Ayecarumba Joined: Nov 17, 2009
• Posts: 6763
August 13th, 2010 at 12:00:42 PM permalink
I think the confusion stems from the "cost" of the opportunity. For the first envelope, there was no cost to the player. They received the \$100 without having to risk anything. However, for the second opportunity, the player must risk the original amount. In this case, by putting in the \$100, 50% of the time the player will receive \$200, and %50 of the time the player will receive \$50. A 50/50 proposition with a \$100 cost.

With cash in hand, and a cost to play the second round, the original parameters do not apply.
Simplicity is the ultimate sophistication - Leonardo da Vinci
weaselman Joined: Jul 11, 2010
• Posts: 2349
August 13th, 2010 at 12:05:18 PM permalink
Quote: Doc

Well, I guess that went right over my head. Your equation is true for any value of p, so I don't see how it makes 50% so special.

If you choose heads or tails with a 50% probability, then you have 50% of winning with any value of p because 0.5*p + 0.5*(1-p) = 0.5
If you use any other strategy (like always picking tails, or always picking heads or anything other that 50/50), then you will only have a 50% chance of winning when the coin is unbiased.
"When two people always agree one of them is unnecessary"
Doc Joined: Feb 27, 2010
• Posts: 7086
August 13th, 2010 at 12:10:08 PM permalink
Actually, I usually flip a coin to decide which way to call a coin toss. :-)
weaselman Joined: Jul 11, 2010
• Posts: 2349
August 13th, 2010 at 1:05:16 PM permalink
Quote: Doc

Actually, I usually flip a coin to decide which way to call a coin toss. :-)

That works ... as long as your coin is unbiased :)
"When two people always agree one of them is unnecessary"
Doc Joined: Feb 27, 2010
• Posts: 7086
August 13th, 2010 at 2:07:18 PM permalink
In an attempt to get away from the coin-flip analogy that I carelessly introduced, I want to go back to discussing the original conundrum and my confusion. Here are some of my more recent rambling thoughts, and I may have stolen/borrowed them from others.

Suppose the problem were slightly reworded. There are two envelopes containing cash. Without either envelope being opened, we are told that one of them contains \$100. We are also told that the other envelope contains either \$50 or \$200, but we are not given any information about the probability of those two possibilities. Can we calculate an expected value of the money in the 50/200 envelope? I don�t think so; what would be the basis?

If I then open an envelope and find \$100 (bringing us back to the original problem), it does not give me any additional reason at all to think that there is \$200 (or \$50) in the other envelope -- it's still that there could be either \$50 or \$200 with unknown probability. In either case, I had a 50% probability of selecting the \$100 envelope, but that does not mean that there is a 50% probability of \$50 vs. \$200 in the other envelope. I still cannot calculate the expected value of the money in the unopened envelope.

This was perhaps weaselman�s point, which I probably initially misinterpreted.
Quote: weaselman

It is not 50% that you picked the higher valued envelope. In general, it just an unknown value.

It is a well-known fallacy to treat any unknown binary value as 50% of each.

If he meant that we can't assume a 50% probability of initially choosing the envelope with more money (which is what his first sentence above sounds like), then I disagree. If he meant we can�t assume that there is a 50% probability that the 50/200 envelope contains \$200 (which is more the way his last sentence sounds), then I agree. I think the EV cannot be calculated with the info we have.

Conclusion for the original conundrum: Pick an envelope and run. You don�t even need to open it until later.
scotty81 Joined: Feb 4, 2010
• Posts: 185
August 13th, 2010 at 2:54:03 PM permalink
Quote: Doc

Suppose the problem were slightly reworded. There are two envelopes containing cash. Without either envelope being opened, we are told that one of them contains \$100. We are also told that the other envelope contains either \$50 or \$200, but we are not given any information about the probability of those two possibilities. Can we calculate an expected value of the money in the 50/200 envelope? I don�t think so; what would be the basis?

This is a vastly different problem than the original problem presented. In this case, we have additional information that makes the choice much easier.

