Riddle. come on math people

[purbeast0]

Originally posted by: 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.

i do see it ... theoretically ... however in the very end, regardless, you are ALWAYS picking 1 of 2 doors, which would be a 50/50 chance.

also, with your little example, you left out one factor ... monte KNOWS which doors have nothing behind it, so he will ALWAYS choose a door that has nothing in it.

this is where the theory seems messed up:

say you have 100 doors, and 1 has a prize behind it. you pick 1 door, and monte removes 98 doors that all have nothing behind it (of course, he knows which one DOES have something behind it). so by this theory, you should switch doors because your odds will then be 99/100 of being right (since the other 98 wrongs were ruled out).

do you really thing switch doors EVERY time you will get the prize 99 times? i doubt it ...

(again, i might not fully understand the original theory so keep that in mind)

 

[DaveSimmons]

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:

The key is that you must decide to switch before the door is opened. Then, as noted by others, you have "moved" your choice from one that includes a single door, to one that includes _2_ doors (one of which will then be opened).

1. You pick door (1 door, 1/3 chance)
2. You decide now that you will switch (2 doors, 2/3 chance).
3. Monty opens one of your 2 doors that is empty / has a goat, you get the other door.

 

[Kelemvor]

You can test this out very simply with a random number generator and a friend.

Have your friend generate a random number (1, 2, 3). Then you pick a random number (1, 2, 3). Your friend tells which one of the numbers you didn't pick is wrong. That leaves your original number or the other number that's left. Just try it and you'll find that that you originally picked the right number 33% of the time and if you switch you are right 66% of the time.

~~~~~~~~~~~~~~
Here's another explanation.

All the doors have the same percentage of being right. Whatever door you pick has 33% chance of being right. You are also 66% likely to be wrong. When Monty removes a door, you picked from the original three so you are still 33% right and 66% wrong. SO you obviously should switch since there'sonly one other option and because of the original doors and what's left, you are still 66% likely to be wrong or 66% likely the that OTHER door is right.

 

[91TTZ]

Since everyone says I'm wrong, I'll just have to agree with the majority and agree that I'm wrong.

The sad part is that I'm unable to figure out why. I must be too dumb.

 

[Kelemvor]

Originally posted by: purbeast0

i do see it ... theoretically ... however in the very end, regardless, you are ALWAYS picking 1 of 2 doors, which would be a 50/50 chance.

also, with your little example, you left out one factor ... monte KNOWS which doors have nothing behind it, so he will ALWAYS choose a door that has nothing in it.

this is where the theory seems messed up:

say you have 100 doors, and 1 has a prize behind it. you pick 1 door, and monte removes 98 doors that all have nothing behind it (of course, he knows which one DOES have something behind it). so by this theory, you should switch doors because your odds will then be 99/100 of being right (since the other 98 wrongs were ruled out).

do you really thing switch doors EVERY time you will get the prize 99 times? i doubt it ...

(again, i might not fully understand the original theory so keep that in mind)

That first statement is wrong. Just becaue there are 2 options doesn't mean that they ahve the same probability. Think of it this way that should clear it up....

You have 3 containers. One is filled with gold, two are filled with manure.
You pick one and you have 33% chance of picking the right one.
Now let's say I take the other two and combine them and then throw out the now empty container. Could be manure + manure, could be gold + manure, could be Manure + gold.
You now have two containers to choose from. Do they have the same percentage of having the gold? No. Yours has 33% chance and the other one, because it has assumed the percentage of one that is now gone, has 66% chance.
End results, You should Always Switch.

Does that make sense?

Base fact is two choices does not mean each is 50%.

