Losing My Marbles

You are probably familiar with Randall Munroe of XKCD fame.  One of his most famous strips is this:

(image © Randall Munroe)

I am usually good about not falling into this trap, but sometimes the bait is too tempting, and I blow an hour or two futilely trying to educate someone – usually with a smile on my virtual face, since I’m a honey-not-vinegar kind of guy.

Yesterday I stuck my foot directly into the bear trap, and (unsurprisingly) it snapped shut. In my defense there was a mathematical riddle just sitting there on the tension plate. What, I was supposed to leave it there?

The riddle was posted innocently to a D&D Facebook group.  It’s an old riddle, long-since solved, but the solution is not intuitive, and there are plenty of smart people out there who insist upon a wrong solution.  Here it is:

“You have three bags, each containing two gems. The first bag contains two blue gems, the second bag contains two red gems, and the third bag contains one blue gem and one red gem.

You pick a random bag and take out one gem.

It is a blue gem.

With what certainty would you guess that the remaining gem from the same bag is also blue?”

That’s it verbatim as it appeared in the Facebook thread. I think traditionally the riddle uses marbles, but this is a D&D group, so we’re probably looking at citrines or tourmalines or something.

The obvious (but wrong) answer is 50% likely.

The less obvious (but correct) answer is 66% likely.

The way to think about it actually quite simple:

  1. You have three bags with two gems each, which means when you pick a single gem, there are six possible ways that you can do it:

a. Draw one of the red gems from the bag with two red gems

b. Draw the other red gem from the bag with two red gems

c. Draw the blue gem from the bag that has one of each color

d. Draw the red gem from the bag that has one of each color

e. Draw one of the blue gems from the bag with two blue gems

f. Draw the other blue gem from the bag that has two blue gems

  1. In the riddle, we know you’ve picked out a blue gem, which means that a, b, and d didn’t happen.
  2. That means there are possible choices you DID make, each with the same likelihood: c, e, and f.
  3. If your draw was e or f, then the remaining gem from your bag is blue. If your draw was c, the remaining gem is red.  Therefore, Q.E.D., there is a 66% likelihood that, given your first gem was blue, that the other gem in the bag is also blue.

I explained this in about four different ways. I even re-taught myself Python so I could write a simple script to simulate the problem.  Sure enough: 66%.  And yet there’s one fellow on this thread who has made a personal crusade out of his certainty that the answer is 50%. (I’ll leave him anonymous; I harbor no ill will towards him.)  He has gone so far as to post YouTube videos showing how the puzzle can be “solved” with (I kid you not) collapsing quantum states of the gem bags. And, no, he’s not trolling the discussion. I’ve been around the block enough times to recognize a bridge under-dweller, and this guy is legit.

He has not convinced me, of course, that 50% is the correct answer.  Because it’s not. What he has convinced me of is that no power on this world or any other will convince him he’s wrong.

The funny thing is, I’m sure he’s thinking the exact same thing about me.

 

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About dorianhart

novelist, game designer, amateur musician, dad
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25 Responses to Losing My Marbles

  1. Dave Korka says:

    Out of curiosity, has he tried it? Have you asked him if he would be convinced by conducting an experiment? You don’t have to run too many trials to testably prove it is not 50/50. I find it interesting that one would argue so much when it is, testably, provably, NOT 50/50. You don’t have to believe the math. Just do the experiment a reasonably large number of times and it will show him that he is wrong. -Dave

  2. Leah Shuldiner says:

    The hard part is wrapping one’s head around the fact that there are six theoretical bags, not three, because it is a two step problem. There’s the rr bag where you pick red1 first. There’s the rr bag where you pick red2 first. There’s the bb bag where you pick blue1 first. There’s the bb bag where you pick blue2 first. There’s the rb bag where you pick red, and there’s the rb bag where you pick blue. Now if he does his experiment with the post-it notes, he would throw away the two rr bags and the one rb bag and be left with three post-it notes, not two. This would change the probability to 60/30, not 50/50.

    We could make a reply YouTube video if you want. 🙂

  3. Anonymous says:

    Dorian- I don’t usually chime in on such discussions, especially when I know how smart the puzzler presenter is (you!). However, I do believe that you got it wrong!! Once you know the drawn color (blue), you know for sure that the other stone is red or blue- probability of 50% being blue. This is the logical approach to the problem. If you were to do an empirical experiment and ran a trial of say 1000000 trials, I would bet that about half would be blue, not two thirds.

