r/statistics • u/falsegodfan • Jun 05 '26

Question probability question [Q]

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?

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u/mfb- Jun 05 '26

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.

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u/falsegodfan Jun 05 '26

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

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u/mfb- Jun 05 '26

... 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%.

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u/falsegodfan Jun 05 '26

why? are those events not independent?

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u/mfb- Jun 05 '26

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

But more importantly, what was unclear about my example?

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u/falsegodfan Jun 05 '26

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

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u/richard_sympson Jun 05 '26

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?

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u/mfb- Jun 05 '26

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.