Viewing a single comment thread. View all comments

uh-okay-I-guess t1_itsc8xy wrote

Reply to comment by eth_trader_12 in How does one figure out what probability is most relevant when deciding how probable something is? by eth_trader_12

Your original post has a list of 7 events (A1 through A7) in increasing order of probability. The difference between P(A1) and P(A7) is many orders of magnitude, with huge jumps from each statement to the next, and there is seemingly no grounds for preferring to study one of these statements to any other. Even if you try to use the principle of including all information, it's not clear where to stop. You'll just worry that you needed to choose A0: Juliet won twice, at 9PM, and the weather was cloudy both times. Compared to P(A1), P(A0) is another order of magnitude lower.

But if you realize that you actually care about, not P(A1), ..., P(A7), but P(R|A1), ..., P(R|A7), then it's easy to see that adding cloudiness does not actually change the result. In a Bayesian formulation of the problem, only relevant information matters. P(R|A0) = P(R|A1). You can ignore the weather.

The point of the last statement in the previous post is that, if you know no particular distinguishing information about Juliet, you can calculate P(R|A) or P(R|J) and it doesn't matter, because you still get the same answer. So as long as you use Bayesian reasoning, you are basically free to pay attention to the specific identity of the person, or ignore it, without affecting your conclusion.

1

eth_trader_12 OP t1_itsfjq6 wrote

That’s what you’re confusing. Looking at the specific probability of Juliet winning the lottery twice does increase the probability that it is rigged, just perhaps not enough. The only reason many consider it still not rigged is because of prior probabiltiies: the vast majority of lotteries in history have been fair; very few have been rigged.

But now imagine as if half of all lotteries were fair and half were all rigged. Let’s assume 10,000 tickets and 10,000 people. Let’s now look at the first case the other commenter mentioned: Juliet won the lottery once. The prior probabiltiies are the same for rigged and fair so they can be ignored. We now look at the likelihoods. The probability of Juliet winning the lottery given a fair lottery is 1 in 10k. The probability of Juliet winning the lottery given a rigged lottery is ALSO 1 in 10k (given others could have rigged it). Fair lottery is equally as likely as a rigged one.

Now, let’s assume Juliet won the lottery twice. The priors are again the same so let’s look at the likelihoods. The probability of Juliet winning two lotteries given chance is (1/10k*1/10k). The probability of Juliet winning two lotteries given that it’s rigged is 1/10k (1 out of 10k people could have rigged it twice). Now, the rigged lottery is MORE likely. Note that if we looked at the more generic description of SOMEONE winning the lottery twice, the likelihood of SOMEONE winning the lottery back to back would be 1…given enough time. But the likelihood of SOMEONE winning the lottery back to back given its rigged..is also 1. Now we must conclude they’re equally likely, but that’s not accurate.

As you can see; looking at specifics seems to work better.

1

eth_trader_12 OP t1_itsfut7 wrote

Sorry I replied to the wrong person but what I said applies to you as well.

In regards to what you said though, “what you care about” is subjective so we’re back to square one.

Ultimately though, I think the more specific information should be taken into account. The example in my other comment highlights that

1