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Topic: MONTY HALL SHOW (Read 14496 times) |
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Arvin Farahmand
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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??
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Mike Peterson
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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
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Russ Arcuri
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on Aug 7th, 2002, 10:38am, 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?? |
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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
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B
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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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JW
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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.
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Arvin Farahmand
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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.
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Jonathan_the_Red
Junior Member
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Posts: 102
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Re: MONTY HALL SHOW
« Reply #6 on: Aug 9th, 2002, 11:26am » |
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on Aug 9th, 2002, 10:38am, 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.
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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.
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My arcade cabinet
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Mark Rowan
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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.
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JW
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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.
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Jonathan_the_Red
Junior Member
Gender:
Posts: 102
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Re: MONTY HALL SHOW
« Reply #10 on: Aug 9th, 2002, 2:53pm » |
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on Aug 9th, 2002, 12:32pm, 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. |
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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.
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My arcade cabinet
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Andrew Liao
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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 =).
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Russ Arcuri
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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.
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tot
Newbie
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Re: MONTY HALL SHOW
« Reply #13 on: Aug 10th, 2002, 3:32pm » |
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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.
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Kozo Morimoto
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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.
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AlexH
Full Member
Posts: 156
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Re: MONTY HALL SHOW
« Reply #15 on: Aug 12th, 2002, 12:46am » |
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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.
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Jeremiah Smith
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Beep!
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Re: MONTY HALL SHOW
« Reply #16 on: Aug 12th, 2002, 11:09pm » |
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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.
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Mongolian_Beef
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Re: MONTY HALL SHOW
« Reply #17 on: Aug 13th, 2002, 8:43pm » |
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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.
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LowIQ
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Re: MONTY HALL SHOW
« Reply #18 on: Aug 14th, 2002, 12:07am » |
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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
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schiefaw
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Re: MONTY HALL SHOW
« Reply #19 on: Aug 15th, 2002, 11:44am » |
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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
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Obgac
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Re: MONTY HALL SHOW
« Reply #20 on: Aug 21st, 2002, 5:19am » |
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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.
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subx
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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.
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Robert Grimsley
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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
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Russ Arcuri
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on Aug 22nd, 2002, 5:51pm, 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. |
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*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.
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zarathustra
Newbie
Posts: 7
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Re: MONTY HALL SHOW
« Reply #24 on: Aug 27th, 2002, 10:17pm » |
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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.
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