Probability 101: Monty Hall
This is a discussion on Probability 101: Monty Hall within the Off Topic & Humor Discussion forums, part of the The Back Porch category; Originally Posted by Pistology
A goat is a goat. If I put three marbles in a can  identical except two are black and one ...

April 24th, 2010 04:10 PM
#76
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Originally Posted by
Pistology
A goat is a goat. If I put three marbles in a can  identical except two are black and one is white, and, blindfolded, you pick a marble and hold it in your fist, and I pull a black marble from the can and show it to you and offer you to exchange the marble in your fist for the one in the can and tell you that I will put $5 on the table if you will put $6 on the table with the total $11 going to you if you hold the white marble and to me if you don't, would you be willing to play the game?
After you choose  whether you chose a goat or the BP  and Monty shows you a goat, by your Youtube / Wikipedia understanding, you give me odds of $6:$5 that you win by switching and play the game 1,000 times. Over the long run, I am up and you are down $500.
If your claim of a 16% (2/31/2) advantage holds, then the results for the same long run are that you are the one up $1,320.
I'd be willing to take odds of 6:5 if you're willing to give 'em.
If I understand this correctly, you're going to give me 5:6 odds on a game that I'm going to win 2/3 (4:6) of the time? You're on. How long are you willing to lose money before you want to stop?
For simplicities sake, let's say we play 900 times (it's easily divided by three). If the probabilities hold (and I am well aware that, in any given event or series of events the "odds" don't always play out exactly according to the math  it is only with infinite events that we will get the EXACT mathematical probabilities), I will win 600 times, earning $3000 (600x5). You will win 300 times, earning $1800 (300x6). I'm up $1200. When can we play?
Try your own experiment  you can mark a penny w/ a marker and put it and two unmarked pennies in the can if you don't have marbles. Draw one without looking, and hold it in your hand. Now look at the other two. If one of them is the marked penny, "show" (i.e. discard) the other one. If they're both unmarked, "show" one at random. Now decide if you want to keep the penny in your hand, or switch to the other remaining one. Do this twenty times, and I'll give you 100 to 1 odds that you end up with the marked penny more often by switching than you do by staying...
P.S. It's not "my" Youtube/Wiki understanding, it's actual, demonstrable, mathematical fact. The entire concept of gambling is predicated on odds and probabilities, and as a semiserious poker player, my understanding of these things is crucial to my pocket book. If you actually want to take the bet you've proposed, your misunderstanding of them will be quite damaging to your pocket book...
A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands  love a woman, build a house, change his son's diaper  his hands remember the rifle.

April 24th, 2010 05:03 PM
#77
Distinguished Member
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Originally Posted by
Hopyard
Now back to the puzzle is there not a choice being left out in Joinin's analysis? Or is what you guys are driving at the business end of "degrees of freedom, n1"???
Actually, going back and reading this comment just gave me another idea towards explaining the "extra information" concept...
Let's consider three Monty Hall strategiesnever switch doors, flip a coin to switch doors, and always switch doors.
We all agree that on our initial choice, we win 33% of the time. If you never switch doors, and choose to ignore the extra info, your odds stay at 33%.
But now you know one door has a goat. If you take this as information only about the goat door you see, you might as well flip a coin between the two remaining doors ("Heads I switch, tails I stand pat.") Now your odds of winning are 50%.
But if you realize that Monty gave you information not just about the door he showed, but also about the door he didn't show, then you switch every time and your odds go up to 67% of winning.
Does that help?
I know I'm over my head when I play with mathematicians and statisticians, but I'm not finding the plain English explanations convincingthough if you guys are in fact real world working statisticians I will have no choice but to bow to your professional
opinion. I guess you have no obligation to convince a "dunce."
I will never call anyone who makes an honest effort to understand something new a dunce. I am a mathematician, and I have taught in a number of areasincluding math and firearms training; I consider the ability to find the right way to explain something to somebody, especially somebody with a different background on a subject, to be one of the top goals of an instructor.
“What is a moderate interpretation of [the Constitution]? Halfway between what it says and [...] what you want it to say?” —Justice Antonin Scalia
SIG: P220R SS Elite SAO, P220R SAO, P220R Carry, P226R Navy, P226, P239/.40S&W, P2022/.40S&W; GSR 5", P6.

April 24th, 2010 07:58 PM
#78
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I'm still lookin' for a quarter.

April 24th, 2010 08:58 PM
#79
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Originally Posted by
kccad
I'm still lookin' for a quarter.
It's behind Door #3!
"Democracy is two wolves and a lamb voting on what to have for lunch; Liberty is a wellarmed lamb contesting the vote."
 Benjamin Franklin

April 24th, 2010 09:58 PM
#80
Senior Member
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The first choice of door is water under the bridge statistically. COMPLETELY meaningless now in probability terms. The NEW choice is statistically independent. You have a choice between two doors that you KNOW has one winner. You have a choice to STAY or a choice to MOVE doors. The only correct answer is 50%, WHEN the question is asked when there are only two doors left. Any other explanation is slight of hand and improper statistics.
If you bring the stats back to the initial choice, then the only answer is 1/3 probability of winning no matter if you switch or not. Monty ALWAYS has the option of picking an empty door whether or not you pick the winner initially so Monty's pick really has no effect.
Here is a simple graphic showing the decision tree of possible outcomes assuming Door 1 is the winner. You can duplicate this for the other three. I get 24 total outcomes, 8 per winning door set (note the video said 9 which is incorrect). Out of the 8, you get 4 winning events, so it's STILL 50%.

