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Old 06-13-2003, 03:31 PM   #16
hot_pastrami
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Er, ok... let me try another way of explaining. This is fun, because it's challenging to explain it, and bends my brain like a pretzel.

I think we can agree that before the player selects a door, each one has a 1/3 chance of being the winning door. When a player selects a door, they basically divide the doors into two groups... the door they have chosen, and doors they have not chosen. The door they have chosen has a 1/3 chance of being a winner, and the doors they have not chosen, as a group, has a 2/3 chance of containing a winner.

So Monty opens one of the doors in the "doors the player did not choose" group. Now, even though you know what is behind one of the doors in that group, the "doors the player did not choose" group still has a 2/3 probability of containing the winning door, you just know not to choose the one that had it's contents revealed. So you go with the remaining door in that group, and use that 2/3 odds to your advantage.

Intuition wants us to break down the problem and say that after a door is opened, and we have one winning door and one losing door before us, we have a 50/50 chance. But it's better to think of the do-I-switch choice as "Was I right on my initial guess?" Chances are, you weren't. It's a 2/3 chance you were wrong, as a matter of fact.

Any better?
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Old 06-13-2003, 05:37 PM   #17
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OK, I follow your logic that the door I chose is 1/3 and the other is 2/3 *IF* you consider the *WHOLE* game. But I still don't understand why the odds are figured on the whole game and not on the final choice.
Can you explain to me why if I win the "Daily Number" and I bet that same number tomorrow, the odds are still 1/1000. Why is the previous day disregarded in this case?

Wait..I see..If you take the day before into account it is 2/2000 which is 1/1000. Nevermind.
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Old 06-13-2003, 06:20 PM   #18
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Well, the problem with the 50/50 intuition is that it assumes that the first choice has no bearing on the second choice. But it does. Consider the two possible scenarios for the initial door selection:

1) You select the door that does contain the prize, which will occur one in three times. Monty can open either of the other two doors at random. His choice is unaffected.

2) You select a door which does not contain the prize, which will occur two in three times. Monty is forced to open the other non-winning door.

Scenario #1 will happen about 33% of the time, and scenario #2 will happen about 66% of the time, right? So, after you make your initial choice, it is safer to bet that #2 happened. So accordingly, it is safe to bet that he was forced to open the other empty door, and therefore it follows that the third door, the one you did not pick, contains the prize. I hope I explained that well, my brain is getting pretty tired.

The reason the Daily Number's odds don't change is that one day's selection has no effect on the other days' numbers, it's always a random draw. Statistics don't always follow intuition... for instance, if you flip a normal coin 100 times and it surprisingly comes up heads every time, what are the odds that it will come up heads again on the next flip? It seems outlandish, like it would become less and less likely for it to turn up heads. But against (my) intuition, the chances are still exactly 50%. That's because each flip of the coin is a random result independent of all the other flips. Wierd.

Anyway, have a good weekend everyone.
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Old 06-13-2003, 06:21 PM   #19
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Good lord I leave for a month and the cellar is suddenly discussing things so far over my head I didn't feel it go by. Well maybe it's not that bad but still way to deep for me.

by the way Hi everyone nice to be around again.
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Old 06-13-2003, 09:19 PM   #20
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Welcome back, Cam.
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Old 06-14-2003, 12:02 AM   #21
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Quote:
Originally posted by vsp
What I do not, and probably will never understand is how the second choice is not _independent_ from the first one. Yes, Monty has information that I do not as to which door has the prize -- but the mere fact that I do not have that information makes it irrelevant to my choosing process. I don't know whether Monty is trying to help me or screw me over by making his offer, and I have no way of finding out short of making a blind choice; all I _do_ know is that between my first choice and my second, the situation has changed and my options are different.
Don't let the Monty-knows-something-I-don't issue trick you. The fact that he knows where the prize is only means that he will never accidentally open the door with the real prize when he's showing you one of the "other" doors.

