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☖ Forums → Discussion Forum → Monty Python and the Yak
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Monty Python and the Yak
2009-09-24, 12:16 PM #41
Originally posted by Pommy:
http://en.wikipedia.org/wiki/Monty_Hall_problem#Bayesian_analysis

There is also a visual diagram further up on the page


I'm familiar with the probability theory. Just because something has the probability of being the winning choice, doesn't mean that it is. I understand that the initial problem is asking for a statistical answer, my point is that the statistical answer isn't always the right one and any 'perceived' benefit from it is purely imaginary. Again, there's a difference between probability and reality. This is the point I'm trying to illustrate.
2009-09-24, 12:16 PM #42
http://www.grand-illusions.com/simulator/montysim.htm
2009-09-24, 12:19 PM #43
Originally posted by Emon:
Knowledge of what's under a cup AFTER choosing does not affect the probability that you chose correctly BEFORE revealing the cup.


Agreed, but that's now how the question was posed. The question is asking "now that you know which one of the two cups that were empty" what is the chance that you picked the right cup. When the question is asked an when/if one of the doors/cups are revealed has an impact on the answer to the question.
2009-09-24, 12:20 PM #44
Originally posted by Alco:
I'm familiar with the probability theory. Just because something has the probability of being the winning choice, doesn't mean that it is. I understand that the initial problem is asking for a statistical answer, my point is that the statistical answer isn't always the right one and any 'perceived' benefit from it is purely imaginary. Again, there's a difference between probability and reality. This is the point I'm trying to illustrate.


The statistical answer suggests what is more or most likely; if you wish to attain a particular outcome, isn't it rational to act in such a way that makes it most likely to occur?
一个大西瓜
2009-09-24, 12:21 PM #45
Originally posted by Alco:
I'm familiar with the probability theory.

Just like how you're familiar with radio waves? :downswords:

Originally posted by Alco:
Just because something has the probability of being the winning choice, doesn't mean that it is.

No ****. You've been arguing that the probability that it's the winning choice is 50%, which is wrong.
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2009-09-24, 12:22 PM #46
It doesn't matter. For the doors (as I'm not keeping up with your hurr cup example), if you pick a door with a yak, and you switch, you will win. Period. Doesn't matter which yak you pick, you still win. The only way you can lose is if you picked the car at the beginning.

That is to say, when you are introduced with the choice of 3 doors, 2 with yaks, you need to look at it as if the 2 yaks are the WINNING doors. Because if you pick either yak, you will win by switching.

If you still think it's 1/2, not 2/3rds, I don't know if you'll ever grasp math.
2009-09-24, 12:24 PM #47
Originally posted by Alco:
Agreed, but that's now how the question was posed. The question is asking "now that you know which one of the two cups that were empty" what is the chance that you picked the right cup. When the question is asked an when/if one of the doors/cups are revealed has an impact on the answer to the question.

Okay, let's try another example.

Say there are one million cups. I pick one at random. Then, 999,998 other cups are revealed, so that the only cups left are the one I chose and the cup with the ball under it. Are you really suggesting that my initial choice had a 50% chance of being correct?

What you're saying only works if I know ahead of time which cups do not have the ball, and then I adjust my choice accordingly.
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2009-09-24, 12:25 PM #48
Originally posted by Darth:
http://www.grand-illusions.com/simulator/montysim.htm


The Monty Hall problem [CENTER] Keep choice: 10 times Wins: 5 cars (50%) Losses: 5 goats (50%) [/CENTER]
[CENTER] Change choice: 0 times Wins: 0 cars (0%) Losses: 0 goats (0%) [/CENTER]
2009-09-24, 12:25 PM #49
Originally posted by Alco:
The Monty Hall problem [CENTER] Keep choice: 10 times Wins: 5 cars (50%) Losses: 5 goats (50%) [/CENTER]
[CENTER] Change choice: 0 times Wins: 0 cars (0%) Losses: 0 goats (0%) [/CENTER]


Law of Large Numbers. Look it up.

