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        riddles >> easy >> MONTY HALL SHOW
        
(Message started by: Arvin Farahmand on Aug 7th, 2002, 10:38am)
Title: MONTY HALL SHOW
Post by Arvin Farahmand on Aug 7th, 2002, 10:38am
Maybe I'm missing something, but it seems like a simple solution.

Since all Monty does is eliminate one of the 3 choices, he is merely increasing the chances of one of the other doors containing the prize (and not a goat) from 1/3 to 1/2.

Initialy all doors have an equal probability of containing the prize - 33%.

All he does is take one of the doors out of the equation. Let's say you pick door 2, and he shows you a goat behind door 1 thus leaving you with the following probability:

1    2   3

G    ?   ?

so we are now dealing with this:

2   3

?   ?

which means your pick, 2 or 3, can be correct 50% of the time. So either way the probability stays the same.

Am I missing something??
Title: Re: MONTY HALL SHOW
Post by Mike Peterson on Aug 7th, 2002, 11:09am
When intuition fails, use brute force trials!

Here's a simple window's script that shows you the results of switching and not switching: (I'll be switching myself  :) )

Const NUM_TRIES = 1000 'Do as many as you like!

Call PlayWithMonty(True) 'Switch doors each time.
Call PlayWithMonty(False)  'Don't switch doors each time.

Sub PlayWithMonty(bSwitch)
     dim prize, pick1, pick2, shown, sResult
            dim lWin, lLoss, try
     
     for try = 1 to NUM_TRIES
           prize = GetRandom() 'Random from 1 to 3
           pick1 = GetRandom()
           shown = GetOtherDoor(prize, pick1) 'Get the door Monty Shows.
           
           if bSwitch then
                 pick2 = GetOtherDoor(pick1, shown) 'pick2 is what we switch to.
           else
                 pick2 = pick1 'Stay with pick1.
           end if
           
           if pick2 = prize then
                 sResult = "WIN"
                 lWin = lWin + 1
           Else
                 sResult = "LOSS"
                 lLoss = lLoss + 1
           end if
           
           'un-comment this if you run under cscript.  (you'll get lots of pop-ups under wscript though)
           'wscript.echo "TRIAL #:" & try & ", PRIZE=" & prize & ", PICKED=" & pick1 & ", SHOWN=" & shown & ",SWITCH_TO=" & pick2 & ", RESULT=" & sResult
     
     next
     
     wscript.echo "TOTAL WINS:" & lWin & ", TOTAL LOSSES:" & lLoss & ", SWITCHING?=" & bSwitch
end sub

function GetRandom()
     Randomize
     GetRandom = Int(3 * Rnd + 1)
end function

function GetOtherDoor(door1, door2)
     dim i
     for i = 1 to 3
           if i <> door1 and i <> door2 then
                 GetOtherDoor = i
                 exit function
           end if
     next
end function
Title: Re: MONTY HALL SHOW
Post by Russ Arcuri on Aug 7th, 2002, 2:21pm

on 08/07/02 at 10:38:32, Arvin Farahmand wrote:
Maybe I'm missing something, but it seems like a simple solution.

Since all Monty does is eliminate one of the 3 choices, he is merely increasing the chances of one of the other doors containing the prize (and not a goat) from 1/3 to 1/2.

[...]

Am I missing something??


Yep, you're missing something. ;)

Think of it this way: Your first choice of three doors has a probability of 1/3 of being correct.  This means that the probability of one of the two unchosen doors containing the prize is 2/3.  Now Monty shows you a goat from behind one of the unchosen doors, eliminating it as a possibility for you.  Since this happened after you made your initial selection, it cannot possibly change the probability of your first choice being correct -- it's still 1/3.  But since Monty has eliminated a bad choice for you from the other two, the remaining unchosen door now has a 2/3 probability of having the prize.

Still not convinced?  Think of it this way: Pretend this game was going to be played 999 times by 999 different people.  How many people are going to pick the prize on the first try?  Approximately 333 of them.  This leaves 666 people who picked one of the goat doors on the first try.

Now Monty must show *everyone* a goat from behind one of the two unchosen doors.  For the 333 people who picked the prize on their first choice, this is easy -- he can open either of the other two doors to reveal a goat.  But for the 666 people who picked a goat on their first try, he has to be careful not to open the door containing the prize -- he opens the other one, revealing a goat.

