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Topic: MONTY HALL SHOW (Read 14407 times) |
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Nyargle
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I think the easiest way to understand the 1/3 - 2/3 versus the 1/2 - 1/2 is to imagine Monty DOESN'T show you what's behind the door. Instead, he lets you pick a door, then he says "OK, you can go with your choice, or you can have both the OTHER doors. If you switch, and the prize is behind either door, you win it"
Since he always shows an unchosen goat, this is effectively what he's doing. Your original choice = 1/3, switching = 2/3
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TimMann
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Re: MONTY HALL SHOW
« Reply #26 on: Aug 31st, 2002, 5:08pm » |
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on Aug 23rd, 2002, 12:29pm, Robert Grimsley wrote:The basic problem has the solution that most posters have mentioned - it's twice as good to switch as to stay.
However, there is an unstated assumption about the basic problem. It assumes the host of the show will offer the 'enticement to switch' either all the time, or on a random basis. Someone asked Monty Hall how this worked back on the real game show "Let's Make A Deal". ... |
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Quite right, and well explained. To be just slightly more specific, the unstated assumption is that Monty's decision about whether or not to offer the enticement is a random variable that's independent of the random variable of whether or not your intial choice was right. (The case where Monty always offers you the enticement is a subcase of that, where his decision is "yes" with probability 1.) Typically when you're given a probability problem and it's not stated whether the different random variables involved are independent, you assume they are, which would make 2/3 the right answer. But (as you note), that independence assumption would be wrong if you were actually on Let's Make a Deal with the real Monty Hall.
The people who insist on 1/2 as the answer can also justify themselves with a different independence assumption: if they assume that Monty's decision as to which if any of the two remaining doors to open is independent of where the prize is. In real life Monty knows which door has the prize and doesn't choose to open that one, but the puzzle (at least as stated here) doesn't say so explicitly. Thus although it's an assumption that doesn't match real life well, someone might make it and then legitimately get an answer of 1/2. Of course, this line of reasoning doesn't mean that all the 1/2 people are right, since most of them don't give a valid argument for their answer. Most of the 1/2 people I've seen will continue to insist on that answer even if you explain that Monty knows which door has the prize and never opens it in this situation.
For more on this puzzle, including quotes from the real Monty Hall, see http://www.wiskit.com/marilyn.gameshow.html and
http://www.dartmouth.edu/~chance/course/topics/Monty_Hall.html.
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| « Last Edit: Aug 31st, 2002, 5:33pm by TimMann » |
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Zaius
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Here's why I feel switching should be right, but yet I can't prove it to myself.
Why I think it should be 50% (and yet have a gut feeling it isn't):
Prize= P, Goat = G.
Ok assume you've chosen a random door. Now Monty is left looking at 2 doors. Here are all the possible combinations (from left to right): GG, GP, PG. I think it is safe to assume that the probability that each of these combinations come up is 1/3 each.
Ok now say he choses the lefternmost door (out of the two remaining ones of course). That rules out PG. So now you know that when he chose that lefternmost door, he basically made it clear that the only remaining door (excluding of course the one you picked) is either a P or a G. I see no reason why P would be more likely to be behind that door than G at this point.
If he choses the righternmost door, you can eliminate GP this time, and you still seem to get an equal likelihood for P or G to be behind the last door (the one neither picked by you or Monty).
Why am I wrong?
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TimMann
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Re: MONTY HALL SHOW
« Reply #28 on: Nov 16th, 2002, 11:30am » |
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You're wrong because you're assuming that in any experiment, once you've gotten down to two possibilities, they must be equally likely.
Suppose you play the game with a deck of cards instead. The game works as follows: The Ace of Spades is the only winning card. You will pick one card from the deck without looking at it and leave it face down. I will sort through the rest of the deck and turn 50 of the remaining cards face up. I will look at the cards as I do so to make sure I don't turn the Ace of Spades face up, but I won't show you the one that I leave face down. I will keep that card.
After we play the game up to this point, you and I will each have one card face down. The only possibilities are that I have the Ace of Spades or you have the Ace of Spades. What is your probability of having won? Do you think it's 1/2?