I've seen several explanations that say something like "but Monty knows where the prize is, so he picks the other one." This is just worded poorly. He isn't trying to help you or screw you, he's just playing the game too, and it has nothing to do with the mathematical odds.
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Old 06-16-2003, 12:52 PM   #22
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Monty's knowledge is critical. If he did not have it, he would sometimes open the door with the prize. In those cases he did not, you would have a 50/50 chance whether you switched or did not.
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Old 06-16-2003, 12:54 PM   #23
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Quote:
Originally posted by hot_pastrami
[b]Well, the problem with the 50/50 intuition is that it assumes that the first choice has no bearing on the second choice. But it does. Consider the two possible scenarios for the initial door selection:

1) You select the door that does contain the prize, which will occur one in three times. Monty can open either of the other two doors at random. His choice is unaffected.

2) You select a door which does not contain the prize, which will occur two in three times. Monty is forced to open the other non-winning door.

Scenario #1 will happen about 33% of the time, and scenario #2 will happen about 66% of the time, right? So, after you make your initial choice, it is safer to bet that #2 happened. So accordingly, it is safe to bet that he was forced to open the other empty door, and therefore it follows that the third door, the one you did not pick, contains the prize. I hope I explained that well, my brain is getting pretty tired.
But... as always, I don't grasp the causal relationship between the two choices.

Let me phrase it this way:

Game #1: Monty shows Joe Contestant three doors. Joe picks Door #1. Monty says "Here's what was behind Door #2," and opens it to reveal a goat. Monty says "Would you like to switch to Door #3?"

Game #2: Monty shows Joe Contestant two closed doors and one open one. The open one contains a goat. (Alternatively, he shows three closed ones, but opens one goat door before Joe makes a choice.) Monty says, "Which one will you choose?"

In my mind, these two games are identical in terms of probability; Game #1 merely has additional non-binding gamesmanship before it's time to choose "for real."

In Game #1, you make an initial choice, but it is in no way binding; the only effect it has (other than suspense for the audience) is to help Monty select which goat door he's going to open. (If you guess the prize door initially, he can open either; if you guess a goat door initially, he will open the other goat door.) Either way, the ambiguity has been removed, and you are now faced with a situation equivalent to Game #2 (two doors that you know nothing substantial about, one prize, one final choice to make).

This doesn't change the fact that there's a 2/3 chance that the prize will be behind one of the two doors you didn't initially choose; that much is simple mathematics. But the situation _after_ Monty opens the door is not the same situation as _before_ Monty opens the door; to claim that the second choice (viewed in a vacuum) isn't a fifty-fifty chance in and of itself is to play clever semantic games with the problem as a whole. That 2/3 chance is tempered by the 100% certainty that one of its two options will be eliminated before the actual choice is made.
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Old 06-16-2003, 01:34 PM   #24
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Quote:
Originally posted by russotto
Monty's knowledge is critical. If he did not have it, he would sometimes open the door with the prize. In those cases he did not, you would have a 50/50 chance whether you switched or did not.
Dammit, that's exactly what I said. Quit trying to confuse the issue.

Vsp thought Monty's knowledge was coming into play in that he is trying to influence the outcome of the game. That is not the case. He simply follows the rules for his role. A trained monkey could do his job and it wouldn't affect the coutcome.
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Old 06-16-2003, 01:38 PM   #25
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Quote:
Originally posted by vsp
But... as always, I don't grasp the causal relationship between the two choices.
Part of the problem is that the scale is so small that intuition misleads us. Here's what basically happens in the three-doors scenario.... the contentant picks one door, then the host opens all but one of the remaining doors. That remaining closed door was unopened because:

A) it contains the prize (likelihood: x-1/x), or
B) you selected the door that contained the prize, so an empty door was randomly opened from those that were left (likelihood: 1/x).

...where x=3 in the Let's Make a Deal context.

So imagine that there are not three doors, but a thousand (x=1000). The contestant picks a door, and like in the three-door game, Monty opens all of the other doors except one... 998 doors spring open to reveal goats. Now, like in the three door game, there are only two closed doors to choose from. But are the chances 50/50? No, because that fact that there are only two doors now does not change the fact that your chances of guessing right were one in a thousand before those 998 doors opened. You can still bet that one of the 999 doors you didn't pick is the winner (likelihood 999/1000, or 99.9%), you just happen to know what is behind 998 of them, so you know not to pick those.

