Three Cards: One Black, One Red, One Half & Half

Just a short one today, all. I’m heading further North. So far from civilization that when I was a boy, it was a cause to celebrate when a cit thirty miles distant opened a Holiday Inn with an—get this— enclosed swimming pool. Such a marvel!

Good probability problems are those that affront the intuition, and the best are simple-sounding but cause fights or furious indignation. The Monty Hall dilemma is one such problem.

Most have heard of the stink made when the self-named Marilyn Vos Savant submitted her correct solution to Parade magazine. Not just civilians, but your actual college mathematics professors wrote in to say Wrong! But wrong they were.

We met something of the same symptom with the “One Boy Born Tuesday” problem. How the hell can knowing which day a kid was born give us any information about his sex?!

This kind of thing doesn’t fit into any of the ordinary slots we set aside for probability knowledge. And that’s what makes it such a wonderful example, because it forces us to carve out new slots once understanding arrives.

The “Three Cards: One Black, One Red, One Half & Half”, and the subject of today’s puzzler, is nothing more than a gentle push on the boundaries. The trick will come in trying to expand it by adding a twist, something I’ll invite you to do.

I have three cards, one of which is all black, one all red, and the last black one side and red the other. Before you on the table is a card whose face is red. Given this information—for we recall that all probability is conditional on stated premises—what is the chance that the other side is black?

Most intuitions suggest that the chance is 1/2, since you have either the all red card or the card that is half and half. But the actual chance is 1/3. Here’s why.

In all probability problems, the only thing to do is to lay out everything that can happen. This can be laborious, but once you do so you cannot go wrong. The mistake people make is to try and jump to the right answer using just their intuition.

What’s everything that can happen? Well, every card has two sides, 1 and 2. You could have seen B1 or B2 of the black card, R1 or R2 of the red card, or CR (combo red) or CB (combo black). You see red, and that can happen three ways: R1, R2, or CR.

If you see R1, then the other side must be R2. Likewise, if you see R2, the other side must be R1. The third and final possibility is you see CR, which makes the other side CB. Thus, the probability—give all the evidence we have—is 1/3.

Not so bad, right? Almost to the point of being dull.

So can we think of a clever way of tarting it up? My guess is that adding seemingly irrelevant, but actually pertinent, information can add the necessary amount of bogglement. Like in the “One Boy Born Tuesday” problem.

How about this? Same initial premise of three colored cards, but this time I say I deal you the card on Tuesday. Given this additional information, what is the probability the other side is black?

I haven’t worked this out—it’s a spur-of-the-moment thought. It might well be that the day of the week is irrelevant. But can we prove it is so?


  1. I would like to revisit the “one boy problem.”

    Suppose that Mrs. Smith has two children, at least one of whom is a son. What is the probability both children are boys?

    Now, suppose you are pondering the mystery of this question, on your way to the store. As you walk down the street you bump into Mrs. Smith, with one of her children in tow. She introduces you to her boy. What is the probability that her child at home is a boy?

    These questions have different answers. So, what is the difference in the question?

    Can we set up the 3 card problem in a way that seems like it is the same question but has a different answer? If an agent has selected a card and placed it in front of you red face up, I don’t think you can say that there is a 1/3 chance that it is black on the other side.

  2. Two one-boy problems for you with two different answers:

    1) Take 100 mothers, each with two children. Remove every mother from the group who does not have at least one boy. Pick one of the leftover mothers at random. What is the probability that she has two boys? (1/3rd)

    2) Take 100 mothers, each with two children. Remove every mother from the group whose first child is not a boy. Pick one of the leftover mothers at random. What is the probability that she has two boys? (1/2th)

    Unfortunately, both of the above cases can usually be described by most forms of the original problem statement. You have to know whether the information “one of the children is a boy” was incidental information from a single observation (ie., a veil on one of the children is lifted and the results are simply reported), or actual sorting information as in case 1. Most people assume (wrongly?) case 2, because case 1 requires a sorting process that is never explained.

    I notice that in your other blog post, you actually state the problem correctly, which is rare. The grammarical construction of your statement makes case 2 very likely.

    The Monty Hall problem, on the other hand, is never stated correctly. You can similarly make two versions of it. The probability depends entirely on whether the talk show host guessed a door at random (making the information incidental), or selected a goat because he knew what was behind what doors.

  3. Well said Tman! I agree with 99% of what you’ve said. The 1% is that I have actually seen the Monty Hall problem well-stated. It takes a lot of text though. It takes the ambiguity out of it, but I suppose it takes the fun out too.