Now assume that EVERYBODY switches.  333 people will be upset, because they had picked the prize the first time but switched to a goat.  However, 666 people will have switched from a goat to the prize.

Hope this helps.
Russ
Title: Re: MONTY HALL SHOW
Post by B on Aug 8th, 2002, 11:39am
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?
Title: Re: MONTY HALL SHOW
Post by JW on Aug 9th, 2002, 8:24am
I believe the answer is 50/50 regardless of whether you switch or not.  You see, all Monty is doing is creating a new problem.  With the original problem you picked 1 door from 3 options so you have a 1/3 chance of picking the correct door.  When Monty eliminates one of the doors (of course this is all assuming he doesn't eliminate the door you choose) he is creating a new problem where you pick 1 door from 2 options so you have a 1/2 chance of winning the prize regardless of the door.  This is true for all of the examples above (even when there are a million different doors) since he gives you the option of switching AFTER he has shown (by default) that the door you initially picked is one of the two remaining doors that could contain the prize.
Title: Re: MONTY HALL SHOW
Post by Arvin Farahmand on Aug 9th, 2002, 10:38am
I see your logic with the 1/3, but I'm still not convinced.

The way I see it, when Monty reveals the door with the goat and asks you to change your choice if you want, he is giving you a new problem.

It may seem like its not a new choice but it is. He is saying ok, now pick one of the two doors. Either you keep the one you already picked (as in re-pick it) or pick the other one. So the problem, IMHO, is that you are picking one of two doors, behind either of which maybe a prize so what is the probability? 1/2.

There are two problems. First, pick one of three doors, then pick one of two doors. For the latter problem the probability is 1/2.

Clearly the goat that Monty reveals is not behind the door that I picked, so I still don't know what is behind the other two doors.

Of course, all of this is assuming that Monty would reveal a goat to me regardless of if my pick is the prize or a goat. Meaning that the fact he is giving me a choice, isn't that there really is a prize behind that door and he wants another oppertunity for me to pick the wrong door.
Title: Re: MONTY HALL SHOW
Post by Jonathan_the_Red on Aug 9th, 2002, 11:26am

on 08/09/02 at 10:38:12, Arvin Farahmand wrote:
Of course, all of this is assuming that Monty would reveal a goat to me regardless of if my pick is the prize or a goat. Meaning that the fact he is giving me a choice, isn't that there really is a prize behind that door and he wants another oppertunity for me to pick the wrong door.


This is precisely why the switching improves your probability to 2/3, because Monty must show you a goat. If Monty opened a door randomly (with the contestant losing if Monty revealed the prize), your odds of success after Monty revealed a goat would indeed be 50/50. But because Monty's choice is limited, you gain information when he opens a door.

There really are many ways to prove this, but I've always liked the "million door" proof that B used. There are a million doors, with one prize. You pick one of these million doors, and then Monty reveals 999,998 goats. Now, keep in mind that whether your initial pick was the prize (highly unlikely) or a goat (very likely), Monty will always be able to show you 999,998 goats... but it's much, much more likely that your initial pick was incorrect and the sole door Monty left alone has the prize behind it.

Still unconvinced? Try it this way. Suppose, utterly certain that switching will have no affect on your chances, you tell Monty right off the bat that you won't switch, no matter what. Knowing that you won't switch, Monty opens both of the doors you didn't choose, instead of just one. What are your chances of walking away with the prize? Answer: 1/3. The strategy "never switch" has a 1/3 chance of success, and therefore the strategy "always switch" has a 2/3 chance.
Title: Re: MONTY HALL SHOW
Post by Mark Rowan on Aug 9th, 2002, 12:01pm
It is 50% The fallacy with the 1/3 argument is thinking of probabilities as absolute values when they are always relative.

If there are 3 unknowns, then yes each door has a 1/3 chance of being correct.  When Monty eliminates one door, the problem changes to 2 unknowns.  The probability is now 1/2 for each door.  There is no absolute property of "chance" that can transfer from the door that has now been removed from the problem.  The 1/3 probabilty was an abstract that only applied to the original problem of 3 unknowns.

