MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
September 9th, 2010 at 7:23:39 AM permalink
I can't take all the credit - I saw a page by Keith Devlin that attacked the problem using Bayes. (But I did work out the calculations myself!)

The real key here is getting past the coin-flip analogy and the intuition that goes with it. Obviously before you flip a coin, the chances of heads/tails are 50/50. But then after you flip it, if the side facing up is an unknown X, the other side is still 50/50 heads/tails. That's not true with the envelopes. Before you pick, the chances of small or large are 50/50. But once you pick an envelope, the chances that the *other* envelope is smaller/larger *than that envelope* are *not* 50/50. That's the assumption that leads to the paradox. The reason it's different is that the range of values for coins is bounded at heads or tails, while the range of values for the envelopes is anything.

Suppose you considered heads = 1 and tails = 2. Then the coin has the same property as the envelopes. However, you know what the coin's range is, so if you get a 2 you keep it and if you get a 1 you switch. For the envelopes the range is infinite, so assuming the envelopes behave like the coin (when they don't) gets you to the improper 50/50 figures and the paradox.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Garnabby
Joined: Aug 14, 2010
• Posts: 197
September 9th, 2010 at 4:57:07 PM permalink
Quote: Wizard

By the way, if we can get to 42 pages, this will become the longest thread in the forum.

I love it!
Why bet at all, if you can be sure? Anyway, what constitutes a "good bet"? - The best slots-game in town; a sucker's edge; or some gray-area blackjack-stunts? (P.S. God doesn't even have to exist to be God.)
mkl654321
Joined: Aug 8, 2010
• Posts: 3412
September 9th, 2010 at 5:23:24 PM permalink
Quote: Wizard

I agree, that is the correct way to look at it. I mentioned this correct way, in other words, in my Ask the Wizard answer.

What is hard to explain is how can the expected gain on a percentage basis (relative to the first envelope) be 25%, but the expected gain on an absolute basis be 0?

By the way, if we can get to 42 pages, this will become the longest thread in the forum.

That's because this is a simple, as in, simplistic problem that is obfuscated by mathematics, when mathematics actually has no place in the discussion. The attempt to bring it in produces a swirling effect where abstractions are batted about, to no purposeful conclusion.

This whole thread seems like a bunch of scientists standing in a circle on a sidewalk, holding a raw egg. They are engaging in a series of discussions about terminal velocity, thickness of the shell, Mohs hardness scale of the concrete, approximate mass/acceleration at 1g of the egg, wind direction and velocity, and the proper height from which to drop the egg. This has consumed hours, when a small child pedals up on a bicycle and inquires what the scientists are doing. "We're trying to figure out what happens when you drop an egg on the sidewalk," one of them replies. The child grabs the egg and drops it on the sidewalk. It breaks. The child shrugs and pedals away. The scientists go back to the university to write research papers.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
Kelmo
Joined: Aug 15, 2010
• Posts: 85
September 9th, 2010 at 8:19:54 PM permalink
Knowledge of the value of x, once chosen, assigns values to x/2 and 2x, one of which does not reside in the final state with x and the contents of the final envelope. Therefore there are two possible states with differing magnitudes. The best we can do is factor in the difference of between these magnitudes.

Chosen Envelope x

Unknown Envelope (State 1)
Probability of State
0.5
Value in Other Envelope
x/2
Weighted Value of Other Envelope
0.5*x/2
Value in all envelopes
1.5x
Equivalent Magnitude of Value Once established
3x /1.5x = 2

Unknown Envelope (State 2)
Probability of State
0.5
Value in Other Envelope
x*2
Weighted Value of Other Envelope
0.5*x*2
Value in all envelopes
3x
Equivalent Magnitude of Value Once established
1.5x / 3x = 0.5

The value of choosing the other envelope should equal to x and both possibilities should contribute.

The value of each possibility should be the weighted value * the Equivalent Magnitude.
If x is the high envelope (State 1), then 0.5*x/2*2 = 0.5x
If x is the low envelope(State 2), , then 0.5*x*2*0.5 = 0.5x
The sum of all values should then be x.
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
September 9th, 2010 at 9:31:48 PM permalink
Quote: mkl654321

That's because this is a simple, as in, simplistic problem that is obfuscated by mathematics, when mathematics actually has no place in the discussion. The attempt to bring it in produces a swirling effect where abstractions are batted about, to no purposeful conclusion.

This whole thread seems like a bunch of scientists standing in a circle on a sidewalk, holding a raw egg. They are engaging in a series of discussions about terminal velocity, thickness of the shell, Mohs hardness scale of the concrete, approximate mass/acceleration at 1g of the egg, wind direction and velocity, and the proper height from which to drop the egg. This has consumed hours, when a small child pedals up on a bicycle and inquires what the scientists are doing. "We're trying to figure out what happens when you drop an egg on the sidewalk," one of them replies. The child grabs the egg and drops it on the sidewalk. It breaks. The child shrugs and pedals away. The scientists go back to the university to write research papers.

An amusing story to be sure, but people had been throwing dice for hundreds of years before Pascal and Fermat asked "wait a minute, why do I win making one wager but lose making the other?" It is precisely that sort of question that mathematics answers (and indeed, is often invented to answer). The entirety of the gambling industry owes its very existence to mathematics entering the discussion.

Many times the interesting question isn't what happens but why.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
mkl654321
Joined: Aug 8, 2010
• Posts: 3412
September 9th, 2010 at 10:42:05 PM permalink
Quote: MathExtremist

An amusing story to be sure, but people had been throwing dice for hundreds of years before Pascal and Fermat asked "wait a minute, why do I win making one wager but lose making the other?" It is precisely that sort of question that mathematics answers (and indeed, is often invented to answer). The entirety of the gambling industry owes its very existence to mathematics entering the discussion.

