[ironborn123]

You are right, but probability estimation of possibly novel phenomena is an ill posed problem.

So either you do not model it at all, or come up with some mental model of what probability you are comfortable with and due to which factors; the factors being more important than the final number.

[layer8]

The exact numbers don’t matter, but your post indicates that you seem to have the wrong mental model of how to adjust your credences. The probability of each factor decreases the overall probability, not increases.

[ironborn123]

Alright lets do some calculations. Feel free to dissect the assumptions.

K event = Kim says Material L is RTSC

D event = DFT says L is RTSC

S event = L is actually RTSC

we have assumed P(S/K) and P(S/D) are each 0.1, though we could have chosen other numbers for them as well.

We want to estimate P = P(S/(K and D))

P = P((K and D)/S)P(S)/P(K and D)

Assuming Kim and DFT are in the business of making positive predictions, they always get it right when L is actually RTSC.

so P(K/S),P(D/S) and P((K and D)/S) are all taken as 1

hence P = P(S)/P(K and D) = P(S)/(P(K)P(D/K)) = P(S/K)/P(D/K) = 0.1/P(D/K)

(similarly, P = P(S/D)/P(K/D) = 0.1/P(K/D))

But ofcourse we dont know P(D/K) or P(K/D). We could check historical data on how often D aligns with K, a messy exercise at best. Say they dont perfectly align, then the conditionals are less than 1,and P>0.1. Even intuitively when K (or D) gets additional support in the form of D (or K), your P should increase, not decrease.

If we assume D and K align on average, then P(D/K) or P(K/D) is 0.5, and we get P = 0.2.

You may estimate everything above differently, thus getting a different P. You can also come up with your own way of modelling this. I came up with my particular estimate to understand how frequently should i follow the news, and care about the whole thing. You should model it according to your usecase.

[dcow]

What? Your model doesn't make any sense if you don’t know how to combine probabilities.

Furthermore, you can’t just say “I understand things semantically different than everyone else talking about probabilities so when I say the probability of two events happening is 20% I actually mean something completely different and my math works this way I invented to describe why my messy naive intuitions about probability don’t match the reality of how probability works”.

[OJFord]

That's not what GP means - even if you assume they're independent variables, 2x 10% probabilities combine to a 1% ('10% of 10%', they multiply, not add) probability of the total outcome, not 20%.

[martindbp]

Huh? I think he's saying OR not AND. Why would two separate pieces of evidence make the whole less probable, that makes no sense, whether you spell out the whole thing more formally or not.

[dcow]

So if I find 10 people that claim, each with a 10% likelihood of being true, that the dark side of the moon has a scar that looks like Pikachu, then it’s 100% true?

That doesn’t make sense either. Seems like someone is falling prey to an intuitive fallacy of probability.

The weird model is just wrong in the first place. The commenter we’re discussing conflates the probability of two people saying something with the probability that what they say is true. So yes, you have to model the situation correctly in the first place.

[ironborn123]

No,even though the 10 pikachu fans assess independently, their estimations will be correlated. As you may already know the overall estimate wont be 100%, rather about 65%. The formula when the assessors are known to be independent, is simpler and has already been indicated by another commenter, P(A1 or A2 or A3 or ..An) = 1 - P(not A1)P(not A2)..P(not An)

On the other hand, if those 10 were engaged in complete groupthink (so no additional information beyond the first guy), the overall estimate would remain 10%.

In general, the answer would lie between 0.1 and ~0.65 depending on how their estimates influence each other.

[martindbp]

Of course you don't add probabilities, but for small probabilities it's an ok estimate. The chance of getting a single heads in N coin flips is one minus the probability of always getting tails, so 1-P(tails)^N. If P(heads) and N are small (e.g 0.1 and N=2), adding the probabilities is a perfectly reasonable estimate, just plug in the numbers and check for yourself.

[OJFord]

Ok yeah I think maybe I misunderstood the original argument. But those are absolutely not independent (which was probably the repliers point, unless we made the same misreading).

[whimsicalism]

dude you so do not know what you are talking about, the GP is talking about basic conditional probability rules lol