Approaching the Reciprocal of the Natural Log Base ("e")

Say the chance of a certain occurrence is 1 / n. What are the odds that the occurrence will NOT take place at least once during n trials?

For instance: you're hitchhiking, and the odds that any particular car will pick you up are 1 / 100. What are the odds that the next 100 cars will pass by without anyone giving you a ride?

As n increases, the likelihood that you won't get a ride approaches 1 / e (about 36.79%). Click here for one of a number of mathematical explanations available online.

In the context described above, your risk of not getting a ride from n cars will be only a little less than 1 / e, unless n is very small. With just 24 cars, the risk exceeds 36%, but it can never reach 37%! Try it out, below, for 1, 2 , 3, ... 1000, ... 9,999,999* cars:

    Enter a number of cars (which will also determine the odds of any particular car picking you up).

    Click button (left) to calculate odds of not getting a ride.



*This particular test gets boring with trials over 2,017,745, because the odds are no longer distinguishable from 1 / e with only seven decimal places. But more decimal places would not always be accurate. My crude method of calculation involves multiplying many fractional numbers ("floats," for you programming geeks), and computers operate with a certain fudge factor when applying arithmetic operations to fractional numbers.