Playing Cards question

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)

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 '19 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 '19 at 23:36

lulu · Answer 1 · 2019-01-24T12:03:35.343

1

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.

edited Jan 24 '19 at 12:03

answered Jan 23 '19 at 22:36

lulu

70,402

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 '19 at 23:39
That's a binomial coefficient. Sometimes written $^{48}C_X$. – lulu Jan 23 '19 at 23:45

score 0 · Answer 2 · answered Jan 23 '19 at 22:22

0

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.

answered Jan 23 '19 at 22:22

Ross Millikan

374,822

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 '19 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 '19 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 '19 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 '19 at 22:39

Playing Cards question

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