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; You have 1/3 of selecting the winning door and 2/3 of selecting a goat. The 2/3 when you select the goat Monty has no choice, ...

April 22nd, 2010 11:10 PM
#16
Member
Array
You have 1/3 of selecting the winning door and 2/3 of selecting a goat. The 2/3 when you select the goat Monty has no choice, because one of the two remaining doors is the winner and he can't open it. So 2/3 times that Monty has no choice the other door is the winner for you. That is 2/3 of winning if you change.
For those who understand, no explanation is necessary.
For those who don't understand, no explanation is possible.
John

April 22nd, 2010 11:42 PM
#17
Senior Member
Array
Originally Posted by
JonInNY
If I pick the winner, he's going to ALSO reveal a goat! So what you are saying does not make sense. No matter which door I pick, he will ALWAYS show a goat!
You're looking at it backwards. Stop looking at the odds of picking the car (1/3) and bet on the better odds of picking a goat (2/3).
Whichever door Monty open's you know the door you just picked has a 2/3 chance of also being a goat.
And that being the case, there's a 2/3 chance of the final door revealing the car, not 50/50 as some have suggested.

April 23rd, 2010 12:03 AM
#18
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You all want the answer? No one has gotten it exactly right yet...
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 23rd, 2010 12:44 AM
#19
Senior Member
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Originally Posted by
OPFOR
You all want the answer? No one has gotten it exactly right yet...
Well I'm pretty darned close, barring other factors that might bias the results.

April 23rd, 2010 01:43 AM
#20
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Ok, here's the answer  skip if you don't want to know yet...
Assume the Big Prize (and where did it become a car?) is behind door number one. There are three original options; you chose 1, 2, or 3.
1) If you chose 1, Monty could show you 2 or 3 (both being goats). If you switch, you lose. If you stay, you win.
2) If you chose 2, Monty could only show you 3 (1 being the BP). If you switch, you win. If you stay, you lose.
3)If you chose 3, Monty could only show you 2 (1 being the BP). If you switch, you win. If you stay, you lose.
Thus, as some have surmised, you have a 2/3 chance of winning if you switch, and a 1/3 chance of winning if you stay. Mathematically, you should absolutely switch. (My apologies to the folks who came up with the right answer  the explanations seemed...off. That may have more to due with the Sam Adams seasonal that's been poured tonight than your explanations, but, hey, I'm the OP... )
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 23rd, 2010 04:50 AM
#21
Member
Array
What are my odds of winning the 1974 Ford Pinto?
http://www.youtube.com/watch?v=dT0J0rcJTLo

April 23rd, 2010 07:13 AM
#22
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Originally Posted by
kccad
I actually have a '74 Pinto.
Gain a 2A vote, take a fencesitter shooting.

April 23rd, 2010 08:49 AM
#23
Senior Member
Array
Originally Posted by
JonInNY
OPFOR, I must be dense today, but I still do not get this. Let's say I choose Door Number 1, ok? This (unknown to me) has the Big Prize.
My odds at this point are 1:3 that I got it right. Correct?
Now, Monty shows me either Door Number 2 or Door Number 3 which WILL have a goat. Correct?
Now I know for certain that I either WILL have a goat, or WILL NOT have a goat. Correct?
The ORIGINAL odds should not matter at this point, because the scenario has changed! Yes, I originally had a 2:3 chance of picking a goat. Since there are now only 2 unknowns, my odds are now 1:2 that I will pick a goat. They are not 2:3 any more!
I do not understand how the odds change to 2:3 in my favor if I switch doors. Can you explain this so a nonstatistician can understand this?
The odds of your door being a goat are STILL 2/3 regardless of the contents of the other two doors.

April 23rd, 2010 08:57 AM
#24
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Originally Posted by
Lewis128
The odds of your door being a goat are STILL 2/3 regardless of the contents of the other two doors.
Yes, I understand that! That's at the beginning BEFORE you know any information at all!
But what I am saying, is that once a given door is known, there are now only two possibilities. The odds of picking EITHER a goat OR a Big Prize at that point is 1:2. I don't understand how that can be any different.
"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 23rd, 2010 09:06 AM
#25
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OK, after more rumination, I believe I understand the concept here. I was stuck on the fact that I might pick the winning door FIRST, in which case switching doors would be a losing proposition.
But I do see that my odds at the outset were 2:3 of a loser.
But what are the actual odds of winning, if, I pick a goat (unbeknownst to me), and he shows me the second goat!?
"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 23rd, 2010 09:34 AM
#26
Distinguished Member
Array
Folks, I lost a fair amount of money in a casino once from this error: please follow this carefully so you don't do the same.
Odds have no memory.
 For starters, you have a 1/3 chance of picking the big prize.
 But once he's opened a goat door, you now have an independent choice.
 "Independence" here means that the two doors don't care if the third one has been opened or not  they hold what they hold.
 Thus you have two closed doors, and you get to choose one.
 The doors don't care if you earlier chose one of them or not: they hold what they hold.
 So you can choose one (stay), or the other (switch).
 If you EARLIER chose this one or that one is irrelevant (again, aside from the theatre of the whole thing).
 You choose one, out of two possibilities = 1/2 or 50% odds of winning the big prize
.
The error comes from considering the 'run' (though I'm not sure that's exactly what folks are explaining, this is the road they're on): if I asked you what the chances were of getting two heads in a row on coin flips, you would rightly say 1/4: only one choice meets the criteria, and you have two possibilities on the first flip and two on the second. BUT each coin flip ITSELF is independent (coins have no memory). So if you get a run of a dozen heads, the next coin flip is STILL 50% AS LIKELY TO GO HEADS AS TAILS. The long odds are for the SERIES, not the individual coin flip.
My error was the red vs. black Roulette version of this same thing: sure, it was unlikely to get black six times in a row... assuming an honest wheel the NEXT spin was still 50/102.
Once you've opened one door, you're at 5050 odds when faced with two doors from which to choose. The doors don't care what you chose or what Monty revealed earlier.