The "always switch" arguments all follow this fallacy, often using examples of "start with x number of doors"  But try this, start with two doors, what are the odds? 50-50.  The probability of a choice between 2 unknowns.  No different than a coin toss.
Title: Re: MONTY HALL SHOW
Post by JW on Aug 9th, 2002, 12:32pm
Having 3 doors or 1,000,000 doors does not change the answer.  When you make your final decision you have two doors and a prize is in one of them.  Therefor your odds are 50/50.  

If you followed your million doors logic you are saying that when you get to 2 doors and you don't switch it is a 1/1000000 chance still - which obviously isnt the case.
Title: Re: MONTY HALL SHOW
Post by Jonathan Chaffer on Aug 9th, 2002, 1:15pm
Yes, it is better to switch.

This is a well known problem, and is one of those where it is very hard to convince intelligent people they are wrong when they are. :) Conditional probabilities stymie many people.

Google is your friend. Some references:
http://www.io.com/~kmellis/monty.html
http://www.cut-the-knot.com/hall.shtml
http://math.rice.edu/~ddonovan/montyurl.html

I'd go into it some more, but these (the top three Google results) all seem to cover it quite well.
Title: Re: MONTY HALL SHOW
Post by Jonathan_the_Red on Aug 9th, 2002, 2:53pm

on 08/09/02 at 12:32:36, JW wrote:
If you followed your million doors logic you are saying that when you get to 2 doors and you don't switch it is a 1/1000000 chance still - which obviously isnt the case.


Um, obviously it is the case. The odds are 1/1,000,000 when you pick the door. Monty opening 999,998 goat-filled doors gives you no new information about the door you picked, because he will always be able to open 999,998 goat-filled doors whether you pick the door with the goat or not.

Here's yet another way to look at it. Switching means you will win if and only if your initial pick was incorrect. Not switching means you will win if and only if your initial pick was correct. The odds your initial pick was correct are 1/3.

I've done my best. Now it's time for me to attempt to profit. Anyone who still believes the odds are 50-50 is invited to play the following game with me:

Take the ace of spades and two red aces from a deck of cards. The ace of spades will represent a prize; the other two goats. Shuffle the three cards and deal them face down. I will select one without turning it over. You look at the other two and turn a red ace face up. I will then switch to the third card -- the one that you didn't turn over. If it's the spade, you pay me $10. If it's a red ace, I pay you $12.

Since according to you, the odds are 50-50, this game is in your favor.  We play until one of us is down $500. Any takers? I'll pay your plane fare.
Title: Re: MONTY HALL SHOW
Post by Andrew Liao on Aug 9th, 2002, 6:27pm
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.

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 =).
Title: Re: MONTY HALL SHOW
Post by Russ Arcuri on Aug 10th, 2002, 12:58pm
JW and Mark Rowan are simply wrong, but it's going to be very difficult, if not impossible, to convince them.  Arvin appears to be wavering; think it through carefully and you'll see I'm right.  Many intelligent people get totally hung up on this one, either unable or unwilling to 'get it.'

There have been numerous excellent arguments put forth to explain this on this thread, and if none of them has convinced them yet, none likely will.  The only way to convince them would be to actually do the experiment with them -- and even that might not work.  I tried this with an engineer friend of mine, and even after the statistics bore me out (we did 30 trials, and in 19 of them he would have been better off switching), he insisted that the trials were simply a statistical anomoly.

There is a big difference between 'straight' probability and conditional probability.  Some people get it, and some people don't.  The fact is that nothing Monty does after you make your selection can change the probability that it was correct when you made it.  

I will make one final attempt, explaining things slightly differently.  Look at it this way: after you make your initial selection, which had a 1/3 chance of being correct, you are then being given a choice between sticking with your first pick or taking BOTH of the other doors.  Because Monty ALWAYS allows you to discard one goat and keep whatever is behind the other door -- in fact, he does it for you.  He's giving you the choice of your initial pick (with a 1/3 chance of being correct) and both other doors (which have a 2/3 chance of one of them containing the prize).  Monty is simply eliminating a goat for you ahead of time.

