0
$\begingroup$

Say I have a hand of X playing cards drawn from a standard 52-card deck, where X can be 1 through 10.

Then say I also have a second deck of the 52 standard playing cards, and draw a card from there. (alternately, say I rolled 1d13)

I need to calculate the odds (at least approximately) that at least 1 card in my X-card hand will A: MATCH the number drawn/rolled, or B: BE WITHIN ONE OF that value. (So if I drew/rolled a 6, B would be satisfied if the cards in my hand included a 5, 6, or 7)

$\endgroup$
  • $\begingroup$ Does ace count as next to both two and king? $\endgroup$ – Ross Millikan Jan 23 at 22:09
  • 1
    $\begingroup$ Hint: In your example, the probability that none of your $X$ cards are among $5,6,7$ is $\binom {40}{X}/\binom {52}{X}$. $\endgroup$ – lulu Jan 23 at 22:09
  • $\begingroup$ Aces are not intended to be within 1 of the king, or vice versa. I’m treating it as a method of generating a number between 1 and 13. $\endgroup$ – Brooks Harrel Jan 23 at 23:36
1
$\begingroup$

Assuming that you mean ranks $\pmod {13}$, so that Aces are next to both $2's$ and Kings:

Then, for part $A$ we note that there are exactly $4$ cards that match the preferred rank. Thus the probability that none of your $X$ cards matches the preferred rank is $$\binom {48}X\Big /\binom{52}X$$ It follows that the answer you want is $$1\,-\,\binom {48}X\Big /\binom{52}X$$

Similarly for $B$ the answer is $$1\,-\,\binom {40}X\Big /\binom{52}X$$

If you don't want Aces to be next to both $2's$ and Kings then $A$ stays the same, but you have to modify $B$ to account for the possibility that the preferred rank only has one neighbor.

$\endgroup$
  • $\begingroup$ I appear to have a gap in my fundamental knowledge- what is that (48 X) notation? (i.e. just the term for it, I can look up rest from there) $\endgroup$ – Brooks Harrel Jan 23 at 23:39
  • $\begingroup$ That's a binomial coefficient. Sometimes written $^{48}C_X$. $\endgroup$ – lulu Jan 23 at 23:45
0
$\begingroup$

When $X=1$ it is easy (assuming the ranks are circular so ace is next to both two and king). You have $1/13$ chance of a match and $3/13$ chance of being within one.

When $X=2$ part A should be easy at $2/13$ except that the two cards you drew might match. The chance of a match is $\frac 3{51}$ because after you draw the first card there are $3$ of the remaining $51$ that match. The chance the die matches one of the two cards is then $\frac {48}{51}\cdot \frac 2{13}+\frac 3{51}\cdot \frac 1{13}=\frac {99}{663}=\frac {33}{221}\approx 0.1493$ Part B has more trouble with overlap. If the two cards match the chance is $3/13$, if they are one apart $4/13$, if they are two apart $5/13$ and if the are further $6/13$. Getting the exact value is a fair mess, and for $X$ larger it is messier yet.

$\endgroup$
  • $\begingroup$ Unless I am misreading, always possible, you can ignore duplicates and such. For part $A$ the probability that no cards match the favored rank is $\binom {48}X/\binom {52}X$, no? So the probability that at least one matches can be found by subtraction. $\endgroup$ – lulu Jan 23 at 22:25
  • $\begingroup$ @lulu: I took the 1d13 to mean that we are only interested in matches in rank. The fact that part B is about being one away supports this. Then it is possible to draw matching cards in the first part. $\endgroup$ – Ross Millikan Jan 23 at 22:34
  • $\begingroup$ Yes...that's how I am reading it as well. There are $4$ cards of the preferred rank, hence $48$ non-matches. Anyway, I posted my version below. Let me know if it is wrong. $\endgroup$ – lulu Jan 23 at 22:36
  • $\begingroup$ @lulu: I see what you mean. I was thinking of matches between the cards drawn from the first deck. Your approach is much simpler. $\endgroup$ – Ross Millikan Jan 23 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.