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John Kieffer, MD MPH's avatar

If I missed this earlier please let me know and I'll go track it down but is it simply intuition that leads us to know that Pr(R_2|B) = Pr(R_2 (R_1 or W_1)|B) leads to the addition of the probabilities? I know that "or" statements equate to addition but I didn't recall that being derived or explained in this series.

William M Briggs's avatar

John,

This would be whenever we did total probability. I can't recall the exact episode from the top of my head. But here, drawing it out completely:

Pr(R_2|B) = Pr(R_2 (R_1 or W_1)|B); since "(R_1 or W_1)", the whole thing given B, is a local truth. And adding a local truth does not change the probability.

That then equals

Pr(R_2R_1|B) + Pr(R_2W_1|B) - Pr(R_2R_1R_2W_1|B)

by the inclusion exclusion formula.

And

Pr(R_2R_1R_2W_1|B) = Pr(R_2R_1W_1|B) = 0

since in Boolean logic "R_2R_2" = "R_2", and it is a local impossibility that you can get both R_1 AND W_1 on the first.

Here's a general resource on the inclusion-exclusion:

https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

I also did in the probability of giant rocks from space post:

https://wmbriggs.substack.com/p/on-the-probability-of-large-rocks

Thanks for the question. Glad you're working through these.

John Kieffer, MD MPH's avatar

Thanks!

Paul Fischer's avatar

P(R3|B)=P(R3(W1W2 or R1W2 or W1R2 or R1R2)

=P(R3W1W2|B)+P(R3R1W2|B)+P(R3W1R2|B)+P(R3R1R2|B)

=P(W1|B)P(W2|W1B)P(R3|W1W2B)+P(R1|B)P(W2|R1B)P(R3|R1W2B)+P(W1|B)P(R2|W1B)P(R3|W1R2B)+P(R1|B)P(R2|R1B)P(R3|R1R2B)

=(n-m/n)(n-m-1/n-1)(m/m-2)+(m/n)(n-m/n-1)(m-1/n-2)+(n-m/n)(m/n-1)(m-1/n-2)+m/n(m-1/n-1)(m-2/n-2)

=(m/n(n-1)(n-2))[(m-1)(m-2)+(n-m)(m-1)+(n-m)(m-1)+(n-m)(n-m-1)]

=m/(n(n-1)(n-2)[-3m+2+nm-n+m+nm-n+m+n^2-nm-n-nm+m]

=m/n(n-1)(n-2)[2-3n+n^2]=m/n(n-1)(n-2)[(n-1)(n-2)]=m/n qed!

Sasha Latypova's avatar

Thank you for the lecture. Are you familiar with Nikolai Kozyrev's "causal mechanics" model? He devised a fairly straightforward way of objectively determining cause and effect by looking at time flow as the energy source for everything.

William M Briggs's avatar

No, don’t know. I’ll look it up. Thanks.