Riddle. come on math people

[CoveX]

Originally posted by: 91TTZ
Update:

I tried that Java simulation 20 times and I think that it is accurate.

By choosing the same door each and every time, I chose correctly 10 of those times. This would back up my theory that it always comes down to a 50% chance, regardless of the door you choose.

Try it another 20 times. You just got lucky.

I wouldn't believe it the first time I read about the problem either. I think the key to understanding it is to think of two scenarios based on the first guess:

1/3 chance: You guessed correct on the first try. Changing the door will mean you lose the prize.

2/3 chance: You guessed wrong on the first try. "Monty" eliminates one wrong door, and switching will ALWAYS give you the prize.

Therefore, it pays off to switch.

Edit: speelingg

 

[LASTGUY2GETPS2]

Originally posted by: desteffy
Suppose there are two envelopes that both contain money and you get to choose one.

You make your choice and then I inform you that one envelope has twice the amount of $$ as the other one.

You are given a choice to either keep the envelope you have, or trade it in for the other, What do you do?


For example if the one you picked first contains $50, then the other one has either $25, or $100 in it.

EDIT:

The real question here is what would be the "best" strategy to do in general for this and why.

Two Words: Game Theory

 

[91TTZ]

Update: I tried it 100 times, and it came to 33 right, which would say that I'm wrong. This of course assumes that the code in that java program is accurate.

 

[Malak]

I'd do the Morrowind trick, kill the guy with the envelopes, and take both.

 

[Jeraden]

But the problem in this post has nothing to do with the Monty Hall problem. There are only 2 choices in this one, and no new details are revealed after you pick the first envelope. You have a 50% chance of getting the higher-amount envelope no matter which you choose and at what point you choose it. This is different from the Monty Hall problem, because in that one, the information given effectively rules out one of the "bad" choices. Since there are only 2 choices here, the information does nothing to skew the odds.

Now, I'm guessing the poster wants us to consider whether its better to switch envelopes, because you have the opportunity to gain $50 at the risk of only losing $25, which in general is a good bet when its 50% odds of that happening. But in this particular case, thats not really true, so your odds are 50% of getting the higher envelope whether you switch or not.

Had the problem been worded that there were 3 envelopes, and you pick 1, and at that point the guy tells you that one of the other envelopes is double and the other is half. At that point it would be wiser to switch envelopes, since in this situation you know you got the middle envelope and the gains outweigh the loss from switching. But with 2 envelopes its 50/50.

 

[Argo]

Doesn't matter.

 

[kranky]

The explanation that finally registered with me is this:
First, you have to understand that if you keep your first choice, you have a 1/3 chance no matter what Monty Hall does.
Once you see that, then you realize that by switching, you get BOTH other doors (=2/3 chance). One of them he shows you is a dud, and you get the other one.

 

[mordantmonkey]

can you look at the envelope you just picked?
in the envelope scenario pick the other one. you risk only half of what you stand to gain.
if your envelope has $50, baisically by betting $25 on 1:1 odds you can triple that bet,. (you get $75 on top of $25 you would have already kept even if you lost.)

 

[91TTZ]

I still don't see the reasoning. Maybe I'm just braindead.

As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:

Close your eyes and click any random door (without looking) and just to get it to reveal a bum door. Forget about your choice, it's irrelevant. He opens the bum door.

Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.

 

[Jzero]

Originally posted by: Jeraden
But the problem in this post has nothing to do with the Monty Hall problem.

That's my fault, I thought it was just the Monty Hall problem told incorrectly.

 

[CoveX]

Originally posted by: 91TTZ
I still don't see the reasoning. Maybe I'm just braindead.

As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:

Close your eyes and click any random door (without looking) and just to get it to reveal a bum door.

Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.

But it has. Based on which two doors you left unselected, the game show host (knowing full well that it's empty), opens one of them. So your previous choice has ramifications.

Edit: As to the OP's problem, it seems to me that you always have a 50/50 chance.

 

[3chordcharlie]

2 envelopes is 50-50, and yes, you have a 50% chance of doubling your money if you switch. Of course there is no solution to the problem based purely on numbers; essentially you should always be willing to switch, as many times as you have the opportunity to do so; in the end, you just have to pick one, and there's no payoff matrix to consider, in dollars. There is one in 'utility' however:

If they show you what's in the envelope, you might change the weighting of your choices; gain 50 / lose 25 is a lot different than gain 50 000 / lose 25 000.

 

[Argo]

Originally posted by: 91TTZ
I still don't see the reasoning. Maybe I'm just braindead.

As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:

Close your eyes and click any random door (without looking) and just to get it to reveal a bum door.

Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.

