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We are in a TV game. We have 8 same looking boxes, each of which has 2 pebbles. Each pebble could be either precious or not precious. We choose a box and the host, without looking inside the box, pulls out a pebble. It comes out to be precious. The TV host then declares that we have exactly 50% chance that the 2nd pebble inside the box is precious, too.

If we know that, in the beginning of the game, the count of precious pebbles is no less than the count of not precious pebbles, which of the statements MUST be true? We assume that the TV host knows the pebbles distribution in the boxes - in what amount of boxes the precious pebbles are 0,1 or 2. The TV host, however, doesn't know the kind of pebbles in each specific box, as he himself doesn't differ the boxes from one another.

  1. If we swapped the boxes, we would have higher chance of picking up precious pebble next.

  2. If we swapped the boxes, we would have same chance of picking up precious pebble next.

  3. If we swapped the boxes, we would have lower chance of picking up precious pebble next.

  4. In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.

  5. In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.

  6. In the beginning of the game, the precious and not precious pebbles have been equal in count.

  7. In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.

  8. The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.

  9. None of the above.

Brandon_J
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Kradec na kysmet
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5 Answers5

6

The answer is:

4, 5, 6, 7 must be true, while 1, 2, 3 are wrong.

Let us denote by $a, b, c$ the amount of boxes of each kind as follows:

PP    PN    NN

a     b     c
First we know that $a+b+c=8$.

Now we use the $50\%$ chance given by the host:

We apply Bayes' formula, where the event PP is "our box contains two precious pebbles" and P stands for "we pick a pebble from our box and it turns out to be precious". We have $$0.5=\Bbb P(P\!P|P)=\frac {\Bbb P(P\!P)}{\Bbb P(P)}$$ $$0.5=\Bbb P(P\!P|P)=\frac {a/8}{a/8+b/8\cdot1/2}$$ So finally: $$b=2a$$

Now the host also said that there are at least as many P's as N's. Therefore:

$$2a+b\geq 2c+b$$ which, given $a+b+c=8$ and $b=2a$, translates to $$a\geq 2$$

Partial conclusion

There is one configuration meeting these conditions, namely 

PP    PN    NN

2     4     2
and there is no other possibility, because $a=3$ would imply $b=6$ and $a+b$ would already be more than $8$. We see that $4, 5, 6, 7$ are all true.

Moving on to questions 1, 2, 3:

If we have a PN box and we swap, then the probability that the next pebble we pick is a P is $$\underbrace{\frac 37\cdot \frac 12}_{PN}+\underbrace{\frac 27}_{PP}=0.5$$ If we have a PP box and we swap, then the probability that the next pebble we pick is a P is $$\underbrace{\frac 47\cdot \frac 12}_{PN}+\underbrace{\frac 17}_{PP}=\frac 37$$ In the first case, the probability stays the same, while in the second it decreases. Therefore 1, 2, 3 are all wrong: it depends.