And as for the 100 doors scenario, Hell yes you should always change. You are 10% right after your first pick. When you remove 98 doors you are still only 10% because you picked before they were removed. Your pick = 1%. Some All the other picks = 99%. Remove 98 bad picks and the one that's left assumes the percentages so you = 1%, other door = 99%.

 

[Kyteland]

Originally posted by: 91TTZ
Since everyone says I'm wrong, I'll just have to agree with the majority and agree that I'm wrong.

The sad part is that I'm unable to figure out why. I must be too dumb.

What part are you having trouble with? I rig slot machines for a living so I do this kind of thing all day. I'll explain whatever you're not clear on.

 

[Yax]

hehehe, we have to go through this argument again.

Monty: chose again after he reveals the losing door, right answer.

double money: if its odd, chose the other envelope, even, keep it. Your chances are much better that way.

dumb answer: if the envelope contains one dollarbills, switch. If it contained two dollarbills keep it.

😀

 

[purbeast0]

Originally posted by: FrankyJunior

Originally posted by: purbeast0

i do see it ... theoretically ... however in the very end, regardless, you are ALWAYS picking 1 of 2 doors, which would be a 50/50 chance.

also, with your little example, you left out one factor ... monte KNOWS which doors have nothing behind it, so he will ALWAYS choose a door that has nothing in it.

this is where the theory seems messed up:

say you have 100 doors, and 1 has a prize behind it. you pick 1 door, and monte removes 98 doors that all have nothing behind it (of course, he knows which one DOES have something behind it). so by this theory, you should switch doors because your odds will then be 99/100 of being right (since the other 98 wrongs were ruled out).

do you really thing switch doors EVERY time you will get the prize 99 times? i doubt it ...

(again, i might not fully understand the original theory so keep that in mind)

That first statement is wrong. Just becaue there are 2 options doesn't mean that they ahve the same probability. Think of it this way that should clear it up....

You have 3 containers. One is filled with gold, two are filled with manure.
You pick one and you have 33% chance of picking the right one.
Now let's say I take the other two and combine them and then throw out the now empty container.
You now have two containers to choose from. Do they have the same percentage of having the gold? No. Yours has 33% chance and the other one, because it has assumed the percentage of one that is now gone, has 66% chance.
End results, You should Always Switch.

Does that make sense?

Base fact is two choices does not mean each is 50%.

yah thanks, that DOES make sense a lot actually 🙂

thx

 

[3chordcharlie]

Originally posted by: 91TTZ
Since everyone says I'm wrong, I'll just have to agree with the majority and agree that I'm wrong.

The sad part is that I'm unable to figure out why. I must be too dumb.

You're not looking at prior probabilities correctly.

You understand that choosing one out of three doors is a 1/3 chance.

Instead of monty removing an incorrect door, imagine having the choice between choosing one door, and choosing two doors. I'm sure you can see that getting to pick two doors is a better chance of winning.

By choosing the strategy 'switch' that's exactly what you're doing. You're getting the opportunity to say 'I think bet the prize was behind one of the two doors I didn't pick'. Which is a good bet.

The fact that one door was removed changes nothing; whether you picked the 'right' door or not, there was at least one door with nothing behind it, that you didn't choose. But it could have been either one (or both). So you're still getting to choose 'the other two' by switching, and that's why it isn't 50-50.

 

[CoveX]

purbeast0: yes, you would win 99 times out of 100.

 

[Kelemvor]

Originally posted by: purbeast0

Originally posted by: FrankyJunior

Originally posted by: purbeast0

i do see it ... theoretically ... however in the very end, regardless, you are ALWAYS picking 1 of 2 doors, which would be a 50/50 chance.

also, with your little example, you left out one factor ... monte KNOWS which doors have nothing behind it, so he will ALWAYS choose a door that has nothing in it.

this is where the theory seems messed up:

say you have 100 doors, and 1 has a prize behind it. you pick 1 door, and monte removes 98 doors that all have nothing behind it (of course, he knows which one DOES have something behind it). so by this theory, you should switch doors because your odds will then be 99/100 of being right (since the other 98 wrongs were ruled out).

