Over the long weekend I was contemplating cuckholdry, that fate worse than butt rape. Various other blogger have discussed cuckholdry and the rate being roughly 10% (or somewhere between 2% and 30%). That’s all well and fine, but I was contemplating what the odds were for the average Renfield with no game and multiple children. It’s not just simply an additive, ie 10% chance for the first child, 20% of being cuckholded at least once with 2 children, etc, etc. Alternatively, if the guy ends up with 10 children, there’s not a 100% guaranteed that at least one child is not his.
I pulled out some reference material and found the following.
P{C of one child} = 0.10 ie 10% then,
P{C1 with multiple children} = 1 – (1-0.10)^n, where n = # of children.
Using this formula I computed the following probabilities and odds of being cuckholded.
2 children = 19% or 4.3:1 odds
3 children = 27% or 2.7:1 odds
4 children = 34% or 1.9:1 odds
5 children = 41% or 1.4:1 odds
6 children = 47% or 1.1: odds
7 children = 52% or 0.9:1 odds
8 children = 57% or 0.7:1 odds
9 children = 61% or 0.6:1 odds
10 children = 65% or 0.5:1 odds
When I was playing poker, if the pot was the right size, I take all of those odds knowing I’d win in the long run. I don’t think I’d take those odds with a typical American woman. Of course these computations assume that all women have the same chance of cuckholding, but I’m sure there are good women out there.
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