On Friday I posted this puzzle….

Caroline, Joan and Sam all go shopping.  They all buy a present, and all of the presents are the same size and wrapped in the same paper.  At the check-out the parcels get mixed up.  What is the probability that……

1) At least one of the three women gets the present she bought?
2) Only one woman gets a wrong present?

If you have not tried to solve it, have a go now.  For everyone else the answer is after the break.

Let’s assume that the presents A, B and C belong to Caroline, Joan and Sam.  There are then 6 ways of distributing the presents:

Caroline, Joan, Sam

1:           A             B          C

2:          A             C          B

3:         B              A          C

4:        B              C             A

5:          C            A             B

6:         C             B               A

Only in cases 4 and 5 does no one get the right present.  As such, there are 4 out of 6 scenarios in which someone gets the right present, and so the chances are two thirds.

2) The probability is zero.  If only one woman gets the wrong present then the other two get the right present, and so the situation simply can’t happen.

Did you solve it? Any other solutions?

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.


  1. Sure, easy enough to solve, following the exact same method using a table.

    However, the stated reason for (2) is a non sequitur. Perhaps a better way to state this is to say that if one woman gets the wrong present, then the intended recipient of that present does not get it either. Thus, a lone error cannot occur.

    1. I hasten to add that this line of reasoning for (2) is the shortcut answer that many folks followed, judging by the Friday comments. With the solutions table at hand (post-hoc) of course it’s simply a matter of scanning the cases.

    2. I don’t see how the stated reason is a non sequitur. If only one woman gets the wrong present, then two women get the right present. But if two woman get the right present, the remaining present can only be the correct present for the remaining woman. Which shows that it is impossible for only one woman to get the wrong present. I know it’s not a formal proof, but I think it’s a perfectly valid explanation of the correct answer.

      What I don’t follow is the relevance of the intended recipient. How does that relate to the answer?

    3. The original reasoning is meant to be a proof by reductio ad absurdum. What’s not spelled out (“missing”) is that two out of three woman having the correct present implies the remaining, single present must in fact be held by the sole remaining woman, which contradicts the assumption that she has the wrong one.

      I argue that this reasoning is more roundabout since you need to consider the entire set. Conversely, if you just consider only one woman having the wrong present and that present’s correct owner you know in general that there must be at least two misattributions already.

  2. I got the right answers (as given) to the question (as stated), but (knowing you) I had assumed that there was a trick.

    I was thinking that the eventual recipients of the presents were not necessarily going to be the same three people who bought them in the first place, i.e. the presents were bought by the three named individuals but were intended to be given as presents to other people outside the confines of the puzzle.

    The ambiguity would have been in the word “get”: Did you mean “get” a present on Christmas Day, or “get” the object at the end of the checkout at the till in the shop?

    1. Ditto.

      I was disappointed in this as a “puzzler”. It doesn’t really require any out-of-the-box thinking.

  3. An alternative way of calculating (1) is to say that the probability of at least one woman having the right present is 1 – probability of no woman having the right present.

    The probability of all woman having the wrong present is the probability of the first having the wrong one (2 in 3) multiplied by the probability of the second having a wrong one (1 in 2) leaving the last woman with a certainty (1) of having the wrong present.

    From that you get 1 – 2/3 x 1/2 = 2/3rds.

  4. There is something known as derangement in Combinatorics which is useful as we can list down the same for more number of people say 10 or more.. of course we can but it will be tedious..
    for 2nd Michael Sternberg is quiet correct..

    1. But that’s invalidated by Question 1, which states that all three are women:

      >> At least one of the three women gets the present she bought

  5. The problem say there are three presents. Then it say the parcels get mixed up. The problem no say there are only three pracels, one per present.

    If a fourth parcel was at the checkout and each women took one parcel and leaving one at the checkout then it is possible with 1/4 chance that just one woman get wrong present,

    Is because wrong present is in fourth pracel.

  6. No twists, no turns, not much of a puzzle this time, could they not have tried to get the most expensive item and had blue spots on their foreheads if they were lying to each other?

  7. The rubric does not require that all three presents necessarily differ in any meaningful or discernible way. If all three presents were essentially identical then the answer to question 1 is “1” in all six possible swaps (if the women themselves can’t tell the difference, then the difference is not discernible and therefore irrelevant). There is also the case to consider where two presents are identical and the third differs. The overall probability considering these cases is left as an exercise for the reader.

  8. > Only in cases 4 and 5 does no one get the right present. As such, there are 4 out of 6 scenarios in which someone gets the right present, and so the chances are two thirds.

    But I took “At the check-out the parcels get mixed up” to mean that it was given that case 1 did not happen. (Which means my answer for question 1 was three fifths.)

    1. The general solution is 1-(1/n), for n at least 2.

      The trick is to count the number of elements of the symmetric group S_n of order n. There are n! ways to order the elements {1,…,n}, but we’ll overcount by a factor of n since the cycle is cyclically permutable. So, there are (n-1)! elements of order n in S_n.

      These are the only elements in which no one gets their own present, so there are n!-(n-1)! ways in which at least one person does, so that the probability is (n!-(n-1)!)/n!=1-(1/n).

      The probability for part 2 is still 0 in the general case, since the stabilizer of the componentwise action of S_n on (1,…,n-1) is trivial. This is obvious, but can also be formally demonstrated by computing the orbit of that element under the action of S_n and using the Orbit-Stabilizer theorem.

  9. same size, same wrapping, but you said nothing about weight. so I’m going to assume that one of the items was incredibly heavy, one was moderately heavy, and one was extremely light. your questions are now rendered moot. have a nice day.

  10. “They all buy a present…”

    So there’s only one present!

    Obviously not the answer, but when you are looking out for twists of logic, this sort of thing stands out to me. But then again, I often tend to over-analyze these puzzles.

  11. If you look at it as a combination rather than a permutation, then there becomes no discernible difference between outcomes for and 5. This puts the probability for question 1 at 80%

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