Monty Hall-esque question

[mykey1961]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.


[/ QUOTE ] This is both wrong, and a very bad example.

Do you mean we would on average win 1/2 the amount wagered? Or do you mean for just those times we bet exactly $50, we would average a $25 win.

Both are wrong.

[/ QUOTE ]

You took this quote out of context. In its original context it read,

[ QUOTE ]
Prior to opening the envelope I think everyone here would say it's 50-50 that it's the smaller envelope. We'd all be willing to gamble on that given favorable odds. We'd certainly bet on it given 2-1 odds. After opening the envelope I doubt many here would want to call off the bet. As far as our bet goes, opening the envelope should not change the 50% chance that it's the smaller envelope. If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.


[/ QUOTE ]

Read in context it should be clear that the "bet" I'm talking about in what you quoted is the wager being discussed in that paragraph. The one that is made after the envelope is picked but before it is opened. The one that is not called off after the envelope is opened. It doesn't matter whether new envelope amounts are chosen each time, or the same ones are used each time, the statement remains true that for THIS wager, with the chance to call it off after seeing the envelope amount, If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.

For example, for THIS wager, if we decide to bet $1000 our EV will be:

(.5)2000 - (.5)1000 = 500

ie. +$25 for every $50 wagered.

Try reading my post again.

PairTheBoard

[/ QUOTE ]


[ QUOTE ]
How is getting 2:1 on half your envelope the best explanation?

[/ QUOTE ]

[ QUOTE ]
I think it's the insight that best satisfies this psychological condundrum.

[/ QUOTE ]

Ok for some reason THIS time you get to choose the amount you wager rather than having to wager 1/2 the amount in the envelope.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.


[/ QUOTE ] This is both wrong, and a very bad example.

Do you mean we would on average win 1/2 the amount wagered? Or do you mean for just those times we bet exactly $50, we would average a $25 win.

Both are wrong.

[/ QUOTE ]

You took this quote out of context. In its original context it read,

[ QUOTE ]
Prior to opening the envelope I think everyone here would say it's 50-50 that it's the smaller envelope. We'd all be willing to gamble on that given favorable odds. We'd certainly bet on it given 2-1 odds. After opening the envelope I doubt many here would want to call off the bet. As far as our bet goes, opening the envelope should not change the 50% chance that it's the smaller envelope. If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.


[/ QUOTE ]

Read in context it should be clear that the "bet" I'm talking about in what you quoted is the wager being discussed in that paragraph. The one that is made after the envelope is picked but before it is opened. The one that is not called off after the envelope is opened. It doesn't matter whether new envelope amounts are chosen each time, or the same ones are used each time, the statement remains true that for THIS wager, with the chance to call it off after seeing the envelope amount, If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.

For example, for THIS wager, if we decide to bet $1000 our EV will be:

(.5)2000 - (.5)1000 = 500

ie. +$25 for every $50 wagered.

Try reading my post again.

PairTheBoard

[/ QUOTE ]


[ QUOTE ]
How is getting 2:1 on half your envelope the best explanation?

[/ QUOTE ]

[ QUOTE ]
I think it's the insight that best satisfies this psychological condundrum.

[/ QUOTE ]

Ok for some reason THIS time you get to choose the amount you wager rather than having to wager 1/2 the amount in the envelope.

[/ QUOTE ]

Have you really actually read and thought about any of my posts mykey? This has been my point all along. If you were allowed to choose the amount you bet you could ignore the envelope amount and your bet would win at 2-1 50% of the time. Switching envelopes amounts to betting at 2-1. Why isn't it the same? Because the amount of your switching-bet is being dictated by the already determined outcome. Try thinking about it.

PairTheBoard

[mykey1961]

[ QUOTE ]
How is getting 2:1 on half your envelope the best explanation?

[/ QUOTE ]

[ QUOTE ]
I think it's the insight that best satisfies this psychological condundrum.

[/ QUOTE ]

[ QUOTE ]
Ok for some reason THIS time you get to choose the amount you wager rather than having to wager 1/2 the amount in the envelope.

[/ QUOTE ]

[ QUOTE ]
Have you really actually read and thought about any of my posts mykey? This has been my point all along. If you were allowed to choose the amount you bet you could ignore the envelope amount and your bet would win at 2-1 50% of the time. Switching envelopes amounts to betting at 2-1. Why isn't it the same? Because the amount of your switching-bet is being dictated by the already determined outcome. Try thinking about it.

PairTheBoard

[/ QUOTE ]


Yes, I've actually read, and thought about your posts.

I agree that getting 2:1 on half your envelope isn't the same as getting 2:1 on X amount of money that you selected the envelope with the lower amount.


But I don't agree that getting 2:1 on half your envelope sheds any light on the paradox.

Once you have opened the envelope and found the $100 you try to calculate the EV of switching.

EV = ($50 + $200)/2 = $125

Using your 2:1 on 1/2 the envelope means

We either lose $50, or win $100.


