eth_trader_12

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

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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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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.

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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.

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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

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eth_trader_12 OP t1_itqty1l wrote

I think the relevant question even in your example of just one lottery is to consider the probability that Juliet winning is small. And there's no problem in that. The probability of Juliet winning given that it was rigged is 1. The probability of Juliet winning normally is very small.

But that's fine. That's the probability of Juliet winning given chance, not the probability of chance given Juliet winning. In order to arrive at the second, we need to look at prior probabilities. Given that the rate at which lotteries is rigged is probably less than the chance of Juliet winning, one would still conclude Juliet won fairly.

An example of something that makes this clear is imagining that every other lottery is rigged. If every other lottery is rigged, would it make sense to look at the probability of merely "someone" winning the lottery. Clearly not. Since the probability of "someone" winning the lottery is very high if not 1 depending on the lottery. But does that mean the probability of it occurring by chance is close to 1? No. Because every other lottery is still rigged.

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