http://www.grand-illusions.com/monty.htm
The Monty Hall Problem
The Monty Hall Problem
Here is something to keep you occupied if that knee is still feeling sore
http://www.grand-illusions.com/monty.htm
http://www.grand-illusions.com/monty.htm
Ales Zazaro
Yeah, the Monty Hall problem is somewhat interesting, but the envelope problem is even more interesting and controversial. (I have heared of job interviews in which the candidate answers correctly but is told they are wrong). Here it is, lets see what kind of discussion this creates:
There are two envelopes containing a positive amount of money, but you do not know how much.
One envelope has twice as much money as the other.
You select an envelope and are shown its contents: it contains 10 dollars.
You now pick one envelope and get to keep the contents.
Which envelope should you choose?
a) The other envelope, because there is a 50% chance it contains 20 dollars and a 50% chance it contains 5 dollars. Thus, on average it contains 0.5*(20) + 0.5*(5)= 12.5, which is more money than in the envelope you looked at.
b) It doesn't matter what envelope you choose, since you had a 50% chance of selecting the correct envelope from the start.
c) No answer/ not enough information
Have fun!
[Edit:] Mods: Maybe this should be in the riddle thread. Sorry!
There are two envelopes containing a positive amount of money, but you do not know how much.
One envelope has twice as much money as the other.
You select an envelope and are shown its contents: it contains 10 dollars.
You now pick one envelope and get to keep the contents.
Which envelope should you choose?
a) The other envelope, because there is a 50% chance it contains 20 dollars and a 50% chance it contains 5 dollars. Thus, on average it contains 0.5*(20) + 0.5*(5)= 12.5, which is more money than in the envelope you looked at.
b) It doesn't matter what envelope you choose, since you had a 50% chance of selecting the correct envelope from the start.
c) No answer/ not enough information
Have fun!
[Edit:] Mods: Maybe this should be in the riddle thread. Sorry!
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slapdash21
- Futureless
- Posts: 4681
- Joined: 29 Sep 2004 14:50
- Location: Beantown, kidd
You should switch the envelope, and take the cash in the other one.
Here is my logic, from a gamblers perspective.
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The easiest way to explain the solution is to make It a two part problem.
You are given information from the guy running the game.
1) He has two envelopes.
2) One evelope has twice as much money in it as the other one.
3) You get to see the cash in one of the envelopes, of your choosing, although who chooses is irrelevant. X = $$$
4) After seeing the cash, one has an opportunity to switch envelopes.
5) After seeing the envelope, I can walk away with X amount of cash.
Okay, that's part one. So, lets put that cash in my wallet, and walk away.
For part two,
On my way out of the event, I'm intercepted by the guy with too much money again. He says, I'll let you toss a coin. if it's heads, I'll pay you 2X money. If you loose, I'll give you 1/2 X amount of money.
So, flipping a coin is an event with a probability of 50:50, and the amount of the bet is X dollars, but because loosing pays 1/2X dollars, we can restate this making the wager amount X as :
"Wager X dollars, and if you win a 50% chance event, you win 4X dollars"
That my friend is a payout of 4:1 for a 50:50 chance. If you play this game enough times, you will make money because you are wagering 1/4 the payout on an event with 1:2 probability of success.
Here is my logic, from a gamblers perspective.
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The easiest way to explain the solution is to make It a two part problem.
You are given information from the guy running the game.
1) He has two envelopes.
2) One evelope has twice as much money in it as the other one.
3) You get to see the cash in one of the envelopes, of your choosing, although who chooses is irrelevant. X = $$$
4) After seeing the cash, one has an opportunity to switch envelopes.
5) After seeing the envelope, I can walk away with X amount of cash.
Okay, that's part one. So, lets put that cash in my wallet, and walk away.
For part two,
On my way out of the event, I'm intercepted by the guy with too much money again. He says, I'll let you toss a coin. if it's heads, I'll pay you 2X money. If you loose, I'll give you 1/2 X amount of money.
So, flipping a coin is an event with a probability of 50:50, and the amount of the bet is X dollars, but because loosing pays 1/2X dollars, we can restate this making the wager amount X as :
"Wager X dollars, and if you win a 50% chance event, you win 4X dollars"
That my friend is a payout of 4:1 for a 50:50 chance. If you play this game enough times, you will make money because you are wagering 1/4 the payout on an event with 1:2 probability of success.
Regarding the envelope scenario,
Let's assume this scenerio was performed on 10 different people. We put the 10 people into two groups of 5. Group A will always stay with the $10. Group B will always take the second envelope. The question is: On average, which team will make more money?
