Freethought & Rationalism Archive

Baloo · 05-08-2003, 05:57 AM

EDIT: Nevermind - I see Lob started a whole new thread with the same one...

Here's one of my favorites:

You're on a game show. The host of the show has shown you 3 curtains: A, B, and C. Behind one of the curtains is a new car - it is up to you to pick the correct curtain. Once you've made up your mind, the host lifts up one of the unpicked curtains, revealing where the car wasn't. You now have a chance to change your mind... the question is, would you?

SULPHUR · 05-08-2003, 06:07 AM

probability has changed from1/3 to 1/2. why change?

Salmon of Doubt · 05-08-2003, 06:51 AM

No, I think I remember this one. You're supposed to change your answer... but I can't remember the mathematical reason why! Go on, tell me

edit... oops, saw it in the other thread. Never mind

SULPHUR · 05-08-2003, 08:06 AM

original probabity 1/3 When the curtain is opened, if you do not change you pick prob is1/6 So pick the other one and it is back to 1/2.

Silent Acorns · 05-08-2003, 10:07 AM

Quote:

Originally posted by SULPHUR
original probabity 1/3 When the curtain is opened, if you do not change you pick prob is1/6 So pick the other one and it is back to 1/2.

I don't think the probability of your initial pick being correct drops to 1/6. It stays at 1/3. The probability of the car being behind curtain number 2, however, climbs to 2/3. This assumes that the host always reveals a loser first and offers you to switch. If he only makes the offer when he knows you have already picked the car, you're being screwed.

SULPHUR · 05-08-2003, 10:41 AM

Yes I can see where I went wrong
thanks

Silent Acorns · 05-08-2003, 10:55 AM

Quote:

Originally posted by Jesse
So the question is, what's the flaw in the reasoning?

I tried looking at it a little differently. Given that there is $100 in envelope #1, that envelope #2 has either $50 or $200, and knowing from symmetry that there is no advantage to switching envelopes, one can calculate that there is a 2/3 chance that envelope #2 contains $50 and a 1/3 chance that it contains $200.

Why should this be the case?

SULPHUR · 05-08-2003, 12:04 PM

Looking at it again I can't see where my reasoning is wrong. The person must change his choice to have the greatest probability

Silent Acorns · 05-08-2003, 01:27 PM

Quote:

Originally posted by SULPHUR
Looking at it again I can't see where my reasoning is wrong. The person must change his choice to have the greatest probability

I agree with you that the best decision is to change curtains. I was only saying that the odds that the original choice was the car stays at 1/3 while the odds of the other curtain goes up to 2/3.

Jesse · 05-08-2003, 02:03 PM

Quote:

Originally posted by Lobstrosity
Could the flaw be that you're taking an average where no average is allowed? Let's take the first case. Suppose we are given that envelope A will always have $100. Next suppose we know that 50% of the time envelope B will be twice what is in envelope A ($200). We also know that if envelope B is not twice envelope A, it must be half what is in envelope A ($50). Given this setup, we expect that if we choose envelope B N times, where N >> 1, we will have approximately N x ($125). Had we chosen envelope A every time, we would have exactly N x ($100), so clearly we benefitted from our knowledge that the value on envelope A was fixed over multiple trials. After all, that's where the expected returns actually makes you money, isn't it?

Now, if you're only given one trial, you have a 50% chance of getting the envelope with the most money. Simply by symmetry one can argue that swapping will not increase your odds of getting the maximal dollar amount for that trial. Multiple trials where you know nothing about the contents of either envelope can just be considered a series of single trials where each time your parameter X has a new value, which voids the idea of averaging the trial results to produce 1.25X. Because these are single trials, I would argue that your consideration of average returns (i.e. 1.25X) is not valid.

Is this anywhere close to the right track?

Sort of. But think of it this way--suppose we do a very large number of trials with all different amounts of money. Then, just look at the subset of trials where X = 100. What would the average return be in these trials? And couldn't we choose any other number and look at the subset of trials where X took that value?

05-08-2003, 06:07 AM	#12
SULPHUR Senior Member Join Date: Nov 2002 Location: hobart,tasmania Posts: 551	car probability has changed from1/3 to 1/2. why change?

05-08-2003, 08:06 AM	#14
SULPHUR Senior Member Join Date: Nov 2002 Location: hobart,tasmania Posts: 551	wager original probabity 1/3 When the curtain is opened, if you do not change you pick prob is1/6 So pick the other one and it is back to 1/2.

05-08-2003, 10:41 AM	#16
SULPHUR Senior Member Join Date: Nov 2002 Location: hobart,tasmania Posts: 551	Its been along time Yes I can see where I went wrong thanks

05-08-2003, 12:04 PM	#18
SULPHUR Senior Member Join Date: Nov 2002 Location: hobart,tasmania Posts: 551	acorn Looking at it again I can't see where my reasoning is wrong. The person must change his choice to have the greatest probability

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05-08-2003, 05:57 AM	#11
Baloo Regular Member Join Date: Jun 2000 Location: St. Louis, MO Posts: 417	EDIT: Nevermind - I see Lob started a whole new thread with the same one... Here's one of my favorites: You're on a game show. The host of the show has shown you 3 curtains: A, B, and C. Behind one of the curtains is a new car - it is up to you to pick the correct curtain. Once you've made up your mind, the host lifts up one of the unpicked curtains, revealing where the car wasn't. You now have a chance to change your mind... the question is, would you?

05-08-2003, 06:51 AM	#13
Salmon of Doubt Regular Member Join Date: Feb 2003 Location: UK Posts: 150	No, I think I remember this one. You're supposed to change your answer... but I can't remember the mathematical reason why! Go on, tell me edit... oops, saw it in the other thread. Never mind

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