Many times the interesting question isn't what happens but why.

I don't disagree, but my point was that the "paradox" being discussed in this thread is due to the failure of mathematical terminology. The child knows that the egg will break. Likewise, the child knows that switching envelopes back and forth can't possibly increase the amount in either envelope.

In other words, there are times when mathematics fails to solve a problem, but words suffice easily.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
Garnabby
Joined: Aug 14, 2010
• Posts: 197
September 12th, 2010 at 6:55:41 AM permalink
Quote: mkl654321

I don't disagree, but my point was that the "paradox" being discussed in this thread is due to the failure of mathematical terminology. The child knows that the egg will break. Likewise, the child knows that switching envelopes back and forth can't possibly increase the amount in either envelope.

In other words, there are times when mathematics fails to solve a problem, but words suffice easily.

Lol. Have none of youze "geniuses" asked, "What if the math (in this case) is indeed correct?" What if it were possible to switch?

None of us is able, even with "free will", to end up doing something otherwise than that we eventually (or hastily) do. Can't say, "I'm going down this path, but to be tricky in the real reality, down that one." Perhaps that new-found (quantum-relativity?) physical ability would be worth something like an EV of 1.25 against inferior opponents.

Certainly, the original problem states nothing about changing the (overall) values of the envelopes in any way or manner throughout.
Why bet at all, if you can be sure? Anyway, what constitutes a "good bet"? - The best slots-game in town; a sucker's edge; or some gray-area blackjack-stunts? (P.S. God doesn't even have to exist to be God.)
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
September 12th, 2010 at 8:47:42 AM permalink
Quote: Garnabby

Lol. Have none of youze "geniuses" asked, "What if the math (in this case) is indeed correct?" What if it were possible to switch?

None of us is able, even with "free will", to end up doing something otherwise than that we eventually (or hastily) do. Can't say, "I'm going down this path, but to be tricky in the real reality, down that one." Perhaps that new-found (quantum-relativity?) physical ability would be worth something like an EV of 1.25 against inferior opponents.

Certainly, the original problem states nothing about changing the (overall) values of the envelopes in any way or manner throughout.

In most cases, problem-solving like this relies upon foundational premises or assumptions. For example, the original problem statement said that one envelope had twice as much as the other. That is taken as a given when addressing the problem. If the reality instead was that after one envelope was selected, the money in other envelope was always stolen by thieves, then it would clearly never be correct to switch. Alternately, if the money in the other envelope was always replaced with exactly half the amount in the opened envelope (a property which would still make the original relationship true), it would still never be correct to switch.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
mkl654321
Joined: Aug 8, 2010
• Posts: 3412
September 12th, 2010 at 12:02:12 PM permalink
Quote: Garnabby

Lol. Have none of youze "geniuses" asked, "What if the math (in this case) is indeed correct?" What if it were possible to switch?

None of us is able, even with "free will", to end up doing something otherwise than that we eventually (or hastily) do. Can't say, "I'm going down this path, but to be tricky in the real reality, down that one." Perhaps that new-found (quantum-relativity?) physical ability would be worth something like an EV of 1.25 against inferior opponents.

Certainly, the original problem states nothing about changing the (overall) values of the envelopes in any way or manner throughout.

We don't have to engage in "what ifs" or other such indulgences in make-believe, "lol", because we assume that the situation being posed will be happening in the real world. In other words, if it WERE possible to increase value by switching, we would not be in this universe, but rather, some other one, and in that universe, no one would have posed the question in the first place ("why, of COURSE you can gain by switching! Everyone knows that!).
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
jvg1981
Joined: Sep 12, 2010
• Posts: 2
September 12th, 2010 at 9:27:13 PM permalink
There's a lot going on in this problem, and Wikipedia goes into a lot of detail.  Here are my thoughts:

1. Quote: Dorothy

It would appear that by switching the man would have a 50% chance of doubling
his money should the initial envelope be the lesser amount and a 50% chance
of halving it if the initial envelope is the higher amount.

This statement is without foundation, as we don't know what distribution the dollar amounts were drawn from.  However, we can make the problem more precise by specifying a distribution.
2. For any dollar amount x, let p(x) denote the probability that the larger amount is x.  Then, if we open an envelope and find x, we can immediately apply Bayes' Theorem to find that the probability that we drew the larger envelope is p(x)/[p(x) + p(2x)] and the probability we drew the smaller envelope is p(2x)/[p(x) + p(2x)].  Thus, given that we've seen x in one envelope, the expected value of the other envelope is xp(x)/{2[p(x) + p(2x)]} + 2xp(2x)/[p(x) + p(2x)] = x{p(x) + 2p(2x)]/[p(x) + p(2x)]}, hence we should switch if p(x) < 2p(2x).  In general, we can't determine if this is true without knowing x.
3. The initial "paradoxical" analysis assumes that, seeing x in one envelope, we are equally likely to see x/2 and 2x in the other envelope.  For this to be true regardless of x, p would have to be a constant function.  This is impossible, though, as Σt p(t) must equal 1.
4. On the other hand, it is possible to choose a distribution p which gives a seemingly paradoxical result.  Wikipedia gives the example p(2n) = 2n-1/3n for every positive integer n and p(x) = 0 otherwise.  Here the expected value is infinite before opening any envelope, and switching envelopes doesn't change that.  After opening an envelope, we can determine our expected gain by switching, and yes we should always switch, but is that really paradoxical?