Your "insight" doesn't address the probability of winning, only the amount at risk, and the payoff when it wins.

The EV calculation from the "paradox" assumes (incorrectly) that the probability of winning is 50%. It might be 50%, or it might not.

[mykey1961]

If I was the one on the game show, and had time to think about the problem:

[ QUOTE ]
Say you have two unmarked envelopes. You're on a new game show, and the host has put money in the envelopes in the following manner: the value of one envelope is twice the value of the other envelope. You have no idea whatsoever about the expected magnitude of the prizes.

You open one of the envelopes, and you see a check for $100. Now, the host offers you the chance to switch to the other envelope. (Again, no game theoretical assumptions about whether or not it is more likely that the game show would have a $200 prize or a $50 prize.) Do you switch?

[/ QUOTE ]

Hmm. If I don't switch, I have $100.

The other envelope contains $50 with probability p, and $200 with probability q.

EV = $50p + $200q = $200 - 150p

EV >= $100 when p <= 2/3

Since I don't know anything about p I can't say if switching is good or bad.

[ QUOTE ]

Simply stepping through a calculation, it seems like you should: the expected value of the unseen envelope is (50+200)/2 or $125.

Now, what if you are presented with the same conundrum, but this time the host didn't show you what was in the first envelope. Should you still switch?

[/ QUOTE ]

I don't know how much I have but my EV for my current envelope is:

EV = $50p/2 + $100p/2 + $100q/2 + $200q/2

EV = $150 - $75p

If I switch my EV would be

EV = $100p/2 + $50p/2 + $200q/2 + $100q/2

EV = $150 - $75p

{quote]It seems like you should; the EV of the other envelope should always be higher through conventional calculation. However, this clearly makes no sense!

Is there a way to explain this seemingly paradoxical fact?

[/ QUOTE ]


The insight to this paradox, in my opinion is that we don't have any information on p so we don't know if switching is right in the first case. But in the second case we know that for any value of p switching doesn't change our EV.

[PairTheBoard]

I like this explanation too.

PairTheBoard

[pzhon]

[ QUOTE ]
[ QUOTE ]
His statement that all of his points have been proven rigorously is dead wrong?

[/ QUOTE ]
No,

[/ QUOTE ]
Good. Note that the first point, that you can't assume the probability is 1/2, specifically negated one argument made in the paradox. Do you disagree with this point? Do you still think there is no way to conclude that one side is wrong when it relies on a fallacy?

[ QUOTE ]
my objection is to what followed immediately afterwards, that:

[ QUOTE ]
but people still argue because the only way these are intuitive is if you have studied this before. Math is hard.

[/ QUOTE ]
People do not argue only because they can't follow the math.

[/ QUOTE ]
First, "counterintuitive" is very different from "can't be followed."

Second, there is a big difference between saying, "People argue because the only way this is intuitive..." and "People only argue because..." One of those is what I said. You objected to the other. What is the real reason you are arguing?

I don't see any real objection to what I said. You are trying to put words in my mouth to say that I am making mistakes you would like me to make.

[ QUOTE ]
Skill in math can be good, but it often induces a special kind of blindness...

[/ QUOTE ]
It is common for people to assume that any strength must be balanced by some weakness. This is wrong, and both tiring and offensive to me. See the past discussion David Sklansky and I had about mathematics majors in the SMP forum. Don't assume I made a mistake just because I was right about something else.

[Pokerlogist]

The explanations presented are fine. Here's another example which is equivalent to the higher level two-envelope situation. It may help conceptualize it.

You are again shown two envelopes A and B. Attached to each one is one secondary envelope. A has envelpe A1 and B has envelope B1. You are told A and B have either $X or $2X in them. Of course the specific dollar amounts are not told to you. You are told that the envelope with $X has a secondary envelope with $2X and the envelope with $2X has a secondary envelope with $X.

If you choose A you can only switch or not to A1.
If you choose B you can only switch or not to B1.

Notice that this puzzle duplicates the classic two-envelope scenario. By first choosing a envelope A or B you are getting 50% chance of $X or $2X . If you switch you will also have 50% chance of either $X or $2X. The chance of getting $X or $2x at the start is 50% and after switching it is still 50%. Seeing the amount in our first choice doesn’t help us. Overall, switching is breakeven. Why? The amounts in A1 and B1 are fixed to the amount in their primary envelopes. They are conditional and dependent on our first choice. Switching doesn't really give us a 50% chance of either doubling or halving the amount of dollars in our orginally chosen envelope. We already determined the amount in the second envelope when we chose the first one.


This is a different situation then when we are specifcally told after picking an envelope that another envelope has an exactly 50% chance of having either twice or half as much as the one we picked. Under these conditions switching is beneficial. Our first choice has not affected the next choice. This scenario is possible only if , after our initial pick, we are given the choice of two other envelopes. One really would have to have twice and one would have 1/2 our originally picked prize money.

Oh well. That's the best I can do with this brain-boggler.