- #1

- 906

- 48

I want to discuss the 50:50 conclusion, which I think is correct and the 2/3 is a fallacy which I will explain.

To summarise the paradox, if you are presented with 3 doors in a competition and there is a car behind one (that you want to win) and a goat behind each of the other two (the booby prize), you pick one door and the chance of you getting the door with the car behind it is obviously 1/3. The host opens one of the other doors to reveal a goat (they know which ones have goats) and asks 'do you want to swap your choice of door'? A cursory inspection of that moment suggests the chance of you now getting the car is 50:50, but a slightly deeper inspection suggests a 2/3 chance if you go with a 'swap' strategy.

One might argue that your chance of picking the door with the car was 1/3, so it therefore remains 1/3. Therefore, all other options (i.e. swapping from that door) must then be 2/3, so it is better to swap, if you want to win.

I have to disagree for the following observation (which I will then back up with a reason the 2/3 is a fallacy); let's say that just at the moment you are about to decide whether to swap doors or not, you go sick and your quiz partner steps in. They are now presented with two doors, which, for sure, one has the car and one has the goat. There is simply no possible way that the probability for them is not 50:50! Yet, somehow the probability for you was different?

This makes absolutely no sense, and the fallacy exists here:- once the host opens one door to a goat, the 1/3 probability that you picked the car in the first 'round' then

*changes*. It is no longer a 1/3 chance that you picked the car, it is now a 1/2 chance. It just is, it is clearly no longer the same probability as before. This is nothing less than Bayesian probability in which the confidence in a given observation changes according to prior observations. In this case, the

*later*observation that there is a goat behind one door

*increases*the probability of your original choice to 50:50. It does not remain the same. This is nothing less than an axiom of science; if you derive a hypothesis that has a very low probability (e.g. your prize car is behind the first door) you then test it, and by making observations that do not disprove that original hypothesis, then the probability of that hypothesis improves.

This is

*precisely*the Monty Hall scenario: The probability your choice of door has a car behind it

*increases*once you get a further non-contradictory observation.

OK, so that deals with the fallacy in words, but this doesn't un-stitch the description of events that lead to a 2/3 'count' of outcomes, if laid out sequentially in some structured matrix. So, where is the fallacy in that?

It is as follows; the 'misdirection' is the focus on the options for what happens when the game player picks a goat, so they could swap and win the car? Right? Well, yes, but what's missing is that there are TWO options that the host can follow if YOU pick the car correctly in the first place! TWO outcomes - they can either pick the 'leftmost' goat or the 'rightmost' goat. These are TWO options, not one, and this is where the 'mathematical' fallacy exists.

See, like this;

Door 1 | Door 2 | Door 3 | | I pick | Host Opens | I stick | I swap |

car | goat | goat | | 1 | 2 | win | lose |

| | | | 1 | 3 | win | lose |

| | | | 2 | 3 | lose | win |

| | | | 3 | 2 | lose | win |

(It doesn't matter what the actual combination is behind the doors, the same would be the case for each combination.)

The point is that if you DO pick the car, then the host has TWO options. One outcome they pick one goat, the other they pick the other one. In a typical explanation of this, this is simply put down as 'the host picks a goat', as if that was one outcome. It isn't, it is two outcomes that look like one. The host can ONLY pick one other option if I pick a goat, but he has TWO options if I pick the car.

There are 4 options for any given combination behind the doors. Two are 'wins' and two are 'loses', whether you fix to one strategy or the other.

It is a 50:50 chance once the host opens a goat door. The question is whether he has opened 'the other' goat door or 'one of' the goat doors. These things are not equal.

This closes the circle between the 'obvious' fact that any 'new' contestant who comes into the competition just as the host opens a goat door is, clearly and obviously, presented with a 50:50 chance, yet the maths didn't say this. The reason is the 'two choice' option the host had is 'hidden' within the definition of the question.

I would welcome a confirmation or rebuttal on this, I have been mulling this over for a couple of weeks and I simply could not reconcile the 'new contestant' scenario with the apparent numbers from the mathematical description. But once you realise the

*host*is actually making one of two choices (in effect, they are pre-selecting two of the contestant's options and reducing them to one) then you realise it is a 50:50 outcome after all. At least, I think this is the case, because the alternative strays so far from intuition it begs us to look for the fallacy in the maths, and I believe this is it.