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'''[[MA 375 Spring 09 HW 5|Back]]''' | '''[[MA 375 Spring 09 HW 5|Back]]''' | ||
| − | + | By a hand containing 5 different kinds, is this including face value, suit and sequence or just face value? | |
| − | + | I think it's just face value, because 5 different suits wouldn't make sense. -Zoe | |
| − | I | + | 13 different kinds and 4 different suits |
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| + | i did (13 choose 5)*4 on top (to represent there are 13 numbers, pick one ad then taking the suits into account - ace of hearts is different from ace of spades) and (52 choose 5) on the bottom to show all the possible hands with 5 cards. | ||
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| + | I think that is mostly right but I believe it should be (13 choose 5)*4^5, in order to account for there being 4 choices for each of the kinds.--[[User:Spfeifer|Spfeifer]] 16:55, 18 February 2009 (UTC) | ||
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| + | I haven't confirmed this solution with anyone, but I found it easier to get the probability of the complement, having a hand with at least two cards of the same type. For the top: Choose any card (52C1), choose another card of that type (3C1) and choose 3 arbitrary cards (50C3)-[[User:jdrummon|jdrummon]] | ||
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| + | Finding the complement, I think, would take a lot longer because you not only have to find the probability of getting a hand with 2 cards the same but getting a hand with 3 cards, 4 cards, and 5 cards the same and them sum these probabilities. Also, using the method you gave above will lead to over count. -[[User:cwithey|cwithey]] | ||
Latest revision as of 09:48, 19 February 2009
By a hand containing 5 different kinds, is this including face value, suit and sequence or just face value?
I think it's just face value, because 5 different suits wouldn't make sense. -Zoe
13 different kinds and 4 different suits
i did (13 choose 5)*4 on top (to represent there are 13 numbers, pick one ad then taking the suits into account - ace of hearts is different from ace of spades) and (52 choose 5) on the bottom to show all the possible hands with 5 cards.
I think that is mostly right but I believe it should be (13 choose 5)*4^5, in order to account for there being 4 choices for each of the kinds.--Spfeifer 16:55, 18 February 2009 (UTC)
I haven't confirmed this solution with anyone, but I found it easier to get the probability of the complement, having a hand with at least two cards of the same type. For the top: Choose any card (52C1), choose another card of that type (3C1) and choose 3 arbitrary cards (50C3)-jdrummon
Finding the complement, I think, would take a lot longer because you not only have to find the probability of getting a hand with 2 cards the same but getting a hand with 3 cards, 4 cards, and 5 cards the same and them sum these probabilities. Also, using the method you gave above will lead to over count. -cwithey