do you really thing switch doors EVERY time you will get the prize 99 times? i doubt it ...

(again, i might not fully understand the original theory so keep that in mind)

That first statement is wrong. Just becaue there are 2 options doesn't mean that they ahve the same probability. Think of it this way that should clear it up....

You have 3 containers. One is filled with gold, two are filled with manure.
You pick one and you have 33% chance of picking the right one.
Now let's say I take the other two and combine them and then throw out the now empty container.
You now have two containers to choose from. Do they have the same percentage of having the gold? No. Yours has 33% chance and the other one, because it has assumed the percentage of one that is now gone, has 66% chance.
End results, You should Always Switch.

Does that make sense?

Base fact is two choices does not mean each is 50%.

yah thanks, that DOES make sense a lot actually 🙂

thx

Welcome.

 

[Yax]

Originally posted by: 3chordcharlie

Originally posted by: 91TTZ
Since everyone says I'm wrong, I'll just have to agree with the majority and agree that I'm wrong.

The sad part is that I'm unable to figure out why. I must be too dumb.

You're not looking at prior probabilities correctly.

You understand that choosing one out of three doors is a 1/3 chance.

Instead of monty removing an incorrect door, imagine having the choice between choosing one door, and choosing two doors. I'm sure you can see that getting to pick two doors is a better chance of winning.

By choosing the strategy 'switch' that's exactly what you're doing. You're getting the opportunity to say 'I think bet the prize was behind one of the two doors I didn't pick'. Which is a good bet.

The fact that one door was removed changes nothing; whether you picked the 'right' door or not, there was at least one door with nothing behind it, that you didn't choose. But it could have been either one (or both). So you're still getting to choose 'the other two' by switching, and that's why it isn't 50-50.

A simple way to put it:
1. you pick a door out of 3.
2. I say to you, do you want to keep your door or trade it for both the other two doors.

3. you say, geez, eventhough I know that the other two doors gives me a 2/3 chance of being right, since one of the other two doors is 100% wrong, I'll just be stupid and keep my door.

 

[purbeast0]

one way it helped me realize the reasoning is with my example of 100 doors.

say you have 100 doors. now you pick 1 door. monte then removes 98 doors that he knows have nothing in them. now by this theory that means 99% chance the correct prize is in the other door, and 1% chance it is in the door you first selected (since you chose 1 out of 100 doors).

at first this didn't make sense to me, but after i wrote it in my previous post and read what i wrote, i realize my previous post was wrong and that i would definitely switch doors in this situation.

 

[vood0g]

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.

thx, your clarification made it clear to me. 😀

 

[91TTZ]

Originally posted by: FrankyJunior
You can test this out very simply with a random number generator and a friend.

Have your friend generate a random number (1, 2, 3). Then you pick a random number (1, 2, 3). Your friend tells which one of the numbers you didn't pick is wrong. That leaves your original number or the other number that's left. Just try it and you'll find that that you originally picked the right number 33% of the time and if you switch you are right 66% of the time.

That made sense to me. Hmmm. I ate something and now I can understand it.

PS- Does anyone else have that problem? If I don't eat every few hours, I get lightheaded and confused. My reasoning ability drops big time. Once I eat, I'm back to normal again. I've been driving before in an unknown area and I easily get lost if I don't have something to snack on. I get confused and angry. What's up with that?

 

[91TTZ]

Originally posted by: Yax
A simple way to put it:
1. you pick a door out of 3.
2. I say to you, do you want to keep your door or trade it for both the other two doors.

3. you say, geez, eventhough I know that the other two doors gives me a 2/3 chance of being right, since one of the other two doors is 100% wrong, I'll just be stupid and keep my door.

Thanks. Now I get it. :thumbsup:

(eating helped also)

 

[chowderhead]

You guys are ALL wrong.
There are no doors, not a single one. Monty Hall reveals curtains. 🙂

 

[Kev]

The Monty Hall problem is stupid.

The way it is worded, you don't have a 33% chance of picking correctly on the first pick, because if he ALWAYS shows you a medium prize, then your initial odds of selecting the big prize on the first try are actually 50%. Or am I crazy/stupid?

 

[3chordcharlie]

Originally posted by: Kev
The Monty Hall problem is stupid.

The way it is worded, you don't have a 33% chance of picking correctly on the first pick, because if he ALWAYS shows you a medium prize, then your initial odds of selecting the big prize on the first try are actually 50%. Or am I crazy/stupid?

If there's three doors, S-M-L, then he can't 'always' show you a medium prize, cause 1/3 of the time, that's the one you already picked...

 

[Kev]

Originally posted by: 3chordcharlie

Originally posted by: Kev
The Monty Hall problem is stupid.

The way it is worded, you don't have a 33% chance of picking correctly on the first pick, because if he ALWAYS shows you a medium prize, then your initial odds of selecting the big prize on the first try are actually 50%. Or am I crazy/stupid?

If there's three doors, S-M-L, then he can't 'always' show you a medium prize, cause 1/3 of the time, that's the one you already picked...

ok then how about instead i say "non-big prize"

doesn't matter whether its small or medium

 

[Jzero]

Originally posted by: Kev

Originally posted by: 3chordcharlie

Originally posted by: Kev
The Monty Hall problem is stupid.

The way it is worded, you don't have a 33% chance of picking correctly on the first pick, because if he ALWAYS shows you a medium prize, then your initial odds of selecting the big prize on the first try are actually 50%. Or am I crazy/stupid?

If there's three doors, S-M-L, then he can't 'always' show you a medium prize, cause 1/3 of the time, that's the one you already picked...

ok then how about instead i say "non-big prize"

doesn't matter whether its small or medium

I dunno which problem you read, but in the one I posted, there is one big prize and the other prizes are of "relatively little value." There's only one "winning" door - the other two are losers.

 

[Kev]

Originally posted by: Jzero

Originally posted by: Kev

Originally posted by: 3chordcharlie

Originally posted by: Kev
The Monty Hall problem is stupid.

The way it is worded, you don't have a 33% chance of picking correctly on the first pick, because if he ALWAYS shows you a medium prize, then your initial odds of selecting the big prize on the first try are actually 50%. Or am I crazy/stupid?

If there's three doors, S-M-L, then he can't 'always' show you a medium prize, cause 1/3 of the time, that's the one you already picked...

ok then how about instead i say "non-big prize"

doesn't matter whether its small or medium

I dunno which problem you read, but in the one I posted, there is one big prize and the other prizes are of "relatively little value." There's only one "winning" door - the other two are losers.

Oh for god's sake. I'll say this one last time.

Since the host will ALWAYS show you a door that IS NOT the WINNING door, you have a 50% chance of choosing the WINNING door from the beginning.

It's just inplausible.

 

[Jzero]

Originally posted by: Kev
Oh for god's sake. I'll say this one last time.

Since the host will ALWAYS show you a door that IS NOT the WINNING door, you have a 50% chance of choosing the WINNING door from the beginning.

It's just inplausible.

I think you've got the order of operations mixed up.
1. You pick 1 door.
2. Host reveals one of the losing doors.
3. You may now stick with your original door or switch to the other door.

 

[weirdichi]

A boat's a boat.. but an envelope's an envelope.. it could even be a boat! You know how much we always wanted one of those!

 

[Kev]

Originally posted by: Jzero

Originally posted by: Kev
Oh for god's sake. I'll say this one last time.

Since the host will ALWAYS show you a door that IS NOT the WINNING door, you have a 50% chance of choosing the WINNING door from the beginning.

It's just inplausible.

I think you've got the order of operations mixed up.
1. You pick 1 door.
2. Host reveals one of the losing doors.
3. You may now stick with your original door or switch to the other door.

I'm a retard. It finally clicked. I don't even know wtf i was thinking