*Estimated read time (minus contemplative pauses): 6 min.*

A strange playing-cards probability question just occurred to me:

There sits before you a standard, well-shuffled deck of 52 playing cards. I remove a card without showing it to you. If you draw a card from the remaining 51-card stack, what’s the probability it’s the 9♣? Easy, you say: 1/52. That’s correct. But suppose I tell you that the first card I removed is the 7♥? Now what’s the probability of drawing the 9♣ from the 51-card stack? Easy again: 1/51. Correct. Rewind the scene. This time, I tell you the discarded card is *not* the A♦. Now what, from your perspective, is the probability of drawing the 9♣? This time you pause, but only for a moment, before producing the correct answer of ……*hover over or click this footnote number to see the answer *^{1}…… Correct again!

Maybe you find this as strange as I do. “Strange” might be the wrong word. Surprising, maybe. At least, I was surprised. My gut response to the last question was that it shouldn’t make any difference to learn what the discarded card is *not*. Why should it? There are 51 cards in the stack. The 9♣ is in the stack, as is the A♦. I don’t know what was discarded. I’ve got 52 cards and don’t know where any particular card is so why not 1/52? So, yeah, maybe strange.

But I also knew this information *must* make a difference. A few seconds’ thought clarified how. Had I been told the discard wasn’t the A♦, 2♦, 3♦… listing every card but the 9♣, then I’d know there was no chance of drawing the 9♣ from the remaining 51 cards. From that perspective, the problem became easy. No longer strange. Maybe I’m just tired.

You obviously found it easy enough, but for those of us who still aren’t sure, or would at least like to exercise our intuitions about conditional probability, here’s an intuitive (I hope!) explanation of what I presume was your thought process—but do let me know if you took a different approach.

Consider again the series of questions posed in the story:

(1) I remove a card from a standard, well-shuffled, 52-card deck. Neither of us sees the card’s face. What’s the probability a card pulled from the remaining 51 cards is the 9♣?

The answer is 1/52. In case that’s mysterious, reflect on the following. Before the card was removed, there was a 1/52 chance of any given card being the 9♣. Probability doesn’t care about what region of space the individual cards occupy, so the probability doesn’t change when you set one of the cards beside the deck or in the other room or if you eat it. You could also spread the cards out side-by-side and there would still be a 1/52 chance of randomly selecting the 9♣ (hopefully that’s obvious). Finally, imagine removing the top card, the second card, third, fourth… until you’re left with just the bottom card. This is the same outcome had you only removed the bottom card from the deck to begin with. Removing the bottom card from a deck of 52 cards clearly has a 1/52 chance of being the 9♣ (or any other card, for that matter).

(2) Same scenario as (1), but this time I flip the discarded card to reveal the 7♥. Now what’s the probability of drawing the 9♣ from the remaining 51 cards?

The answer is 1/51. We know what the discarded card is, so we can eliminate it from the denominator.

(3) Reshuffle the deck and start over. I remove a card. I look at it and tell you it is *not* the A♦. What, from your perspective, is the probability a card pulled from the remaining 51 cards is the 9♣?

I’ll lead up to the answer. First, I’d like to emphasize that I’m asking what the probability is from *your* perspective. I’ve seen the card, so from my perspective the probability of drawing the 9♣ from the remaining 51 cards will either be 0 (due to the card I’m looking at actually being the 9♣) or will be 1/51 (due to the card I’m looking at *not* being the 9♣). But the probability is different from your perspective because your evidence is different than mine.

This is true of the discarded card as well. It started, for both of us, with a 1/52 chance of being the 9♣. I’ve learned what that card is, so for me it’s either 0 or 1 that it’s the 9♣. Your current evidence isn’t strong enough to take you to a 0 or 1, but you do have a some new evidence that recommends a probability update. Namely, your evidence has gone from the discarded card possibly being any card in the deck to its being any card *but* the A♦. In other words, there were 52 cards it could have been, but now there are 51. So the probability, from your perspective, of the discarded card being the 9♣ goes from 1/52 to 1/51. From this, we can work out your probability that the next card pulled from the remaining 51-card stack is the 9♣. To see how, let’s dissect an easier example.

I mentioned above the significance of telling you that the discarded card is not the A♦, 2♦, 3♦… listing all but the 9♣. You’d then know that the probability of the discarded card being the 9♣ is 1, and the probability of the 9♣ coming out of the 51-card stack is 0. Notice that those two probabilities add up to 1. Now imagine what happens if I tell you it’s not the A♦, 2♦, 3♦… listing all but *two* cards—say, the 9♣ and the 3♥. Now you know that the discarded card has a 1/2 probability of being the 9♣. This means, intuitively, that there is also a 1/2 probability that the 9♣ is in the 51-card stack. This is also borne out by the basic idea that the probability of the 9♣ being the discarded card OR being in the 51-card stack must be 1. In other words, the respective probabilities must add up to 1 (just as it did in the first example in this paragraph). This is a key idea here.

We’ve established that there’s a 1/2 probability the 9♣ is in the 51-card stack. Every card in that stack has the same probability of being the 9♣. So, we need to break that 1/2 probability into 51 equal parts. In other words, each card has a probability of (1/2)/51 = 1/(2×51) = 1/102. That it’s so low—i.e., about .0098 or .98%—makes sense, as it should get us close to the probability you have when knowing with a probability of 1 that the discarded card is the 9♣ (which, again, translates to a 0% chance of drawing the 9♣ from the 51-card stack).

Notice that the above, in which we break up the probability into however many parts, is essentially what we’re doing with all the questions here. When you pull a card from a 52-card deck, there’s a probability of 52/52 that the 9♣ is in the deck. We break that up into 52 equal parts to find the probability of the 9♣ (or any other card) being pulled: (52/52)/52 = 1/52. And when we discard one card without looking at it, there is a 1/52 chance that the discarded card is the 9♣, and there’s a 51/52 chance it’s in the remaining 51-card stack, which we break into 51 equal parts to figure out the probability of pulling that (or any other) card from the stack: (51/52)/51 = 1/52.

All we have to do now is apply the above insights to question (3). We’ve learned that the discarded card is not the A♦ and have established that this translates to a 1/51 probability of that card being the 9♣. There’s still a probability of 1 that the 9♣ is either the discarded card or one of the cards in the remaining 51-card stack. So we subtract 1/51 from 1 to get a 50/51 probability that the 9♣ is in the 51-card stack. Finally, we evenly disburse that probability (i.e., 50/51) among the 51 cards in the stack by breaking it into 51 equally sized parts: (50/51)/51 = 50/51² = 50/2601.^{2}

That final number appears less strange when we look at decimal estimates. For pulling the 9♣ from the stack, we started with a 1/52 ≈ .01923 probability, and then used the evidence that a single discarded card is not the A♦ to update that number to 50/2601 ≈ .01922. That’s a decrease of about .000007. So, as expected, learning what the discarded card isn’t decreases, if only slightly, the probability for the 9♣ coming from the stack; well, unless we learn that the discarded card isn’t the 9♣, in which case our probability for pulling it from the stack *increases* by about .000377 from 1/52 ≈ .01923 to 1/51 ≈ .01961.

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#### Footnotes:

- 50/2601
- Mental math tip for quickly squaring 51: visualize it as (50 +1)(50 + 1) and then FOIL: (50)(50) + 2(50) + (1)(1) = 2500 + 100 + 1 = 2601.