Group A: They get $50 for sure, that's easy to see. $10 * 5 = $50
Group B: Since we are assuming averages, there is a 1 in 2 chance that you will get $20 when switching. So out of 4 people, 2 will get $5 and 2 will get $20. Totaling up to $50. But wait, there is still one more person that needs to go. That means most of the time Team B will either get $55 or $70. Which is more than Team A will get. For more concrete evidence, I'll show all the possibilities for Team B.
|5|5|5|5|5| - $25
|5|5|5|5|20| - $40
|5|5|5|20|20| - $55
|5|5|20|20|20| - $70
|5|20|20|20|20| - $85
|20|20|20|20|20| - $100
So, out of the 5 possible scenarios for Team B, only 2 of them produce less money than Team A. So, 3/5 times Team B will beat Team A.
Although since we are discussing money, people see money in different ways. If you see getting $5 or $10 as losing and getting $20 is winning then you might as well switch. If you see getting $5 is losing and getting $10 or $20 as winning then you might as well quit while your ahead and take the $10. But if you're like me and see getting $5, $10, or $20 as winning, you might as well switch cause you have nothing to lose.
If it were me, I'd switch envelopes for sure. Plus I'm from Vegas and winning $5 for "losing" a bet is a win in itself.
Let's assume this scenerio was performed on 10 different people. We put the 10 people into two groups of 5. Group A will always stay with the $10. Group B will always take the second envelope. The question is: On average, which team will make more money?
Group A: They get $50 for sure, that's easy to see. $10 * 5 = $50
Group B: Since we are assuming averages, there is a 1 in 2 chance that you will get $20 when switching. So out of 4 people, 2 will get $5 and 2 will get $20. Totaling up to $50. But wait, there is still one more person that needs to go. That means most of the time Team B will either get $55 or $70. Which is more than Team A will get. For more concrete evidence, I'll show all the possibilities for Team B.
|5|5|5|5|5| - $25
|5|5|5|5|20| - $40
|5|5|5|20|20| - $55
|5|5|20|20|20| - $70
|5|20|20|20|20| - $85
|20|20|20|20|20| - $100
So, out of the 5 possible scenarios for Team B, only 2 of them produce less money than Team A. So, 3/5 times Team B will beat Team A.
Although since we are discussing money, people see money in different ways. If you see getting $5 or $10 as losing and getting $20 is winning then you might as well switch. If you see getting $5 is losing and getting $10 or $20 as winning then you might as well quit while your ahead and take the $10. But if you're like me and see getting $5, $10, or $20 as winning, you might as well switch cause you have nothing to lose.
If it were me, I'd switch envelopes for sure. Plus I'm from Vegas and winning $5 for "losing" a bet is a win in itself.
Ben Skaggs
Amateurs practice until they can get it right.
Professionals practice until they can't get it wrong.
No, I don't play soccer. Yes, there are competitions. 4 years. Lots of practice.
Amateurs practice until they can get it right.
Professionals practice until they can't get it wrong.
No, I don't play soccer. Yes, there are competitions. 4 years. Lots of practice.
So I meant to post this a long time ago, but I am now stuck with slow internet for a couple of weeks.
Basically, there is no answer to this problem without more information. Unless you have some prior expectations about the amount of money in the envelopes, there is no mathematical solution. This is because there is no well-defined sample space to work with (whenever you have probabilities you need a sample space first). There is a temptation to say that the sample space consists of the following two events (which are each then given 50% probability, and averaged)
1) You picked the envelope with less money, so the envelopes contain $10 and $20
2) You picked the envelope with more money, so the envelopes contain $5 and $10
However, as soon as money is put into the envelopes, one event is true with certainty (known to the person with the envelopes but not to you), so you can't try to average between events. Since you can't know what event is true by looking in just one envelope, the problem has no solution unless you know something extra. For example, if you knew that the amount of money in the envelopes was random, with some distribution that you also knew, then the problem could be solved. On a side note, the idea of Bayesian statistics is to assume some "prior" distribution to solve such problems.
Anyways, this problem is confusing as hell and troubled me for a long time when I first heared it. To top it all off, if you pick an envelope and DON'T look at how much it contains, then there IS an answer to the problem, and it clearly doesn't matter if you switch or not.
Basically, there is no answer to this problem without more information. Unless you have some prior expectations about the amount of money in the envelopes, there is no mathematical solution. This is because there is no well-defined sample space to work with (whenever you have probabilities you need a sample space first). There is a temptation to say that the sample space consists of the following two events (which are each then given 50% probability, and averaged)
1) You picked the envelope with less money, so the envelopes contain $10 and $20
2) You picked the envelope with more money, so the envelopes contain $5 and $10
However, as soon as money is put into the envelopes, one event is true with certainty (known to the person with the envelopes but not to you), so you can't try to average between events. Since you can't know what event is true by looking in just one envelope, the problem has no solution unless you know something extra. For example, if you knew that the amount of money in the envelopes was random, with some distribution that you also knew, then the problem could be solved. On a side note, the idea of Bayesian statistics is to assume some "prior" distribution to solve such problems.
Anyways, this problem is confusing as hell and troubled me for a long time when I first heared it. To top it all off, if you pick an envelope and DON'T look at how much it contains, then there IS an answer to the problem, and it clearly doesn't matter if you switch or not.