Arnaud Mortier
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  • Precious pebbles are >= Not precious; Chance for 2nd precious pebble in our box is 50%, as stated by the host – Kradec na kysmet Mar 09 '19 at 14:35
  • @Kradecnakysmet is that an admission that statement 8 is false? – Weather Vane Mar 09 '19 at 14:37
  • Yeah statement 8 is false, if you draw it, it starts looking more clear :P – Kradec na kysmet Mar 09 '19 at 14:44
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    You write "it depends", and therefore conclude that statement 3 is wrong. I disagree. It depends on what's in your chosen box, but you don't know what is in that box, and probability theory is the way to deal with that unknown information. When you make the choice to switch boxes, and know you have a 50:50 chance of having NP or PP in your box, then you can calculate the exact probability of picking a precious stone from the box you switch to. It is $\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{3}{7}=\frac{13}{28}<\frac{1}{2}$. Therefore statement 3 is correct. – Jaap Scherphuis Mar 09 '19 at 22:44
  • @JaapScherphuis I agree with the general concept that Probability theory is here to deal with unknown info. But in the present case, the OP asks "which of these statements MUST be true", in capital bold letters. Imo this can only mean "in all cases, no matter what the other pebble in your box is". – Arnaud Mortier Mar 09 '19 at 22:51
  • Statement 3 is incorrect in every possible case – Kradec na kysmet Mar 09 '19 at 22:56
  • @Kradecnakysmet Not if your box contains PN, in that case the chance for the next pebble to be P is $3/7$ which is less than $0.5$. But since your question asks "which statement MUST be true" the only thing that matters is that there are situations where 3. is false. – Arnaud Mortier Mar 09 '19 at 23:11
  • Check again, 3 isn't ever true – Kradec na kysmet Mar 09 '19 at 23:20
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    Also, since I deleted my answer, here's our chat discussion, Arnaud. – Brandon_J Mar 09 '19 at 23:29
  • @Kradecnakysmet If your box is PP, and you swap, you have 1) a 4/7 chance of getting a PN box and then 1/2 chance of getting P, and 2) a 1/7 chance of getting the other PP box, after which you are certain of getting P. $$4/7\cdot 1/2 + 1/7\cdot 1< 0.5$$ – Arnaud Mortier Mar 09 '19 at 23:33
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    Arnaud Mortier, I agree with @JaapScherphuis. Statement 3 is correct. It is not about "what will happen" which "depends". It is about comparing probabilities of a specific event in two scenarios. And, if you follow your own calculations, you should see that if we swap the boxes the probability of picking up a precious pebble diminishes". – ppgdev Mar 10 '19 at 00:09
  • Arnaud Mortier, what would you say if in the same setting you were asked to calculate and compare the probabilities of the following two events. First, the probability of pulling a precious pebble out the same box (no swap). Second, the probability of pulling a precious pebble out of a new box (swap). My reading of the first three statements is that you are asked to determine if they are true or false based on the comparison of the probabilities of these two events. – ppgdev Mar 10 '19 at 01:40
  • @ppgdev I see your point. Both readings make sense to me. It's not clear what was originally intended. – Arnaud Mortier Mar 10 '19 at 01:54
  • What you are doing feels to me a bit like the joke from Anchorman: "60% of the time, it works every time". :-) – Jaap Scherphuis Mar 10 '19 at 05:58
  • @JaapScherphuis Not at all, the question is "does it work all the time?" and the answer is "only if your box is of such type (which you don't know obviously, but that's not the point)" – Arnaud Mortier Mar 10 '19 at 09:57
  • @JaapScherphuis Anyway, it's up to the OP to state what their intention was, and the OP's comments seem to indicate that there are indeed different cases. – Arnaud Mortier Mar 10 '19 at 09:58
  • This answer was accepted, but it still has two crucial falsehoods. Firstly, Bayes' theorem counts permutations, but the TV host is making a statement about the probability of the pebble combinations. The order in which we draw the pebbles does not impact whether we chose a PP-box, it only alters when we find out if we did. Imagine you only had two boxes, one PP-box and one PN-box. The probability of choosing the PP-box is 50%, but using the Theorem the same way as in this answer would have the host say you have a 66% chance of drawing another precious pebble by not swapping, which is false. – Afrodeity Mar 29 '19 at 15:09
  • Secondly, the only way the statements 1,2 and 3 can be false is if there are different possible cases with different box types distributions. However, since this answer claims there's only one possible configuration, an average probability must exist. – Afrodeity Mar 29 '19 at 15:11
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    @Afrodeity I'm not sure I understand any of your comments. Bayes' theorem doesn't count permutations, it's a very simple conditional probability formula, and there is clearly no mistake in the way I use it. As for your second comment, the question has been raised before, there are indeed two ways to read the problem and accordingly two ways to answer. Which one is correct only depends on the OP's intention. If the answer was accepted, it means that the question needed to be read the way I did. – Arnaud Mortier Mar 29 '19 at 15:33
  • So, I spent some more time thinking about your method and I have to admit that reading the host's words the way you did is also valid. However, your second part is still wrong, because you compare the [probability of success by switching if the first pebble was precious] to the [probability of the second pebble being precious if the first one was], when instead you should be comparing it to the [probability of success by not switching if the first pebble was precious].Essentially, you compare it to 50%, when the actual value is less. I edited my answer to provide details in this regard. – Afrodeity Mar 29 '19 at 18:49
  • @Afrodeity For me the second and third $[...]$ in your comment are one and the same. I'm guessing that it's again a matter of how you read the question that is asked, however I can see no other way here. – Arnaud Mortier Mar 29 '19 at 18:56
  • My bad, I meant [2] is about picking the right pebble, [3] is about picking the right box. Our choice is between 6 boxes, not between 8 pebbles. The host is guaranteed to draw a precious pebble first (if there is one), so you should imagine he looks at both pebbles to make his decision. Now the chance of the first pebble in a PN-box being precious is the same as that of a PP-box: 100%. By eliminating the NP-permutations, we're not making the choice of a PN-box less likely (or PP more likely) because at the time of choosing, we don't know which box is or isn't a PN-box. – Afrodeity Mar 30 '19 at 08:01
  • @Afrodeity An NP box and a PN box are the same, the pebbles aren't ordered. – Arnaud Mortier Mar 30 '19 at 09:38
  • Let us continue this discussion in chat. – Afrodeity Mar 30 '19 at 17:16
2

The selection procedure ensures that each of the 16 stones has an equal chance of being picked by the host.

