It’s the Friday Puzzle!


A fun but tricky one this week….

Imagine that 4 cards have been removed from a pack of 52 cards and placed face down in front of you.  If you were to choose any two of the four, the chances of getting two red cards is 50%.  What are the chances of picking a black pair?

As ever, please do NOT post your answers, but do say if you think you have solved the puzzle and how long it took.  Solution on Monday.

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.


72 comments on “It’s the Friday Puzzle!

  1. mirfaan says:

    I had to draw all the possible combinations to work it out, but still under 2 minutes. Sneaky sneaky.

  2. the urban server says:

    Maybe thirty seconds in my head. Nice.

  3. Falsedog says:

    Got it quite quickly. Nice start to the day!

  4. I think I got it in seconds… but then I think the question is geared up to create self doubt

  5. Carl says:

    Couple of minutes. Once again, I think the wording of the question leaves something to be desired, but at least the answer is unambiguous this time (I think).

  6. Nick says:

    It took a few seconds to confirm the obvious answer.

    Then there’s the matter of whether there’s sufficient ambiguity in the wording of the question to permit a different answer. On one reading, the procedure used to remove the four cards will determine the required probability. It could thus be argued that since we aren’t told the procedure, we can’t determine the probability. On another reading, we are meant to regard the procedure as irrelevant because the cards “have been removed”. There is some element of ambiguity, but I’m unsure whether it’s sufficient to reasonably permit a different answer.

    • Bob O'H says:

      If you’re worrying about this, you’ve got the wrong answer.

    • Nick says:

      I’m sure I have the correct expected answer, but it’s still interesting that how the four cards were drawn can, under a different interpretation, give different answers. (I’m sorry if that seems a little cryptic, but I’m trying not to give too much away!)

    • Steve Jones says:

      I’m struggling to see how the way the original four cards were drawn has any impact on the answer. From what I can see the original 4 cards could have been drawn at random or carefully and deliberately chosen, and it makes no difference to the chance of a random draw of any two of the four giving a pair of blacks once it’s known that there’s a 50% chance of any such pair being both red.

      However, I’m also not sure why Richard says this problem is tricky – perhaps I’ve missed something.

    • Jimbo says:

      @Steve, how many red and black cards do you think there are? Calculate the probability of picking 2 red cards from those 4.

    • Steve Jones says:


      It’s not a matter of how many red & black cards I think there are, it’s a matter of how many there must be for a 50% probability of randomly drawing a pair of red cards from the four. There is only one such combination.

    • Nick says:

      @Steve Jones: The question is: How do you get from (presumably) (a) a regular pack of 52 cards to (b) 4 cards such that the probability of drawing (without replacement) 2 cards is 1/2? There’s more than one way to get from (a) to (b), and more than one answer. But I’m guessing the wording “4 cards *have been* removed” (my emphasis) is meant to indicate that we should start not from (a) but from (b). In that case, the answer is unique.

    • Steve Jones says:


      The question makes no sense at all from the point of view of starting from (A). The probability of choosing two red cards at random from 4 cards, also drawn at random from a full pack of 52, is 25/102 (and the same probability as choosing two black cards of course). So it would never match the 50% criterion to consider the probability from that point.

      The question clearly is about the selection of two cards at random from (B), and there’s only one config of red & black that gives a 50% chance of drawing 2 red cards from the 4. It matters not at all how we got from (A) to (B). As it ia the clear implication it’s the same (B) (or else the 50% information is completely irrelevant), then there is only one answer to this problem that makes any sense.

    • Nick says:

      @Steve Jones: The question doesn’t say the 4 cards are drawn *at random* from a full pack. The method of drawing the cards can indeed influence the answer, and to a significant degree. The 50% information is still relevant.

      If the question is only about the selection of 2 cards from (b), why mention the full pack?

    • Tort says:

      @Nick It can’t influence the answer. It makes no difference how the cards are drawn. The question indicates the probability of choosing two red cards from the four in front of you, it makes no difference the probabilities that went into selecting those cards just as it makes no difference when you are flipping a coin that you have already flipped heads 3 times, the chance of the next flip being heads is 50-50. @Steve Jones has it right.