I doubt this explanation will convince you either, but it was worth a shot.  For those still unconvinced, I will be happy to liberate you from your money in a manner similar to how Jonathan_the_Red offered to.
Title: Re: MONTY HALL SHOW
Post by tot on Aug 10th, 2002, 3:32pm
My attempt to convince someone.

Since everyone agrees that the odds of the inital pick is 1/3, and since the probability of either of the closed doors honding the prize is 1, the odds of the other door must be 1 - 1/3 = 2/3.  What Monty does is not to give a new problem, he just moves all the remaining 2/3 odds to behind a one single door.  If you believe that the odds are 1/2, that must mean that your initial pick chance was 1 - 1/2 = 1/2.

The 1/2 odds would be true if you are presented just two closed doors out of three.  Your advantage is that you know that one of the doors was picked randomly from three doors, whereas the other door was eliminated by Monty to not contain a goat, thus giving the remaining closed door 2/3 chance being the correct one.

Extend the above examples to hundred or million doors, and it should be more obvious.  I remember when I was initially presented this problem, my gut feeling answer was 50-50, but extending the number of doors to extreme, it was immediately obvious.
Title: Re: MONTY HALL SHOW
Post by Kozo Morimoto on Aug 11th, 2002, 1:39am
Yes, the easiest way to see it, is to run a few thousand runs of a Monte Carlo simulation of the game.  Monte Carlo sims won't "prove" that its 2/3 to switch, but it will definitely support the 2/3 to switch answer.

Also, a lot of people miss the point that Monty NEVER opens the door with the prize.  His choice of a door is ALWAYS biased.

If, however, Monty's choice is purely random between the remaining 2 doors and he doesn't show you what's behind his door, then yes, switching won't alter the odds.  But his choice is NOT random, and he ALWAYS shows what's behind his door.

Another way to look at it is, you are given a choice of picking 1 door out of 3 (like the game show) or you are given a choice of picking 2 doors out of 3.  Which would you choose?  2 doors out of 3 would be obvious answer.  You know from the outset that when you pick 2 out of 3, one of the 2 won't have the prize.  And Monty will always pick THAT door out of the 2 that you picked.  You will still stick with the initial choice of the 2 doors 'cause you already knew that one of them had the donkey.
Title: Re: MONTY HALL SHOW
Post by AlexH on Aug 12th, 2002, 12:46am
The one problem with this well known puzzle is that the phrasing is often a bit too loose and the colorful use of Monty Hall is actually misleading. If the rules of the game are that Monty must always reveal a goat and offer you a choice then the always switch strategy is the way to go and pays off 2/3 as has been explained. However, on the actual Let's Make a Deal show Monty had options about what to do only one of which was revealing a door. In such a case the switch strategy can have payoff as low as 0.

This problem is actually used in experiments on people to analyze various behavior aspects. Not that many people realize at first that the switch strategy is right, but in the course of the experiment they can learn by observing the outcomes.
Title: Re: MONTY HALL SHOW
Post by jeremiahsmith on Aug 12th, 2002, 11:09pm
I believe that Marilyn vos Savant, who is supposedly* the world's smartest woman and has a riddle-ish column that sometimes appears in newspapers and the like, had a lot of problems with this one. People would tell her she was wrong when she said the odds were 2/3. But she's right, as are most of the people in this thread. Marvin Gardner had a similar problem with his Scientific American column, when he had a similar question.

* I'm not denying that she's smart, but it might be a bit of a stretch to say she's the smartest.
Title: Re: MONTY HALL SHOW
Post by Mongolian_Beef on Aug 13th, 2002, 8:43pm
the answer is 2/3 if you switch 1/3 if you dont. this is a very famous problem with a very famous solution. its amazing theres still debate.

think about it this way:

if you stick. you only get the prize if its behind the door you picked originally (you bet you were right on the first try)

if you change. you only fail to get the prize if its behind the door you picked originally. (you bet you were wrong on the first try)

what are the odds you were right on the first try? 1/3.

what are the odds you were wrong on the first try? 2/3.

its not 50-50 because that assumes you have no knowledge whatsoever about either of the two doors. the knowledge you DO have is that the door you didnt pick originally (the one you would switch to) was once part of a bigger subset of doors (2 doors not 1 door) that have been all opened except for one. thus as part of this bigger subset it has a greater chance to be correct.  
Title: Re: MONTY HALL SHOW
Post by LowIQ on Aug 14th, 2002, 12:07am
I use this method to explain to my friends. Think in EXTREME situations.

ie suppose you have 1,000,000 doors with 1 of them having the $ and the rest having goats. Now suppose after you picked yours, the contestant removes away 999,998. (this leaves two doors unpicked, the remaining one and the one you originally picked).

