Door 2 does have a 2/3 chance of winning. As others have said, Monty has effectively merged the unchosen doors into one door, so contestant 2 should choose that door. The fact that he/she is not the same person as contestant 1 makes no difference
The key to it is being allowed a choice after Monty has taken two doors (each of which might have been the car) and choosing which one doesn’t have the car and eliminating it. His choice provides you with information that you can then use to improve your odds.
In your first example, both contestants are choosing without the benefit of any input from Monty. It therefore has nothing to do with the Monty Hall problem.
In your second example, Monty is not taking two doors and telling you which one isn’t the car: he is constrained to choose the door that the two contestants have not. He is not making any choice, so his choice tells you nothing.
You misunderstood my second example.
It is identical to the original Monty scenario. But “after” Monty reveals a losing door, he invites a 2nd contestant up.
He tells the 2nd contestant to choose either contestant 1’s door or the remaining door. I maintain contestant 2 has a 50-50 chance of picking the winning door.
This contradicts what’s been said about door 2 having a 2/3 probability of being the winner.
Read my water and fish and filtration example.
I’m really just repeating what’s been said a dozen ways, but let’s try putting it another way. You say that contestant 2 has a 50-50 chance. That would be true if he has precisely the same information about both remaining doors, but he doesn’t. He knows nothing about door 1 except that it is one of the original three choices (1/3 chance). He knows that out of 2 and 3 (ie 2/3 of the original possibilities) the one Monty didn’t open is the only one that might be a winner. And because 2/3 of the possibilities have been narrowed down into one door, it has 2/3 odds of being the winner.
No, we’re getting out of synch. By “second example” I meant the second example in your post #38.
But why aren’t the odds re-assessed once more information comes into play?
Sounds like you’re saying my “clueless” contestant who wanders into the studio after Monty reveals door 3 has a 50-50 shot at guessing which of the 2 remaining doors is a winner, but contestant 1 doesn’t.
Now you’ve changed the setup 
If contestant 2 doesn’t know which door contestant 1 originally chose, then yes he has a 50-50 chance. If he does know, then he knows that the other door is the “combined” door and should choose it.
I believe this is wrong, and that OldGuy and CurtC are correct.
Game: Monty Hall problem, “Monty’s random choice” variant
Rules: You chose one of 3 doors. Monty then opens one of the other two at random.
Case to analyze: Monty’s randomly-chosen door reveals a zonk.
Question: Given the above, what are your chances of winning by switching doors?
Let ‘P’ represent a door with a prize and ‘Z’ be one with a zonk. Here are the three possibilities after you have chosen and before Monty opens a door (with your choice shown at the left): PZZ ZPZ ZZP
Next, Monty will open either remaining door, at random. Let ‘O’ denote an opened door. Each of the 3 cases above has two possibilities, leading to six total cases: POZ PZO ZOZ ZPO ZOP ZZO
But we are told that the door Monty opened is a zonk, so we know we aren’t looking at either #3 or #6 from the list of six cases - we are dealing with one of the other four. Here are the choices they offer (opened door removed): PZ PZ ZP ZP.
Thus, switching wins half the time.
As noted above, the original Monty hall problem resolves to “Would you like to keep the door you first chose, or trade it for both the other doors.”* This variant is not the same.
(* To which should be added the minor qualification: “But you may keep only the best prize you find behind either door.”)
Please be patient with me. I just went over this with a colleague - he agrees with you all.
I see this playing out as 4 scenarios (assumption for clarity is that I always choose door 1.
p = my original pick w = winning door s = shown door
Scenario 1:
1pw 2s 3 - if I swap I lose
Scenario 2:
1pw 2 3s: - if I swap I lose
Scenario 3:
1p 2w 3s: If I swap I win
Scenario 4:
1p 2s 3w: If I swap I win.
This is 50-50 to me. My friend (and probably you all) feel that 1 & 2 collapse into 1
sceanrio, but I see no reason why they shouldn’t be distinct cases.
You are, in effect, counting the “Bar 12” in the Don’t Pass EV calculation, which leads to misleading results.
There is only 1 possible action for Monty to take in any variation of the puzzle: open all remaining zonks. You are making that one possibility worth two, which is fallacious. Consider: how does your logic hold up with the deck of cards scenario, where you pick one and Monty uncovers 50 losers?
Same response as above.
Scenarios 3 and 4 are each as likely as scenarios 1 and 2 combined. When you pick the winning door straight off (i.e. scenario 1 or 2), sometimes Monty will open door 2, sometimes door 3, but the chances that you got the winning door is 1/3.
I can explain that by postulating the puzzle in a slightly different situation. I offer a game of three-card monte to you. We bet $100. I show you the ace, then scramble the cards so fast that there’s no way you can keep up with it. When I stop, you pick your card. At that point, I (knowing which card is the ace) reveal a non-ace card and ask if you’d like to switch. Should you?
I guarantee you that if you switch, I will get your $100. My own policy is to offer the switch only if you have picked correctly the first time (why else would I offer it? To make the game more interesting? Yeah right).
