Timeline for combinatorics: deck of cards and suits
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 22, 2022 at 6:35 | vote | accept | MeTAZOIS | ||
| Sep 22, 2022 at 0:44 | comment | added | JMoravitz | A final note about enumeration vs probability., I had used binomial coefficients here with the intention of ignoring the order in which cards are drawn, treating all hands containing the same set of cards as "the same" since it does not matter when calculating the probability so long as you are consistent how you do so. If you did want the enumeration and you did care about the order of the cards, just multiply by $6!$ to reintroduce order of cards as mattering to the count. | |
| Sep 22, 2022 at 0:40 | comment | added | JMoravitz | @user2661923 no, I had addressed it as I intended. I read your comment as questioning the rephrasing in terms of "at least one not present" in addition to questioning whether it should have been enumeration or probability. The first part is answered in the first comment, the second part addressed to the both of you in the second comment. Further clarification to the OP about inclusion exclusion correctly using subtraction in the second comment. | |
| Sep 22, 2022 at 0:37 | comment | added | JMoravitz | @MeTAZOIS The enumeration problem and the probability problem are one and the same, the probability problem just takes it one step further and takes the ratio. Ignore the division if you want the enumeration. As for why you see some subtraction going on, that is simply how inclusion-exclusion works. And yes, inclusion-exclusion works just as well for enumeration as it does for probability. You alternately add and subtract increasing intersections, adding all singles, subtracting intersections of two, adding of three... | |
| Sep 22, 2022 at 0:37 | comment | added | user2661923 | @JMoravitz My previous comment was suggesting that the original poster was not trying to compute a probability, but was instead trying to enumerate. | |
| Sep 22, 2022 at 0:35 | comment | added | JMoravitz | @user2661923 "at most 3 suits are present" and "at least one suit is not present" are equivalent statements. | |
| Sep 21, 2022 at 23:39 | comment | added | MeTAZOIS | It's actually an enumeration problem, it's probably my problem describing the question unclearly, but I don't see why you wrote $4 * {39 \choose 6}$ to minus the later part, I thought it was going to be the sum of those 3 identities to obtain the total number of possible arrangements. | |
| Sep 21, 2022 at 23:23 | comment | added | user2661923 | I am unsure whether the original poster is presenting a Probability problem, or an enumeration problem. I didn't see anything in the posting that asked for the probability that not all $(4)$ suits are involved. | |
| Sep 21, 2022 at 22:55 | history | answered | JMoravitz | CC BY-SA 4.0 |