Timeline for how many different bridge hands have exactly 2 six-card suits?
Current License: CC BY-SA 3.0
12 events
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| May 11, 2015 at 13:40 | comment | added | JMoravitz | @Jimfy I will again reiterate, "two six-card suits" means that there is a suit with six cards, a second suit different from the first with six cards, and a third suit different than both of the previous two with one card. | |
| May 11, 2015 at 13:27 | comment | added | JMoravitz | @Jimfy no it should not. That almost answers the question of "how many bridge hands have two suits with at least 6 cards or one suit with at least 12 cards." which is a very different question. (I say almost because the C(13,6)C(13,7)C(4,2) term should not have a C(4,2) but rather a C(4,1)C(3,1) instead since you can clearly distinguish between the suit with 7 cards and the suit with 6 cards even without knowing what the suits are). Only the third term ( C(13,6)C(4,2)C(13,6)C(26,1) ) is needed. All other terms are counting cases which do not match the problem statement. | |
| May 11, 2015 at 4:25 | comment | added | JimfyWinsy | If the question is asking:How many different bridge hands have exactly 2 six-card suits?Should the answer be:C(13,12)C(4,1)(39,1)+C(13,13)C(4,1)+C(13,6)C(4,2)C(13,6)C(26,1)+C(13,6)C(13,7)C(4,2)? | |
| May 10, 2015 at 18:19 | comment | added | wythagoras | Yeah, you are right. +1 to your answer. | |
| May 10, 2015 at 18:18 | history | edited | wythagoras | CC BY-SA 3.0 |
In the 4th case, it does matter which is the first and which is the second suit.
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| May 10, 2015 at 18:04 | comment | added | JMoravitz | Teachers and tutors alike are prone to the same human mistakes as the rest of us. The tutor seems to have been confused as to the specific interpretation of the question or the specifics of answering a question like this in general. Even with his (flawed) presumed interpretation of the problem, he has overcounted, in which case your alternate answer seems to (almost) fix his mistakes (you still have a $\binom{4}{1}\binom{3}{1}$ in the third summand that should be a $\binom{4}{2}$). I would again reiterate that in problems like these "six-card suit" implies "exactly six-card" not "at least." | |
| May 10, 2015 at 17:58 | comment | added | wythagoras | @JMoravitz Indeed you are right, I edited it, but it doesn't explain why the tutor got the answer thet JimfyWinsy gave. The 40 implies that there is chosen form the remaining 52-12=40 cards, so including those form the same suit. | |
| May 10, 2015 at 17:56 | history | edited | wythagoras | CC BY-SA 3.0 |
per the comment
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| May 10, 2015 at 17:42 | comment | added | JMoravitz | @wythagoras the condition seems quite clear. A hand with one twelve-card suit is definitely not a hand with exactly two six-card suits. Also, read my answer as to why you should not use $\binom{3}{1}$ for selecting the second suit. Instead you should have selected both sextuple suits simultaneously. You have as a result doublecounted. | |
| May 10, 2015 at 17:16 | history | edited | wythagoras | CC BY-SA 3.0 |
added 324 characters in body
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| May 10, 2015 at 17:03 | comment | added | JimfyWinsy | But one of my tutor give me such an answer:C(4,1)*C(13,12)*C(40,1)+C(4,2)*C(13,6)*C(13,6)*C(40,1) | |
| May 10, 2015 at 16:43 | history | answered | wythagoras | CC BY-SA 3.0 |