If Monty’s policy is to offer the switch only when you’ve chosen correctly on your first guess, then you win by switching 0% of the time.
No, there are really only three scenarios. This is where you end up with incorrect percentages.
As others have said, these first two are really only one scenario. You pick 1 and the prize is behind door 1, and if you swap you lose. That’s the scenario. Which door contains the prize doesn’t matter. This happens 1/3 of the time.
Scenarios 3 and 4 also each happen 1/3 of the time. Once you grasp this distinction, everything should fall into place.
You pick the right door 1/3 of the time. You pick the wrong door 2/3 of the time. Concentrate on that, everything else follows from it.
Doesn’t matter, I still had a 33% chance at picking the winner to begin with. If I switch I have a 66% chance. Just because the situation falls into the 33% category doesn’t mean I didn’t still have a 66% chance with switching when I made my initial selection.
That’s what I don’t get. The odds seem to be different based on how you calculate. I have a similar example from craps: Doey/Don’t with no odds. This means 1 unit on the Pass and 1 unti on the Don’t Pass every time, never taking any odds. What is the EV of this system?
Well, the EV of the Pass and Don’t Pass averaged together is roughly -1.404%.
Then again, all resolutions are a push except the Bar 12, when I lose the Don’t. The only possible resolution is a loss, so I might think the EV is -100%.
However, when you lose, you can only lose 1 unit, and the aggregate bet is 2 units, so clearly the EV can’t be -100%, but if played to an arbitrary duration, you are GUARANTEED to be left with a single unit, regardless of how many units you brought to the table.
You seem to be arguing that the EV of this craps bet should be considered -100%, or at least close to it. My understanding is that this is an incorrect application of probability.
Can you help me understand what this means?
Why is it not possible to have the variant I (and others) have described, where Monty opens one of the unchosen doors at random?
If the 50 losers are uncovered at random (unlikely, but obviously possible) then either of the remaining cards is equally likely to be the winner.
I’m reminded of the Martingale. After two heads in a row, bet a bunch on tails, as three heads in a row only happens 12.5% of the time. But in reality the next flip is still 50-50. Why?
Because two heads in a row only happens 25% of the time in the first place, so you’ve already realized the low-percentage event you’re trying to capitalize on. Having been painted into a 25% total probability corner, you still have the original 12.5% chance for a third head, but there’s only 12.5% chance left, which is the chance for a tail. So you’ve got a 12.5% chance for heads, a 12.5% chance for tails, and a 75% chance of never have gotten here in the first place.
Gladly, but I have to run right now. Look for a response either later this afternoon or late tonight/early tomorrow morning. (Unless a helpful poster wants to explain the relevant aspects of the line bets in craps.)
If he opens the prize, then there is no problem. If he opens a zonk, it is then the Monty Hall problem, and the randomness/motivation of his choice affects nothing; it’s still 1:2.
Patently false. Try a computer simulation and see for yourself.
Ellis Dee - your knowledge of odds and your calculations are correct. However, you have missed the key assumption when trying to understand these calculations. [Now that I re-read this post I put together, a better example would be to illustrate what has been repeated to saoirse]
Let me try to simplify things:
A) If Monty randomly opens a door after you picked, without any further motivation on his part, then you standing pat is a 50-50 chance of winning; likewise, choosing the last remaning door has a 50-50 chance of winning (as others have already correctly pointed out).
Conversely,
B) If Monty only opens a door after you picked and always reveals a zonker, then the best way to maximize your odds is to always choose to pick the last remaining door.
Well I’m now in the “I get it camp”. Re-examining the “3 possibilities” diagram I see that the two scenarios I claimed were distinct are in fact the same (probability wise) as the other two scenarios.
Bring it on Monty!
There we disagree. A vital aspect of the original Monty Hall problem (OMHP?) is the fact that Monty knows what’s behind the unchosen doors, and always opens one that offers a zonk.
Well, “patently” means “obviously” and I’d submit that even if false, this thread indicates that it’s less than obvious. But I believe my statement is true.
We are presuming a pack of 52 different cards, of which one (say, the Ace of Spades) has been declared the winner. I choose one. Remaining cards are then chosen at random and revealed. If this reaches the point where 50 cards have been exposed and the winning card has not yet been seen, how does the chance that my card is the winner compare to the chance that the one unturned card is?
I say they are equal. Indeed, at any point in this game, all unturned cards - mine and all the others not yet exposed - have an equal chance of being the winner. (If the selection process is truly random, how can it be otherwise?) Most trials end before 50 cards have been turned. The probability that the Ace of Spades is the final card is one in 52 - the same as for my original choice.
When the OMHP is extended to a deck of cards, things are different: I pick a card and then we have Monty carefully checking each card before deciding to expose it. The process of choosing cards is anything but random. Every trial reaches the point where 50 cards are exposed. But we know this is nearly always because Monty has encountered the winner before the end and has chosen to suppress it.