# The Paradox of the Second Ace

*This post is inspired by a Futility Closet article. Do visit them and subscribe to their excellent podcast!*

Suppose you’re dealt a bridge hand ((13 cards)), and someone asks whether you have any aces; you check, and yes! you find an ace. What’s the probability you have more than one ace?

This is a slightly messy combinatorics problem, but one that can be solved with a bit of elbow-grease.

### Dud man’s hand again

The probability of getting no aces is simple: there are 48 non-aces in the pack, so it’s $\frac{48}{52} \times \frac{47}{51} \times \frac{46}{50} \times … \times \frac{36}{40}$, or $\frac{48!39!}{52!35!}$. There’s quite a lot of cancelling in that, and it works out to be $\frac{39\times 38 \times 37 \times 36}{52 \times 51 \times 50 \times 49} = \frac{19 \times 37 \times 9}{17 \times 25 \times 49} = \frac{6327}{20825}$, about 30%.

The probability of a single ace is a bit harder, but can be figured out as the probability of getting 12 non-aces ((let’s call them ‘duds’)) followed by an ace, multiplied by the number of ways the ace and duds could be arranged. If we replace the final dud in the previous product (the $\frac{36}{40}$) with the probability of getting an ace ($\frac{4}{40}$), we have the probability of 12 duds and an ace in that order; there are 13 equally-likely orders, so the probability is $\frac{9139}{20825}$, or about 44%.

That’s actually enough to solve the problem: 70-odd percent of hands have at least one ace in, 44% have exactly one ace in, so 26% have more than one. Given you have at least one ace, you’ll have several aces about $\frac{26}{70}$ of the time, or around 37%. More precisely, it’s $1-\frac{9139}{20825-6327} = \frac{5359}{14498}$. ((Phew, we agree with FC.))

### Win some, lose some, it’s all the same to me

The follow-up discusses a similar idea: suppose you have the ace of spades. What’s the probability you have at least one other ace?

This is, surprisingly, a less tricky problem. Given we have the ace of spades, what’s the probability that the remaining twelve cards are duds? It’s very similar to the all-duds problem from before, only picking 12 consecutive duds out of the 48 available. You get $\frac{48!39!}{51!36!} \approx 44\%$. (This is exactly the same as the any-one-ace probability, but for slightly different reasons).

That means, 56% of the time you have the ace of spades, you have at least one other ace in your hand!

### But whyyyyy?

It’s a bit counter-intuitive that you end up with a better chance of having an extra ace *if you know the kind of ace you have*, but it does make some sense: after all, only one in four of the single-ace hands contain the ace of spades, but all of the four-ace hands, three-quarters of the three-ace hands and half of the two-ace hands do.

That is to say, when you don’t care about which ace you have, the hands are split 44-26 between one ace and several. Since only a quarter of the one-ace hands contain the ace of spades, but at least half of the multiple-ace hands do, the ratio becomes 11:13+ - which is 54% even just assuming that every multiple-ace hand has two aces in.

I think that’s neat.