Also might want to look up "Sample Size" while you're at it.
2009-09-24, 12:28 PM #50
Originally posted by Cool Matty:
It doesn't matter. For the doors (as I'm not keeping up with your hurr cup example), if you pick a door with a yak, and you switch, you will win. Period. Doesn't matter which yak you pick, you still win. The only way you can lose is if you picked the car at the beginning.

Alco might not realize that the host knows which door has the car, and isn't going to pick that one (he does not want to give it away to the player, the player must guess). Thus, if the player picks a yak, the host must show the yak. At this point, if the player switches, he will win.

However, if the host forgets which door has the car, it becomes a 50% chance.
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2009-09-24, 12:30 PM #51
Originally posted by Alco:
The Monty Hall problem [CENTER] Keep choice: 10 times Wins: 5 cars (50%) Losses: 5 goats (50%) [/CENTER]
[CENTER] Change choice: 0 times Wins: 0 cars (0%) Losses: 0 goats (0%) [/CENTER]

The Monty Hall problem [CENTER] Keep choice: 1300 times Wins: 428 cars (32%) Losses: 872 goats (68%) [/CENTER]
[CENTER] Change choice: 1300 times Wins: 871 cars (67%) Losses: 429 goats (33%) [/CENTER]
2009-09-24, 12:32 PM #52
HEY LOOK AT THAT

The Monty Hall problem [CENTER] Keep choice: 1000 times Wins: 334 cars (33%) Losses: 666 goats (67%) [/CENTER]
[CENTER] Change choice: 1000 times Wins: 679 cars (67%) Losses: 321 goats (33%) [/CENTER]
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2009-09-24, 12:42 PM #53
Originally posted by Darth:
Law of Large Numbers. Look it up.

Also might want to look up "Sample Size" while you're at it.


I know, I just thought it was funny the first time that I ran it that it came back as 50%. Interesting link. For sample size of 100, it would show 2/3 win ratio for changing choices about 70% of the time. For the smaller sample size of 10 it would give a 2/3 ratio only about 50% of the time. However, the question doesn't suggest that you will have 100 of attempts.

Doing three single attempts, I won 2/3 of the time by keeping. However doing three sets of three single attempts did reveal ~ 2/3 win ratio for changing choices.

If the host must always choose a Yak to be revealed (as the game illustrated), then yes I would agree with the 2/3 win strategy by switching as the HOSTS CHOICE does influence the outcome.
2009-09-24, 12:45 PM #54
Originally posted by Alco:
However, the question doesn't suggest that you will have 100 of attempts.

...
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2009-09-24, 12:45 PM #55
I was going to post :carl: there, but I was too :carl: to be able to do it
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2009-09-24, 12:46 PM #56
It was meant as sarcasm.
2009-09-24, 1:03 PM #57
Star Wars: TODOA | DXN - Deus Ex: Nihilum
2009-09-24, 10:59 PM #58
Originally posted by Detty:
Therefore, if you always follow a strategy of switching, your 2 in 3 chance of originally picking a yak gives you a 2 in 3 chance of winning the riches instead. If you follow the strategy of not switching, the best you can hope for is your original 1 in 3 chance of guessing correctly.


But one of the choices was eliminated, the original choice is no longer 1 in 3, it's now 1 in 2.
You can't judge a book by it's file size
2009-09-24, 11:07 PM #59
Originally posted by Emon:
Alco might not realize that the host knows which door has the car, and isn't going to pick that one (he does not want to give it away to the player, the player must guess). Thus, if the player picks a yak, the host must show the yak. At this point, if the player switches, he will win.
.


ah, see, now that makes sense.
I was running under the assumption it was a fair game, but if the host does have prior knowledge then yes, switching makes sense.
Unless it's a bluff >_>
You can't judge a book by it's file size
2009-09-24, 11:12 PM #60
I like to think of the problem as a game. Remember: Monty, with the power to see behind the doors, must always reveal a single goat in the second round.

