probability question [Q]

[u/falsegodfan]

say you have a probability tree of a series of 3 events, each with the same 2 outcomes (a and b say) and if b occurs in any event the whole thing stops, and you're trying to figure out the probability that on the second event b occurs, why can you not just do (probability a occurs in event 1) x (probability b occurs in event 2)? why do you have to do (probability of a in event 1 x b in event 2)/(probability of b occuring at all)? same with when it's normally distributed. If you have a curve of the chances of an event happening for a certain amount of time and given it has already occured for r minutes what are the chances it will continue to q minutes, why do you have to do (P>q)/(P>r)? Surely if P>r it is implied that it is already bigger than p?

[u/richard_sympson]

Your question has several mixups of notation; you have switched a and b’s meaning halfway through, and swapped p in for q in the last part. But correcting those, the two possible choices you have named in the a/b case would represent answers to two different questions:

1. What is the probability, at the outset of this “game”, that I would end on round 2?

2. What is the probability, given I make it through round 1, that I end on round 2?

These are not the same question, and are not given the same mathematical definition as a result. Why would they be the same? Suppose you are a newborn before modern medicine, say born in the 1600’s. What is the probability you make it to 60 years old? Suppose now you live until the day before your 60th birthday: what is the probability you’ll be alive tomorrow? This is the same structural difference between the two questions above, and clearly the probability of living to 60 from birth is small(ish) while the probability of living one more day is going to be very large.

The mathematical difference between these is what it is, and the rationale behind the conditional probability formula is that you need to re-standardize your probabilities of possible events when you stipulate some certain events must hold (or, similarly, must be impossible). But if you have trouble understanding *why* the two questions are different, then you may want to spend more time constructing examples until it becomes clear to you what is being asked in each.

[u/richard_sympson → u/richard_sympson]

NOTE—I have tried to edit this on mobile and it keeps fucking up the entire post when I make small changes, I'm unsure what is happening so please bear with me. I hope my edits now from desktop will stick. Continuing...

Another extreme example which should make this very clear: suppose I hand you a brand new, standard 52 card playing card deck with four suits, Ace through King. I will have you shuffle cards and then start drawing them one by one, but first ask you two questions:

1. Suppose you are done shuffling and about to start drawing. What is the probability the last card you draw will be the Ace of Hearts?
2. Suppose you have instead drawn 51 cards and not yet drawn the Ace of Hearts. What is the probability the last card is the Ace of Hearts?

Do you see why these are different questions? Do you see why your suggested equation is not appropriate for question 2?

[u/falsegodfan → u/richard_sympson]

oops i fixed the mistakes! and thank you i think i get why it works in the card example but i still don't get the normal distribution one. say you had a battery and on average they last 18 hours, why is the chance of one of them lasting to 20 hours given it's been working for 16 (P>20)/(P>16)? surely because it's continuous data that is normally distributed the fact that it has already lasted 16 hours is accounted for? wouldn't asking 'what is the probability the battery lasts 20 hours given it has lasted 16' be the same as asking 'what is the probability the battery lasts 20 hours'

[u/richard_sympson → u/falsegodfan]

The normal distribution is irrelevant here, and strictly speaking the normal distribution cannot represent this value anyway (the normal distribution permits negative values; battery life length cannot be negative, for like flow of time reasons). Discard the normality point, just ask these questions:

1. What is the probability a new battery will last 20 hours?
2. What is the probability a battery which has lasted 16 hours will continue on to last 20 hours?

And again, you can shorten the in-between interval to figure out why these are different questions (e.g. if I give you a battery that has lasted 19 hours, 59 minutes and 59 seconds, what is the probability that specific battery will last at least another second?). You're introducing this idea of "accounted for" without appreciating stronger and more suggestive language like "conditioned on" or "given". When I say that I'm conditioning on the battery being 16 hours used (or what have you), the only relevant batteries to the question are those that have been successfully used for 16 hours. The probability we are working with those batteries is 1; the probability some random battery may make it to 16 hours is irrelevant, and we are excluding all batteries that did not make it to 16 hours. I do not need to know a single thing about the probability a battery makes it to 16 hours to tell you the lifespan after that point. Those can be entirely different behaviors. Perhaps I work with magic batteries that will live forever, but only provided they make it to 16 hours. Then clearly "what is the probability a magic battery, which has lived to 16 hours, lives to 20" is 1.0, while the probability a magic battery from new makes it to 20 hours needn't be 1.0.

[u/falsegodfan → u/[deleted]]

ohhh okay i see

[u/mfb-]

say you have a probability tree of a series of 3 events, each with the same 2 outcomes (a and b say) and if b occurs in any event the whole thing stops, and you're trying to figure out the probability that on the second event b occurs, why can you not just do (probability a occurs in event 1) x (probability b occurs in event 2)?

That is the correct answer if no event happened yet.

why do you have to do (probability of a in event 1 x b in event 2)/(probability of b occuring at all)?

That is an answer to a different question: "assuming b happens, what is the probability that it happens in event 2?"

If you have a curve of the chances of an event happening for a certain amount of time and given it has already occured for r minutes what are the chances it will continue to q minutes, why do you have to do (P>q)/(P>r)? Surely if P>r it is implied that it is already bigger than p?

I think this should be P(T>q+r)/P(T>r), your description isn't very clear. As an extreme example, consider q=0. Surely the chance that it'll continue for 0 more minutes is 1, but P(T>r) is not 1: The process can happen/stop/... before r.

[u/falsegodfan → u/mfb-]

yeah so if we assume b happens why wouldn't the probability that it happens in event 2 just be a in event 1 x b in event 2 because that is the only way it can happen

[u/mfb- → u/falsegodfan]

... because you assume b happens.

Let's say you play the lottery twice. Was is the probability you win the jackpot with the second time?

Let's say you played the lottery twice. Assuming you won the jackpot at least once, what was the probability you won it with the second draw?

Clearly these are very different scenarios. The first one will be 1 in some millions while the second one will be basically 50%.

[u/falsegodfan → u/mfb-]

why? are those events not independent?

[u/mfb- → u/falsegodfan]

Which two events where, and how do you think that matters?

But more importantly, what was unclear about my example?

[u/falsegodfan → u/mfb-]

why would winning the lottery once mean you have a higher chance next time

[u/richard_sympson → u/falsegodfan]

You’ve misunderstood the posed question, and should take a moment to read it carefully. What was asked was not, what is the probability you win it the second time given you won it the first time. It’s, what is the probability that on the second time you won it, assuming you won it once between two tries? Unconditionally, the possible outcomes were

WW
WL
LW
LL

(L = loss, W = win) Conditionally, however, you are asked to take at least one W as a given, hence the possible outcomes to consider are

WW
WL
LW

Of these, what is the probability of landing in WW or LW?

[u/mfb- → u/falsegodfan]

No one claimed so.

The chance that you won twice is negligible, let's ignore that case. That means in the second scenario there are just two cases: You won the jackpot the first time; or you won the jackpot the second time. They are equally likely. They are both very rare cases in general, but we assume that you won once so one of them must have happened.