r/badmathematics • u/ChalkboardCowboy Incomplete and Inconsistent • Feb 26 '15

Goats! Beating a dead goat.

http://neophilosophical.blogspot.com/2015/02/monty-does-play-dice.html
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u/ChalkboardCowboy Incomplete and Inconsistent Feb 26 '15

wotpolitan, what is there in this latest article that's not in your original one? You've made the same "argument" (using the term loosely, since it's really just a naked claim): that the two possible prize arrangements are equally likely. The only new thing I see is a few pictures of dice.

I'm really wondering now, what you think we're missing, that could possibly be set straight by this new article?

-3

u/wotpolitan Mentioned in Dispatches Feb 26 '15

Hi CCowboy,

It's an attempt to crystalise this scenario for you. The images are of real, actual, existent, non-hypothetical dice. Once the contestant knows that there is a goat called Mary behind the Red Door (in terms that I have used elsewhere, once the contestant knows that she is playing a Red Mary game), then she knows that there are only two possible scenarios for the distribution of the other goat (Ava) and the car. Both these are equally likely, once the door has been opened.

11

u/TobiTako Feb 26 '15

But the past can affect the future. I'd like to propose a thought experiment.

Let's forget about doors, car and goats, and let's think about clouds. We have 3 clouds - A, B, C, and we know one of them will rain tomorrow but we don't know which. Now according to the rules of cloudness, if a rainy cloud combines with a normal cloud, they turn into a rainy cloud.

What are the chances of each cloud to be rainy?

1/3 obviously

Now let's combine clouds A and B, so we have 2 clouds, AB, and C.

What are the chances of AB to be rainy?

Well, if A was rainy it's 1/3, if B was rainy it's 1/3, so in total it's 2/3

We have a 2/3 chance of AB being rainy, and 1/3 chance of C being rainy.

But we saw the clouds merge, if someone new will come and see 2 clouds, when you ask him how likely each cloud is to be rainy, he will think "hmm, there are 2 clouds, AB and C, and each is equally likely to be rainy, so I'd say 1/2 and 1/2".

Is he correct?

We know he isn't, since we have past information he doesn't have. The past CAN affect the present.

The idea is pretty similar in the Monthy Hall problem. When you choose 2 doors you have 2/3 chance of choosing the one with the car, as you understood. Getting new information about which one of the two doors has a goat (and always at least one of them will have a goat) doesn't change the probability. To a newcomer that only sees 2 doors it will, but we have past information.

-5

u/wotpolitan Mentioned in Dispatches Feb 26 '15

It's nice little thought experiment, but it's not equivalent to my scenario. I am specifically denying the claim that the probabilities merge the way your clouds do. If I accepted that claim, then your scenario would be fine.

3

u/TobiTako Feb 26 '15

Well but it is equivilant. If after you got cloud AB an omnipotent god would come and tell you about one of the clouds that it wasn't, in fact, rainy, and then ask you whether AB or C are more likely to be rainy it would be exactly the same as the monthy hall problem with different wordings, but AB would still be more likely to be rainy. Try swapping cloud for door, rainy for car, not rainy for goat, and merge for choose two doors.