April 23rd, 2010 10:11 AM
#27
Senior Member
Array
Originally Posted by
Paymeister
Folks, I lost a fair amount of money in a casino once from this error: please follow this carefully so you don't do the same.
Odds have no memory.
 For starters, you have a 1/3 chance of picking the big prize.
 But once he's opened a goat door, you now have an independent choice.
 "Independence" here means that the two doors don't care if the third one has been opened or not  they hold what they hold.
 Thus you have two closed doors, and you get to choose one.
 The doors don't care if you earlier chose one of them or not: they hold what they hold.
 So you can choose one (stay), or the other (switch).
 If you EARLIER chose this one or that one is irrelevant (again, aside from the theatre of the whole thing).
 You choose one, out of two possibilities = 1/2 or 50% odds of winning the big prize
.
The error comes from considering the 'run' (though I'm not sure that's
exactly what folks are explaining, this is the road they're on): if I asked you what the chances were of getting two heads in a row on coin flips, you would rightly say 1/4: only one choice meets the criteria, and you have two possibilities on the first flip and two on the second. BUT each coin flip ITSELF is independent (coins have no memory). So if you get a run of a dozen heads, the next coin flip is STILL 50% AS LIKELY TO GO HEADS AS TAILS. The long odds are for the SERIES, not the individual coin flip.
My error was the red vs. black Roulette version of this same thing: sure, it was unlikely to get black six times in a row... assuming an honest wheel the NEXT spin was still 50/102.
Once you've opened one door, you're at 5050 odds when faced with two doors from which to choose. The doors don't care what you chose or what Monty revealed earlier.
But the doors DO have memory unless the contents are randomly switched again. Nobody's moving goats and cars around, the second goat and car are still behind the same door they started at in the beginning. It's no longer a random chance, like roulette or a flipped coin.
No matter which door is opened, the remaining doors "remember" what was behind them in the first pick.
So the odds are Goat1 1/3, Goat2 1/3, Car 1/3. You have a combined chance of 2/3 to pick a goat. Revealing the other goat does not change the odds that you have a goat because the doors are not reset after you pick.
The examples of roulette and coins do not hold because the act of spinning the wheel or flipping the coin reintroduces the random element. The doors remain set, therefore the 1/3 vs 2/3 odds remain throughout.
So when you make your initial pick, you presume you are going to pick a goat 2/3 of the time, which means 2/3 of the time ONE of the other two remaining doors will reveal a car.
Once Monty opens the door revealing a goat, you know there is a 2/3 likelyhood that the remaining door hides the car.

April 23rd, 2010 10:19 AM
#28
VIP Member
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Originally Posted by
Lewis128
So when you make your initial pick, you presume you are going to pick a goat 2/3 of the time, which means 2/3 of the time ONE of the other two remaining doors will reveal a car.
Once Monty opens the door revealing a goat, you know there is a 2/3 likelyhood that the remaining door hides the car.
I am losing you on this last paragraph. It seems as though you have too many "2/3" possibilities. Can you write this out a different way so us nonmath people can understand what you are saying.
I still don't understand why it's in my best interest to switch doors.
"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 23rd, 2010 10:25 AM
#29
Senior Member
Array
Originally Posted by
JonInNY
I am losing you on this last paragraph. It seems as though you have too many "2/3" possibilities. Can you write this out a different way so us nonmath people can understand what you are saying.
I still don't understand why it's in my best interest to switch doors.
Turning the random mess in my head into a coherent explanation is not always easy for me.
Teaching is not my strong suite, that would be my sister. I'll ask her and reply again later. She's at work... teaching.
Found a Youtube.com that explains it. It may help.
http://http://www.youtube.com/watch?v=koPBkK_Rak
There's also a Wikipedia link, but it's written at college grad level and makes my brain hurt. Didn't finish college, I ran out of $$ about the time my motivation kicked in.
Math and Business classes were fun to me, but "required" undergrad classes depleted my funds and under employment wiped them out entirely. But I'm veering way off topic. Here's the link.
Monty Hall problem  Wikipedia, the free encyclopedia
Last edited by Lewis128; April 23rd, 2010 at 10:15 PM.

April 23rd, 2010 11:12 AM
#30
Member
Array
Originally Posted by
ppkheat
I actually have a '74 Pinto.
That's okay, one of my cars is an '81 AMC Eagle, so I guess we are both pretty much in the same boat.
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