A real poser this time

Originally posted by richjohnson
I don't want it to be expressed in any particular way, but I am curious as to how one would properly express the probability of the envelope not chosen being the one with more money.

The probability that the envelope not chosen is the one with more money is expressed as 1/2. That is, there are two possible states (I chose the big one, I chose the small one) and you have no information that can be used to discern which you are in.

Always 1/2, regardless of whether you open the first envelope or not. The actual values contained in the envelopes are irrelevant to the analysis, as is their ratio. If they were in a 10:1 or 3:2 ratio, a proper analysis would yield an identical conclusion about the probability that you have picked the small one: 1/2.

Originally posted by DoctorScribbles
The probability that the envelope not chosen is the one with more money is expressed as 1/2. That is, there are two possible states (I chose the big one, I chose the small one) and you have no information that can be used to discern which you are in.

Always 1/2, regardless of whether you open the first envelope or not. The actual values contai ...[text shortened]... uld yield an identical conclusion about the probability that you have picked the small one: 1/2.

OK, then is that not the same as saying I've got a 50% chance of improving my situation and a 50% chance of worsening my situation if I switch envelopes?

Originally posted by richjohnson
OK, then is that not the same as saying I've got a 50% chance of improving my situation and a 50% chance of worsening my situation if I switch envelopes?

It leads to that conclusion, yes.

Originally posted by DoctorScribbles
It leads to that conclusion, yes.

I take it then, the problem arises when the relative values of the envelopes are compared. It seems that the proper expectation value for the money in each envelope is 1.5*A, where A is the amount of money in the smaller envelope. Thus, when you open the envelope and see $20, you don't know if A=$10 or A=$20. Can probabilities be assigned to these two possibilities?

Originally posted by richjohnson
I take it then, the problem arises when the relative values of the envelopes are compared. It seems that the proper expectation value for the money in each envelope is 1.5*A, where A is the amount of money in the smaller envelope. Thu ...[text shortened]... A=$20. Can probabilities be assigned to these two possibilities?

Yes. After observing 20 in one envelope, with probability .5, A=20, and with probability .5, A=10.

As we let A denote the amount in the smaller envelope, let B denote the amount in the larger envelope. From the two cases above, respectively, we deduce that with probability .5, B=40, and with proability .5, B=20.

Now in the first case, our expectation for either envelope would be (20 + 40)/2=30=1.5*A. And also in the other case, our expectation would be (10 + 20)=30=1.5*A. So, as you claimed, in all cases, the expectation equals 1.5*A.

You'll note that this is quite different and contradictory to the following flawed analysis: Given that we see 20 in one evelope, the other envelope contains either 10 or 40 with equal likelihood, and thus our expectation after switching is (10+40)/2=25>20.


Do you see that that flawed analysis mixes and matches the best of both cases for the value of A, just like operands of addition and subtraction are carelessly mixed and matched in the missing dollar problem? If any analysis of this problem relies on performing an operation on the pair of operands 10 and 40, it is flawed. Those values are completely unrelated.