    • Nick says:

      @Tort: Here’s an example of how the selection method can influence the answer. I don’t think I need to give a *** SPOILER ALERT *** because the solution is so different from the expected one, but avert your eyes if you wish…

      Suppose the selection protocol is:
      (a) select a card at random from the pack, and then
      (b) select three more cards of the same colour as (a).

      Now, the chance of picking two red cards from these four is 50%, the same as the chance of picking two black cards. Rather different from the expected solution.

    • Anonymous says:


      Suppose that in (a) you got a black card. By (b), all 4 cards must be black.
      So, the probability of picking 2 red cards from 4 black cards is 0.
      But they are telling us that the probability of picking 2 red cards from the 4 cards is 0,5.

    • Nick says:

      @Anonymous: There’s only a 1/2 chance that a black card is selected in (a), and thus a 1/2 chance that 2 black cards will be picked from the 4. Similarly if you replace “black” with “red” throughout.

      If you were to follow this protocol 100 times, roughly half the time you’d pick 2 red cards from the 4.

  7. StewE17 says:

    Looks obvious at first glance but took a few minutes to work it out properly.

  8. Goliath says:

    “A fun but tricky one this week…”

    And I am looking for the trap because I got an answer in a few minutes but am now unsure whether it’s right or not.

    Well, let’s see on Monday if I get a good start into the new year.

  9. dcmonk79 says:

    I think I’ve got it!!

    Don’t quite know how it happened, it just popped into my head. I normally have no chance of getting The Friday Puzzle right!

  10. benjaminsa says:

    Agree with Nick, this was fairly ambiguous.

  11. Stray Cat says:

    Cute. Yeah, I think I got it.

  12. JimC says:

    Got an answer after a couple of minutes that I’m quietly confident about. I think the information that you have a 50% chance of drawing a red pair from the four cards is sufficient to reduce any ambiguity.

    • Confused says:

      Does black pair mean two black cards or two black cards of the same face value? i.e. the 8 of spades and the 8 of clubs. If the the former the answer is easy. If the latter I can’t see anyone solving this in “30 seconds”

    • JimC says:

      If my answer is correct, your question is irrelevant.

    • Anonymous says:

      @confused. From the question I’d say two black cards of any value. I’m not going to explain why, but the question makes it impossible to make it mean “black cards of the same value”.

    • Bob O'H says:

      @Confused – the ‘two pairs’ problem isn’t that difficult – the first card just has to be black, and that defines what the second card has to be.

    • Jimbo says:

      @JimC, who said anything about a red pair?

    • JimC says:

      @Jimbo – to clarify, by “red pair” I meant two red cards, not a pair in the poker player’s sense of two cards of equal value. Loose wording on my part.

  13. AMWhy says:

    Aside from the annoyingly incorrect non-puzzle answer I’m sure people will give of ‘picking’ meaning you can look at the cards, actually looking at the puzzle itself I have an answer in about a minute.

  14. Bob O'H says:

    Oh duh. About a minute, about 5 secs once I had realised it wasn’t a semantic question.

    Nice one.

  15. M says:

    Took me about 10 seconds to understand the question, and another 5 to answer it.

  16. SofARMaths says:

    Couple of minutes. Another minute to confirm with tree diagram.. Good one.

  17. Yat says:

    Got stuck for ten seconds reading “If you were to choose any two of the four, the chances of getting two red cards is 50%” before I understood this was just an additionnal information about the 4 cards. This information immediately eliminates all but one possibility about the number of black cards, without calculations.
    then I read the question, which was totally obvious at this point.

    • Berhard says:

      it took me some few tens of seconds more, to have the same conclusion, but i made a small doolde just for fun to proof it…

  18. Paul Durrant says:

    Ten seconds or so. Another twenty seconds double-checking.

  19. Lazy T says:

    I reached over for a pencil and realised the solution so I drew a smiley face.

  20. ers says:

    10-15 seconds.

  21. Pumblechook says:

    Shamefully, took me about five minutes.

  22. Tom (iow) says:

    Got it after a few minutes, assuming it means from the same four cards and not from a fresh four.

  23. Charles White says:

    Solved the Friday puzzle about two red cards. Easy.

