Statistics riddle problem continued.

[Sir Kenyth]

This is a little math riddle that was being discussed previously and submitted by Barry:

quote:

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OK, heres a probability one for you. You are playing a game show and the host asks you to pick your prize. There are three doors you can pick and one of them has a prize behind it. The other two are empty (which, I admit, would be a prize to some people but not to most).
He asks you to guess so you pick one. He then opens one of the other doors to reveal nothing.
He asks you if you are still sure.
On the basis of pure probability should you stick or change?

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This is a tricky little problem. Your fist instinct would be to say theres a 50/50 chance for the remaining two doors. Two doors, one choice, 50/50 chance. It's not that simple though. There's another factor to take into consideration. The trick is the fact that the game guy KNOWS which door the prize is behind. His choice of an empty door is NOT RANDOM. He has two doors to pick from and he knows which is empty. Therefore, since he HAS to pick an empty door, the door he DOES NOT PICK has a higher chance of being a prize door than your TOTALLY RANDOM guess. It's only because he knows where the prize is, see? His indirect choice has a higher probability than your direct choice because he's informed and you're not. Get it?

The thinking is a bit abstract on this one.

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NOTE: It appears I was wrong and that the original 50/50 argument holds. Read Legolas' reply further on and see how he mapped out the possible paths. 250 and I are bantering uselessly through most of this thread, but you might get a kick out of it!

[250]

3 doors are too simple

lets look at 10 doors

1 has the prize, the other 9 are empty

now I ask you to give the probability for "switch" choice (meaning, pick one door, then switch to another)

for every 10 times, you might get the prize for the first try, so 1/10 is a constant

then if you pick a bad door: the chance is 9/10
now here is the KEY:
when we calculate chance, we see the total possibility as a whole "1", in this case, the "1" is replaced by "9/10"

so the next pick you can get the prize out of the 9 door is: 1/9
but this 1/9 is 9/10's. which is 9/10 * 1/9 in a total of 1

so the total possibility is:
1/10 + 9/10 * 1/9 = 2/10

understanding that, look back to the original problem (which is more simpler)

1/3 + 2/3 * 1/2 = 2/3

P.S. whoever pick the prize shouldn't be count into consideration

[Sir Kenyth]

quote:

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Originally posted by 250:
3 doors are too simple

lets look at 10 doors

1 has the prize, the other 9 are empty

now I ask you to give the probability for "switch" choice (meaning, pick one door, then switch to another)

for every 10 times, you might get the prize for the first try, so 1/10 is a constant

then if you pick a bad door: the chance is 9/10
now here is the KEY:
when we calculate chance, we see the total possibility as a whole "1", in this case, the "1" is replaced by "9/10"

so the next pick you can get the prize out of the 9 door is: 1/9
but this 1/9 is 9/10's. which is 9/10 * 1/9 in a total of 1

so the total possibility is:
1/10 + 9/10 * 1/9 = 2/10

understanding that, look back to the original problem (which is more simpler)

1/3 + 2/3 * 1/2 = 2/3

P.S. whoever pick the prize shouldn't be count into consideration

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250, you're trying to prove the mathematical part. I'm explaining the somewhat abstract reasoning. I'm not even sure that this math applies. Either of the remaining doors are a switch from the original one the gamemaster opened. I think the logic you're presenting is that subsequent tries always have a higher probability for a given task. The straight probability is only for a one try deal. If you try a 50/50 task four times, there's a better chance of you winning it at least once. Statistical concepts are somewhat abstract and fairly malleable. That's why statisticians can pretty much come up with whatever numbers the person who hires them wants.

The first try of ten doors is 1 in 10. The second try can be seen as 1 in 9 or 1 in 5 depending on what logic you use. Since it is a subsequent try, I beleive that 1 in 5 would be the correct answer. It's strange to think that the exact same situation can produce different probabilities. Here's a brain twister. If another person came along and had a choice of the nine doors, would the probability be 1 in 9 because they are on their first try?

[250]

nah uh

do you even understand what I wrote?

EDIT: I am not "trying to prove the mathematical part", I am proving it

[Sir Kenyth]

quote:

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Originally posted by 250:
nah uh

do you even understand what I wrote?

EDIT: I am not "trying to prove the mathematical part", I am proving it

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I added some more to the top post. I understand what you wrote. It's looking at it from one point of veiw. Always remember that "proof" with statistics is a fickle thing.

Here's a brain twister. If another person came along and had a choice of the nine doors, would the probability be 1 in 9 because they are on their first try, or would it be combined with the previous choice by a different person to still make a 2 in 10 probability? What if someone covered the first chosen door so the second person couldn't see it and didn't know about it?

[250]

no no no, how do you get the idea of "1 in 5" ?

if the second person comes along, that'd be a DIFFERENT problem, mind you.

[Sir Kenyth]

The same physical situation can have different probabilities depending on how you reason things. That's why simulators and studies are used.

[Sir Kenyth]

quote:

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Originally posted by 250:
3 doors are too simple

so the total possibility is:
1/10 + 9/10 * 1/9 = 2/10

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Quoted from you. 2/10 simplifies to 1/5.

[Waluin]

(startrekgeek)
Infinite diversity in infinite combinations
(/startrekgeek)

[250]

nope, you are changing the conditions

by reasoning, we dont change conditions

out assumption only stays true if ALL the conditions that the assumption is based upon stays constant. you add one more thing in, the whole thing changes

so you admit 2/10 is the answer

in fact, it is 2/n for all doors (n is whatever constant)