Q. Should you switch?

if put this way, you would definitely switch because the probability is not 1/2 and in fact is 999,999/1,000,000. The same theory applies to hte scenario if its 3
Title: Re: MONTY HALL SHOW
Post by schiefaw on Aug 15th, 2002, 11:44am
The answer is to switch. I didn't want to belive it either, but I proved it to myself through a computer simulation. They had this riddle on Car Talk and I thought they were nuts. It doesn't seem to make sense, but it works in simulation.

Tony
Title: Re: MONTY HALL SHOW
Post by Obgac on Aug 21st, 2002, 5:19am
I the answer can be both. If the door chosen by the quizmaster was randomly chosen (thereby also 1/3 chance of the already picked door been chosen) then nothing changes and assuming an unopened door was chosen, you are at 50% probability. Otherwise its a 2/3 probability.
Title: Re: MONTY HALL SHOW
Post by subx on Aug 22nd, 2002, 5:51pm
The probability is more than half, but it's not the full 2/3, because once you remove the other 2 there's still a larger chance that yours is right as well, and your given a chance to select BOTH doors equally, and at 3 doors, the number isn't enough (imo) to justify ALWAYS taking the other door, and having 2/3 ratio of getting it right, it would be somewhat below that.  

Even with the 1,000,000 problem, the chances that the other one is right is not the full amount, but slightly less (albeit, with a number that large, MUCH smarter to take the other door).

I know it sounds a little messed up, but think about it, but yes it still is the "better" choice to take the door not opened.
Title: Re: MONTY HALL SHOW
Post by Robert Grimsley on Aug 23rd, 2002, 12:29pm
The basic problem has the solution that most posters have mentioned - it's twice as good to switch as to stay.

For those that are having trouble with this concept, the most important aspect of the problem has to do with randomness.  The host of the show knows what is behind each door, and he is making NON-RANDOM decisions based on his knowledge.  Some of the assumptions one normally makes get thrown out the window due to this non-randomness.

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".  He said he was more likely to make 'enticement to switch' offers like this when the contestant had, in fact, chosen the best prize (once again, the all-knowing host is making non-random decisions).

Now one must factor this into the equation.  Some of the possibilities:

1) The host is making 'enticement to switch' offers completely randomly or all the time.  All the logic that has been previously discussed applies.  It is twice as good to switch.

2) The host only makes 'enticement to switch' offers when the contestant has chosen the best prize.  This one is real simple.  You have the best prize, always stay put.

3) The host makes 'enticement to switch' offers more often to contestants with the best prize, but sometimes does this with contestants who have the lesser prize to avoid being predictable  (this was the case in real life on the show).  There is a range of possibilities depending on what percentage of the time Monty is offering the enticement.  If he is offering it twice as often to contestants with the best prize, I believe this will exactly negate the 2-to-1 benefit of switching in case (1) above, so we're back to square one - it makes no difference if you switch or stay.  If he is making 'enticement to switch' offers at an even greater rate to contestants with the best choice already (say 75 percent of the time), now it becomes best to stay put.  If he is making 'enticement to switch' offers at a lower rate than 2/3 of the time (say 60 percent) now the best strategy swings back to switching, although the percentage improvement from this strategy is less than the 2-to-1 ratio of case (1) above.

Whew.

Someone at the Game Show Network should review the tapes and find out what the real percentage is regarding these switch offers.

Overall, if this show gets revived, I don't think understanding all this is going to do any future contestants much good, as I suspect the reality is pretty close to the 'it all washes out to even' scenario.

grimsler@bellsouth.net
Title: Re: MONTY HALL SHOW
Post by Russ Arcuri on Aug 27th, 2002, 11:54am

on 08/22/02 at 17:51:27, subx wrote:
The probability is more than half, but it's not the full 2/3, because once you remove the other 2 there's still a larger chance that yours is right as well, and your given a chance to select BOTH doors equally, and at 3 doors, the number isn't enough (imo) to justify ALWAYS taking the other door, and having 2/3 ratio of getting it right, it would be somewhat below that.  