Your first decision has three possible outcomes. Two of those outcomes are goats.

Monty must reveal a goat, but two thirds of the time he is forced to reveal the only remaining goat, since you've already eliminated the other one.

This means that most times, switching after Monty has been forced to reveal a goat ensures that you are switching to the door Monty was unable to open, the car.
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2009-09-25, 4:37 AM #61
Another way of looking at it:

label the three doors A, B and C. A is the one with the car/riches/prize behind it.

You pick one of the doors, each with probability 1/3.

If you pick door A, Monty can reveal door B or C, each with probability 1/2.

If you pick B or C, Monty must reveal the other of B / C, with probability 1

So you have four possible outcomes:

1) P(You pick A, Monty reveals B) = 1/3 * 1/2 = 1/6
2) P(You pick A, Monty reveals C) = 1/3 * 1/2 = 1/6
3) P(You pick B, Monty reveals C) = 1/3 * 1 = 1/3
4) P(You pick C, Monty reveals B) = 1/3 * 1 = 1/3

Switching makes you lose in cases 1 and 2 with probability 1/6 + 1/6 = 1/3
Switching makes you win in case 3 and 4 with probability 1/3 + 1/3 = 2/3


In other news, the plane takes off and 0.9 recurring equals 1.
2009-09-25, 6:09 AM #62
Originally posted by JediKirby:
Remember: Monty, with the power to see behind the doors,


Yeah, see, that's the all important bit that wasn't mentioned that wasn't mentioned in the original post.
You can't judge a book by it's file size
2009-09-25, 6:34 AM #63
Yeah nice one Pommy, Monty Python and the Yak :downswords:

:P
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2009-09-25, 6:44 AM #64
It's easier to explain it like this... imagine instead of 3 doors there are 10. 9 have a goat behind them, and one has the car. You have a 1/10 chance of picking the car.

You choose Door #2. Monty then opens doors #3-10, revealing goats (he knows where the goats are). This leaves doors #1 and #2. The probability that you chose the car door is still 1/10 - you made the choice when there were 10 options.

But now that the other doors have been eliminated, you know that Door #1 has a 9/10 chance of being the winner - because unless you chose correctly the first time (1/10), the door you picked hides a goat (9/10).
2009-09-26, 10:47 AM #65
This title stinks considering this is first time I ever saw this problem and Pommy didn't provide all of the correct information required.
2009-09-26, 12:11 PM #66
When I was taking statistics, I remember the professor saying that a lot of students were being asked this question in job interviews and getting it wrong.
If you think the waiters are rude, you should see the manager.
2009-09-26, 1:49 PM #67
This makes my head hurt.
2009-09-26, 2:38 PM #68
Originally posted by Michael MacFarlane:
When I was taking statistics, I remember the professor saying that a lot of students were being asked this question in job interviews and getting it wrong.

Yeah, I can see that. The thing is, even experienced statisticians get it wrong. It has nothing to do with knowledge about statistics, in a sense. Our minds are just generally bad at this type of problem, no matter how smart you are.
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2009-09-26, 2:51 PM #69
I think some people also refuse to pay attention to explanations and insist that it's 50/50. I know several VERY intelligent people that still think this problem is bull****.
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2009-09-26, 3:11 PM #70
I thought this was interesting.... IMO Alco is just looking at it differently.

The coin flip I think is easier to understand. You flip it, it's head's. The next flip, obviously it's still a 50/50 chance of being heads or tails.. the probability of it being heads though goes down. Flip it again and it's heads, the probability goes down again... because the chances of getting heads so many times consecutively is a low probability.

Statistics take the whole equation into account... Alco is taking each question and showing the probability. Each coin flip is 50/50 chance, thus it is a 50/50 chance. Irregardless of previous flips.