Once his pick is revealed to be precious, it is still equally likely to be any one of the precious stones in the boxes. For the host's statement to be true, there must be the same number of precious stones with a precious stone partner as there are with a non-precious partner.
This means that for every box with two precious stones (which obviously both have a precious partner) there must be two boxes with a precious stone partnered with a non-precious stone.
This leaves two possibilities:

1xPP, 2xPN, 5xNN
2xPP, 4xPN, 2xNN

We are given that the number of precious stones is at least as many as the non-precious ones, so we must be in the second case:

2xPP, 4xPN, 2xNN

The other boxes have 6, or 7 precious stones in them, depending on whether the current box is PP or PN. You do not know which of these is the case, though you do know they are equally probable because the host said so. There are therefore on average 6.5 precious stones in the other boxes, and when switching boxes you have a probability of 6.5/14, or less than 50% to pick a precious stone next. Therefore statements 1 and 2 are false, 3 is true.
Statements 4-7 are easily seen to be true, and 8 and 9 are false.

Jaap Scherphuis
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    2xPP + 4xPN doesn't make it 50% chance – Kradec na kysmet Mar 09 '19 at 15:24
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    @Kradecnakysmet Actually, the more I think about it, the more I believe that is exactly what's going on. Given that the first pebble pulled was precious, there's twice the amount of chance that the box had two precious pebbles in it. 2xPP + 4xPN has 8 ways of pulling a precious pebble out. 4 of those ways will have another precious pebble in the box, and 4 will not, giving you a 50% chance. – Amorydai Mar 09 '19 at 17:52
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    @Kradecnakysmet yes it does. See my answer for another way to look at it. – Arnaud Mortier Mar 09 '19 at 21:24
  • @Jaap I agree with everything except the last part. I don't think you should think in terms of average. See the last part of my answer. – Arnaud Mortier Mar 09 '19 at 21:27
  • Solving it like that you now have 33,33% for 2nd precious pebble – Kradec na kysmet Mar 13 '19 at 00:08
  • @Kradecnakysmet No. The probability of having a PP box is twice the probability of a PN box, because half the cases where a PN box is chosen get eliminated when it is revealed that the precious stone was picked from it. It is exactly the same as this three card problem, except that I have made the number of NP boxes larger to compensate. – Jaap Scherphuis Mar 13 '19 at 06:27
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There are two arrangements that satify the given facts:

A. 4 boxes with two precious pebbles, 4 boxes with one of each
B. 3 boxes with two precious pebbles, 3 with one of each, 2 with two non-precious

Because

There are at least 8 precious pebbles
Half of the boxes with a precious pebble contain another, half do not.

So for each statement:

  1. If we swapped the boxes, we would have higher chance of picking up precious pebble next.

    FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is higher.

  2. If we swapped the boxes, we would have same chance of picking up precious pebble next.

    FALSE: in case A it is either more or less.

  3. If we swapped the boxes, we would have lower chance of picking up precious pebble next.

    FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is lower.

  4. In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.

    FALSE: in both solutions.

  5. In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.

    FALSE: this is only true in one solution.

  6. In the beginning of the game, the precious and not precious pebbles have been equal in count.

    FALSE: there are more precious pebbles in both solutions.

  7. In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.

    FALSE: in both solutions.

  8. The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.

    FALSE: there are two arrangements that satisfy the condition.

  9. None of the above.

    TRUE:

So my answer is

Statement 9 is the only true statement.

Weather Vane
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These are all the statements that MUST be true:

1. If we swapped the boxes, we would have higher chance of picking up precious pebble next.

There are three kinds of boxes:

  • PP-box: Two precious pebbles inside
  • PN-box: One precious pebble, one regular pebble inside
  • NN-box: Two non-precious pebbles inside

The host might be saying that I. of all combinations containing a precious pebble, half contain a second, or he might be saying that II. of all permutations starting with a precious pebble, half contain a second.

If he means I. then half of all boxes with one precious pebble contain another, so there is an equal amount of PP- and PN-boxes. The only possible box distributions with at least half of all pebbles being precious are A: 3PP, 3PN, 2NN and B: 4PP, 4PN

If he means II. then, since there are twice as many permutations in PP-boxes than PN-boxes for the first pebble to be precious, there must be half as many PP-boxes as PN-boxes to maintain an even 50% ratio. The only possible box distribution with at least half of all pebbles being precious is C: 2PP, 4PN, 2NN.

When we choose a box at the start and don't swap it afterwards, we are choosing 1 of 8 possible combinations.

