I remember looking at a brain scan of a patient who had metastatic cancer. There were 4 nodules, all in the same hemisphere. I thought, “What are the odds of that!”, and speculated that the patient had carotid narrowing on one side, serving to “protect” that hemisphere.
Then I thought, “What really are the odds of that?”
Well, the first met (metastasis) is clearly going to be on the same side as itself, so 1 met is 100% unilateral. (1.0)
The second met has 50% chance of going to same side. (0.5*1.0)
With the third met, the odds are 25%. (0.50.51.0)
So, the odds of four mets all going to the same side is 12.5%. Not really that unusual.
I think the math of the draw is similar to second step here. (1/75*1.0).
I would have guessed that the odds of getting paired up with same girl 3 years running would be way less than 1:5000.
To be paired up (both ways) with the same partner for three years would be one chance in 75^5. First, you’ve got to pick each other the first year. That’s 1 in 75, as we’ve established. Then, the next year, E-DUB has to pick her again: That’s another factor of 1 in 75. And she has to pick him again: A third factor of 1 in 75. And then the third year, they’d also have to pick each other, another two factors of 1 in 75.
That simulation is not right, it is indeed way less than that. If we allow that it can be any pair, it’s a 63% chance in the first year (see septimus above). But if you now require it to be the same pair for each subsequent year, it’s (1/75)^2 per year. That’s slightly off because the 63% allowed for 2 or more hits in the first year, but it’s close enough. It comes out to more like 1 in 10s of millions for 3 years running.
So, about 1: 2,373,046,875 or better than one chance in 2.4 billion. More likely than all the fast molecules running to left side of the room. But less likely than picking the winning Powerball ticket (around one in 300 million).