I have experienced such uncanny-seeming coincidences as well. But when that happens, I try to recall that we often underestimate the probabilities of coincidences, and that can make the coincidence seem more “canny”.

For example, suppose you have two clock-like devices that each display a number between 1 and N, inclusively, with the number displayed changing once per second, just like the seconds display on a clock. The displayed number is always selected randomly. So, for any given display second T, the probability that a given number M will be displayed during T is unaffected by the numbers that have been displayed at times prior to T. For any number M between 1 and N, and any display second T, the probability that M will be displayed during T is therefore 1/N.

Since there are two of these devices, A and B, then at any time T, the numbers displayed on A and B either match or they do not. There are N^{2} possible combinations of numbers for two dials. The number of possible matches is N, and the number of possible mismatches is N^{2} - N. The probability of a match is thus N/N^{2} = 1/N. The probability of a mismatch is 1 - 1/N = (N^{2} - N)/ N^{2} = (N – 1)/N.

Now suppose the two devices are started at the same time and allowed to run indefinitely. What is the probability that, after the device has run for two seconds, there will have been two consecutive mismatches? The answer is:

(N – 1)/N x (N – 1)/N = (N-1)^{2}/N^{2}

What is the probability that, after the device has run for five seconds, there will have been five consecutive mismatches? The answer is:

(N – 1)/N x (N – 1)/N x (N – 1)/N x (N – 1)/N x (N – 1)/N = (N-1)^{5}/N^{5}

In general, for any number k, the probability that there will have been k consecutive mismatches after the device has run for k seconds is (N-1)^{k}/N^{k}, or [(N-1)/N]^{k}. Since N-1 is less than N this fraction decreases monotonically as k increases.

Now, how large does k have to be for the probability of k consecutive mismatches to have decreased to 1/2? In other words, how large does k have to be for the probability of k consecutive mismatches to equal the probability of at least one match? The answer is the solution for k in the equation [(N-1)/N]^{k} = 1/2, which is

k = log _{(N-1)/N} (1/2).

We can now calculate the answer for different values of N. If N = 1000, so each of the devices is capable of outputting only 1000 random numbers, k is 692.8 seconds. For reference, there are 57,600 seconds in a 16 hour “waking day”, so 692.8 seconds is just a bit over 1/100th of a waking day. Now we can look at the answers for a few other possible values of N. In the table below, the first column is the value of N, the second column is the number of seconds k for which the probability of a match has grown to 1/2 or fifty-fifty, and the third column is the number of waking days corresponding to the number of seconds in the second column.

N |
Seconds |
Waking Days |

10,000 |
6931.1 |
.12 |

100,000 |
69314.4 |
1.2 |

500,000 |
346,573.2 |
6.02 |

1,000,000 |
693146.8 |
12.03 |

So, even if both devices can output a million numbers, and even if they operate completely independently and randomly, changing the number displayed once per second, there is a fifty-fifty chance that if the devices are allowed to run for just over 12 waking days, at least one match will occur.