    The probability of selecting 2 of the same color is 66%!

  4. Caryn says:

    Anonymous above is Caryn Hart!

  5. shusda says:

    Dorian, I’m afraid I’ll have to agree with Caryn on this one. The way that I would think about it is a bit different from yours, but equally simple:
    1. You have three bags with two gems each, which means when you pick a single gem, there are four possible ways that you can do it:
    a. Draw one of the red gems from the bag with two red gems
    b. Draw the blue gem from the bag that has one of each color
    c. Draw the red gem from the bag that has one of each color
    d. Draw one of the blue gems from the bag with two blue gems
    2. In the riddle, we know you’ve picked out a blue gem, which means that a and c didn’t happen.
    3. That means there are two possible choices you DID make, each with the same likelihood: b and d.
    4. If your draw was d, then the remaining gem from your bag is blue. If your draw was b, the remaining gem is red.  Therefore, there is a 50% likelihood that, given your first gem was blue, that the other gem in the bag is also blue.

    • Leah Shuldiner says:

      The way I think of it is that your 4 options (a, b, c, d) don’t have equal probability. For example, you are twice as likely to draw a blue gem from a bag with two blues that you are to have drawn a blue gem from a bag with one blue and one red. Given that, and given that we know you picked a blue gem, you are twice as likely to have picked the bag with two blues as you are to have picked the bag with one blue and one red.

      I would list your possibilities as follows.
      a. draw a red gem from the 2 reds bag
      b. draw the OTHER red gem from teh 2 reds bag
      c. draw the blue gem from the bag with one of each
      d. draw the red gem from the bag with one of each
      e. draw the blue gem from the bag with 2 blues
      f. draw the OTHER blue gem from the bag with 2 blues

      • shusda says:

        Then let me put this another way: Say you HAVEN’T yet drawn a marble from any of the three bags. In that case, the probability of drawing two marbles of the same color is indeed 66% (or odds of 2 to 1), since two of the bags have marbles of the same color, and only one bag has marbles of different colors. However, in Dorian’s riddle, you HAVE drawn a marble, and it is blue. That means that you can eliminate the bag with two red marbles, leaving just two bags. That reduces the probability to 50% (or odds of 1 to 1).

    • Marc LeBlanc says:

      In order for (b) or (c) to happen, you have to first pick the right bag (1/3 chance) and then pick the right gem (1/2 chance). There is a one in six chance of that happening.

      In order for (a) or (d) to happen, you just have to pick the right bag, so the chance is 1/3.

      So (a) and (d) are more likely than either (b) or (c).

  6. Marc LeBlanc says:

    Consider this game:

    1) Dorian draws a random gem from a random bag.
    2) if the gem is red, game over. No big deal.
    3) If the gem is blue, He draws the other gem in second gem is the bag.
    4) If the second gem is red, Dorian pays you $6.
    5) If the second gem is blue, You pay him $5.

    If you believe the probability is 50%, you should be happy to play this game 2000 times; you will win about about 1000 times and lose about 1000 times, and make a net $1000.

    Dorian, please cut me in on your winnings.

  7. Leah Shuldiner says:

    Here’s another way to think of it. This is a two step game. First I pick a bag and second I pick a marble. The probability that I pick any one bag is 1/3.

    Given our scenario, where I’ve picked a blue marble, here are the probabilities:

    I had a 1/3 chance of picking the blue/blue bag. Once I pick that bag, I have a 2/2 chance of picking a blue marble. 1/3 x 2/2 = 2/6 chance.

    I had a 1/3 chance of picking the blue/red bag. Once I pick that bag, I have a 1/2 chance of picking a blue marble. 1/3 x 1/2 = 1/6.

    Therefore, I am twice as likely to have a blue marble remaining in the bag as a red one.

  8. Leah Shuldiner says:

    Also, Dorian, I hate you because I have done nothing else with my day other than think of this problem! 😉

  9. Marc LeBlanc says:

    Yet another way to think about it:

    The three bags are:

    – (100 red gems)
    – (1 blue gem, 99 red gems)
    – (100 blue gems)

    I pull a random gem from a random bag. It’s blue. Intuitively, does it seem likely as not that it came from the bag with the 1 blue gem?

