OK, here is the new puzzle…..

My friend Albert is trying to open a safe.  The following numbers are on the safe door….

77 – 49 – 36 – xx

…the next number in the sequence will open the safe.

What number should Albert use to get inside the safe?

Yesterday I posted this puzzle….

Erica has two children. One of them is male. What is the probability of her other child also being male?

If you haven’t tried to solve it, have a go now. For everyone else, the answer is below….

Let’s imagine that Erica has two children. Assuming that they were not born at the exact same time, there are four possible combinations….
Male Male

Male Female

Female Male

Female Female

This will give rise to 50% males and 50% females. However, we know that one of Erica’s children is male, therefore she cannot fall into the last category, thus leaving us with…..
Male Male

Male Female

Female Male

All of these options are equally likely, so the chances of her other child being male is one third.

Any other answers? If so, post them below.

Follow the blog for more daily puzzles. See you tomorrow!


    1. I thought the same. By virtue of the first child being male, Female Male should be off the board. Should be 50/50.

    2. The birth order of the children. There are two different ways someone can have two children of the opposite sex from one another.

      Male Female refers to the elder child being male and the younger child being female. Female Male refers to the elder child being female and the younger child being male.

      In contrast, there is only one way to have two male children.

    3. @Tom: If the first child was male, you’d be correct, but we aren’t told WHICH child is male, only that one of them is. Maybe it’s the eldest child that’s male, or maybe it’s the youngest child.

  1. Dear both,

    Here’s today’s puzzle but alas, no solution to the fishy problem. Mum, I think your answer is a very good one!

    How are you with numbers. See below….

    Lots of love,

    Ann xoxoxox


  2. I am so stupid. I thought that if you said ONE was male, three the other would be female. 🥰

    1. The birth order of the children matters. Your answer of 50% would be correct if we were told WHICH child were male, but we’re not given that information in the question.

      Here’s another way to think about it: Of all families with two children, what’s the probability that both children are male? Answer: 25% (50% chance that child 1 is male x 50% chance that child 2 is male). So if you choose a two child family at random, there’s a 25% chance both kids will be boys. Similarly, there a 25% chance both kids will be girls, or a 50% chance there will be one of each.

      So, if we weren’t told anything else about Erica’s kids, other than she has two of them, we would assign a 50% probability that either of them is male, or a 25% probability that both of them are.

      If we were told the eldest child were male, we would assign a 100% probability that the eldest is male, but we still wouldn’t know anything about the youngest child, so our probability that the youngest is male would stay at 50%. This would imply a probability of them both being male of 50%. Similarly if we were told the youngest child were male.

      However, if we’re not told which child is male, only that one of them is, it might be the eldest or it might be the youngest. We therefore can’t assign a 100% probability that the eldest is male, or a 100% probability that the youngest is male. But what we can do is rule out the possibility that both children are female. In other words, we know this particular two child family is not one of the 25% of two child families with two girls. Our potential pool of possible families has therefore shrunk to 75% of the total pool of two child families (those with two boys and those with one boy and one girl).

      We already know that two child families where both children are boys make up 25% of the total. But now our denominator (our potential pool) has shrunk to 75% of the total. This gives us a probability of a two boy family as: 25/75 = 1/3 = ~33.3%.

  3. Love the daily puzzle! thank you!

    However, I have to disagree with the probability solution. As one child has been born already and is male there can only be a %50 probability of the next child being male.

    If you are asking about the probability of any person having 2 male children before any one has been born then your answer is correct but that is a different question.

    1. Hi John,
      The problem is the ambiguity of the question. It doesn’t say that the other child is yet to be born, nor even that the child, whose sex was revealed, was the first child.
      So when you say that is a different question, it’s actually not! Your interpretation was potentially the different question 😊
      Of course, if probability only refers to events that haven’t happened, rather than events that have happened, but haven’t been revealed yet, then you would be right. Perhaps, that is the strict mathematical definition (it was too long ago for me to remember), but the
      argument would then be that this puzzle is a generalist one using the word in it’s general sense.

    2. Hi John
      I should say that I do agree with your 50% though.
      No-one is caring the order of the children really.
      The chance the first mentioned child is the first AND a boy is 2/3. The second is either a boy or a girl and that is 1/2 (as you rightly say). Overall that gives the 1/3 figure. The same applies to the other way around giving a 1/3. That leaves 1/3 for the BB combination, which isn’t the question as we have already been told that first part, so the 2/3 was removed from the equation.
      So I agree that you are right, for the right reason, but that birth is irrelevant 😊

  4. Dear Richard,

    Living the puzzles What about the length of the fish though?

    Driving us crazy 🤯 Thanks Gary

    Sent from my iPhone


    1. Gary,

      This is where your secondary school algebra would have helped you in a few seconds!