  24. MaciejK says:

    It took me more time that it should have taken because I thought that it said that you have 50% chance of selecting two red cards in general (which is clearly wrong). Then I realized that it is simply an information about the way the 4 cards have been selected.

    It is also the opposite situation to the one we had some weeks ago with the blue stones. While the puzzle is perfectly unambiguous this time as well, the procedure does matter and we are told in the puzzle what it was.

    • Anonymous Aussie says:

      Maybe Richard thought this one was “tricky” as he didn’t get the right answer to the blue stones problem

  25. Scott R. says:

    10 seconds to solve it, and another 5 to find the second (impossible) solution

  26. Blue Sky says:

    So there are equal number of both black and red cards in a deck? 🙂 Soo….obvious.

  27. About 30 seconds in my head.

  28. Tim Greenwood says:

    A few seconds for the instinctive answer. A coupleofminutes to prove it to myself.

  29. Anonymous says:

    done – about 30 seconds with a piece of paper. Nice teaser though.

  30. I got the answer almost immediately, and took a few more seconds to check it mentally.

  31. Berber Anna says:

    Once I figured out the first bit, a few seconds to reach my conclusion. First bit took me a few minutes and a check of the comments to see if more people were thinking in this direction. Now to see if I’m correct.

  32. ken gilhooly says:

    Hi think got red/black card prob in c 5 mins with aid of P&P diagram..

  33. physicalist says:

    Got it. A minute worrying about how to interpret it. Another minute writing out some possibilities to sort my suspicions. And then a few seconds confirming that I had the right answer.

  34. I don’t know how to do the math, but I do know what the tricky part is. Is that worth anything?

  35. Jerry says:

    First, I needed to find a paper and pencil. Second, I needed to think a little. Third, I found the surprising answer. Total, say, five to seven minutes.

  36. themenniss says:

    Solved it in about 30 seconds on paper. it’s either really simple or i’m wrong.

  37. I think it’s easy – though I am assuming the clue is in the phrase “the chances of getting two red cards is 50%”.

  38. matthew naylor says:

    I read a few comments to be sure it wasn’t tricky wording. People will generally assume that there’s an even chance of getting either colour so that’s the only ‘trick’ cos actually you just need to know how to calculate probability and try out the possible configurations in your head.
    Couple of minutes or three including reading comments.

  39. Gail Hatton says:

    Not entirely sure if there is an answer as we know there are two red cards in the four removed, however four red cards could have been chosen, so I don’t know the answer to this one. Nagging me now, nice one from the Wiseman.

  40. Dave Rickey says:

    About four seconds, because I could tell right away there was something odd with the initial statement of the odds.


  41. John Loony says:

    Got it! About 1½ minutes, although I had to write down and work through all the combinations first. Another (unusually) satisfying one, in that I know that I have got the correct answer without having to worry about double meanings or dubious interpretations or dodgy definitions or any of the usual mind-bending bourgeois degeneracy which Comrade Wiseperson usually inflicts upon us.

  42. Anon says:

    I thought that the Bourgeois degeneracy was the best bit Mr Loony

  43. I think I got the answer. Yeah, it to me more than 5 minutes.
    Seems to me that how the cards were selected is at random. Otherwise, the method should have been specified.

  44. Phil Watts says:

    Depends if there’s a difference between selecting two cards of the same colour (one after the other) or a pair of the same colour (two cards simultaneously).

  45. Ed says:

    Less than 15s to understand the question and arrive at the solution w/o any calculation. Then 15s to confirm my answer after reading the comments here.

  46. Ronald Lett says:

    Took me a few seconds to realize all the information before the probability of getting two red cards was just a magician’s trick of providing irrelevant extraneous information. Very cute. =)

  47. anibal morbo says:

    explicaciones cientificas de la astrologia se encuentran como:
    “month of birth.behaviour.etiology”
    en la web en: “Figure 19. Trends in percent of
    > forked vs. spiked antlered yearlings for 217 deer in the
    > —Genetic/Environmental Interactions in White-tailed
    > Deer“.” se relaciona la frecuencia del alelo genetico y la actividad solar anual, lo que pudo confundir a los astrologos

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