So over on Puzzling.SE, a user asked this question. DISCLAIMER: I'm not asking you guys for the answer to the question, but rather an answer to the discussion that ensued about evaluating the logic of a certain answer to the question. I believe that a key part of it isn't correct; the answer-er (obviously) believes that it is, so we started a discussion in the chat.

The problem is that we are going in circles on our discussion - we agree on a lot, but we have these two little areas that we can't see straight on. Could you guys shed some light on which one of us is correct, and why?

NOTE: Please do not address the full question and its answer. If possible, address only the "key part" that I mentioned earlier.


In the original question discussed above, we are told that on a game show, there are 8 boxes, each containing 2 stones. Some (at least half) are precious, and the rest are not precious. The participant selects a box, from which the host selects a stone at random. The stone turns out to be precious. At this point the host (a perfect logician for our purposes) announces that there is a 50% chance that the other stone in the box is precious. The host does know how many boxes there are with two precious stones, one precious stone, and zero precious stones. He does not know which boxes they are.

Now, given that information, both the other answer-er and I conclude that the two possible distributions of precious stones among the boxes is 00111222 and 11112222. I have since realized that this is false. However, with those distributions, our discussion still stands:

We begin to differ on the question "If we swapped the boxes, is it true that we would always have a higher/lower/same(choose one or none) chance of picking up precious pebble next, given the possible distributions". I argue that the probability for this is always higher; he argues that it isn't always higher.

Are either of us correct?

  • $\begingroup$ Could you put all the relevant information here? Even with the links, it's a lot of work just to understand what your question is in the first place. $\endgroup$ – The Laconic Mar 9 '19 at 21:01
  • $\begingroup$ @TheLaconic edited. $\endgroup$ – Brandon_J Mar 9 '19 at 22:20
  • $\begingroup$ When you say "selects a box and a stone", does that imply that we know what the stone is? $\endgroup$ – jbowman Mar 9 '19 at 22:33
  • $\begingroup$ Could you explain exactly what calculation the host is carrying out when they declare there is a "50% chance" about the other stone? Their meaning is obscure. $\endgroup$ – whuber Mar 9 '19 at 22:44
  • $\begingroup$ For the benefit of the OP, I'll just mention that this question appears to be a variation on the famous Monte Hall problem, for which there is a well-developed mathematical/statistical literature and solution. $\endgroup$ – Ben - Reinstate Monica Mar 9 '19 at 22:46

The probability is less.

First, the only collection of boxes that can satisfy both the conditions given - half or more of the stones are precious, if you draw a precious stone there's a 50-50 chance the other stone is precious - is one where the eight boxes have $(0, 0, 1, 1, 1, 1, 2, 2)$ precious stones respectively. Bayes' Theorem should convince us that the number of boxes with one precious stone must be twice that of the number of boxes with two stones for the "50-50" chance criterion to hold, and the only other arrangement where this is true is $(0, 0, 0, 0, 0, 1, 1, 2)$, which is ruled out by the "at least half the stones are precious" criterion. (If you are having trouble seeing this, refer to the Monty Hall problem linked to in comments above.)

If you switch, you have a 50-50 chance of switching from a box with two precious stones to a box drawn from the set of boxes with $(0, 0, 1, 1, 1, 1, 2)$ precious stones and a 50-50 chance of switching from a box with one precious stone to a box drawn from the set of boxes with $(0, 0, 1, 1, 1, 2, 2)$ precious stones. In the first case, the probability that you choose a precious stone is $0 \cdot 2/7 + 1/2 \cdot 4/7 + 1 \cdot 1/7 = 3/7$. In the second case, the probability is $0 \cdot 2/7 + 1/2 \cdot 3/7 + 1 \cdot 2/7 = 1/2$. The overall probability of selecting a precious stone is therefore $1/2 \cdot 3/7 + 1/2 \cdot 1/2 = 13/28 < 1/2$.

Note that with the arrangement of stones you've provided, e.g., $(0,0,1,1,1,2,2,2)$, the probability that I select a precious stone from a box with two precious stones is NOT 1/2. You might be inclined to say, for example, "the probability of choosing box 3 is equal to the probability of choosing box 6, so...", but that's not the probability we are interested in. The probability of choosing box 3 AND selecting the precious stone from it is half that of choosing box 6 and selecting a precious stone from it, because if we select box 3, we only have a 50% probability of also selecting the precious stone, but with box 6, we have a 100% probability of selecting the precious stone. Therefore, if we select a box from either box 3 or box 6, and select a precious stone, the probability is 2/3 that we have chosen the box with two precious stones - and this logic applies to all arrangements with equal numbers of boxes with one and two precious stones.

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  • $\begingroup$ I realize now that the 00111122 is the only one that satisfies the criteria. Our discussion was on the ones we then thought were correct. Anyway, thanks for helping us out, and sorry for my poor explanations. $\endgroup$ – Brandon_J Mar 9 '19 at 23:38
  • $\begingroup$ No worries, Monty Hall type problems are notorious for the difficulty of explaining them clearly. $\endgroup$ – jbowman Mar 9 '19 at 23:39

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