Exercise from probability

Question

In normal deck of cards each of the cards have some rank (2 have rank 2, 3 have rank 3, ... , J have rank 11, Q have rank 12, K have rank 13 and A have rank 14). 3 cards are drawn. We have the random variable X which gives us the absolute difference of the first 2 drawn cards. Using the random variable X express the probability that the rank of the 3rd card will be between the first two cards exclusively (ex. if the first drawn card is 7 and the second one is 10 what is the probability that the third one will be 8 or 9)?

Answer 1

Denoting this event by $E$ you have:$P\left(E\right)=\sum_{x=2}^{12}P\left\{ E\mid X=x\right\} P\left\{ X=x\right\} =\sum_{x=2}^{12}\frac{4\left(x-1\right)}{50}P\left\{ X=x\right\} $.

After drawing the first $2$ cards there are $50$ left and $4\left(x-1\right)$ of them have a rank between the first two cards.

This is not a full calculation of the probability, but it is an expression of it that makes use of rv $X$.

Edit:

To compute the probability it is better to choose for another route.

Let $R_{1}$ denote the ranking of the first, $R_{2}$ the ranking of the second and $R_{3}$ the ranking of the third card that is drawn.

Let $N$ denote the cardinality of the set $\left\{ R_{1},R_{2},R_{3}\right\} $.

Then:

$P\left(E\right)=P\left\{ E\mid N=1\right\} P\left\{ N=1\right\} +P\left\{ E\mid N=2\right\} P\left\{ N=2\right\} +P\left\{ E\mid N=3\right\} P\left\{ N=3\right\} $.

Here:

$P\left\{ E\mid N=1\right\} =0=P\left\{ E\mid N=2\right\} $, $P\left\{ E\mid N=3\right\} =\frac{1}{3}$ and $P\left\{ N=3\right\} =\frac{48}{51}\times\frac{44}{50}$.

So we end up with:

$P\left(E\right)=\frac{1}{3}\times\frac{48}{51}\times\frac{44}{50}=\frac{352}{1275}$

Answer 2

A different point of view:

How many pairs (a,b) are there for $|a-b|=k$?

if k>0 then {a,b} gives two sets. also if the least element is given this set is uniquely determined. There are 13-k values for the least one. therefore there are 2(13-k) such pairs.

How many representations in actual cards does each pair have? 16 since each one of the two can be ace,spade,diamond or heart.

so for each k there are 32(13-k) ways to get the first two cards with that space in between. that means that the third card can be any out of 4k cards.

therefore for each k there are $32(13-k)(4k)=128(13k-k^2)=1664k-128k^2.$

So now we just have to add this for all positive values of k with $k\leq13$.

but $\sum_{k=1}^{13}1664k-128k^2=\sum_{k=1}^{13}1664k-\sum_{k=1}^{13}128k^2=1664\sum_{k=1}^{13}k-128\sum_{k=1}^{13}k^2=1664(\frac{14*13}{2})-128(\frac{13(14)(27)}{6})=46592$

That means that there are 46592 triples that work.

Therefore the probability you want is $46592$ over the number of triples which is just $54*53*52=148824$ and $\frac{46592}{148824}\approx 0.31306778476$

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The part of this you are asking is given a k(in your case X)what is the probability:

its $1664k-128k^2$ as explained above.