Any better? I know my explanations aren't the greatest.

EDIT: I had the little formulas reversed in my A/B scenarios... oops
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Old 06-16-2003, 01:48 PM   #26
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Quote:
Originally posted by vsp
But... as always, I don't grasp the causal relationship between the two choices.

...

But the situation _after_ Monty opens the door is not the same situation as _before_ Monty opens the door; to claim that the second choice (viewed in a vacuum) isn't a fifty-fifty chance in and of itself is to play clever semantic games with the problem as a whole.
The causal relationship you are looking for is this: the door Monty opens is affected by the door that you choose.

You can't view the second half in a vacuum, because your initial choice determined which door he would open (or, 1/3 of the time, it at least narrowed down the choices).

In Game 2, the odds are indeed 50/50, because the door that Monty opens was chosen between the two losers at random. But in Game 1, 67% of the time Monty <B>is forced by the rules of the game</B> to open the only remaining losing door.
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Old 06-16-2003, 02:39 PM   #27
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Quote:
Originally posted by hot_pastrami
So imagine that there are not three doors, but a thousand (x=1000). The contestant picks a door, and like in the three-door game, Monty opens all of the other doors except one... 998 doors spring open to reveal goats. Now, like in the three door game, there are only two closed doors to choose from. But are the chances 50/50? No, because that fact that there are only two doors now does not change the fact that your chances of guessing right were one in a thousand before those 998 doors opened. You can still bet that one of the 999 doors you didn't pick is the winner (likelihood 999/1000, or 99.9%), you just happen to know what is behind 998 of them, so you know not to pick those.
It still doesn't entirely make logical sense to me. x can be 3, 1000 or 3.4985767E+12, but if (x-2) are eliminated before you make a choice, you're still choosing between two options in the end, one of which is an unknown and the other one of which is (998 losers + one unknown). You make a choice earlier in the game, of course, but that choice doesn't determine whether you win or lose.

If I was applying for a job with 999 others, and only two of us were brought back for a second interview (myself being one of them), I'd be a hell of a lot more confident after the reduction than before...
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Old 06-16-2003, 03:04 PM   #28
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Originally posted by vsp
If I was applying for a job with 999 others, and only two of us were brought back for a second interview (myself being one of them), I'd be a hell of a lot more confident after the reduction than before...
Well, that's not an effective parallel, since it's a decision based on information about all of the choices, and there is no random element.

Say you're outside with some friends, staring at a beautiful starry sky with a new moon. Your friend tells you that she's chosen one star from the sky, and wants you to guess which one she is thinking of. You contemplate the question for a moment, then, from the thousands of visible stars, you indicate a star in the western sky. She responds with, "Ok, I was thinking of the one you pointed out, or..." she points to another, near the south horizon... "that one." Which is more likely? There are only two stars now to choose from, so you might think the chances are 50/50, but in order for it to be the star you pointed out, you would had to have correctly guessed from all the stars in the sky.

What are the chances that you guessed right, and she just picked some random star as an alternative? And what are the chances you guessed wrong, and she's pointing out her real star as an alternative?

Another example... say you entered a contest for a $1 million prize. The company conducting the contest is bringing every entrant into their office, so you go wait your turn in line. Once you get in, they tell you that the winner of the $1 million was either you, or some lady named Blarda in Columbus, OH. You have to pay them $1000 to find out, or keep your money and risk losing the $1 million. Do you think you ave a 50/50 chance of being the winner, or do you just keep your $1000?
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Old 06-16-2003, 03:22 PM   #29
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Originally posted by hot_pastrami
Well, that's not an effective parallel, since it's a decision based on information about all of the choices, and there is no random element.
I don't know about that -- I've been involved in some seriously random-seeming job interview processes...
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Old 06-16-2003, 05:02 PM   #30
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What are the chances that you guessed right, and she just picked some random star as an alternative? And what are the chances you guessed wrong, and she's pointing out her real star as an alternative?
Chick's head games have no place in a discussion mathmatical odds.:p
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