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Friday puzzler -- pick an envelope

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Friday puzzler -- pick an envelope

#1

Friday puzzler -- pick an envelope

Alex Y

This week and next are two subtly different versions of the same puzzle, with very different answers.

You are shown a machine that creates random real numbers. However, you do not know how these random numbers are distributed, e.g, they could be roots of Pi (rhubarb being my favorite ;) ) powers of e with an arbitrary sign, etc. As an example, the puzzlemaster gets the machine to create a number, which he shows to you, and in this case, you notice that it is a negative number with three digits to the left of the decimal place and four digits to the right.

Now, without your being able to see, the puzzlemaster uses the machine to generate two additional numbers, which he seals in envelopes. Using a coin toss, he chooses one of the envelopes to give to you to open. After you have opened it and looked at the generated number inside, he says that you are to pick the envelope with the larger of the two numbers -- either the one you were given or the unopened envelope.

Is there a strategy that lets you improve your odds of success to greater than 50%? If so, what is that strategy, and what odds do you achieve?

Re: Friday puzzler -- pick an envelope

#2

Re: Friday puzzler -- pick an envelope

Bill Earl

Not quite the 'Full Monty' there.

Re: Friday puzzler -- pick an envelope

#3

Re: Friday puzzler -- pick an envelope

Alex Y

:) Yes, this does look a little like the MH problem, and also a little like the two-envelope paradox, but it is neither.

Actually, I think this is easier than the Monty Hall problem, since it avoids the ambiguity about the actions MH takes.

But stay tuned for next week's, which is NOT MH.

Re: Friday puzzler -- pick an envelope

#4

Hint

Alex Y

Use all the information at your disposal.

Re: Friday puzzler -- pick an envelope

#5

Re: Hint

Bill Earl

You have no knowledge of the distribution, so you can't infer anything from the two numbers you have seen. And the puzzlemaster's choice of envelope was random, so you can't infer anything indirectly from his choice. I don't think you can do any better than a coin-flip on this one.

Re: Friday puzzler -- pick an envelope

#6



Alex Y

You have no knowledge of the distribution, so you can't infer anything from the two numbers you have seen. And the puzzlemaster's choice of envelope was random, so you can't infer anything indirectly from his choice. I don't think you can do any better than a coin-flip on this one.

I think you can. I'll post mow later today. Hopefully I am thinking of this correctly.

Re: Friday puzzler -- pick an envelope

#7

Re: Hint - Question

Larry Barrett

Is it always the case that the puzzlemaster first shows you a random number, then presents two envelopes to you? In other words, do you always see one random number, then see the contents of one envelope, then make a choice about the envelopes?

Re: Friday puzzler -- pick an envelope

#8

Re: Hint - Question

Alex Y

I don't know about always...but that's what happened in the case in question! :)

I think you may be onto something!

Re: Friday puzzler -- pick an envelope

#9

Answer?

Alex Y

I'm fairly confident of this answer, but I've been confident and wrong before, so can someone confirm or refute this, please?

I believe that you have a 2/3 chance of being right with a strategy of comparing the number in the envelope you were given to the initial random value you were shown, and sticking with the envelope you have if it is larger than the sample, or switching if it is smaller.

Rationale:

No matter the distribution, with three applications of this machine, there will be a largest (L), smallest (S) and middle (M) value. The possibility of those values for:

The sample, the envelope opened, and the other envelope are:

1) S,M,L

2) S,L,M

3) M,S,L

4) M,L,S

5) L,S,M

6) L,M,S

Since the process is random, there is an equal likelihood of each of these possibilities. (We don't know the distribution, so the relative sizes of S,M, and L cannot be known, but for instance, the largest of the three could equally likely appear in any of the three spots.)

Now, if the envelope opened has a smaller number than the sample, we can rule out cases 1, 2, and 4 and in the remaining three cases, switching pays off in 3 and 5

Similarly, if the envelope opened has a larger number than the sample, we can eliminate cases 3,5, and 6, and see that it is best to stay put in two of the three remaining cases.

QED (I think!)

Re: Friday puzzler -- pick an envelope

#10

Re: Answer?

Bill Earl

Hmmmm. Your answer presumes something about the distribution of the numbers. What says that the only possibilities are H, M & L? Why can't it be LLS, SSL or even MMM? Why can't two (or more) of the numbers be the same?

Even with an even distribution this is possible. But consider the case where the distribution is a 50% probability of the number being -abc.efgh and a 50% probability of it being +ijk.lmnop. This case reduces to the equivalent of a coin flip.

Now re-cast the question as, witnessing two coin flips, is there any strategy for predicting the third?

Re: Friday puzzler -- pick an envelope

#11

Re: Answer?

Alex Y

Hmmmm. Your answer presumes something about the distribution of the numbers. What says that the only possibilities are H, M & L? Why can't it be LLS, SSL or even MMM? Why can't two (or more) of the numbers be the same?

Good catch. I should have specified that the process used by the machine gave a "diffuse" distribution where there was an infintesimally small chance of any single result. (Maybe there is a better way of saying that?) Another fix would be to state that the machine generates random real numbers uniformly distributed within a range that you do not know. Then there will be a L, M, and H with probability approaching 1.

I agree with your example of a distribution of random real numbers that only included two such numbers with equal probability.

Thanks for "keeping me honest"!

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