Odds of $X$ number of items getting picked twice or three times in a row

Question

There is a list of $63$ items, from which $20$ items are picked in month $1$. What is the chance of $1$ item from the $20$, being picked again in the month $2$?(month $2$ also picks 20 items)

Then further, what is the chance that $2$ items from the $20$ from month $1$ are picked again in month $2$.

Then further, what is the chance that $3$ items from month $1$ are picked again in month $2$..... etc... in theory all the way to the probability of picking all $20$ items identically from month $1$ to month $2$ out of a total of $63$.

My goal is to able to do this for any number, such as the chance that $9$ items from month $1$ are picked again in month $2$.

From my understanding, the chance of getting 1 item the same from $20$ items picked from a total of $63$ from month $1$ to month $2$ is:

$$1 - \frac{_{43}C_{20}}{_{63}C_{20}}$$ or $99.99$%

Answer 1

It seems to me that what you're after is the probability mass function of a hypergeometric distribution.

The hypergeometric distribution gives the probability of k successes in n draws, without replacement, from a finite population of size N that contains exactly K successes.

The pmf of a hypergeometric distribution is well known:

$$ P ( X = k ) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$$

In your case, the population size $N$ is 63. The 'successes' that you are after are the $K=20$ balls that were drawn in the first month. Here, $n = 20$, the number of draws in the second month. You want to find out the probability that exactly $k$ balls that were drawn in the first month were also drawn in the second month, and we can use the formula above to answer that question.

So, for example, the probability that exactly one item is the same (the other 19 are different) in the second month is given by:

\begin{align*} P(X = 1) &= \dfrac{\binom{20}{1} \binom{43}{19}}{\binom{63}{20}}\\ &= \dfrac{169600600}{142894872351}\\ & \approx 0.001186890734 \end{align*}

This may seem quite small, compared with the value you calculated in the statement of your question above. A little bit of consideration, however, will lead to the observation that it would in fact be quite rare for there only to be a single item drawn in both months. In fact, what you have calculated in the statement of the question is the probability that there is at least one item selected in month two that was also selected in month 1; i.e. $P(X \geq 1)$.

Hope this helps

Answer 2

Hint

What you have computed is the probability of getting at least 1 item from the 20.

Can you divide the items into two types : 20 selected ("special") and 43 not selected ("ordinary") in month 1, and work out ?

Hint 2

If you have not heard of the hypergeometric distribution, suppose you want to compute the probability of 3 "special" and 17 "ordinary" selected, does the formula below make sense ?

$$\frac{{20\choose 3}\cdot{43\choose {17}}} {63\choose 20}$$