Is it important the host looked behind the doors?
If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.
The overall point is that the host is not picking at random, and is thus affecting the outcome in a statistically reliable way.
WLOG assume the car is in door A (The rest will be the same by symmetry.)
You pick A, B, or C.
The host picks one of the other two doors to show you.
The universes can be described by your choice, followed by the host's random choice. There are 6 possibilities: AB, AC, BA, BC, CA, CB.
Since we're not in a universe where the host chose a car, we eliminate BA and CA.
We are left with four possibilities: AB, AC, BC, CB.
AB = You chose the correct door. (probability 1/4)
AC = You chose the correct door. (probability 1/4)
BC = You chose the wrong door. (probability 1/4)
CB = You chose the wrong door. (probability 1/4)
Since the host's choice was random, it's 50/50 as to whether the car is behind the door you chose or the remaining door.
This same reasoning breaks down when the host knows what's behind the doors because you're no longer eliminating a couple of the universes and it's a 2/6 vs 4/6 situation.
If you're playing blackjack, and you hit on a 19, but you have x-ray vision and you know a 2 is at the top of the deck, you're not playing the same game as you would be if you didn't have that knowledge.
Same with the Monty Hall problem. It is a critical distinction whether opening the door has a 0% chance, or a 33% chance, of showing you a car.
I thought as you think. I wrote a simulation to show I was right. I was wrong.
I have had this interchange several times. Invariably it goes one of two ways. They have endless reasons why they have to be right and they don't need to write a goddamn simulation, or they tell me they wrote the simulation and they have learned they were wrong.
This is basically just a different way of saying that Monty looks behind the door to be sure to only reveal goats.
Yes, the probability of wins will be different in these two scenarios, but it doesn’t affect the conclusion: when Monty reveals a goat, you should always switch.
Edit: if you believe I am the confused one after this comment, I will go make the simulation as you suggest.
In one, Monty reliably reveals a goat. In the other, Monty picks randomly between the other two doors.
We are, in parallel universes, playing both of those games. We know which. The host has just revealed a goat.
I read fouronnes3 as saying that these two situations are the same. They are not.
As I said, the outcome per game will be different (since you suddenly have an additional opportunity to lose), but the math around whether or not to switch when shown a goat remains unchanged.
To put it another way, if Monty is choosing randomly, half the time where you would win by switching, instead the game just ends/isn't counted, but the same is not true of the case where you win by not switching. From a bayesian point of view, Monty randomly revealing a goat should increase your belief that you picked the car the first time.
That said, switching is not worse than not switching unless Monty is biased towards revealing the car instead.
1) I roll a 3-sided die (or a 6 sided, wrapping) and keep it covered.
2) You pick a number, 1-3
3) I flip a coin. If it's heads, I pick the lower available number; if it's tails I pick the higher.
4) I peak at the die. If it's my number, we reveal and start over.
5) I (always, at this point) offer you a wager: if the die shows your number (so you would have lost if you switch), you pay me $7; if the die doesn't show your number, I pay you $5.
Is it the same game if I flip the coin secretly, peek at the die, announce my number (picked algorithmically in the obvious way), and then pay out according to whether switching would win (as above)?
Because (assuming I've explained these games as I intend... it's getting late) I would play the former with you not the latter (but I would play the latter if we switched the payments around - I chose 5 and 7 because 7/12 is halfway between 1/2 and 2/3).
No, it does not.
>> if (montys_choice === car_door) { i -= 1; continue; }
This discards the game where Monty chose a car.
>> do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door);
This makes Monty never choose a door with a car.
Neither one counts the game where Monty opened a door with a car as a win for either staying or switching.
var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice); if (montys_choice === car_door) { i -= 1; continue; } if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins);
var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins);
The subset of games where they close different doors is smaller than all possible games.
The probability that the host opens a door with a goat is always the same. (Again, under jtsiskin’s assumption that he picks randomly!)
As before, switching is essentially saying "I think I got it wrong on my first try", which is a good bet.
Monty opening a door with a goat doesn't change the probability of your initial door being right: it's always 1/3. You're not choosing between two doors; you're choosing between "I think I got it right the first time" (1/3) vs "I got it wrong the first time" (2/3) -- the second is more probable.
The game where there are two simultaneous participants, both free to open any door (including both picking the same door) only muddles things and offers no insight. It's effectively two simultaneous but independent games. How they are going to split the price if they both win, anyway? ;)
And if they switch both have probability 2/3 of getting the car? Is this some form of car sharing?
Maybe you should try to play with three cards and one ace. Pick one card (you are the first contestant). Pick a second card (you are also the second contestant). Flip the remaining card. If was the ace start again.
Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
Yes.
> Where does the car go if no one gets it?
I don't understand your question. What does this have to do with probabilities?
> And if they switch both have probability 2/3 of getting the car?
Yes. Remember, because both can pick the same door, this is effectively two independent games running simultaneously.
They each have a 2/3 probability of getting the car because, when they first chose, they had each 1/3 of getting it right. Each contestant's choice and corresponding probability of winning is independent of the other's.
> Is this some form of car sharing?
