Comments on: Two Envelopes

By: Seb

Seb — Thu, 08 Nov 2007 16:06:50 +0000

For me R = IR = the real numbers. So “a random number X with a positive probability of being between every two distinct elements of R” is just a continuous random number (say, a normal distribution). This is easier to imagine than a random distribution on Q!

By: yaniv

yaniv — Thu, 08 Nov 2007 16:01:29 +0000

Got you (I shouldn’t have used the term “number”, my bad ).

And, yeah, generalizing the riddle to the reals is possible (that is the second part of the first paragraph of my comment). My point was that the statement “a random number X with a probability positive over R” should be replaced by “a random number X with a positive probability of being between every two distinct elements of R” (this is probably what you meant).

By: Seb

Seb — Thu, 08 Nov 2007 15:21:46 +0000

When I hear number I think “real”, not integer! so my solution works even if the two numbers are reals (even negative ones). It will therefore also work for N, Z, Q and other subsets of R.

By: yaniv

yaniv — Thu, 08 Nov 2007 15:00:00 +0000

Well, I think there is a slight imprecision in the statement: “with a probability positive over R”. What you actually want to do is generate a number that has a positive probability of falling between every 2 distinct positive integers (for example, let y be geometrically distributed with any parameter p between 0 and 1, and let x be y + 1/2).
This solution easily extends to the case where what is written in the envelopes are not integers but elements of some arbitrary ordered set, on which you can define a probability distribution that gives a positive probability to a dense subset (dense in the sence that between every two elements of the original set there is an element of the subset – note that with this wording, Z+1/2 is dense in Z).

Another solution to the riddle: say that the number x you are seeing is the lesser of the two with a probability of 1/2+1/x (or some other formula of the sort).

By: Seb

Seb — Thu, 08 Nov 2007 10:13:56 +0000

I draw a random number X with a probability positive over R. If X is smaller than the number I have on the envelope, I will say that it’s because it’s the larger, if not I will say it is the smaller. I will be right 1/2+a/2 of the times, with a being the probability that the number X is between your two envelopes.

By: Seb

Seb — Thu, 08 Nov 2007 09:33:23 +0000

There is anothe great riddle with two envelopes that you might know.

In one envelope I put some amount of money, and in the other one I put double that amount (you don’t know the amount). I shuffle the envelopes, and you select (randomly) one of them, which you open, and you observe the amount, say $10. Now, the other envelope holds $5 with probability 1/2 and $20 with probability 1/2. So the average value for the other envelope is $12.5, higher than the one you have. I offer now the possibility to change and take the other envelope. Would you do so? why, or why not?

Now, imagine you don’t open the envelope when you first choose it. Does anything change? (this envelope has x inside, and the other one has on average 1.25x. Or not?)