A has 3 PP-boxes, so our success rate without swapping is 3 in 8

B has 4 PP-boxes, so our success rate without swapping is 4 in 8

C has 2 PP-boxes, so our success rate without swapping is 2 in 8

Statements 1, 2 and 3 are true if the success rate of switching is always greater/equal/less respectively for all cases.

Since the meaning is unclear, statements 4,5,6 and 7 are indeterminable. Statement 8 is false.

In case A, 9 of the 16 pebbles are precious and of those, 6 are in PP-boxes.

This means there is a 6 in 9 chance we picked the PP-box at the start. Switching leaves a choice between 2PP-, 3PN- and 2NN-boxes, so the chance of drawing another precious pebble right away is

$$\frac{2\cdot 2 + 3\cdot 1 + 2\cdot 0}{7\cdot 2} = \frac{7}{14}$$

Likewise, we have a 3 in 9 chance we picked the PN-box instead. This time there are 3PP-, 2PN- and 2NN-boxes to switch to, so the probability is

$$\frac{3\cdot 2 + 2\cdot 1 + 2\cdot 0}{7\cdot 2} = \frac{8}{14}$$

In total, swapping boxes in case A has a success rate of:

$$\frac{2}{3}\cdot\frac{7}{14} + \frac{1}{3}\cdot\frac{8}{14} = \frac{11}{21} \approx 0.524$$

In case B, 12 of the 16 pebbles are precious and of those, 8 are in PP-boxes.

This means there is an 8 in 12 chance we picked the PP-box at the start. Switching leaves a choice between 3PP- and 4PN-boxes, so the chance of drawing another precious pebble right away is

$$\frac{3\cdot 2 + 4\cdot 1}{7\cdot 2} = \frac{10}{14}$$

Likewise, we have a 4 in 12 chance we picked the PN-box instead. This time there are 4PP- and 3PN-boxes to switch to, so the probability is

$$\frac{4\cdot 2 + 3\cdot 1}{7\cdot 2} = \frac{11}{14}$$

In total, swapping boxes in case B has a success rate of:

$$\frac{2}{3}\cdot\frac{10}{14} + \frac{1}{3}\cdot\frac{11}{14} = \frac{31}{42} \approx 0.738$$

In case C, 8 of the 16 pebbles are precious and of those, 4 are in PP-boxes.

This means there is an 4 in 8 chance we picked the PP-box at the start. Switching leaves a choice between 1 PP-, 4 PN- and 2 NN-boxes, so the chance of drawing another precious pebble right away is

$$\frac{1\cdot 2 + 4\cdot 1 + 2\cdot 0}{7\cdot 2} = \frac{6}{14}$$

Likewise, we have a 4 in 8 chance we picked the PN-box instead. This time there are 2 PP-, 3 PN- and 2 NN-boxes to switch to, so the probability is

$$\frac{2\cdot 2 + 3\cdot 1 + 2\cdot 0}{7\cdot 2} = \frac{7}{14}$$

In total, swapping boxes in case C has a success rate of:

$$\frac{1}{2}\cdot\frac{6}{14} + \frac{1}{2}\cdot\frac{7}{14} = \frac{13}{28} \approx 0.464$$

Results:

Case A Success Rates: Not switching - $\frac{3}{8} < \frac{11}{21}$ - Switching
Case B Success Rates: Not switching - $\frac{4}{8} < \frac{31}{42}$ - Switching
Case C Success Rates: Not switching - $\frac{2}{8} < \frac{13}{28}$ - Switching

Conclusions:

Statement 1 is true, therefor 2,3 and 9 are false.

Afrodeity
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RULE: Once you pick a precious stone (P), there is exactly a 50% chance for the other stone in the box to be P.
Conclusion1: You must have the same number of PP as PN. .... As pointed out by Brandon_J, this is not right. :( The problem is that I had a different chance of picking the P from a PN box, than from a PP box. It's like (without looking at it) flipping a coin with 2 heads, vs. a normal coin. If I get heads, I can retroactively adjust my chances of which coin I was holding. I would have lost a bet on this one!

w3c_ee
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  • Conclusion1 is incorrect. See @ArnaudMortier 's answer and my computer simulations from this discussion. – Brandon_J Mar 28 '19 at 01:26
  • OK, now i am getting your point. I didn't randomly pull the first stone. I was more likely to pull it from a PP box. OK. – w3c_ee Mar 30 '19 at 04:58
  • Yep. It's super difficult to wrap your mind around (I actually have a deleted answer on this question that says exactly what you were originally saying, so I understand). You might want to try looking at the Monty Hall Paradox if you're looking to understand this concept better (it helped me). – Brandon_J Mar 30 '19 at 12:28