April 24th, 2010 10:32 PM
#81
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Rusty, you're not big on reading the whole thread before posting, huh?
A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands  love a woman, build a house, change his son's diaper  his hands remember the rifle.

April 24th, 2010 10:32 PM
#82
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Originally Posted by
JonInNY
It's behind Door #3!

April 24th, 2010 10:53 PM
#83
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re: Rustn...
Originally Posted by
Rustynuts
The first choice of door is water under the bridge statistically. COMPLETELY meaningless now in probability terms. The NEW choice is statistically independent. You have a choice between two doors that you KNOW has one winner. You have a choice to STAY or a choice to MOVE doors. The only correct answer is 50%, WHEN the question is asked when there are only two doors left. Any other explanation is slight of hand and improper statistics.
If you bring the stats back to the initial choice, then the only answer is 1/3 probability of winning no matter if you switch or not. Monty ALWAYS has the option of picking an empty door whether or not you pick the winner initially so Monty's pick really has no effect.
I was firmly of the same opinion until I did the experiment JoininNY suggested.
It isn't really obvious that OPFOR is right, but he is. Here's why. If you don't switch you are stuck with having made only one choice where the odds of being wrong are 2/3. By switching on the second chance, the nature of the problem gets changed.
The problem becomes, "calculate the probability of two successive choices getting a result."
That is a slightly different question than the probability of what happens when you make one choice and one choice only (what happens if you stay put with your original guess).
And yes, the chances for each outcome on each toss are indeed independent, but that fact actually misleads to the wrong conclusion in this particular problem.
There is no slight of hand here or improper statistics. Just do JoininNY's experiment and it will become obvious. If you read the whole thread you will see that I was almost adamantly in your camp on this thread until I spent the better part of a night tossing around in bed going nuts over the problem, trying to prove to OPFOR that he was wrong, and in the process convinced myself that he is right.

April 24th, 2010 11:29 PM
#84
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It's hard to prove a mathematical fact wrong, isn't it?
Look, I certainly didn't come up with the Monty Hall problem, or its solution  it's been around for quite a while. The answer is what it is  it's been proven time and time and time again, I'm just relaying that answer here. It's exactly the same as if I said that 2+2=4, and you tried to convince me that it wasn't...
Congrats to you, Hopyard, and the others who were able to get beyond the "obvious" (though wrong) answer and see the truth (even if it isn't obvious, it's true) of the correct answer. Now, if only there were a mathematical way to PROVE that my brand/caliber/style is the best!
A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands  love a woman, build a house, change his son's diaper  his hands remember the rifle.

April 25th, 2010 10:52 AM
#85
Senior Member
Array
Originally Posted by
OPFOR
....Now, if only there were a mathematical way to PROVE that my brand/caliber/style is the best!
Ah, but that's subjective. There are no proofs for subjective opinion, only objective facts.
But you're right, I like my brand/caliber/style the best also.

April 25th, 2010 11:53 AM
#86
Senior Member
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OK I give, but i still don't like it!

April 25th, 2010 11:54 AM
#87
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Originally Posted by
highvoltage
Ah, but that's subjective. There are no proofs for subjective opinion, only objective facts.
But you're right, I like my brand/caliber/style the best also.
Ah, but there IS proof of that. here it is!

April 25th, 2010 02:57 PM
#88
Senior Member
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Originally Posted by
Rustynuts
Ah, but there IS proof of that. here it is!
Not really, subjective drawings by 6th graders don't hold water!

April 25th, 2010 03:41 PM
#89
Moderator
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To the non believers:
I'm going to give you a choice of three doors. Behind one door is a big prize, behind the other two doors nothing. Now do you want to pick one door and one door only, or do you want to pick two doors out of three? Obviously if you pick two doors one of them will be empty (and I will show you which) but you will still win the prize. What is you choice?
Pick one door with odds of one in three?
Pick two doors with odds of two of three?
OK you've obviously picked two doors for a chance of two out of three. Even if I show you that one is empty (we already knew one had to be) your odds are still 2/3 that ONE of these two doors held the prize.
The odds are that 66.6% of the time you will win if you pick two out of three doors. It is 100% that one of your two doors will be wrong and we will now remove the wrong door. Now your remaining door carries the full pecentage value of the incorrect door equal to 33.3%. It also carries it's own third 33.3%. 33.3% + 33.3% = 66.6% (2 out of 3).
The single door still carries it's original chance of 33.3% (1 out of 3).
Two doors remain from the original 3 with two different sets of odds.
"To my mind it is wholly irresponsible to go into the world incapable of preventing violence, injury, crime, and death. How feeble is the mindset to accept defenselessness. How unnatural. How cheap. How cowardly. How pathetic." Ted Nugent
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