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I'm a bit confused about something.

Let's say I want to cross a road to my friend's house. The probability of the road to be opened is $p$, and $1-p$ to be closed. So I know the probability for one direction at a time. What would it be if I would like to go and return on the same day? note: Let's assume that the road was opened/closed for the whole day.

My answer is $p^2$ (or $(1-p)^2$), depends on the status of the road.

Ofek Pintok
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    Ask yourself, what is the probability of return if you can't cross in the first place? – Phil H Jun 17 '18 at 22:44
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    What to do you mean? If the road is open or closed all day, it will either be open with start and open when you return or closed when you start an return. So it's p that it'll be open both times and 1-p it will be closed both times? Is your question why isn't it $p^2$ or $(1-p)^2$? – fleablood Jun 17 '18 at 22:45
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    If it's closed, how can you cross it to make the return? – Phil H Jun 17 '18 at 22:46
  • @PhilH so for the closed state, it remains $(1-p)$ because the prob. of the return is depended on the status of the road.. But what about the opened state? is it also depended on the first status of the road? – Ofek Pintok Jun 17 '18 at 22:50
  • Same thing with the road open, my point was that your answer made no sense if you considered the question I asked.. – Phil H Jun 17 '18 at 22:54
  • I'm confused by the wording "The probability of the road to be opened is $p$, and $1−p$ to be closed. So I know the probability for one direction at a time." Which I read as you have one of these probabilities when you leave and one when you depart, i.e., the road could be open or closed in either direction. However the last line confuses everything: "Let's assume that the road was opened/closed for the whole day. " as it seems to contradict your initial statement. – Daniel Buck Jun 17 '18 at 23:09

5 Answers5

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If it is given that the status of the road is the same for the entire day, then the answers of "$p^2$ or $(1-p)^2$" don't make sense. With this interpretation, only the status of the road matters, but it does not matter at all how many times you want to cross the road. With probability $p$ the road is open for the entire day — so this is the probability that you can visit your friend, or visit him and come back home, or visit him and come back home and visit him again, or … (you get the idea). And the probability that you can't cross the road is just $1-p$, regardless of how many times you want to cross it.

zipirovich
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If the road is open it will be open all day and the probability that it will be open all day is $p$. And if the road is closed it will be closed all day and the probability of that is $1-p$.

So the probability you can go and return is the probability the road is open all day is $p$ and the probability you can't is the probability the road is closed all day is $1 - p$.

So I guess you question is why don't we get a paradox that the probability of GOING and RETURNING isn't product of Prob GOING $\times$ Prob RETURNING $= P^2$?

Well $P(A \text { and } B) = P(A)\times P(B)$ only if the events are independent of each other. That is certainly not the case here.

To figure the probababilities of dependent events is $P(A \text{ and } B) = P(A)*(P(B|A))$ ($P(B|A)$ means the probability of $B$ given that we know $A$ happened. In this case we know that if the road was open and went there, then the Probability that we return is $100\%$. So $P(A\text { and } B)= p*1 = p$.

And $P(\text { not } A \text { and not} B) = P(\text {not }A)*P(\text { not }B|\text { not }B) = (1-p)*1 = 1-p$.

And by the way $P(A \text{ and not } B) = P(A)*P(\text{ not }B|P(A)) = p*0 = 0$ and $P(\text{ not } A\text { and }B) = P(\text{ not }A)*P(B|\text{ not A}) = (1-p)* 0 = 0$.

So everything works out in the end.

fleablood
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If the road doesn't change state for the whole day $$Pr[state=open,~on ~return|state ~on ~first ~trip]=1,$$ if $state=open$ on first trip and zero otherwise, so the overall probability is $p$.

kodlu
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Let the probability the road is open be $P(O)=p$, and the probability the road is closed be $P(C)=1-p$.

If the road is closed you can't make the journey at all and this is with $P(C)=1-p$.

If the road is open you can make the journey and this is with $P(O)=p$. Now dependent on whether the road is open or closed when you leave gives your final probability: you are stranded if the road is closed with a probaility of $p(1-p)$, otherwise it is open and you return home with probability $p^2$. (I think this is what you meant when you said that you "know the probability for one direction at a time".)

Daniel Buck
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I think answer is p,because who comes back and who went are the same,so he knows the road is open or closed,when the probabality of being open is P when he wants to come back anything changes,so thats just one job,so answer is P

sobhan zahedi
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