Does this make me a hurr? I do see the logic in the Monty Hall question... To me though, it doesn't seem like it changes whatever door the money is behind.
2009-09-26, 3:52 PM #71
Originally posted by Citizen86:
The coin flip I think is easier to understand. You flip it, it's head's. The next flip, obviously it's still a 50/50 chance of being heads or tails.. the probability of it being heads though goes down. Flip it again and it's heads, the probability goes down again... because the chances of getting heads so many times consecutively is a low probability.


this is wrong... so wrong...
(or you're not explaining yourself very well)
2009-09-26, 4:21 PM #72
The doors are not like the coin flips at all in any way shape or form.
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2009-09-26, 5:06 PM #73
I cheated because we did this problem in my statistics class last semester so I already knew the answer. But I enjoyed reading the thread. Sounds exactly like the conversation we had in class till we ran it through the computer simulation.
Life is beautiful.
2009-09-27, 6:01 AM #74
Originally posted by Citizen86:
I thought this was interesting.... IMO Alco is just looking at it differently.

The coin flip I think is easier to understand. You flip it, it's head's. The next flip, obviously it's still a 50/50 chance of being heads or tails.. the probability of it being heads though goes down. Flip it again and it's heads, the probability goes down again... because the chances of getting heads so many times consecutively is a low probability.

Statistics take the whole equation into account... Alco is taking each question and showing the probability. Each coin flip is 50/50 chance, thus it is a 50/50 chance. Irregardless of previous flips.

Does this make me a hurr? I do see the logic in the Monty Hall question... To me though, it doesn't seem like it changes whatever door the money is behind.


I'm not sure which sentence you mean to be true.. Each coin flip is an independent event, and doesn't affect subsequent coin flips (other than maybe in some mechanical way, not a statistical way). It may well be unlikely to get 10 heads in a row, but if you've had 9 heads in a row the outcome of the next coin toss is still 50/50.
"The trouble with the world is that the stupid are cocksure and the intelligent are full of doubt. " - Bertrand Russell
The Triumph of Stupidity in Mortals and Others 1931-1935
2009-09-27, 8:58 AM #75
Right, that's what I was trying to say... statistically if you flip heads, the chances of heads coming up again goes down. If you flip heads again, the chances go down again even more, and so on. If you look at the coin flip as an individual coin flip though (like the first time), you could say it's still a 50/50 chance. In a mechanical way, I guess as you say Mort-Hog.
2009-09-27, 9:42 AM #76
Originally posted by Citizen86:
Right, that's what I was trying to say... statistically if you flip heads, the chances of heads coming up again goes down. If you flip heads again, the chances go down again even more, and so on. If you look at the coin flip as an individual coin flip though (like the first time), you could say it's still a 50/50 chance. In a mechanical way, I guess as you say Mort-Hog.


This isn't right. It's independent probability; the probability of the next flip coming up heads is 50% whether you look at it as part of a series of flips or as a single flip.
If you think the waiters are rude, you should see the manager.
2009-09-27, 10:04 AM #77
Yeah, it's a bit confusing

The probability that you get x heads in a row is (1/2)^x, but the probability that your next flip will be a heads will be 1/2 regardless of what x is
一个大西瓜
2009-09-27, 10:11 AM #78
Michael, that was the point I was making... Alco was looking at it as a single flip regardless of previous flips, but statistics looks at it as a series of flips. It is confusing, and I don't think either are wrong... the more you flip heads, the less chance the next flip will be heads... but it's still going to be either heads or tails...
2009-09-27, 10:13 AM #79
Originally posted by Mort-Hog:
(other than maybe in some mechanical way, not a statistical way)


This strikes me as an unusual way of saying "our idealized statistical model (which assumes that coin flips don't affect each other at all) may not be a perfectly accurate description of reality". You make it sound as if "mechanical" and "statistical" effects are phenomena of the same kind.
2009-09-27, 10:14 AM #80
Originally posted by Citizen86:
the more you flip heads, the less chance the next flip will be heads


THIS IS WRONG
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