It does in this case. When you picked the first door you had 33% chance of winning. Now MOnte elimated the bad door, but he can't elimanate the door you already chose, so that door still has 33% chance of winning. However, the other door now has 63% chance of winning because it represents both doors.

Think about it this way: Mary, Joe and Frank each chose an equal pile containing 33 apples. But then Frank left and gave all his apples to Mary. It'd make sense for Joe to switch piles with her.

 

[rgwalt]

Originally posted by: 91TTZ
I still don't see the reasoning. Maybe I'm just braindead.

As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:

Close your eyes and click any random door (without looking) and just to get it to reveal a bum door. Forget about your choice, it's irrelevant. He opens the bum door.

Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.

Monty has given you knowledge that changes the situation. He has revealed one wrong answer. The probability that you picked wrong is 2/3, and after a wrong door is revealed, you will get the prize by switching.

R

 

[Kelemvor]

Exactly what Argo said. each door has 33% chance. So the door you have has 33% chance of being the Big Prize door. Whatever the rest of the coices are have a combine percentage of 66%. When you remove one of the other two, your door still has 33% because it was taken from the original 3. So now the last door has a 66% chance.

We did this in a programming clasee back in high school. You shoudl always switch.

But the OP makes no sense because with only 2 to start with they each have 50%....

 

[purbeast0]

Originally posted by: Argo

Originally posted by: 91TTZ
I still don't see the reasoning. Maybe I'm just braindead.

As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:

Close your eyes and click any random door (without looking) and just to get it to reveal a bum door.

Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.

It does in this case. When you picked the first door you had 33% chance of winning. Now MOnte elimated the bad door, but he can't elimanate the door you already chose, so that door still has 33% chance of winning. However, the other door now has 63% chance of winning because it represents both doors.

Think about it this way: Mary, Joe and Frank each chose an equal pile containing 33 apples. But then Frank left and gave all his apples to Mary. It'd make sense for Joe to switch piles with her.

no i agree with 911TTZ.

think about this ... say you do not pick any door at all, and he opens 1 door. you are now left with a choice between 2 doors to choose.

that is the same exact thing as you picking one, and then he reveals one of the dud doors. either way he's going to remove one, leaving you with a choice of one or the other doors.

i do not follow the explanation either

 

[purbeast0]

Originally posted by: rgwalt

Originally posted by: 91TTZ
I still don't see the reasoning. Maybe I'm just braindead.

As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:

Close your eyes and click any random door (without looking) and just to get it to reveal a bum door. Forget about your choice, it's irrelevant. He opens the bum door.

Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.

Monty has given you knowledge that changes the situation. He has revealed one wrong answer. The probability that you picked wrong is 2/3, and after a wrong door is revealed, you will get the prize by switching.

R

and after he reveals one of the wrong ones, your odds are now 1/2. thats the thing i do not understand. either way, your NEW odds are 1/2, regardless of what door you picked at first becuase he always eliminates a dud door.

 

[91TTZ]

Originally posted by: kranky
The explanation that finally registered with me is this:
First, you have to understand that if you keep your first choice, you have a 1/3 chance no matter what Monty Hall does.
Once you see that, then you realize that by switching, you get BOTH other doors (=2/3 chance). One of them he shows you is a dud, and you get the other one.

I still don't get that one.

Let me preface with 2 given principals:

1. We can all agree that with odds, there is no prior knowledge, the odds are the same each time.

2. Your final choice is the only one that matters.

Keeping those 2 things in mind, Each and every time he asks you for your final choice, you have a 50% chance of winning because there are only 2 doors to choose from.

 

[purbeast0]

Originally posted by: FrankyJunior
Exactly what Argo said. each door has 33% chance. So the door you have has 33% chance of being the Big Prize door. Whatever the rest of the coices are have a combine percentage of 66%. When you remove one of the other two, your door still has 33% because it was taken from the original 3. So now the last door has a 66% chance.

We did this in a programming clasee back in high school. You shoudl always switch.

But the OP makes no sense because with only 2 to start with they each have 50%....

aaah okay i see how it makes sense in THAT explanation, however regardless, i still see the door you pick as having a 50% chance at THIS POINT because there are only 2 doors left.

but i do understand the theory now.

EDIT:

but as the guy above me stated (which i also agree with) is that either way, you are choosing between 2 doors in the end so your choice is still a 50/50 choice.

 

[mordantmonkey]

trying to do to much at once. thread is too fast for me.

 

[Argo]

Originally posted by: purbeast0

Originally posted by: Argo

Originally posted by: 91TTZ
I still don't see the reasoning. Maybe I'm just braindead.

As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:

Close your eyes and click any random door (without looking) and just to get it to reveal a bum door.

Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.