Even with the 1,000,000 problem, the chances that the other one is right is not the full amount, but slightly less (albeit, with a number that large, MUCH smarter to take the other door).

I know it sounds a little messed up, but think about it, but yes it still is the "better" choice to take the door not opened.
*Sigh.*  The odds are EXACTLY 1/3 for non-switchers and 2/3 for switchers.  The odds for switching ARE the "full 2/3."

With the 1,000,000 doors scenario, the odds are 1/1,000,000 if you don't switch and 999,999/1,000,000 if you do.  Those are exact odds, easily calculated -- not guesses or estimates -- just like the 1/3 and 2/3 probabilities when you only have three doors.
Title: Re: MONTY HALL SHOW
Post by zarathustra on Aug 27th, 2002, 10:17pm
For non believers, heres a way to test the problem with 52 doors:
Shuffle a deck of cards.  Remove a card but dont look at it (you want it to be the ace of spades).  That is your first choice.  Now have someone else look through the deck and remove the ace of spades, or if its not there, then remove one other card.  That represents what monty did when he removed all the door with the goats.  Ok so now you still shouldnt havent looked at eigther card yourself.  Where is the ace of spades?  You'll learn prety quick that the ace of spades is almost always the card you didnt pick because your first choice has only a 1/52% chance of being correct, not a 50% chance.  However it is possible to "reset" the odds to 50% if you shuffel up the 2 remaining cards so you don't know which is which.  Hope that helps someone.
Title: Re: MONTY HALL SHOW
Post by Nyargle on Aug 28th, 2002, 6:53pm
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
Title: Re: MONTY HALL SHOW
Post by TimMann on Aug 31st, 2002, 5:08pm

on 08/23/02 at 12:29:52, 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".  ...


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.
Title: Re: MONTY HALL SHOW
Post by Zaius on Nov 16th, 2002, 1:55am
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?
Title: Re: MONTY HALL SHOW
Post by TimMann on Nov 16th, 2002, 11:30am
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?
Title: Re: MONTY HALL SHOW
Post by Zaius on Nov 16th, 2002, 12:30pm
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
Title: Re: MONTY HALL SHOW
Post by Rodrick Crider on Nov 21st, 2002, 12:41pm
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:
--------------------------------
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.
--------------------------------
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.
Title: Re: MONTY HALL SHOW
Post by KAuss on Nov 25th, 2002, 11:21pm
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  ;D)

If you get this explanation, then you can easily see the 3 doors one.
Title: Re: MONTY HALL SHOW
Post by jeu on Dec 1st, 2002, 12:39pm
Test
Title: Re: MONTY HALL SHOW
Post by jeu on Dec 1st, 2002, 12:43pm
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
Title: Re: MONTY HALL SHOW
Post by towr on Dec 1st, 2002, 2:09pm
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..
Title: Re: MONTY HALL SHOW
Post by Patient of Dr Whitey on Dec 2nd, 2002, 11:32pm
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.
Title: Re: MONTY HALL SHOW
Post by ghjgj on Dec 10th, 2002, 1:39pm
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 ;)
Title: Re: MONTY HALL SHOW
Post by Paul Brockman on Dec 13th, 2002, 11:23am
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.
Title: Re: MONTY HALL SHOW
Post by Chronos on Dec 13th, 2002, 2:24pm
No, that's the probability of winning altogether if you switch as often as you don't.  But why would you do that?
Title: Re: MONTY HALL SHOW
Post by Bryan Cannon on Dec 20th, 2002, 7:27am
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.
Title: Conditional Probability
Post by Kevin on Jan 15th, 2003, 3:56am
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.
Title: Re: MONTY HALL SHOW
Post by Frank on Jan 27th, 2003, 2:34pm
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?
Title: Re: MONTY HALL SHOW
Post by rmsgrey on Apr 15th, 2003, 11:23am
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)...
Title: Re: MONTY HALL SHOW
Post by Boody on Apr 16th, 2003, 5:52am
Lot of posts for an easy riddle.
:o

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.
Title: Re: MONTY HALL SHOW
Post by Rejeev on Oct 7th, 2004, 9:24pm

on 08/08/02 at 11:39:11, 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?