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Zaius
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Ya I understand now how it is 2/3 to switch. The best explanation was by Nyargle:
I think the easiest way to understand the 1/3 - 2/3 versus the 1/2 - 1/2 is to imagine Monty DOESN'T show you what's behind the door. Instead, he lets you pick a door, then he says "OK, you can go with your choice, or you can have both the OTHER doors. If you switch, and the prize is behind either door, you win it"
Since he always shows an unchosen goat, this is effectively what he's doing. Your original choice = 1/3, switching = 2/3
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Rodrick Crider
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I have met several critics of the monty hall problem. I have come up with two easy ways of explaining the solution, and I have not yet found someone who I could not convince with these explanations. Someone may have already posted these explanations, but I didn't read through all the posts. Here it goes:
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Explanation # 1)
If you don't switch doors, you are betting that you made the correct choice the first time. Probability is 1/3.
If you do switch the doors, you are betting that you did not make the correct choice the first time. Probability is 2/3.
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Explanation #2)
If you don't switch doors, you are betting that you made the correct choice the first time. Probability is 1/3.
If you do switch the doors, you have exhausted all the possible outcomes, therefore the probability is 1 - 1/3, which equals 2/3.
-------------------------------
If neither of those explanations can convince someone of the correct answer, then you shouldn't be having this conversation with them.
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KAuss
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Re: MONTY HALL SHOW
« Reply #31 on: Nov 25th, 2002, 11:21pm » |
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easy to explain... the CATCH to this question here is the fact that the host KNOWS which door you picked.....
(The following examples assumes monty CANNOT open the door you choose if monty is knowledgable of your pick)
use the 1 Million door example. But this time pick in your head. Monty now opens doors without knowing which one you picked... he opens 999,998 doors... What are your chances that the doors he opened will be the one you picked?
Your chances of your door being picked and opened as wrong is 999,998 / 1 Million and the chance to pick the last 2 wrong or correct door will be 2 / 1 Million... (as you see, the chance to pick it at first try will be very slim)
now do the same, but let monty know which door you picked, now what are the chances of him picking your door? ZERO, so your chances of KNOWING you're wrong is ZERO. Which brings us to the numbers of you picking the wrong proven door at first 0 - 999.998, the chance of you picking one of the last 2 doors? 1 / 2... Now think back, did monty not open this door be cause he was not allowed to? Or because you picked the right one? With the first example, you can clearly see if monty WAS ALLOWED to pick your door, monty probably would of. Now if you still think sticking to that door is right, then why not pick in your head instead of picking and letting monty know?
The catch is monty not being able to open the door you choose. If he was allowed to, then you would of lost already, and not even get close to repicking.
The trick to convincing someone is not to show them, but merely ask them what they don't understand... Half the time, they won't even understand what they don't understand... (note half the time is an estimate  )
If you get this explanation, then you can easily see the 3 doors one.
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jeu
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Re: MONTY HALL SHOW
« Reply #32 on: Dec 1st, 2002, 12:39pm » |
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Test
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jeu
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Re: MONTY HALL SHOW
« Reply #33 on: Dec 1st, 2002, 12:43pm » |
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Heres a simple C program that proves the solution
#include <iostream.h>
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>
#include <unistd.h>
#include <math.h>
#define ITERATE 100000 /* No of Times to Iterate */
enum { EMPTY, FILLED };
/**
* Retuns a random integer between min and max
*
*/
int rndtime(int min, int max)
{
timeval tv;
gettimeofday(&tv, NULL);
srand(tv.tv_usec);
return (int) ((rand() % (max - min) ) + min) ;
}
int main() {
int count = 0, switchsuccess = 0, initsuccess = 0;
while (count++ < ITERATE) {
int room[3] = {EMPTY,EMPTY,EMPTY}; // All Rooms are empty intial
ly
int actual = rndtime(0,3); //Randomly fill gold inot one of th r
ooms
room[actual]=FILLED;
int montyshow=-1;
int initpick = rndtime(0,3); // Simulate ur initial pick
while(1) { // Simulation of Monty' pick
montyshow = rndtime(0,3); // Pick a random room
// This randomly picked room should not be the either th
e one
// that you picked or the one that actually has the gold
if(montyshow != actual && montyshow != initpick) break;
}
// Simulate a switch: Choose the room that is different from the room
// that you picked initially and also differnet from the one th
at Monty SHowed you
int switchpick = 3 - initpick - montyshow;
// Lets see how many times a switch is a success
if(switchpick == actual) switchsuccess++;
if(initpick == actual) initsuccess++;
}
// Initially a the room that you picked had a 1/3 chance of winning;
// Monty then picked a room that has a zero chance of winning;
// which means that the other room has a chance of 2/3 to one
// Which means the number of times you siwtch should give the correct re
sult 66.66 %
//
// Runs this program to see waht the result is
cout << "success rate by switching = " <<((float) (float)switchsuccess/(
float)ITERATE) << endl;
cout << "success rate by not switching = " <<((float) (float)initsuccess
/(float)ITERATE) << endl;
return 0;
}
Success rate is approximatley .66 when you siwtch and .33 otherwise
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towr
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Re: MONTY HALL SHOW
« Reply #34 on: Dec 1st, 2002, 2:09pm » |
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I wonder why no one tried it with a simple table yet..
