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 hand1, 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-aces2 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 $1-\frac{9139}{20825-6327} = \frac{5359}{14498}$.3

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.


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. 13 cards []
  2. let's call them 'duds' []
  3. Phew, we agree with FC. []


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