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.
Title: Re: MONTY HALL SHOW
Post by rfh on Oct 19th, 2004, 4:03am

on 08/09/02 at 18:27:02, 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 =).
Title: Re: MONTY HALL SHOW
Post by b_ecstasy on Oct 28th, 2004, 2:14pm
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.
Title: Re: MONTY HALL SHOW
Post by Yobgal on Jan 12th, 2006, 11:04pm
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?


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?


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?


In those questions, the answers are 1/3 and 1/2, respectively. I hope this explanation is more clear for some people.
Title: Re: MONTY HALL SHOW
Post by Zero on Jan 14th, 2006, 2:09pm
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...
Title: Re: MONTY HALL SHOW
Post by towr on Jan 15th, 2006, 7:11am

on 01/14/06 at 14:09:31, 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...
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.
Title: Re: MONTY HALL SHOW
Post by sojibby on Jun 1st, 2006, 4:23pm
There are three doors labled 1, 2, and 3.  For argument's sake, let's say that the car is behind door one and there are goats behind door 2 and door 3.  We'll call them goat A and goat B (it makes a difference).

There are 8 scenarios here:

SCENARIO ONE

Contenstant choice:  1 (car)
Host opens:  2 (goat A)
Contestant choice:  switch (lose car)

SCENARIO TWO

Contestant choice:  1 (car)
Host opens:  2 (goat A)
Contestant choice:  no switch (win car)

SCENARIO THREE

Contestant choice:  1 (car)
Host opens:  3 (goat B)
Contestant choice:  switch (lose car)

SCENARIO FOUR

Contestant choice:  1 (car)
Host opens:  3 (goat B)
Contestant choice:  no switch (win car)

SCENARIO FIVE

Contestant choice:  2 (goat A)
Host opens:  3 (goat B)
Contestant choice:  switch (win car)

SCENARIO SIX

Contestant choice:  2 (goat A)
Host opens:  3 (goat B)
Contestant choice:  no switch (lose car)

SCENARIO SEVEN

Contestant choice:  3 (goat B)
Host opens:  2 (goat A)
Contestant choice:  switch (win car)

SCENARIO EIGHT

contestant choice:  3 (goat C)
Host opens:  2 (goat A)
Contestant choice:  no switch (lose car)


These are all the possible scenarios, 8 in total.

In half of them she switches and in half of them she wins.

Therefore the probability is a sensible 1/2.
Title: Re: MONTY HALL SHOW
Post by rmsgrey on Jun 1st, 2006, 6:17pm

on 06/01/06 at 16:23:35, sojibby wrote:
In half of them she switches and in half of them she wins.

Therefore the probability is a sensible 1/2.

Here's a rather less interesting game. I give the contestant a blank piece of paper and then ask her to choose whether to switch it for the pile of 10 $10 bills on the table or keep the piece of paper, nominal value rather less than a cent. There are two scenarios:

SCENARIO ONE

Contestant choice: switch (get $100)

SCENARIO TWO

Contestant choice: no switch (get nothing)

In half of them she switches and in half of them she wins, so the probability of her leaving with $100 is 50% if she tosses a coin to make her decision.

If, instead, she makes her choice based on analysing the situation, she will choose to switch every time and leave with $100 100% or the time...

Since the analysis is supposed to determine whether the contestant should switch or not, then you should keep it at four scenarios and jsut decide for each scenario whether switching is good or bad:

Scenario one:

Contestant choice: 1 (car)
Host choice: 2 (goat A)
Shouldn't switch

Scenario two:

Contestant choice: 1 (car)
Host choice: 3 (goat B)
Shouldn't switch

Scenario three:

Contestant choice: 2 (goat A)
Host has no choice: 3 (goat B)
Should switch

Scenario four:

