Choose at least N from large groups

Question

Having 40 aces and 60 kings in a deck of hundred (100), what is the probability of a sample of 50 cards containing 25 or more aces?

Is it required to do sum over binomial distributions for 0 to 24 to archive to the probability or is there "math" way obtaining result (distribution like binomial but for limited non-independent groups).

Raw power solution would lead to something like this

$\mathbb{P} = 1 - \sum^{24}_{i = 0} \frac{permutations\ of\ i\ aces}{total \ permutations} =\ ??$

Is it solvable by hand?

Answer 1

There is a pencil-and-paper method which avoids summing binomial probabilities, provided you have a table of cumulative Normal distribution probabilities and are happy with an approximation.

The exact distribution of the number of aces is a Binomial distribution with parameters $n=50$ and $p=0.4$, which has mean $\mu = np = 20$ and variance $\sigma^2=np(1-p) = 12$. This distribution can be approximated by a Normal distribution with the same mean and variance. We want to find $P(X \ge 25) = 1 - P(X \le 24)$. Since the Binomial distribution is discreet and the Normal distribution is continuous, we apply a "correction for continuity" and find $P(X \le 24.5)$. We have $$X \le 24.5$$ if and only if $$\frac{X-\mu}{\sigma} \le \frac{24.5 - 20}{\sqrt{12}} = 1.30$$ and $(X- \mu)/\sigma$ has (approximately) a Normal distribution with mean zero and variance one, so we can look up the cumulative probability in a table, with the result that $$P \left( \frac{X-\mu}{\sigma} \right) \le 1.30\approx 0.9032$$ so the probability of having 25 or more aces is about $1-0.9032=\boxed{0.0968}$.

This result compares fairly well with the exact result of $$1- \sum_{i=0}^{24} \binom{50}{i}(0.4)^i (0.6)^{50-i} =0.0978$$ for the Binomial distribution but requires less computation, again assuming you have a table of Normal probabilities available.

Answer 2

I don't think there is another way than the following.

Let 50 be the sample size and if you draw an ace, it's a succes. If not, it's a fail. The probability you draw an ace is 40%. With these figures you can model it as a binomial distribution and you can find the answer to:

$ P(x\geqslant25)$ , where $x$ is the amount of times you draw an ace in this sample of 50. You can find this via the CDF of the binomial distribution:

$\sum_{k=25}^{50} \binom{n}{k}p^k(1-p)^{n-k}$

, where $N = 50$ , $k = 25,26,...N$, $p =0.40$

or via excel:

$1-BINOM.DIST(24,50,0.4,TRUE)$

As a result: the probability of a sample of 50 cards containing 25 or more aces is roughly 9.78%

Answer 3

You must be careful here on how the sampling is done.

a) With replacement , i.e. you shuffle the deck, draw a card and take note of it for your sample,
put it back, reshuffle and proceed to a new draw.
In this case the probability of extracting an ace is constant for each draw, and you can
use the binomial distribution as already suggested.

b) Without replacement, i.e. you shuffle the deck and extract your sample , sequentially or randomly choosing
out of the cards remaining in the deck.
In this case, instead, the probability of drawing an ace is not constant along the draws, as it depends on the result
of all the precedent ones.
As the space of equi-probable events you shall consider that of all possible outcomes of the shuffling, i.e. the permuations
of the deck, and count the aces appearing in the first $s$ draws, which will become your sample of $s$ elements.

Refer to this Wikipedia article on this subject.

Therefore, given a deck with $a$ aces and $b$ kings , a sample of $s$ cards, and denoting by $p(k;a,b,s)$ the probability
of having $k$ aces in the sample,then

in the case with replacement you have $$ \eqalign{ & p_{\,r} (k;a,b,s) = \cr & = \left( \matrix{ s \cr k \cr} \right)\left( {{a \over {a + b}}} \right)^{\,k} \left( {{b \over {a + b}}} \right)^{\,s - k} = \cr & = {1 \over {\left( {a + b} \right)^{\,s} }}\left( \matrix{ s \cr k \cr} \right)a^{\,k} b^{\,s - k} \cr} $$

while in the case without replacement it is $$ \eqalign{ & p_{\,o} (k;a,b,s) = {{\left( \matrix{ a \cr k \cr} \right)\left( \matrix{ b \cr s - k \cr} \right)} \over {\left( \matrix{ a + b \cr s \cr} \right)}} = \cr & = {{a!} \over {k!\left( {a - k} \right)!}}{{b!} \over {\left( {s - k} \right)!\left( {b - s + k} \right)!}}{{s!\left( {a + b - s} \right)!} \over {\left( {a + b} \right)!}} = \cr & = {{a!b!} \over {\left( {a + b} \right)!}}{{s!} \over {k!\left( {s - k} \right)!}}{{\left( {a + b - s} \right)!} \over {\left( {a - k} \right)!\left( {b - s + k} \right)!}} = \cr & = {1 \over {\left( \matrix{ a + b \cr a \cr} \right)}}\left( \matrix{ s \cr k \cr} \right)\left( \matrix{ a + b - s \cr a - k \cr} \right) \cr} $$ which is a Hypergeometric Distribution.

The two probabilities become close when the sample size is small wrt the population ($s<<a,b$) but are quite different for large sample sizes.

Take a small example, which is easy to verify by drawing a sketch, with $a=2, \; b=4, \; s=3$, together with that in your post ($a=40, \; b=60, \; s=50$): the result are as depicted below

Concerning the question if it is possible to compute by hand
$$ \eqalign{ & P(25 \le k) = \sum\limits_{25\, \le \,k\, \le \,40} {p_{\,o} (k;40,60,50)} = \cr & = {1 \over {\left( \matrix{ 100 \cr 50 \cr} \right)}}\sum\limits_{25\, \le \,k\, \le \,40} {\left( \matrix{ 40 \cr k \cr} \right)\left( \matrix{ 60 \cr 50 - k \cr} \right)} \cr} $$
I do not see any practical way, if not resorting to the approximation via the Normal
distribution, as suggested by @akward. Only that the variance in the second case is quite less
than the Binomial distribution.