# What are the odds that 1 suit shows up at least 3 times in the first 8 cards chosen?

(We keep inventing betting games...and I'm curious if I'm giving the right odds :)

In a normal, shuffled, deck of cards,

What are the odds that [Lets say, Hearts] shows up at least 3 times in the first 8 cards chosen?

Cards are not returned to the deck between each pick.

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What are the odds there at most two hearts in the first 8 cards? – JavaMan Jul 21 '11 at 1:17
Please note that ncmathsadist and Byron Schmuland are looking at two different problems. Byron Schmuland solves the Hearts problem, while ncmathsadist shows how to approach calculating the probability that at least one suit appears $3$ or more times. There is no simple relationship between the two problems. – André Nicolas Jul 21 '11 at 4:04

Let's warm up with the easier question of getting exactly three hearts.

The number of ways to choose 8 cards from a deck of 52 is the binomial coefficient $52\choose 8$. The number of ways of
getting exactly three hearts is ${13\choose 3}\times{39\choose 5}$. Here "13" is the number of hearts (of which we take 3) and "39" is the number of non-hearts (of which we take 5).

The probability of getting exactly three hearts is, therefore, the ratio $${{13\choose 3}\times{39\choose 5} \over {52\choose 8}}.$$

For the probability of getting 3 or more hearts, you repeat this pattern replacing 3 with 4,5,6,7 and 8, then add the results. The probability you want is

$${{13\choose 3}\times{39\choose 5} \over {52\choose 8}}+{{13\choose 4}\times{39\choose 4} \over {52\choose 8}} +{{13\choose 5}\times{39\choose 3} \over {52\choose 8}}+{{13\choose 6}\times{39\choose 2} \over {52\choose 8}} +{{13\choose 7}\times{39\choose 1} \over {52\choose 8}}+{{13\choose 8}\times{39\choose 0} \over {52\choose 8}}.$$

When you do the calculations, this turns out to be $118943/378350 = 0.31437$.

As ncmathsadist points out, for calculation purposes, it is easier to consider the opposite problem of getting two or fewer hearts, and subtracting from 1. The probability you want can also be written

$$1- {{13\choose 0}\times{39\choose 8} \over {52\choose 8}}-{{13\choose 1}\times{39\choose 7} \over {52\choose 8}} - {{13\choose 2}\times{39\choose 6} \over {52\choose 8}} .$$

Problems of this kind, where you sample with replacement, give hypergeometric probabilities. If you had replaced the card after each draw, you would use binomial probabilities instead.

http://en.wikipedia.org/wiki/Binomial_probability_distribution

http://en.wikipedia.org/wiki/Hypergeometric_distribution

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This is $1 - P(\hbox{2 cards of each suit is drawn})$. Now try some fiddling with binomial coefficients to get to the answer.

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