The Monty Hall Problem.
This guy’s methodology is completely flawed.
When he “re-spins” the wheel, the spinning should be relegated to the areas where the door Monty opens actually contains the car, that is when the contestant is not correct with his guess. This model splits those “re-spins” between those times and the times when the contestant has actuall guessed correctly.
Are you talking about the first wheel or the second, Bing? Isn’t it true that Monty never opens the door that contains the car, so your situation would never occur?
The problem is with the second wheel:
(1) The only way Monty Hall can ever pick the door with the car is if the contestant DOESN’T. (the red parts of the wheel)
(2) If Monty Hall picks the other door without the car, the contestant switches w/o a problem.
(3) If Monty Hall picks the door with the car, he will re-choose. Again, this will only happen if the contestant doesn’t have the car.
(4) Here’s where it gets dicey: When Monty re-chooses, he only has one door from which to choose. This is the same as (2). The website’s second wheel, however, nullifies these probabilities entirely.
Thus the site only balances when the contestant chooses the door with the prize (1/3) against when the contestant doesn’t AND Monty Hall doesn’t either (2/3 * 1/2 = 1/3).
The missing third (the black pieces on the wheel) are treated as neutral when they really indicate situations where switching is better.
Oh, I get what you’re saying. “The problem says only that Monty opened a door with a goat behind it so we interpret this to mean that if the car is revealed then the game is over and the next contestant plays the game”
Under that condition, then their analysis makes sense. It still doesn’t seem, however, in line with the spirit of the problem.
I thought this problem was pretty simple. When the contestant makes the original choice, there’s 1/3 odds. Now Monty shows him another non-winning door. (Whether or not the original choice was a winner is irrelevant, as that is not revealed until after.) So now the contestant has 1/2 odds of getting it right, if he rechooses. It’s the choice between 1/3 odds and 1/2 odds. Which would you pick?
That’s not right. The problem is easiest thought of in this manner:
If the contestant switches every time, he/she only loses when they have chosen the correct door on the first guess. There is a one in three chance of that.
Interesting: I just read a bit about this problem in The Curious Incident of the Dog in the Nighttime.
Ok, wait a minute. I thought I had this one figured out years ago, but evidently not.
Take 1 (Monty Hall according to Phil): Monty can always show you a door that doesn’t have the car behind it, and he always will. If you switch every time, you lose only when you guessed right the first time. Since there’s a 1/3 chance of guessing right the first time, there’s a 2/3 chance of winning if you switch every time.
Take 2 (what I thought before): Monty always can, and always will, show you a door with no car behind it. The car is either behind the door you pick first or the door you could choose after Monty throws one out. That’s a 50/50 proposition.
One third of the time the contestant guesses right to begin with. In that case, Monty provides no information: the doors he chooses from are equivalent in every way. Two thirds of the time the contestant guesses wrong, and in those cases Monty essentially tells the contestant the answer. So, (1/3)*0 + (2/3)*1 = 2/3. The expected value of the certainty Monty provides is 2/3, so it stands to reason that the contestant should be able to get the right answer 2/3 of the time if he/she follows the proper strategy. Phil summed up the strategy: switch every time.
Another way to think about this is like so: assume two people jointly pick an initial door. Monty then opens a door that has a goat behind it. If one of the players sticks with the initial door and the other takes the remaining door, one of them is certain to get the car.
We know that the initial pick has a 1/3 chance of being right, so player 1 has a 1/3 chance of getting the car. Since one of the two players has to get the car, player 2 (who always switches) has to have a 2/3 chance of getting the car (1/3 + 2/3 = 1).
Say you choose door A. There’s a 1/3 probability that door A wins, and a 2/3 probability that either door B or door C wins.
When Monty asks if you want to switch, what he’s really asking is if you want to stick with door A, or whether you’d rather see what’s behind *both* doors B and C.
Opening one of the doors for you effectively gives you a way to choose *both* doors, in a way that’s in accordance with the rules of the game.
I like to think of it as “collapsing the wavefront”: it’s the same 2/3 probability for door-B-or-door-C, but it’s all been collapsed into being represented by the door that Monty didn’t open and you can choose.
(I’m not a physicist, and I don’t know what those words really mean. :-) )
Or, to simplify that slightly, I hope we can all agree that the probability of you having chosen the correct door in the first place is 1/3, and doesn’t change no matter what Monty says. After he opens a door, there is only once choice left, and the probabilities have to sum to 1, so the probability of the remaining door being the one must be 2/3.
Another exploration of the Monty Hall problem.
Yeah, John; my wife and I finally came around to the “do you want your one door or would you rather have both of the other two doors” point of view this afternoon. We’d already convinced ourselves that switching gives a 2/3 chance of winning, but we needed something to give us the warm fuzzies about why. That was what finally did it.
I’m dropped back by to post that insight, but you beat me to it. :-)
I can talk philosophy all night long, but my brain just starts to fizz when I read this stuff.
This is how I made sense of this.
Monty always shows you what’s behind a loser door. There are three possible situations about what’s *really* behind the doors:
Door 1 Door 2 Door 3
Reality A: Car Goat Goat
Reality B: Goat Car Goat
Reality C: Goat Goat Car
Say you choose Door 1. If the reality A and you switch, you lose. But if the reality is either B or C, you win. Thus, you have twice the likelihood of winning if you switch.
hmmm formating broke down above - just move the door 1, 2, and 3 lines over a few spaces so they line up with the columns…
This thread is closed to new comments. Thanks to everyone who responded.