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That paper is a far cry from "LLM's always give the wrong answer"!

A paper just published in Nature says that AGI is here, but we don't want to admit it...

Do you mean this in the trivial sense that she cannot see her own reflection, or are you saying that if the test could be modified to...

Indeed. She would have failed the rouge test.

Fortunately, it also works for cards😉

PS note that we can simply add the expected value for each colour: $$E = E(N, R) + E(N, B) + E(N, G)$$

Inspired by the posts by @Gavran we can show that for ##R## red balls and ##N## balls in total, the expected number of red balls at the...

## P(X=g) ## - probability that we have g green cards remaining ## P(X=b) ## - probability that we have b blue cards remaining ## P(X=r)...

I also calculated it to be 1.515 using a different but related method. What the OP is asking for is modeled by the negative...

That would be the same as drawing N cards without replacement that match the first color.

That is very far off. The correct answer is 1.515 Your concept of “green survives” appears to be the problem.

I put the precise calculations into a spreadsheet (it only took a few minutes) and got 1.514988104

Alternatively, you could note that the probabilities above are related to each other by: Probability that precisely the last (r+1)...

Here's some pseudo-code: Specify number of balls of each colour R = 40, B = 35, G = 25, N = R + B + G Initialize expectation value E =...

If you can revolutionise physics in your spare time, why are you wasting your talents on software development?