this post was submitted on 23 Dec 2025
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Who ever could it be? (piefedimages.s3.eu-central-003.backblazeb2.com)
submitted 5 days ago by RmDebArc_5@piefed.zip to c/tumblr@lemmy.world
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[–] tensorpudding@lemmy.world 14 points 5 days ago (1 children)

If each friend in a group of four had an independent and identical 25% chance of being gay, the probability that at least two of them are gay is about 26%, I like those odds.

[–] carotte@lemmy.blahaj.zone 2 points 5 days ago (2 children)

are you sure it’s 26%? i’m arriving at around 57%, but maybe there’s a flaw in my reasoning

[–] tensorpudding@lemmy.world 6 points 4 days ago* (last edited 4 days ago) (1 children)

Its possible I made an error, I used 1 minus the cdf of the binomial distribution with n = 4 and p = 0.25 evaluated at 1, calculated from https://binomialdistributioncalculator.com/binomial-cdf-calculator/

[–] carotte@lemmy.blahaj.zone 5 points 4 days ago

huh. i did the same thing, but by hand…

oh wait no i see my mistake. you’re right, it’s about 26%. my bad 😅

[–] MissingInteger@lemmy.zip 4 points 4 days ago

@tensorpudding@lemmy.world is correct:

P(at least 2 gay) = 1 - P(0 gay) - P(exactly 1 gay)
= 1 - 0.75^4 - 4×0.25×0.75^3
= 1 - 0.31640625 - 0.421875
= 0.26171875
≈ 26%