I was discussing this the other day, and wanted clarification..

The answer that seems obvious, but I *think* is incorrect, is that the the chances of finding a block in less than 10 minutes are the SAME as the chances of finding it after 10 minutes. So 50%.

But I have seen stack exchanges about 'Poisson Distribution', 'probability density' and 'exponential functions' ..

And then e^-1 pops out at ~36%.. for the chance AFTER the mean. More than 10 minutes.

And 1-e^-1 at ~64%.. for the chance BEFORE the mean. Less than 10 minutes.

Can't see how these figures were come by.. (the first e^-1)

Can someone explain it - thank you!

The "time to next event" in a Poisson stream of rate lambda (= on average lambda events per unit of time) has an exponential distribution, given by:

P(Delta_T) = lambda * exp(- Delta_T * lambda)

This means that the probability for the time to the next event to be between Delta_T and Delta_T + dt equals

P(Delta_T) * dt

(it is a probability DENSITY)

https://www.probabilitycourse.com/chapter11/11_1_2_basic_concepts_of_the_poisson_process.phpAs such, the total probability for Delta_T to be smaller than 1/lambda (which is your question) amounts to:

integral P(Delta_T) dDelta_T taken from Delta_T = 0 to 1/lambda

which gives us: 1 - 1/e = 0.632...

So there's 63% chance to find it in less than the average time (10 minutes).

The reason why it amounts to nevertheless the average time, is that even though there's only 37% chance to be longer, it can be REALLY longer sometimes.

EDIT:

If you want a more intuitive picture, your question is probably "how can the average be at value X, and not have half of the probability on the left, and half on the right" ?

The point is that the average value is not the MEDIAN value (which has, exactly, as a definition: half the probability left, and half, right).

Think of it this way: think of a metal object, a kind of bar, but such that its thickness is exponential. It is thick on one side, and its thickness diminishes exponentially to a very, very thin needle on the other side. Now, the "average" is the centre of mass, the spot where you have to hold this bar so that it remains in equilibrium. Well, that centre of mass is not in the spot where there's an equal mass on the left and the right, but rather, the "short" part will have a bigger mass than the "long" part, simply because the long part has a bigger lever arm.

You can feel that by thinking about this still the following way: suppose you put on the two extremities of a ruler, a weight of 1 kg and a weight of 200 g. In order to keep the ruler in equilibrium on your finger, you'd need to hold it much closer to the 1 kg weight than to the 200 g weight. So at that "centre of gravity", you still have more weight on the side of the 1 kg, but its lever arm is shorter ; and you have less weight on the 200 g side, but its lever arm is longer.