That wouldn’t make very good television. You have to pad the game out to an hour. You might as well play Deal or No Deal with two suitcases!
Losing whether he switches or not? Reread what Uncertain wrote.
I hated this scene in the movie 21. This same question was asked and the “smart” kid immediately jumped on it by saying you should obviously switch doors. However, in 21, it wasn’t established that Monty has to open a door. In fact, Monty can choose to only open a door when the player has correctly picked the car. In that case, your odds of winning go from 100% to 0% if you switch. The fact that the pivotal scene establishing the bona fides of the protagonist contained that sort of error bugged me throughout the movie.
I love the subheading in Cecil’s original column.
THE LAST YOU’LL EVER HAVE TO READ ABOUT THIS
The irony. It burns!
Well don’t leave the irony plugged in, and make sure you let it cool after ironying your clothes.
Powers &8^]
If Monty chooses to open another door (and offer a choice) based on something other than 100% of the time or total randomness, then of course the situation is completely different.
friedo said:
At this point we had ceased to talk about television and were discussing how to illustrate the principle. You brought up the need to change the numbers, so I showed the logical way to do that.
One thing that makes the logic problem so confusing is the basis of three doors/cards. It becomes less difficult to understand with 52 options, same rules.
ZenBeam said:
Uncertain was quoting morodin5. Uncertain only posted two conditions and asked what morodin5 was suggesting the third situation was. morodin5 did not list those specific cases, he asked why the odds fall the way they do.
Reading between the lines, Uncertain was asking morodin5 what he thought happened to the other 1/3 probability once Monty shows the door. morodin5 is confused why that probability is assigned to swapping to the other door rather than split evenly between the two doors.
The alternative case that DSYoungEsq cites would be swapping to the door that Monty revealed. “Monty, I’ll take the goat.” That’s not usually considered an option.
And of course, the answer to my presentation of morodin5’s question stated above is, Monty is offering you the option of keeping your original door at 1/3 probability, or swapping to both other doors, each with 1/3 probability of having the car, “oh and by the way, one of them definitely has a goat, see?”
Evil Economist
Cecil already discussed this:
Typically the problem is stated in such a way that the assumptions above are explicit, regardless of what Monty really did in the TV show. I didn’t see the movie 21, so can’t comment on how they presented the problem.
My point was this. Saying that you have a 1/3 chance of winning if you don’t switch is the same as saying that 1/3 of the time you would win if you didn’t switch. Similarly, a 1/3 chance of winning if you do switch means that 1/3 of the time you would win if you did switch. So what happens the other 1/3 of the time? You lose either way? Nope, can’t happen.
I’ll put it another way. Either you win if you do switch (and lose otherwise) or win if you don’t switch (and lose otherwise). The probabilities must add to 1. They can’t both be 1/3.
I just realized it will soon be twenty years since I first argued with people about this. I think I’ll take my pension and get out of here.
I thought DSYoungEsq had missed the “if and only if” part of Uncertain’s post. Regardless, I don’t see how your interpretation of what DSYoungEsq said fits with what Uncertain wrote.
Uncertain said:
Right, I got that. The problem is that morodin5 was not trying to argue that. He is confused why one option stays at 1/3 probability and the other option becomes 2/3 probability, rather than both remaining options splitting the probability from the door that was eliminated. If both choices are 50%, that sums to 1 as well.
ZenBeam said:
DSYoungEsq said that the remaining 1/3 of the time you lose, which is consistent with not being in the 1/3 chance of winning if/only if switching and not being in the 1/3 chance of winning if/only if not switching. You didn’t win and you didn’t win, so you lost. I was pointing out that Uncertain was looking at 2 doors, but there was actually a 3rd door - the one Monty reveals. If you pick it, you’re assured to get a goat and not a car. Ergo, that is a guaranteed lose case.
One third of the time Monty lands on his edge.
–
John W. Kennedy
My heart belongs to Dada.
The only real question is why people have so much problem explaining this. If Vos Savant is so smart, why does she have so much trouble getting the point across?
A third of the time, you pick the winning door to start with. 2/3 of the time, you leave Monte with a winning and losing door. Whenever that happens, Monte is nice enough to get rid of the loser for you and offer you the winner. If you picked the winner 1/3 of the time and he is offering you the winner the other 2/3 of the time, switching makes sense.
The odds of any door being a winner are always 1/3. Showing you or not showing you what is behind any door does not change that. But Monty is not offering you a random door.
Let’s assume we have three marbles. Two blue ones are losers and a red one is a winner. There are two events. To model the scenario where you would always switch doors, you can view it as eliminating a random marble in your first event, leaving you with two marbles. 1/3 of the time, you eliminated the red marble in the first event. You cannot win the second event.
But what is the second event? 2/3 of the time, there is a red and blue marble left and somebody removes the blue for you. That leaves you with a 100% chance of picking a red marble.
So your probability is 1/3 * 0 * 0 + 2/3 * 1 * 1 = 2/3.
If we model the scenario where you pick a door and keep it, 1/3 of the time you get the red marble. There is no second event. The probability of winning is 1/3.
Another simple way of looking at it is that you pick a door and then either switch to a new one or don’t switch. If you don’t switch, there’s a 1/3 chance that you win. That means there is a 2/3 chance you win if you pick the only alternative.
If Monte picked a door randomly, then he’d be eliminating the possibility of winning 1/3 of the time. The rest of the time, you’d be moving to a different random door if you switched. The odds would not change. It would be like having a bag with the three marbles and each of you picking one. There would be a 1/3 chance you picked the red, 1/3 chance that Monte picked the red, and 1/3 chance that the red would be in the bag. If you switch, you gain nothing. But you cannot apply the same rules of probability once a non-random event is introduced.
This is the part that trips most people up, because the probabilities get too complex to be intuitive. You have to either know enough probability theory to be able to do the calculations, or you have to try an experiment several times and see the empirical results. That’s why it’s so hard to explain in text.
Powers &8^]
So, not being a fan of the show… Does he always pick another door and offer the choice? Or not always? Obviously “only when you picked the right door first” is too easy and a giveaway.
As far as the actual game show went, Monty was not constrained. He didn’t have to offer you the chance to switch. He typically did because that creates the drama, the tension.
Here’s a more challenging question, which also has practical applications: How can we make this interminable discussion stop? I call this the Monty Halting Problem.
Right. But of course, like the plane-treadmill question, the question being asked is more theoretical than practical.
Powers &8^]
Shhhhh! Don’t bring up the treadmill. 
Hey, if we are going to discuss something controversial here, we might as well go the whole nine yards. 