Here are some of the relevant sections from Satoshi's paper:
http://bitcoin.org/bitcoin.pdf

"11. Calculations
We consider the scenario of an attacker trying to generate an alternate chain faster than the honest chain. ... The race between the honest chain and an attacker chain can be characterized as a Binomial Random Walk. The success event is the honest chain being extended by one block, increasing its lead by +1, and the failure event is the attacker's chain being extended by one block, reducing the gap by -1. ... The probability of an attacker catching up from a given deficit is analogous to a Gambler's Ruin problem. ...

p = probability an honest node finds the next block
q = probability the attacker finds the next block
qz = probability the attacker will ever catch up from z blocks behind

Given our assumption that p > q, the probability drops exponentially as the number of blocks the attacker has to catch up with increases. With the odds against him, if he doesn't make a lucky lunge forward early on, his chances become vanishingly small as he falls further behind ... the attacker's potential progress will be a Poisson distribution ... To get the probability the attacker could still catch up now, we multiply the Poisson density for each amount of progress he could have made by the probability he could catch up from that point.

Converting to C code..

```
include <math.h>
double AttackerSuccessProbability(double q, int z)
{
double p = 1.0 - q;
double lambda = z * (q / p);
double sum = 1.0;
int i, k;
for (k = 0; k <= z; k++)
{
double poisson = exp(-lambda);
for (i = 1; i <= k; i++)
poisson *= lambda / i;
sum -= poisson * (1 - pow(q / p, z - k));
}
return sum;
}
```

Running some results, we can see the probability drop off exponentially with z.
"

Well I have a coinbase transfer that's been pending for nearly 24 hours and it has 19 confirmations. Wtf. I could've just used a credit card for all this hassle – user609926 – 2018-01-06T09:43:30.443