I am willing to grant that Cecil may be right in that, since Monty knows what is behind the doors, the statistics are not just random because Monty is make the decision about when/what door to open.
However, Mailyn makes no such claim (at least in the text that is provided) and so gives not reasonable argument for the statement “…but the second door has a two-thirds chance.”
That statement, given the information presented is completely incorrect.
Unless I’m misreading who’s being quoted, it appears Marilyn vos Savant did say that the host knew what was behind the doors: “You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat.”
Yes, but she makes no claim as the effect that it has on the outcome. It would be fair to say that it has an effect (which, again, she does not explicitly do) but it is a giant leap to claim that “…the second door has a two-thirds chance.” That could only true if she were to show, statistically, that Monty makes that the case. She very clearly seems to be implying that the number of doors determines the probability, not Monty’s decisions.
Further, any claim that says the the second door has two-thirds the chance would mean the Monty and the shows producers and extremely stupid as it wouldn’t take the viewers/participants long to figure that out. It is to the shows disadvantage to have the odds be very far from 50/50. It removes the tension and costs extra money.
As this thread demonstrates, people do not have a good instinctive grasp of probability. If people really could figure out the odds of something happening and act accordingly, we wouldn’t have lotteries and casinos.
Cecil goes into detail about the actual running of the show, and how it does not really conform to the “Monty Hall Problem”. Monty himself has said as much in some interview somewhere. The “Monty Hall Problem” was set up as a probability demonstration using a well-known game show situation to frame the problem, but for the demonstration to work, the problem originator (Steve Selvin) had to set constraints. Those constraints are sometimes not understood by the person repeating the problem, so sometimes you might encounter the problem misstated or not explicitly making the statements.
That said, Marilyn’s presentation makes account for the host’s selection process with the remark that the host knows what is behind the doors. In other words, he is not guessing when he reveals a door, but purposely revealing a non-winning door. That is not stretching or assuming, that is a valid reading of the words presented. That is further proven with the Answer and the further example of the million doors, where she says:
Bolding and underlining added for emphasis. It’s clear from that she is using the host’s selection process as part of the evaluation of the probability.
Good point. Given how counterintuitive the situation is, and how people continue to argue against the facts, I can speculate that the situation would provide a 2/3 advantage to the house. People will assume 50:50, and given other predilictions, will tend to keep the door they’ve already chosen over swapping to a different door with the same odds (as demonstrated by the Mythbusters). So Monty could in fact run strictly by the rules of always offering a switch and selecting his reveal, and most contestants would be dumb enough to not switch, thus the odds would be in the game show’s favor.
The situation doesn’t change much if Monty doesn’t know what’s behind the doors. If he opens a door and reveals a goat, you switch to the third door (the one neither you nor he chose). If he opens a door and reveals a new car, then you switch to the one he just opened, which after all is known to contain a new car. Your net odds of winning the car remains the same, 2/3.
What’s really critical, though, and which Marilyn didn’t specify, is the stipulation that he always opens a door at all. Maybe he always does, in which case the “standard” answer is correct. Maybe he’s perverse about it, and offers you the choice only if you got the right guess the first time, but sticks you with your original goat whenever you guessed wrong, in which case the “standard” answer is the worst advice possible. Maybe he uses a combination of the two tactics to try to psych you out and manipulate you into always making the wrong choice (this is in fact the way the real Monty operated).
Amazingly, such a mathematician as Paul Erdos refused to believe the correct (Marilyn’s) answer. And computer/math columnist Brian Hayes didn’t believe until he ran a Monte Carlo simulation.
Of course the answer will depend on the details, but nearly everyone seemed t agree that the host knows the correct door, always opens an incorrect door and always gives you the option to change. Under those circumstances, assume you always switch. Then whenever your initial choice is correct (1/3 of the time) you will lose and whenever your initial choice is wrong (2/3 of the time) you will win. It’s that simple.
Well, if the host didn’t know what was behind the doors, and the question was: You select door #1. The host opens door #2 and you see the car. Do you switch to door #3?" … isn’t very interesting, is it?
