Board 8 > I have a (likely stupid) probability question for the math folks here.

(+) Yeah! (+)

I definitely learned how to calculate this sort of thing at some point, but that knowledge has long since departed me.

Say we have a deck of 20 cards. 4 of those cards are red and the other 16 are blue.

Six individual people are dealt 3 cards each, so 2 cards are not dealt at all. What is the probability that an individual person has been dealt at least one red card? What about if there are five people being dealt 4 cards each?

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NBIceman posted...

Say we have a deck of 20 cards. 4 of those cards are red and the other 16 are blue.

Six individual people are dealt 3 cards each, so 2 cards are not dealt at all. What is the probability that an individual person has been dealt at least one red card?

1 - (16/20)*(15/19)*(14/18) = 0.5087719298 or about 50.8%

What about if there are five people being dealt 4 cards each?

1 - (16/20)*(15/19)*(14/18)*(13/17) = 0.6243550052 or about 62.4%

I don't think the numbers 6 or 5 are actually relevant here, since everybody's chances are congruent. Would be the same as just saying "if 1 person drew 3 cards, what are the chances" or "if 1 person drew 4 cards". Whether the other cards get dealt out to others or just stay in the deck shouldn't matter I don't think?

Maybe I'm missing something, idk algebra 2 was a long ass time ago but I don't think multiple people drawing from the deck would change the odds

Basically you invert the question to be, "what is the chance of an individual getting all blue cards" then subtract that from 1 since they are opposite scenarios or whatever

Yeah, I was definitely overcomplicating that. Many thanks.

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