samuelaplc

2022-09-06

I'm trying to understand what kind of probability distribution I need to use in order to calculate a very simple example using a deck of cards.

Assume that there is a standard deck of cards (52 cards):

Let X be the number of non-hearts until I get 13 hearts (without replacement). What would be the distribution I need to use?

Assume that there is a standard deck of cards (52 cards):

Let X be the number of non-hearts until I get 13 hearts (without replacement). What would be the distribution I need to use?

procjenomuj

Beginner2022-09-07Added 8 answers

Step 1

Place the cards in a row. If we only discern hearts and non-hearts then there are $(}\genfrac{}{}{0ex}{}{52}{13}{\textstyle )$ arrangements. Now fix $n\in \{13,\dots ,52\}$ and notice that there are $(}\genfrac{}{}{0ex}{}{n-1}{12}{\textstyle )$ arrangements such that the last heart is placed on spot n. So denoting N as the spot of the last heart we come to:

$P(N=n)={\textstyle (}\genfrac{}{}{0ex}{}{n-1}{12}{\textstyle )}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{13}{\textstyle )}}^{-1}$

Now realize that $X=N-13$ so that $P(X=k)=P(N=k+13)={\textstyle (}\genfrac{}{}{0ex}{}{k+12}{12}{\textstyle )}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{13}{\textstyle )}}^{-1}$

This for $k=0,1,\dots ,39$

Step 2

Equality $\sum _{n=13}^{52}P(N=n)=1$ leads to the observation that $\sum _{n=12}^{51}{\textstyle (}\genfrac{}{}{0ex}{}{n}{12}{\textstyle )}={\textstyle (}\genfrac{}{}{0ex}{}{52}{13}{\textstyle )}$. More generally it can be shown (e.g. by induction) that $\sum _{k=r}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{k}{r}{\textstyle )}={\textstyle (}\genfrac{}{}{0ex}{}{n+1}{r+1}{\textstyle )}$.

So a more general setting here would lead to a combinatorial proof of this equality.

Place the cards in a row. If we only discern hearts and non-hearts then there are $(}\genfrac{}{}{0ex}{}{52}{13}{\textstyle )$ arrangements. Now fix $n\in \{13,\dots ,52\}$ and notice that there are $(}\genfrac{}{}{0ex}{}{n-1}{12}{\textstyle )$ arrangements such that the last heart is placed on spot n. So denoting N as the spot of the last heart we come to:

$P(N=n)={\textstyle (}\genfrac{}{}{0ex}{}{n-1}{12}{\textstyle )}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{13}{\textstyle )}}^{-1}$

Now realize that $X=N-13$ so that $P(X=k)=P(N=k+13)={\textstyle (}\genfrac{}{}{0ex}{}{k+12}{12}{\textstyle )}{{\textstyle (}\genfrac{}{}{0ex}{}{52}{13}{\textstyle )}}^{-1}$

This for $k=0,1,\dots ,39$

Step 2

Equality $\sum _{n=13}^{52}P(N=n)=1$ leads to the observation that $\sum _{n=12}^{51}{\textstyle (}\genfrac{}{}{0ex}{}{n}{12}{\textstyle )}={\textstyle (}\genfrac{}{}{0ex}{}{52}{13}{\textstyle )}$. More generally it can be shown (e.g. by induction) that $\sum _{k=r}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{k}{r}{\textstyle )}={\textstyle (}\genfrac{}{}{0ex}{}{n+1}{r+1}{\textstyle )}$.

So a more general setting here would lead to a combinatorial proof of this equality.

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