There are 12 windows, and behind each is either one of eight red cards or one of four white cards. The player must guess what is behind each window. The player also knows the distribution of the colors, so he won’t, for example, guess that five or more windows hide white cards.
What is the probability of a player guessing 12 correctly?
According to ChatGPT, the odds are 0.202%.
I have no idea if it’s right, but that’s what I got from the AI. It says 1 in 495 odds of guessing right. I tried this with four separate models of ChatGPT, just to re-verify, and all four said the same thing.
Yeah, AI is really dumbing down my ability to think.
It doesn’t matter which order you guess, so the odds of guessing white correctly are (4/12)*(3/11)*(2/10)*(1/9) then 8 certainties which is 0.00202 or 1 in 495 attempts. It will have the same difficulty regardless of what order you guess.
[Moderating]
An answer from an AI that you have no idea whether it’s right adds nothing to a discussion. If you can’t add actual knowledge to a thread, just don’t answer.
If it’s behind a window, surely the player can just look thru the window and see the result.
So the probability would seem to be 100% (except for blind players).
Assuming I understand the situation correctly, and all the cards of the same color are identical, there are 495 different ways of distributing the cards in the windows, and thus the probability of guessing correctly is 1/495.
One way to think of this is that you just need to select which of the 12 windows get the 8 red cards (so that automatically the other four get the white). The number of ways to choose 8 out of 12 is C(12,8) (“12 choose 8”), which is 495.
Another way to think of it is counting the number of permutations (orders or ways of arranging) of 12 objects, 8 of which are identical and 4 of which are also identical. This number is 12!/(8!4!) = 495.
Does it matter if he’s given information as he’s choosing? I.e. He chooses red for window #7, and is told that is correct.
No, every individual sequence, like WWWWRRRRRRRR or RRWRRWRRWRRW is independently random, and therefore has exactly the same chance of getting it right. Your only choice is whether to frontload the hard choices or leave them for the end, the overall probability doesn’t change.
Not unless, when being told you’re incorrect, you get some sort of opportunity to fix your mistake.