Probability Paradox

RealEstateBroker · 2006-07-24 19:22:11

A college professor asks for two volunteers. Mary and John come forward.

The professor hands each of them an envelope. He tells them that there is money in each envelope. He explains that one envelope has twice as much money in it as the other, and that he will give the students the option of either keeping their envelope, or trading it for the other envelope.

Mary looks in her envelope and sees that there is a $10 bill. She thinks to herself: "If I trade it for John's envelope, there is a 50/50 chance that I will end up with $5 and a 50/50 chance that I will end up with $20. That means that I will either lose $5 or else I will gain $10." She reasons that this is the same as getting two to one odds on a 50/50 bet. In fact, if she were to get this opportunity 100 times, she would expect to lose 50 times and to win 50 times, thus ending up with a large, net gain. So she agrees to exchange envelopes.

At the same time, John goes through the same thought process, and he also agrees to exchange envelopes.

So the question is: How can they both be right?

George,Y · 2006-07-28 12:37:28

Is your question well known?

RealEstateBroker · 2006-07-28 12:49:38

Even though I do not have the slightest clue about what the symbols mean, so that George Y's response is totally undiscernable to me: Nonetheless there is something about it that rings true. I am sure that George Y has given us the right answer, but I am frustrated that I simply do not read the language.

I promise that I have a logical mind, and I can follow a reasoned explanation, so if George Y would be willing to take the time to explain the answer, it would not be lost on me. Any chance George would be willing to walk us through it, or at least say what each of those symbols refer to?

Zhylliolom · 2006-07-28 12:55:29

I think you're thinking that his signature is his answer?

George,Y · 2006-07-28 16:51:14

My signature explains how Linear Least Error Approximation works, which is in matrix domain.

Sorry I haven't got an answer yet.

All_Is_Number · 2006-07-30 02:47:23

RealEstateBroker wrote:

A college professor asks for two volunteers. Mary and John come forward.
The professor hands each of them an envelope. He tells them that there is money in each envelope. He explains that one envelope has twice as much money in it as the other, and that he will give the students the option of either keeping their envelope, or trading it for the other envelope.
Mary looks in her envelope and sees that there is a $10 bill. She thinks to herself: "If I trade it for John's envelope, there is a 50/50 chance that I will end up with $5 and a 50/50 chance that I will end up with $20. That means that I will either lose $5 or else I will gain $10." She reasons that this is the same as getting two to one odds on a 50/50 bet. In fact, if she were to get this opportunity 100 times, she would expect to lose 50 times and to win 50 times, thus ending up with a large, net gain. So she agrees to exchange envelopes.
At the same time, John goes through the same thought process, and he also agrees to exchange envelopes.
So the question is: How can they both be right?

Let X = total money in each envelope.

There are two possible cases. Mary will receive (1/3)X and John receives (2/3)X

or

Mary receives (2/3)X and John receives (1/3)X

If Mary receives (2/3)X by trading she can expect to gain (2/3)X or lose (1/3)X.

If she receives (1/3)X, by trading she can expect to gain(1/3)X or lose (1/6)X.

The reality is that the potential gain and loss is always (1/3)X of a total X.

Every time Mary loses, she has overestimated the total amount in the envelopes and the potential winnings by a factor of two. Every time she wins, she has overestimated her potential loss by a factor of two.

A better way for Mary to examine the probability would be to assume that since any given time, her envelope contains either ((1/2)+(1/6))X or ((1/2)-(1/6))X, on average her envelope will contain (1/2)X, and by trading or not each participant risks (1/6)X at even odds, such that the "pot" on any given transaction equals (1/3)X. Her risk analysis is not skewed by looking at the problem in this manner.

The potential winnings for either player in any round are a function of the total amount X, and not the amount in either individual envelope.

George,Y · 2006-07-30 13:32:51

Great! you fix the amount of money as a constant and eliminates its variability to get only one random variable on which to chose!

RealEstateBroker · 2006-08-01 10:50:58

It looks like All_Is_Number has got it, and presented the solution in a way which even I can understand. George, Y is right about simplifying the amount by making it a constant. When I was trying to figure this out, I kept over-complicating the problem.

Funny that I thought George's signature was the solution! Reminds me of a real-life "who's on first?" mix-up which I was involved in recently. I had asked for the password of a protected site which I have the right to visit, and was told that the password is secret. I replied: "I understand that the password is secret, but what is it? You guessed, they had neglected to put quotes around "secret," so I thought they were telling me that the password is secret, but they actually meant to tell me that the password is, "secret."

I appreciate the friendly ambiance on this site. No one would even think of implying that I am stupid for not realizing that the equation is George's signature on mathisfun. It seems that people who are interested in the discussions are welcome no matter what their level of understanding.

Math Is Fun Forum

#1 2006-07-24 19:22:11

Probability Paradox

#2 2006-07-28 12:37:28

Re: Probability Paradox

#3 2006-07-28 12:49:38

Re: Probability Paradox

#4 2006-07-28 12:55:29

Re: Probability Paradox

#5 2006-07-28 16:51:14

Re: Probability Paradox

#6 2006-07-30 02:47:23

Re: Probability Paradox

#7 2006-07-30 13:32:51

Re: Probability Paradox

#8 2006-08-01 10:50:58

Re: Probability Paradox

Board footer