Playing Cards question

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)

• Does ace count as next to both two and king? – Ross Millikan Jan 23 at 22:09
• Hint: In your example, the probability that none of your $X$ cards are among $5,6,7$ is $\binom {40}{X}/\binom {52}{X}$. – lulu Jan 23 at 22:09
• 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. – Brooks Harrel Jan 23 at 23:36

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.

• 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) – Brooks Harrel Jan 23 at 23:39
• That's a binomial coefficient. Sometimes written $^{48}C_X$. – lulu Jan 23 at 23:45

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.

• 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. – lulu Jan 23 at 22:25
• @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. – Ross Millikan Jan 23 at 22:34
• 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. – lulu Jan 23 at 22:36
• @lulu: I see what you mean. I was thinking of matches between the cards drawn from the first deck. Your approach is much simpler. – Ross Millikan Jan 23 at 22:39