Where the prize is, and what door you choose is independant, unless you're clearvoyant. So every combination is equally likely.
choice\prize| A | B | C
------------------------------------
A | 0 | 1 | 1
------------------------------------
B | 1 | 0 | 1
------------------------------------
C | 1 | 1 | 0
------------------------------------
1 denotes when you should switch to win.
Clearly switching gets you the prize 6/9 = 2/3 of the time..
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| « Last Edit: Dec 1st, 2002, 2:11pm by towr » |
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Patient of Dr Whitey
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I dont really understand why this is difficult at all. Simply choose the door that the host showed you. You are then GAURANTEED to win the goat. Do this 3 times and you can use them in the 2 condoms 3 "women" riddle.
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ghjgj
Guest
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sticking will support your guess. meaning you are confident in yourself.
so you could think of it like this.
would you rather bet £1 your lottery ticket is the big jack pot winner or £1 your ticket will lose? if its lose switch to the win side 
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Paul Brockman
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May I be so profound to say that both sides of this arguement are right. It all depends on what question you ask. The orginal question asks what is the probabilty of winning for each case, to swich or not to swich. The answer is that the odds are 2/3 and 1/3 respectivly. Now, ask a different question. What are your chances of winning the game all togeher? In this case the odds are 1/2. Proof: Go back to the computer simulations that prove the 1/3, 2/3 odds. If you did 99 iterations where you kept the same door and 99 iterations where you swiched probability dictates that you should get 33 wins during the first 99 iterations and 66 wins during the second 99 iterations. This means that you won 99 out of 198 times, 1/2.
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Chronos
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Re: MONTY HALL SHOW
« Reply #38 on: Dec 13th, 2002, 2:24pm » |
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No, that's the probability of winning altogether if you switch as often as you don't. But why would you do that?
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Bryan Cannon
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Of course, this will just be yet another way of stating what's already been said, but it has been the most effective (and simple) way for me to explain this.
There are only three possibilities of how this plays out:
1) You choose the right door -> Monty shows a goat -> you switch and lose.
2) You choose wrong door #1 -> Monty shows the other wrong door -> you switch and win.
3) You choose wrong door #2 -> Monty shows the other wrong door -> you switch and win.
Therefore, 2 out of 3 times, you win by switching.
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Kevin
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This problem can be easily solved with conditional probability.
Let
s1 = "Picking the winning prize initially"
s2 = "Winning the prize after switching"
Then,
P[s1] = 1/3
P[s2 | s1] = 0
P[s2 | not s1] = 1
Therefore,
P[s2] = P[s2 | s1]*P[s1] + P[s2 | not s1]*P[not s1] = 2/3
So your chance of winning the prize after switching is 2/3. You can perform similar operation to determine that you have 1/3 chance of winning if you stay with your first choice.
If the Monty does not show one of the doors, P[s2 | not s1] now equals to 1/2 instread of 1. In this case, your chance of winning the prize after switching, P[s2], is 1/3.
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Frank
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You know there is a lot easier explanation for this. The answer is 1 in about 10,000 will win regardless if they switch or not. I mean come on. Do you really think there is a million dollars sitting behind a door. No!!! Not only that, but there is only one goat! After the contestant picks a door they move their one goat behind one of the other doors and open that door. After they close that door they move the goat between the other two doors. After the contestant decides to keep the door they picked the same goat is moved behind that door. If the contestant switches doors then thy move the goat behind the door that the contestant switched to. Either way YOU LOOSE!!! If it were 2/3, 1/2, 1/3 or even 1/10 how many shows do you think there would be before they went bankrupt giving away $1,000,000 every few days?