It does in this case. When you picked the first door you had 33% chance of winning. Now MOnte elimated the bad door, but he can't elimanate the door you already chose, so that door still has 33% chance of winning. However, the other door now has 63% chance of winning because it represents both doors.

Think about it this way: Mary, Joe and Frank each chose an equal pile containing 33 apples. But then Frank left and gave all his apples to Mary. It'd make sense for Joe to switch piles with her.

no i agree with 911TTZ.

think about this ... say you do not pick any door at all, and he opens 1 door. you are now left with a choice between 2 doors to choose.

that is the same exact thing as you picking one, and then he reveals one of the dud doors. either way he's going to remove one, leaving you with a choice of one or the other doors.

i do not follow the explanation either

Your example is flawed. First of all you picking a door and you not picking one is totally different scenarios. In the one case Monte has a choice of 2 doors to open, while in the other case he can either have a chocie of 2 or a choice of one. I'm tired of explaining it, but if you guys don't see this perhaps probability isn't your strong suit.

 

[Kyteland]

Originally posted by: 91TTZ
I still don't see the reasoning. Maybe I'm just braindead.

As we all know, odds do not change based on the previous pick. If you flip a coin that has a 50% chance of landing on heads, the next flip will still have a 50% chance of being heads even if the previous 5 flips were also heads. With that in mind:

Close your eyes and click any random door (without looking) and just to get it to reveal a bum door. Forget about your choice, it's irrelevant. He opens the bum door.

Ok, now here's the real choice. You have 2 doors left, and the prize could be behind either one. Each time you try it, the scenario will be the same- you're always left with 2 doors, one of which contains the prize. Your previous pick has no bearing on this current pick.

What you're missing is that the doors are never reshuffled after you take away the wrong ones.

Take this example. I want you to pick the ace of spades from a deck of cards. You pick one card leaving 51 other cards. I remove all 50 non-ace of spades from that pile leaving one card and you are asked to stay or switch. What should you do? switch.

Here's the math. 1/52 times you pick the ace of spades on your first try and it doesn't matter the cards I remove from the remaining 51. 51/52 times you don't pick the ace of spades and the only option that leaves me is to remove the 50 non-ace of spades. In those 51/52 times the remaining card after I remove 50 *must* be the ace of spades, so it is to your advantage to switch.

The confusing thing is that just because I remove something doesn't change the probability distribution. When I separate a deck of cards in to a pile of 1 and a pile of 51, the pile of 51 has a 51/52 chance of having the ace of spades. If I am forced to leave the ace of spades in the pile then it desn't change the probabilities if I remove 50 other cards. There is still a 51/52 chance that it is there.

Now, if I took your original card back and randomized them behind my back, *then* it becomes a 50/50 chance again.

Now I see why you have such a hard time with the 0.99.. argument.

Edit: edited for speeling.

 

[Argo]

People who think that switching is like chosing from two doors (hence 50-50 chance) are wrong. It's not chosing from two doors, sicne there are prior factors affecting the percentages. Prior factors being that when Monte chose one door he couldn't choose YOUR door, he could only chose from two of the others. Plus, he could only chose the LOSING door.

 

[3chordcharlie]

Originally posted by: purbeast0

Originally posted by: FrankyJunior
Exactly what Argo said. each door has 33% chance. So the door you have has 33% chance of being the Big Prize door. Whatever the rest of the coices are have a combine percentage of 66%. When you remove one of the other two, your door still has 33% because it was taken from the original 3. So now the last door has a 66% chance.

We did this in a programming clasee back in high school. You shoudl always switch.

But the OP makes no sense because with only 2 to start with they each have 50%....

aaah okay i see how it makes sense in THAT explanation, however regardless, i still see the door you pick as having a 50% chance at THIS POINT because there are only 2 doors left.

but i do understand the theory now.

EDIT:

but as the guy above me stated (which i also agree with) is that either way, you are choosing between 2 doors in the end so your choice is still a 50/50 choice.

No, it's a 1/3 versus 2/3 choice.

Thre is a 66% chance you were 'wrong' the first time.

If you you were wrong (66%), and you 'switch' you now have a 100% chance of winning.

If you were right (33%) and you switch, you now have a 0% chance of winning.

Therefore it is a 2/3 chance of winning by switching; there is no 50-50 anywhere in the problem.

Your error is assuming an equal probability of the remaining doors being the right one; but the probability isn't equal, so the choice is not 50-50.

 

[mordantmonkey]

you try to think of it like the coin thing which is not the same... the doors are dependent, there is only one prize. you start with 1/3 chance of being right and 2/3 being wrong. after one of the doors are opened your original choice was still 1 in 3 chance of being right, but now the ONLY other door (the one you didn't choose) has a 2/3 chance of being right.