      I’m guessing it’s ok to post the answer as it’s the day after.

      The extra length is half the total length.
      Let’s call the extra length X. The total length of the fish is then 2X.
      The length is also given as 30 + X
      2X= 30 + X
      And so X = 30
      The fish is 60 (2X).

      That’s if I remembered the question right!

  5. Hi Rich,

    I’m struggling to see where the ambiguity is:

    > Erica _has_ [so I suppose we have to agree that 2 children exist right now rather than one being born, one yet to be born] two children.
    > One of them _is_ male [i.e. one child exists, is male]. What is the probability of her _other_ child [i.e. the child who is not definitely male] also being male?

    As phrased this can only be %50 probability surely? even if the child referred to in the statement “one of them is male” is not known to be the first or second born I don’t see any ambiguity as the next statement refers to the “other” child.


    And yes I do have too much time on my hands.


    1. Oops we were posting at the same time. I don’t think you are wrong! See my post under my initial one.
      Best wishes, Rich

  6. Hi John,
    My answer to your post here, referring you to my second post above, when we appeared to be writing at the same time hasn’t appeared yet. Perhaps, I’m writing too many too soon. So what colour is the apple? 😂

    Seriously, I hope you saw it as I’m not disagreeing with your logic.

  7. Hi Rich,

    The apple is clearly red, this has never been in question and is obvious given that the monk is a Franciscan and not a Benedictine, the clue was in the name.

    Best wishes,


    1. Hi John,
      I am astonished at how quickly you got that, given that the monk didn’t even realise he was in the question until you pointed it out. I’m very impressed.
      Hope you have a great day (given the restrictions).
      Best wishes

  8. The step “All of these options are equally likely” is wrong: once you use information to remove the [F,F] case, you have to compute the new probabilities either using Bayes’s rule or from state-event pairs (if you’re a frequentist).

    This problem is isomorphic to flipping two coins, seeing one heads and asking what the probability of another heads.

    This is a math problem, so one should do the math, not hand-wave: http://sitacuisses.blogspot.com/2019/07/a-family-has-two-children-one-is-boy.html

    1. José, you’re right: it’s the same probability as flipping two coins, and then, when seeing that at least one of them is heads, asking what the probability of both of them being heads then is. And the maths does work out at ⅓.

      You can try it!

      Get two coins and flip them. If you get at least one heads, note whether the other coin is also a head.

      A quarter of the time you’ll get 2 tails, so not note anything.
      A quarter of the time you”ll get 2 heads, so note ‘yes’.
      And the other half of the time you’ll get 1 of each, note ‘no’.

      And at the end you’ll have twice as many ‘no’s as ‘yes’es.

      The page you link to addresses a subtly different question: in the ‘Frequentist Approach’ it correctly points out that if you pick a boy at random from all the 2000 boys in the survey, then there’s a 50% chance of their sibling also being a boy. But that’s from picking a _boy_ at random. So a family with 2 boys is twice as likely to be picked as a family with just 1 boy.

      Here we didn’t start with a boy. We started with a family. And of the 3000 families with a boy, only 1000 of them have 2 boys.

    1. I think spoiling others fun by publishing the answer on the day of release wasn’t really in the spirit of this page.

  9. Each birth is independent of any other. The possibilities on a birth are 50% male, 50% female. Any subsequent birth is also 50% males. 50% female.

  10. @Ally Why can’t I respond directly to your posts?

    Anyway, if not knowing the order of the births is relevant than distinguishing between female male and male female is redundant. The only thing that makes any sense given the explanation provided by Richard is that prior to determining the gender of either child what is the probability you end up with male male, which would be 1 in 4.

    1. No idea why you can’t reply directly to my posts Tom.

      After thinking on this for another day and reading the other responses, particularly the one by José Camões Silva, I’m now pretty sure I’m wrong anyway.

      José Camões Silva is right. Where both I and Richard Wiseman have gone wrong is by assuming “All of [the remaining] options are equally likely” (MM, MF, FM) after eliminating the FF option. This is incorrect. As José states, once you use information [the observation that one of Erica’s children is male] to remove the [F,F] case, you also have to compute the new probabilities for the remaining possibilities. You can’t just divide the prior probabilities by the new number of possibilities as I have done.

      See the link to José’s blog, or here: https://betterexplained.com/articles/an-intuitive-and-short-explanation-of-bayes-theorem/ for the full explanation.

      The prior probability of two boys in a randomly selected family of two children is 25%.
      The probability of a ‘true positive’ (i.e. observing one child is a boy in a family with two boys) is 100%.
      The probability of a ‘false positive’ (i.e. observing one child is a boy in a family without two boys) is ~33.3% (1/3).

      If you stick the numbers into Bayes Theorem you end up seeing that the correct answer is actually 50%, as many peoples intuition has already told them.

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