll set the bottom card over on that table; no matter what, you might as well pick it in the end. – Lopsy Jan 15 '15 at 21:18

We show that for $r$ red cards and $b$ black cards your chance of winning is $\frac{b}{r + b}$ and that this is true for all possible strategies. We do this by induction on $T = r + b$, the total number of cards left. This means we need to show that the chance to win is $\frac{b}{T}$.

For $T = 1$ you must, according to the rules of the game, call "bet". If $b = 1$, only a black card remains and your chance of winning is $1$. If $b = 0$, only a red card remains and your chance of winning is $0$. Your chance to win is $\frac{b}{T}$ in both cases.

Now for the induction step, we assume that your chance to win with $b$ black cards remaining in $T - 1$ cards is $\frac{b}{T - 1}$ (for any $b \leq T - 1$). We calculate the chance to win with $b$ black cards remaining in $T$ cards (for any $b \leq T$).

If you choose to call "bet" now, your chance of winning is clearly $\frac{b}{T}$. If you wait for the next card instead then your chance of winning = (chance of red on next card $\times$ chance of winning with 1 less red) $+$ (chance of black on next card $\times$ chance of winning with 1 less black). The chances to win after removing a card are given by our induction hypothesis. $$\frac{r}{T} \times \frac{b}{T - 1} + \frac{b}{T} \times \frac{b - 1}{T - 1}\\ = \frac{rb + b(b - 1)}{T(T - 1)}\\ = \frac{b(r + b - 1)}{T(T - 1)}\\ = \frac{b(T - 1)}{T(T - 1)}\\ = \frac{b}{T}$$ Regardless of whether you bet now or wait, your chance of winning is $\frac{b}{T}$.

This completes the induction, and so your chance of winning is always $\frac{b}{T} = \frac{b}{r + b}$. For the values given in the problem ($r = 26$, $b = 26$), this chance is $0.5$.

I agree the answer is

50%

But the way I thought of it was:

There are two types of strategies: one, wait after $x$ number of cards or wait after $x$ number of red cards (symmetry for black).

First strategy:

If I wait for one card and it's red, I now have a 25/51 chance of getting black on the next shot. If the card is black, I have a 26/51 chance of getting black. You gain or lose 1/102 chance with a 50% chance, which doesn't help. You can apply this for any $x$ number of cards.

Second strategy:

If I wait for the first red card, it's still the same thing as above only the position I stop at isn't decided ahead of time but through chance. If I wait for 10 reds, the chances that I find this before the 20th card and after the 20th card are the same, which correspond to a higher or lower chance of winning.

Other than these two strategies, there are no other (that I can think of) distinctive possible strategies other than cheesy ones like "go through all 51 in hopes the dealer messes up and shows you the next card".

• There are other strategies. Like "Wait until the number of black cards left in the deck is at least 20% more than the number of red cards left. (If this never happens, take the last card.)" – Lopsy Jan 15 '15 at 9:26
• I guess depending on what percentage you use, it changes the math. If you wait until the number of black cards are 5% more, you are more likely to hit that condition raising your chances, and if you happen to hit the last card, it is that much less likely to be a black card. I'm not sure how to prove this and I don't think the other answers specifically counter this either. Theosza's answer is basically my strategy one with variables, Jean's is basically "it's common sense" and disregards feedback strategies. – Quark Jan 15 '15 at 17:20
• The other answers don't specifically counter this. Because they don't need to: they work no matter what the strategy is. If you're not sure why another answer works, leave a comment on it, we're happy to explain better. – Lopsy Jan 15 '15 at 17:25
• So you're implying theoza's answer covers that theory but my strategy one doesn't? Can you specify where this distinction arises from? – Quark Jan 15 '15 at 17:32
• The first sentence of theoza's post says that their solution works "for all possible strategies." Your post considers two possible strategies. – Lopsy Jan 15 '15 at 19:06

Wait for him to drop almost all cards. Now there a three different possibilities:

• all 26 red cards are already on the table: you would have 100% chance to win since there are only (at least 2) black cards left.

• you can see that already 25 of your black cards are on the table at some earlier time: you have to wait for the end anyway because there is one black card among reds with a lower chance to hit. Not sure about the exact chance to win here.

• if there were one red and one black card left you would have to bet on the last card but one; giving you a 50% chance anyway - nothing to lose (Y).