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rmsgrey
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Re: MONTY HALL SHOW
« Reply #42 on: Apr 15th, 2003, 11:23am » |
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Someone mentioned conditional probabilities, so (taking the unconventional version of the problem where you read Monty's script before the show and saw that it said "open an unchosen door to reveal a goat"):
P(Monty reveals goat|right initial choice)=1
P(Monty reveals prize|right initial choice)=0
P(Monty reveals goat|wrong initial choice)=1
P(Monty reveals prize|wrong initial choice)=0
This compares to the "normal" situation where Monty lets you pick which door is opened second (the script didn't say which door to pick):
P(Monty reveals goat|right initial choice)=1
P(Monty reveals prize|right initial choice)=0
P(Monty reveals goat|wrong initial choice)=0.5
P(Monty reveals prize|wrong initial choice)=0.5
Without bothering to write out the full calculations, I think that highlights where the two cases differ (which is the usual problem for people who can't see why it's not the 0.5 they were taught)
In the latter case, seeing a goat gives you new information (it's more likely your first choice was right than if you hadn't seen the goat behind the second door because you had the chance to be wrong). In the former, seeing a goat doesn't tell you anything (you already knew there would be a goat there...)
By the way, I've used the 52 cards Ace-of-spades variant a few times myself. Running it ten times is usually pretty convincing - particularly if you mention that the chances of a 50-50 chance coming up all in favour of the card not first selected 10 times in a row are about 0.1%... of course, if one does come up you have to use 2% as your figure, which is a little less impressive. If it comes up twice, you start suspecting that the cards may be marked somehow... Speaking of which, one of my friends (I hope hypothetically) proposed a chat-up technique involving getting his target to pick a card from a pack, and then tell her with full confidence that it was the ace of spades. He reckoned the pay-off from the (extremely) rare success would make the trick's expected value positive (assuming she hadn't heard about it in advance)...
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Boody
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Re: MONTY HALL SHOW
« Reply #43 on: Apr 16th, 2003, 5:52am » |
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Lot of posts for an easy riddle.
Because of irreductible people, I propose to make a vote for the good solution.
I vote for the "switching give us 2/3 chances to win" solution.
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Rejeev
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Re: MONTY HALL SHOW
« Reply #44 on: Oct 7th, 2004, 9:24pm » |
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on Aug 8th, 2002, 11:39am, B wrote:| Or, think of it this way: instead of three doors, suppose there are one million. You pick a door, Monty opens all but one of the remaining doors. The prize is either behind the door you picked or the door he didn't open. Which door do you think has the prize? |
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Very good reasoning! Another angle of viewing this problem is: changing the choice after monty open one door is equvalent to: we win if the price is behind any of the two doors. Consider the a dry run: suppose we choose door 1. first and then we change the choice after monty reveals.
Case1: price is in Door 1:- we loose
Case2: price is in door 2:- Monty opens door3 and we choose door2 and wins
case3: price is in door3:- Monty opens door2 and we choose door3 and wins
That is we win 2 out 3 cases.
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rfh
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on Aug 9th, 2002, 6:27pm, Andrew Liao wrote:the odds are 2/3 if you switch, 1/3 if you dont.
here's the simple logic: if you pick one door arbitrarily your odds are 1/3 for being right (one door out of three). that means 2/3 chance you're wrong (2 other doors). however monty gets rid of the wrong door from the 2/3 (2 other doors) so when you pick to switch youre essentially picking the option thats 2/3 (the 2 other doors). Think of it as getting to pick 2 doors. RESTATED: its 2/3 odds your initial guess was wrong. thus if you pick one of those 2 doors your chance would be 1/2 of 2/3. but since monty says which is wrong. its the full 2/3."
This is essentially the probablity compression argument
which is frequently used but not correct. The correct
way to state it is as follows. Pick a door (say #1).
1/3 of time that door wins if you stay. By always switching you will not get the prize that 1/3 of the time but you will pick up the prize the 1/3 of the time it is behind 2 and the 1/3 of the time it is behind 3. So out of all the trials you
will get the prize 2/3 of the time it is offered. Since the prize
is offered every round you will win 2/3 of the time.