Sorry but the horse is still kicking. Cecil’s solution didn’t sit well with me at all so I got my maths pen out of its dusty draw and I’m pretty sure Adam and Anna are right. So here goes;
The way Cecil went about it is all wrong, he’s based his original four possibilities on events that are are impossible in the bounds of the problem, so they’re not of any use.
Here’s an accurate way to veiw the possibilities;
One of the children is the definite female let’s denote her as D(F), the other child has a 50/50 chance of being either (F) or (M) I’ll denote this one as C. There are four possibilities.
Child A is D(F) and child B is C(F)
Child A is D(F) and child B is C(M)
Child A is C(F) and child B is D(F)
Child A is C(M) and child B is D(F)
Each of these possibilities has a 1/4 chance and 2 of them mean one child is a male. so the odds are 1/2 of a child being a male. No paradox involved. The misleading part is the original four options given in the articles solution.
A better way of describing the problem with Cecil’s four options is that there never was a choice 4, nothing to eliminate. But choice 2 is twice as likely as choices 1 and 3.
The flaw here is that child A is defined as the one who entered the room – who is observed to be female. That eliminates #4, leaving only three possibilities, and in two of them, both children are girls.
Or, you would have to state it like this:
Child A is D(F) and child B is C(F) and child A enters first.
Child A is D(F) and child B is C(M) and child A enters first.
Child A is C(F) and child B is D(F) and child A enters first.
Child A is C(M) and child B is D(F) and child A enters first.
Child A is D(F) and child B is C(F) and child B enters first.
Child A is D(F) and child B is C(M) and child B enters first.
Child A is C(F) and child B is D(F) and child B enters first.
Child A is C(M) and child B is D(F) and child B enters first.
Since it was a girl who entered the room first, then we eliminate #4 and 6, leaving 6 possibilities, and only two of those (#2 and #8) include a male child.
Hi there. I think you’ve mixed up the two situations here, Cecil was saying if you knew the first child into the room was a girl then there’s a 50/50 chance of the second being male or female, on the other hand the chance of one child being male (ignoring the first to enter the room part that he introduces after the main puzzle) is 2/3 in his view. Here’s the whole dialogue.
I need to think more about the second question and Cecils answer of 1/2 chance of the second child being a male but I’m confident about the answer to the first question.
OK, on to the second question. I’m saying that if the first child is know to be a girl then there’s a 1/3 chance of the second being a boy.
The three possibilities are;
first child second child
D(F) C(F)
D(F) C(M)
C(F) D(F)
There’ll all equally likely and only one of the three has a male as the second child so 1/3. Knowing the gender of the first child is female raises the possiblity of the second being male from 1/4 to 1/3 because it reduces the odds of the second child being the definite female D(F) from 1/2 to 1/3.
No matter how I look at it, the 2/3 number seems arbitrary at best or making very strange assumptions about the motivation of the show.
If Monty were to choose EVERY time to reveal a losing door ONLY when the contestant had CHOSEN the winning door, then what are your odds of winning if you choose the “other” door?
Of course the show would not do that because people would get wise. So, the odds will contantly change depending on how the host bluffs. Ultimately though it will be approximately 50/50 at best since, assumedly, the show is always going to be better at playing the game than the contestant.
Here’s a way to consider it. Follow the standard set-up: one prize, three doors, you pick a door, Monty knows where the prize is.
But suppose Monty is feeling lazy. He doesn’t want to walk over and open a door. So he just says “Okay, here’s my offer. You can keep what’s behind the door you picked. Or you can have what’s behind both of the other two doors.”
So which do you pick? Do you think you have better odds of finding the prize with one door or two doors?
But that is not what is happeneing. In the scenario presented, one choice is eliminated. Once that is the case, there are only 2 options. Just as with coin flipping, regardless of what occurs before, the chances of a win between 2 items is 50/50. The only change to those odds are based off of the host choosing to bluff and how. It just does not add up to a 2/3 chance regardless of how you look at it.
Even with the scenario you just presented you only end up with a 2/3 chance. that is a best case scenario and not how the game is played so Marilyn’s claim doesn’t seem to be resonable.