In the original problem, we didn't know the amounts in advance, so we had no way of knowing if we had just opened the fixed amount envelope, or the +100%/-50% envelope. In this case, we do. We know for sure that one envelope contains \$100, and the other envelope contains either \$200 or \$50.

If we open an envelope and it contains \$200, we take it.
If we open an envelope and it contains \$50, we switch. The other envelope contains \$100.
If we open the envelope with \$100, we should switch because the expected value of the \$200/\$50 envelope is \$125 (\$200 + \$50 / 2). So, we are trading a sure \$100 for an expected value of \$125.

In the original problem, the expected value of one envelope is X, and the expected value of the second envelope is 125% of X. IF we knew which envelope we were opening (either X or X*1.25), then we could make a decision. As it stands, there is not enough information to make an informed decision and I can see no advantage to switching based upon the original wording.

The classic Monte Hall problem also has this same kind of additional information injected into the problem. The additional information in that problem is that the door exposed after the initial choice was not random, but rather it was known in advance that that door did not contain the prize.
Prediction is very difficult, especially about the future. - Niels Bohr
Wizard Joined: Oct 14, 2009
• Posts: 23117
August 13th, 2010 at 3:12:23 PM permalink
Quote: weaselman

Yes, 50% is special, because p+(1-p)=1.

Here is what I think he means. Let's say Doc has a 50% chance of calling tails, and the probability the coin lands on tails is t. Let x be the probability Doc wins

x=.5*t + .5*(1-t)

Next multiply both sides by 2

2x = t + (1 - t)

We see from the weaselman equation that p + (1-p) = 1, so it must be true for t as well...

2x = 1

x=1/2.

Then again, we just could get the same answer by canceling the .5t and -.5t terms, leaving x=0.5.
It's not whether you win or lose; it's whether or not you had a good bet.
Doc Joined: Feb 27, 2010
• Posts: 7086
August 13th, 2010 at 3:31:50 PM permalink
Quote: scotty81

This is a vastly different problem than the original problem presented. In this case, we have additional information that makes the choice much easier.

In the original problem, we didn't know the amounts in advance, so we had no way of knowing if we had just opened the fixed amount envelope, or the +100%/-50% envelope. In this case, we do. We know for sure that one envelope contains \$100, and the other envelope contains either \$200 or \$50.

If we open an envelope and it contains \$200, we take it.
If we open an envelope and it contains \$50, we switch. The other envelope contains \$100.

To this point I agree with you -- I was not clear enough in my statement. Restate it this way: In the original problem, you do not originally know the amounts in either envelope. Once you open one envelope and find \$100, then you know that the other envelope contains either \$50 or \$200, but you have no basis for knowing what the probability of either of those two possibilities is. I contend that you are then in exactly the same situation as in my modified problem where you are told the amounts before the envelopes are opened.
Quote: scotty81

If we open the envelope with \$100, we should switch because the expected value of the \$200/\$50 envelope is \$125 (\$200 + \$50 / 2). So, we are trading a sure \$100 for an expected value of \$125.

This is where I disagree with you completely. You divided by 2. What basis do you have for believing that there is a 50/50 chance of \$200 and \$50? As weaselman pointed out, the fact that there are only two possibilities does not suggest that they are equally likely. I contend that you have been given no basis whatsoever for dividing by 2. It could as well be that there is a 25% chance of \$200 and a 75% chance of \$50. In reality, it is not even a probabilistic event -- it is deterministic and was determined by the person who put the money in the envelopes. I don't think we have any information as to which they would have chosen or what the probabilities are for how they would make their choice.
Quote: scotty81

In the original problem, the expected value of one envelope is X, and the expected value of the second envelope is 125% of X. IF we knew which envelope we were opening (either X or X*1.25), then we could make a decision. As it stands, there is not enough information to make an informed decision and I can see no advantage to switching based upon the original wording.

I agree with your conclusion. I'm just not sure I follow your line of thought in this paragraph. I don't follow what you mean by "the expected value of one envelope is X", and I contend that there is no basis for the 1.25 factor. But you draw your conclusion without regard to what this factor is -- it could as well be 3.5, and you would still realize that you have no justification to switch. So we come to the same final conclusion.
Doc Joined: Feb 27, 2010 