  4. You have three cards: one has the word “boy” on both sides, the second the word “girl” on both sides and the third has “boy” on one side and “girl” on the other. Mrs Smith shows you one side of a card and it says, “boy”. What’s the odds it says, “boy” on the other side? Well, the other side could say “girl” or it could say “boy” or it could say, “boy”. So, 2/3. Right? Well, yes. If she introduced you to one son he could be either of her two sons if she has two sons.

    Except that we already decided that a two-child family presents four possibilities so why only three cards?

  5. One source of error isn’t just the result of jumping to the intuitive, but the confusion of details. Monty Hall is the classic example (just see the comment above). There may in fact be more than one potential process at work, which alter the likelihood of potential outcomes, and we are not told or able to determine what the likelihood of either process being at work.

    A variation is to lead a reader to formulate an incorrect understanding of the process.

    In the simple posing of the question for your card trick, your language helps to entice a reader to :

    one card is ALL red
    one card is ALL black
    one card has red ON ONE SIDE and black ON THE OTHER.

    It is easy for a reader to make the mistake of drawing an image of playing cards. And hence assuming mistakenly that your card that is “red all over” is “red all over one side with a common pattern on the other”

    To make such an error alters that possible outcomes and lead sto the erroneous 1/2 answer – for in that case there would be 2x black sides, 2x red sides and 2x “back” sides. 4 possible elimated, leaving the other 2; of which one case has black on the other side, the other has the common pattern.

    Language that entices people to use erroneous prior assumptions like this is a good way to trick them and hence make money. Many good confidence tricks would use this approach.

  6. A variation is to lead a reader to formulate an incorrect understanding of the process.

    You beat me to it – I was about to say the exact same thing.

    Then there’s the issue of how it works. Are we just told you have 3 cards and you’re just showing us 1, or are all 3 cards on the table? Is 1 on the table and 2 behind your back? How do we know you’re not cheating and telling us a false premise?

    It reminds me of Marlyn’s column where there’s something like… “You’re taking a test. You get 3 pts for every right answer. 1 pt for every wrong answer (including blank I think) and -5 pts for every guess.” Everyone wrote in to say she was wrong that you should never guess and she stressed that “No, you get punished for guessing, you shouldn’t do it” without realizing that what everyone was having a problem with was: How does the test distinguish between a guess and a wrong answer? (IIRC she even said the test was multiple choice, making it a false premise even more)

    I’m willing to bet that’s the biggest issue a lot of people have with stats. (which didn’t Briggs point this out with game theory also?) If the premise is true, yes they’ll agree that such and such is correct, but in the details, the premise seems to fall apart. The best example ever is this webcomic. When it comes to cons, your chance of winning is always 0%.

  7. These sorts of problems have always made me crazy – although the explanations make sense logically, they’ve never “felt” right to me – up till now (thank you, Briggs). And I think part of the problem is that there are in at least some of these scenarios two probability situations that are distinct logically and mathematically, but which (at least with me) can become conflated psychologically (if that makes any sense).

    Here’s how I got to a comfort level with the “three card problem” presented by Briggs – perhaps if there are some students who have a tough time dealing with it, this approach might help.

    The two probabilities involving the three cards (one all red, one all black, one mixed) are 1) what’s the likelhood of a red face showing up after you lay down one of those cards? (that’s 1 out of 2, right?), and then what’s the probablility of that red face having a black face reverse side (that’s the annoying 1 out of 3).

    So I took the original statement of the problem and modified it a tad: what if I had one card, red on both sides; one card with one face red, one black; and a million all black cards? Now imagine shuffling and flipping those cards (the tough part once you move past the gedanken experiment),
    drawing one at random, and slapping it on the table. The chances of seeing a red rectangle are almost infinitesimal – but still finite.

    Yet if you do get a red face, you know that card is one of only two possible cards in the deck. However, if you want to figure the odds of the other side being black, you’ve got to think of them not as two possible cards out of a million and two, but (since both faces are equivalent, except for (possibly) color) what’s actually four “cards” out of two million and four “cards”.

    And those four “cards” are black showing, red reverse; red showing, black reverse; red showing, red reverse; and (other) red showing, (other) red reverse. Now the “card” on the table obviously can’t be the black obverse, red reverse, which leaves only the other three as options. And of those other three, only one has a black reverse. So 1 out of 3 – Yesss!

    Now I’ll go see if I can make sense of the complicated versions of this thing.

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