That does not mean that switching to, say door 2 is 2/3.
Suppose for example monty never opens door 3 when the
prize is behind 1. Switching to door 2 become 100% whenever door 3 is open even though the contestant still only gets the prize 1/3 of the total times it is offered by switching to door 2.
its particularly obvious if you make a decision chart yourself and draw out possibilities:
the right door must be 1,2, or 3
stick: you pick door 3. so odds = 1/3 because the right one could be 1, 2 or 3. the fact that monty tells you one is wrong doesn't matter because you assumed that in your choice anyway.
change:you pick door 3 but switch. then your odds go up to 2/3 because now if its behind 1 or 2 youll get the prize since monty gets rid of the choice (from 1 or 2) thats wrong.
sorry for being so verbose =). |
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b_ecstasy
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Haha, that's funny. Yes, the answer is 2/3. I have two ways of looking at it, and they've both been explained, but I'll try to visually demonstrate one of them more clearly.
Suppose x = prize (x marks the spot!) and o = goat.
There are three scenarios:
X O O
O X O
O O X
Okay, we know that we will choose the prize 1/3 of the time on the first try, so assume that we choose the first door in all three scenarios. We will get the prize 1/3 of the time, as the probability has shown in the long run. I'm just short circuiting the process. Okay, so the first door in each scenario is chosen, and Monty opens a door. - denotes open.
X - O
O X -
O - X
So two out of the three scenarios, you should switch.
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Yobgal
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Re: MONTY HALL SHOW
« Reply #47 on: Jan 12th, 2006, 11:04pm » |
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I'm a wordsmith, not a mathematician. I won't begin to claim any sort of math knowledge, so I think I can sympathize with the people claming 1/2. Since I haven't seen any attempt at this explanation on this thread, I will try. The final question is simply worded to leave out information.
Quote:| What is the probability of winning the car if she stays with her first choice? What if she decides to switch? |
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What if it was asked in these words?
Quote:
There are three closed doors. One of the three doors hides a car. She doesn't know what any of the three doors hides. She chooses one of the three doors. What is the probability that she wins the car?
There are three closed doors. She knows that a goat is behind the first door. One of the three doors hides a car. She chooses one of the three doors. What is the probability that she wins the car? |
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She began with three doors. She chose one door. She now has the option of selecting a different door of the three doors. One door has been opened, but it is still there. Common sense that you can't possibly get one out of two when there are three choices. Compare that puzzle to this:
Quote:I have three coins on a table, each completely covered and by itself. Exactly one of those coins is a nickle. What are the odds that you select a nickel?
You incorrectly choose one of those coins. I put that coin in my pocket. I now have two coins on a table, each completely covered and by itself. Exactly one of those coins is a nickel. What are the odds that you select a nickel? |
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In those questions, the answers are 1/3 and 1/2, respectively. I hope this explanation is more clear for some people.
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Zero
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Re: MONTY HALL SHOW
« Reply #48 on: Jan 14th, 2006, 2:09pm » |
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oh my gosh! I can't believe this was so complicated!
I have to admit I was a hardcore 50% fan when I first saw the riddle.. but you folks made it sound so confusing I actually thought about it
The millions of doors example makes it so obvious. What are the chances you picked the prize? 1 in a million! What are the chances Monty has the prize? Well if YOU don't have it, he does!
----------
I think this is the same kind of rationale problem as when people say... I've flipped a coin 9 times and it's come up tails. Assuming it's a fair coin, what are the chances of it coming up heads? Technically it's 50%... but considering recent history I'd bet my money it's coming up heads next time...
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towr
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Re: MONTY HALL SHOW
« Reply #49 on: Jan 15th, 2006, 7:11am » |
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on Jan 14th, 2006, 2:09pm, Zero wrote:| I think this is the same kind of rationale problem as when people say... I've flipped a coin 9 times and it's come up tails. Assuming it's a fair coin, what are the chances of it coming up heads? Technically it's 50%... but considering recent history I'd bet my money it's coming up heads next time... |
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It wouldn't be a fair coin if it was influenced by it's history, though.
And a sequence of 9 heads should happen once every 512 runs, on average.
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