That's what I asked :) Car sharing makes little sense, prize-wise, which is why this makes no sense as a contest. But in regards to probabilities, it makes no difference: yes, both contestants should switch.
> Maybe you should try to play with three cards and one ace.
There are only 3 doors, so you should only use 3 cards.
> Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
I don't understand your question. Probabilities are computed independently for each contestant. There's no "simultaneous preference".
Player #1 has picked door A
Player #2 has picked door B
The host has opened door C and there was a goat
At this point:
What are the chances of winning of player #1 if he keeps his pick of door A? And if he switches to door B?
What are the chances of winning of player #2?
Player #2 also had a 1/3 prob of having picked the right door. Therefore, she has a 2/3 prob of winning if she switches.
In summary, both should switch. This is independent of whether they chose the same door between themselves. In other words, in a game where you're not forced by the other player's choices, your probability of winning is independent of the other player's. Your strategy should always be to switch (you can verify this empirically!).
The probability of player #1 getting the car if he keeps his door, which is equal to the probability that the car is behind door A, is 1/3
The probability of player #1 getting the car if he switches, which is equal to the probability that the car is behind door B, is 2/3
The probability of player #2 if he keeps his door, which is equal to the probability that the car is behind door B, is 1/3
The probability of player #2 getting the car if he switches, which is equal to the probability that the car is behind door A, is 2/3
In summary, the car is:
Behind door C with probability 0
Behind door A with probability 1/3
Behind door B with probability 2/3
Behind door B with probability 1/3
Behind door A with probability 2/3
You don't have probabilities "behind doors", you have probabilities of "chose right the first time". The probabilities of players #1 and #2 choosing right are clearly independent.
But don't trust me: try it!
So you have Player A who chose door 1. The host flips the goat in door 3.
You have Player B who chooses door 2. For them the host flips open the goat in door 1.
Player A should switch, and wins the car. Player B should switch, but had the car and loses it. The right move in both their cases was to switch.
Really, its no different than playing two separate games. Unless, like you said, you somehow change the rules and limit someones choices, in which case Im not sure why its a part of the conversation, other than being introduced through confusion.
https://news.ycombinator.com/item?id=24713352
If the players pick different doors, the host will open the third door. That prevents him from being sure to get a goat, in the same way that opening a door at random in the single-player case prevents him from being sure to get a goat.
For this reason, both players should be playing "blind", without seeing the other's choices or the door Monty opens for the other player. Anything else doesn't make sense for the game.
As it does removing the host's ability to choose a non-winning door in the variant where "the host had just picked at random from the two doors, and it happened to show a goat". That doesn't work the same either because it's not the same game.
After player #1 has picked door A, player #2 has picked door B and door C has been opened revealing a goat:
a) if player #1 stays with door A his probability of winning is ___
b) if player #1 switches to door B his probability of winning is ___
c) if player #2 stays with door B his probability of winning is ___
d) if player #2 switches to door A his probability of winning is ___
e) the probabilty that the car is behind door A is ___
f) the probabilty that the car is behind door B is ___
g) the probabilty that the car is behind door C is ___
a) The probability of winning if staying: 1/3
b) The probability of winning if switching: 2/3
c) The probability of the car being behind door the player chose is the same as a): 1/3
d) The probability of the car being behind the door Monty didn't open is 2/3
e) The probability of the car being behind the door Monty did open is zero: we know he chose a goat!
Note that the game doesn't work if it's not run independently for both players, because if they both choose goats, Monty would be forced to open a door hiding a car, ruining the game (it makes no sense at that point to either stay or switch). In order for the game to work the way I imagine you want, we must assume one player always picks a goat and the other a car... which cannot be guaranteed.
Rules2: "One player picks a door, the second picks a door (maybe the same), the host opens a door that has not been picked (if there are two he picks at random). Then the players can either keep their choice or switch to the other closed door."
Scenario2: "Player #1 picked one door, player #2 picked another door, the third door was opened and there was a goat behind it."
I agree it's different from the original problem, that's the point!
Rules0: "The player picks a door, the host opens a door that has not been picked with full knowledge that there will be a goat behind it. Then the player can either keep his choice or switch to the other closed door."
Scenario0: "The player picked one door, the host opened another door and there was a goat behind it."
Rules1: "The player picks a door, the host opens another door selected at random. Then the player can either keep his choice or switch to the other closed door."
Scenario1: "The player picked one door, the host opened another door and there was a goat behind it."
However the solution of Problems 1 and 2 is the same. In both cases the rules allow for the car being unveiled when the host opens the door (it happens with probability 1/3). In both cases under the scenarios proposed there is no point in switching doors.
They are both different to the original problem where everybody knows beforehand that the door opened will hide a goat.
[0] The problem I proposed at https://news.ycombinator.com/item?id=24707305 is also equivalent:
Rules1': "The player picks a door, and opens another door selected at random. Then he can either keep his choice or switch to the other closed door."
Scenario1': "The player picked one door, then opened another door and there was a goat behind it."If the host picked a door at random, and it happened to show the prize, then the game would be ruined. That would make some terrible television. Therefore, he can't be picking a door at random.