    Now remove 1 gem (red, red, and blue respectively) from each bag and play again. Keep removing 1 gem from each back until you get back to the original game.

  10. shusda says:

    Let me put this another way: Say you HAVEN’T yet drawn a marble from any of the three bags. In that case, the probability of drawing two marbles of the same color is indeed 66% (or odds of 2 to 1), since two of the bags have marbles of the same color, and only one bag has marbles of different colors. However, in Dorian’s riddle, you HAVE drawn a marble, and it is blue. That means that you can eliminate the bag with two red marbles, leaving just two bags. That reduces the probability to 50% (or odds of 1 to 1). (N.B.: I also posted this comment further up the string as a reply, but I’m not sure anyone will see that one.)

    • Marc LeBlanc says:

      If you have drawn a marble, and it’s blue, you can ALSO eliminate the case where you drew the red marble from the mixed-color bag. If you only eliminate the draw-from-red-bag cases, you haven’t gotten them all.

    • Leah Shuldiner says:

      Yes, but given that you drew a blue marble, you are twice as likely to have drawn the bb gag as the br bag, no?

  11. Adam Strong-Morse says:

    Okay, so now I’m stuck trying to figure out if this puzzle is the same as the Monty Hall problem (https://en.wikipedia.org/wiki/Monty_Hall_problem) or just similar. I think it’s just similar, not a restatement, but I’m not positive. 🙂

    Also, this reminds me very much of the principle of restricted choice in bridge–a principle that I know to be true, but find very counterintuitive nonetheless. https://en.wikipedia.org/wiki/Principle_of_restricted_choice

    • Anonymous says:

      I think it is the same as the Monty Hall problem. Instead of three doors, there are three bags. You’ve chosen a bag, and you want the bag with two blues. Of the two remaining bags, I show you the one with two reds so now you know the two remaining bags have a blue. Do you switch, or stay pat?

      • Adam Strong-Morse says:

        So, it’s at least mildly different from the Monty Hall problem. First, in the Monty Hall problem, there are actually three doors, and all three matter to the problem. In this problem, there are ostensibly three bags, but in fact there are only two relevant bags. The problem would be exactly the same if there were two bags, one with two blue gems and one with a blue and a red gem, and you drew out a blue gem. The bag with two red gems is a red herring (although it may help some people with the analysis to reach the correct answer. In the Monty Hall problem, also, new information is being introduced by a functionally omniscient third party; here, you’re just making an observation of the outcome of the choice you made. So it’s a variant on the same theme, but not identical.

        Also, in your framing, the conclusion is exactly opposite from Dorian’s framing. If you start with three bags, and want the bag with two blues, and before you draw you pick a bag, and then I reveal the bag with two reds, you should always switch. In Dorian’s framing, having drawn a gem and found that it is blue, you should never switch (because you are twice as likely to have drawn from the two-blue bag as from the red and blue). (Also, note that this produces the weird contrary to the normal Monty Hall pattern, you can pick a bag and have the bag that you picked revealed to be the two-red bag, because the two losing bags are different, and by hypothesis a specific one of the losing bags will be revealed.)

        So I don’t think they’re the same, although they’re definitely related.

  12. shusda says:

    One last thought on this, and then I promise I’ll shut up:
    In my mind, THIS slightly reworded question SHOULD be answered 66%:

    “You have three bags, each containing two gems. The first bag contains two blue gems, the second bag contains two red gems, and the third bag contains one blue gem and one red gem.

    You pick a random bag and take out one gem.

    With what certainty would you guess that the remaining gem from the same bag is the same color?”

    • Marc LeBlanc says:

      Yes!

      It’s the exact same problem. Imagine your friend says “I pulled a blue gem, what color should I guess the other one is?”

      Blue, right?

      You just said beforehand that the probability of getting a single-color bag is 2/3. Knowing that the gem is blue just tells you WHICH of those single-color bags you likely got.

  13. PalmerEldritch says:

    This is Bertrand’s Box Paradox first posed by Joseph Bertrand in his 1889 work ‘Calcul des probabilités’. And the answer is indeed 2/3. The forerunner of The Monty Hall Problem, Three Prisoner Problem and many other variations of the same.

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