Contestant choice: 3 (goat B)
Host has no choice: 2 (goat A)
Should switch


The catch here is that the four scenarios are reached by different processes - in each of the first two scenarios, there are two choices made; in the last two, only one choice each. As an example, if you roll an ordinary six-sided die, and reroll it if it comes up a six the first time, you get 11 possible outcomes: 1, 2, 3, 4, 5, 6&1, 6&2, 6&3, 6&4, 6&5, 6&6, with a 6 coming up on the first roll in 6 of the cases. If you try it about 35 times over, if the 11 cases were equally probable, you'd expect to see each one about 3 times. I would expect it to actually come up more like each of 1, 2, 3, 4, 5 coming up about 6 times each, and the other 6 possibilities coming up about once each - getting a 6 on the first roll is as likely as getting any other value, and that probability is then shared among the 6 scenarios that share that initial 6...
Title: Re: MONTY HALL SHOW
Post by Icarus on Jun 1st, 2006, 7:35pm
The point of rmsgrey's post, sojibby, is that you are assuming that your 8 scenarios are all equally likely. This is not true.

Like you, I will label the door with the car as "1", and the doors with the goats "2", and "3".

There is no reason for the contestant to favor one door over the other when making her initial choice, so we have 3 equally likely cases:

Case 1: (probability = 1/3) She picks door 1.
Case 2: (probability = 1/3) She picks door 2.
Case 3: (probability = 1/3) She picks door 3.

In Case 2, the host will always pick door 3.
In Case 3, the host will always pick door 2.
Both situations have probability 1/3 of happening.

In Case 1, the host is free to pick either of door 2 or door 3, and has no reason to favor one door over the other, so we assume that he picks randomly, and each would have equal chance of being picked (actually, it does not matter how he makes his decision, but for simplicity's sake, assume he does so at random). So we have 2 subcases:

Case 1a: She picks 1, he picks 2. Probability = (1/3)(1/2) = 1/6
Case 1b: She picks 1, he picks 3. Probability = (1/3)(1/2) = 1/6.

Your 8 scenarios come from adding her decision on switching to these 4 cases. But note that two of the cases, 2 and 3, are twice as likely to occur as the other two, 1a and 1b.

Let's look at the "switches" scenarios:

Scenario 1: (Case 1a - probability 1/6)
She chooses 1; He chooses 2; She switches - Loses car.
Scenario 3: (Case 1b - probability 1/6)
She chooses 1; He chooses 3; She switches - Loses car.
Scenario 5: (Case 2 - probability 1/3)
She chooses 2; He chooses 3; She switches - Wins car.
Scenario 7: (Case 3 - probability 1/3)
She chooses 3; He chooses 2; She switches - Wins car.

If she switches, she loses the car 1/6 + 1/6 = 1/3 of the time, but she wins the car 1/3 + 1/3 = 2/3 of the time.

Consider now the scenarios where she does not switch:

Scenario 2: (Case 1a - probability 1/6)
She chooses 1; He chooses 2; She stays - Wins car.
Scenario 4: (Case 1b - probability 1/6)
She chooses 1; He chooses 3; She stays - Wins car.
Scenario 6: (Case 2 - probability 1/3)
She chooses 2; He chooses 3; She stays - Loses car.
Scenario 8: (Case 3 - probability 1/3)
She chooses 3; He chooses 2; She stays - Loses car.

Here the probabilities are just reversed (as they have to be): 1/6 + 1/6 = 1/3 of the time she wins, 1/3 + 1/3 = 2/3 of the time she loses.

Better to switch.
Title: Re: MONTY HALL SHOW
Post by sojibby on Jun 4th, 2006, 5:29am
It's really common sense when you think about it isn't it.

She has a 2/3 chance of picking a goat - from which a switch would result in a win.

Can't believe I was so blind to that.

Thanks for making me think!
Title: Re: MONTY HALL SHOW
Post by rmsgrey on Jun 4th, 2006, 3:56pm

on 06/04/06 at 05:29:57, sojibby wrote:
It's really common sense when you think about it isn't it.

She has a 2/3 chance of picking a goat - from which a switch would result in a win.

Can't believe I was so blind to that.

Thanks for making me think!


Thenk-you for actually stopping to think - a lot of people who start out with a particular answer in mind are so convinced by their initial reasoning that they refuse to accept any argument that contradicts it.


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