Where am i thinking wrong,i should be in a wormhole...?

[KSakuraba]

I remember once reading that it is possible that a wormhole could appear right in front of you and take you to another dimension...(a bit unprobable though...[img]/images/graemlins/crazy.gif[/img]). So lets say that the probablity of a wormhole appearing in front of me at moment t is p. So the probability would be the same p at the moment of t+1 second,right? If we split the 1 second into N parts,we would have same probability at the moments of t,t+(1/N),t+(2/N),...,t+((N-1)/N),t+1? Now when the N goes to infinite, we have infinite amount of samples, which each have the same probability of the worm hole appearing.
So would'nt 1-((1-p)^N) go to 1 when p<1 and N goes to infinite?? I know im lost here big time(largely because im not in a wormhole [img]/images/graemlins/wink.gif[/img]) but can't find where...anyone who can help me?

[Spicymoose]

0 times infinity is sometimes 0.

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So lets say that the probablity of a wormhole appearing in front of me at moment t is p

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The problem is in the term 'moment' Nothing happens if time does not progress. You need to think in terms of time intervals instead, as there is 0 probability of something happening in a 0 time interval.

Let's say that for a time interval, T = 1 second, there is a W% chance of a wormhole happening. If you half the time interval, W also halves to W/2 per interval. Therefore, in every T= 1 second, you have 1/2W*2 = W% chance of the event happening, as before. Cut the time interval to 1/1000, you have 1/1000*W*1000 = W% chance of it happening in 1 second. Exactly as before.

Also, just to clarify, 1-((1-p)^N) in not correct, as p is a function of N (as N increases, p for each interval decreases.)

If P is the probability in one second, the equation is

1- (1-(P/N))^N