Timeline for If I deal 7 cards what is the probability that there is at least one card from each suit.
Current License: CC BY-SA 4.0
19 events
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| Feb 19, 2023 at 19:32 | comment | added | true blue anil | @Thomas Andrews: Ok, thanks | |
| Feb 19, 2023 at 19:25 | comment | added | Thomas Andrews | No, the second value, $\binom{4}1\binom{39}7$ can't be seen as counting anything that is useful here. In inclusion/exclusion, the individual sets $|A_1|,\dots,|A_n|$ count things, but if these are all the same, the value $n|A_1|$ doesn't count anything. | |
| Feb 19, 2023 at 8:57 | comment | added | true blue anil | @ThomasAndrews: I understand that neither exactly one missing nor at least one missing correctly reflects the situation. But if I want to write something linguistically rather than in set cardinalities, does one missing pass muster ? | |
| Feb 17, 2023 at 6:57 | history | edited | true blue anil | CC BY-SA 4.0 |
Verbiage pruned
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| Feb 16, 2023 at 23:10 | comment | added | Thomas Andrews | If $N$ is the random variable counting the suits not in your $7$ cards, you can consider $e=\frac{4\binom{39}{7}}{\binom{52}{7}}$ as the expected value of $N.$ So $e=P(N=1)+2P(N=2)+3P(N=3).$ But you are saying $e=P(N>0)=P(N=1)+P(N=2)+P(N=3).$ | |
| Feb 16, 2023 at 23:04 | comment | added | Thomas Andrews | $[\text{all combos}]-[\text{at least one suit missing}]$ is the exact count you want. There would be no reason to have any terms after that. But your second terms is not $[\text{at least one suit missing}],$ it is an over-count of that value. | |
| Feb 16, 2023 at 23:00 | comment | added | Thomas Andrews | Interpreted your way, that answer would be wrong, too. But the "particular suit(s)" phrase, when read carefully, can be interpreted to be correct. Here, they subtract the hands which don't have clubs, the hands which don't have diamonds, etc. $\binom{39}{7}$ counts each of those. But if you subtract $4\binom{39}7$ you are not subtracting the number of hands with at least one suit missing, but a number bigger than that - you are attempting to count that, but you are over counting that number, hence the need for correction terms. | |
| Feb 16, 2023 at 22:49 | comment | added | true blue anil | @Thomas Andrews: (Similar Q) "Count the number of five card hands. Remove the number of five card hands missing at least a particular suit and which suit that was (among possibly missing more). Add back the number of five card hands missing at least two particular suits, subtracting missing at least three particular suits, etc... You can end at that point since it is impossible to be missing four or more suits at a time here" What's wrong with what I wrote ? math.stackexchange.com/questions/4328530/… | |
| Feb 16, 2023 at 22:29 | comment | added | Thomas Andrews | He wrote $+$ signs for the sum of the counts, and used $\cup$ for the union. The point is $|A_1|+|A_2|+|A_3|+|A_4|$ is not the same as $|A_1\cup A_2\cup A_3\cup A_4|.$ The latter is the number of "at least one missing," but the former, which is $\binom41\binom{39}7,$ does not count the cases of "at least one missing." | |
| Feb 16, 2023 at 21:49 | comment | added | true blue anil | @ThomasAndrews: I can't argue with you on technicalities. If the wording requires minor changes to be technically correct, you are free to correct it. I'll look at my dusty notes tomorrow ! | |
| Feb 16, 2023 at 21:20 | comment | added | true blue anil | You wrote + signs instead of U signs which confused me.And the formula doesn't end at the top line, see all the lines. In the way I have done it, from all combos, first I have subtracted one suit without caring whether other suit(s) have got subtracted or not , and so on, and so forth, so it is at least one suit missing. | |
| Feb 16, 2023 at 21:17 | comment | added | Thomas Andrews | @trueblueanil you've confused $\cup$ for $\cap.$ And I agree, $\binom{4}{1}\binom{39}7$ does not count "at least one missing." Otherwise, you wouldn't need the other terms. It overcounts "at least one suit missing," which requires the later terms for correction. | |
| Feb 16, 2023 at 20:58 | comment | added | Mike | No it does not. If a hand has all 4 suits, then it does not fall into [Hands with no Diamonds] $\cup$ [Hands with no Clubs] $\cup$ [Hands with no Spades] $\cup$[Hands with no Hearts] | |
| Feb 16, 2023 at 20:52 | comment | added | true blue anil | @Mike: I don't understand: What you have written amounts to the null set ! | |
| Feb 16, 2023 at 20:43 | comment | added | Mike | I do agree that it is Inclusion-Exclusion as done here, but I also have an issue with how the top line of this is written. It isn't for example [At least one suit missing] ; it is instead [Hands with no Diamonds] $+$ [Hands with no Clubs] $+$ [Hands with no Spades] $+$ [Hands With no Hearts]. As in Thomas Andrews' answer below, we ultimately want the cardinality of [Hands with no Diamonds] $\cup$ [Hands with no Clubs] $\cup$ [Hands with no Spades] $\cup$ [Hands With no Hearts], and then we subtract this from ${{52} \choose 7}$. | |
| Feb 16, 2023 at 20:37 | comment | added | true blue anil | Glad to have been of help ! $\;\;$ :) | |
| Feb 16, 2023 at 20:36 | comment | added | Aidan | Thank you so much this makes a lot more sense! | |
| Feb 16, 2023 at 20:34 | history | edited | true blue anil | CC BY-SA 4.0 |
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| Feb 16, 2023 at 20:27 | history | answered | true blue anil | CC BY-SA 4.0 |