uh-okay-I-guess t1_itrh646 wrote
Let's say you let J be the event that Juliet wins two state lotteries back to back, and let A be the probability that anyone wins two state lotteries back to back. Let R be the event that the lottery is rigged. What you really want to know is the posterior probability that the lottery is rigged, and this could be P(R|J) or P(R|A).
I'm interpreting your question as asking: how should I choose whether to care about P(R|J) or P(R|A)?
Fortunately, if we know nothing about Juliet, it doesn't really matter. See, you can calculate P(R|J) = P(J|R)P(R)/P(J), and you can also calculate P(R|A) = P(A|R)P(R)/P(A). I claim these calculations will produce very similar values when we know nothing about Juliet. P(R) is of course common to both. Canceling it from both expressions leaves P(J|R)/P(J) in the first expression and P(A|R)/P(A) in the second.
In the second expression, both the numerator and denominator are larger. There are at least a thousand people who play the New Jersey lottery^([citation needed]), so we expect P(A) > 1000P(J). But similarly, if the lottery is rigged, it could be rigged in favor of any of those one thousand people, so we also expect P(A|R) > 1000P(J|R). If we genuinely do not think any of those people is more likely to be favored by the rigging than any other, then the ratios will be exactly the same and our calculated probabilities will be the same too. In this case it doesn't matter whether we care about P(R|J) or P(R|A).
Remember, this only works if Juliet is an arbitrary person. If we do know that Juliet is the daughter of a crooked New Jersey politician, then maybe we think that if the lottery is rigged, it's quite likely to be rigged in her favor. In that case, we might say P(A|R) = 10P(J|R), while still believing that the lottery is probably not rigged and P(A) = 1000P(J). Then P(R|J) is going to be 100 times bigger than P(R|A). In this case it would be wrong to use P(R|A), because it's throwing away important information (Juliet's crooked connections).
In summary, in both cases we should really condition on the more specific event (i.e. we care about P(R|J)), because that takes into account all the available information. But luckily for our sanity, when we have no special information about Juliet, P(R|J) = P(R|A). So even though you want to condition on all the available information, it's fine to ignore information that means nothing to you.
eth_trader_12 OP t1_its2ozt wrote
I completely agree with you except the last statement. P(R|A) given the same principle of more information that you just said assumes that "all the information" we have is that someone at some time won two lotteries twice. As in, if you knew that someone won two lotteries at some point or another, then yes, P (R|A) would suffice.
But in this case, we know that Juliet won. Hence, we calculate P (R|J), even if we don't know anything else about Juliet
Coomb t1_its9nxp wrote
>I completely agree with you except the last statement. P(R|A) given the same principle of more information that you just said assumes that "all the information" we have is that someone at some time won two lotteries twice. As in, if you knew that someone won two lotteries at some point or another, then yes, P (R|A) would suffice. > >But in this case, we know that Juliet won. Hence, we calculate P (R|J), even if we don't know anything else about Juliet
Why? Why do we do that when as far as we know Juliet is no different from anyone else? There is no more reason to assume the lottery is rigged because a particular individual whose name you know won twice in a row if you know nothing at all about that individual. If you don't know them from Adam, then Juliet could just as easily have been Adam or Bobby or Charles or Doug. Knowing literally nothing other than her name is the same as knowing nothing at all about her, unless the lottery is rigged for anyone named Juliet and you know that.
eth_trader_12 OP t1_itsfkc8 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.
pucklermuskau t1_itv0hxz wrote
what difference does it make that you know her name?
eth_trader_12 OP t1_itvas30 wrote
It doesn’t. You missed the entire point of the example. Even if I didn’t know her name, I would know that a specific person won it, and the math would be the same.
The math would be different only if I knew that someone at some time won two lotteries, not at a specific time. The time is what’s relevant here
pucklermuskau t1_itvbekr wrote
in all examples that are being discussed, those that you're strangely dismissing, we are assuming that the same person has won twice in a row. you're misunderstanding the argument.
[deleted] t1_itvxksv wrote
[removed]
uh-okay-I-guess t1_itsc8xy wrote
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.
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.
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
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