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#1
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Re: probability homework question
Hi Bruce - Thanks for your response. I had already resolved to stay out of your hair from now on.[img]/images/graemlins/crazy.gif[/img][ QUOTE ]
and the number of ways to get the other 2 cards <font color="red">4*4</font>. [/ QUOTE ]If you have chosen the specific lower and higher rank, that's true. Otherwise there are, for threes, for example, 2*4=<font color="blue">8</font> lower cards, excluding aces, and 9*4=<font color="blue">36</font> higher cards, also excluding aces. And then we have to add in the aces. Where Z can be anything but a three or an ace in A333Z, there are <font color="blue">44</font> cards that can be the Z, not just <font color="red">4</font>. Buzz |
#2
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Re: probability homework question
[ QUOTE ]
If you have chosen the specific lower and higher rank, that's true. Otherwise [/ QUOTE ] Original problem: [ QUOTE ] What is the probability of getting 5 cards dealt (normal deck) and three of the cards are the same denomination (trips), and the other two cards are one above and one below...example: 8 8 8 9 7 [/ QUOTE ] (emphasis added) |
#3
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Re: probability homework question
Hi Bruce - I didn't solve it for the specific example, but for the general case when hero has trips with one card higher and one card lower. For trip eights, it wouldn't only be 88897, but
88896 88895 88894 88893 88892 8889A and then 888T7 888T6 888T5 888T4 888T3 888T2 888TA etc. and then there would be trip sevens, and they're a bit different from trip eights, and then trip sixes which are also a bit different, etc. I did it for all of them. For all cases. It always depends on how you look at things, I guess. I'll get out of your hair now. Thanks. Buzz |
#4
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Re: probability homework question
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Hi Bruce - I didn't solve it for the specific example, but for the general case when hero has trips with one card higher and one card lower. [/ QUOTE ] I understand. I didn't just solve it just for the example either, I solved it for the question which I believe was asked, which is for any 3-of-a-kind where the other 2 cards are one RANK above and one RANK below. While the words RANK were not present in the problem statement, I am pretty sure that is what was intended as I simply don't believe that your interpretation, which is "the other two cards consist of one higher and one lower card", is correctly expressed in English by the words "the other two cards are one above and one below". Gene Fish, please tell us which interpretation you intended. |
#5
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Re: probability homework question
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one RANK above and one RANK below [/ QUOTE ]Hi Bruce - I see another way to read the problem that I had not even considered before. 222A3, 33324, 44435, etc. (but not 44436 or 44425 or 44426 etc.) just exactly one rank above the trips and exactly one rank below the trips. Interesting. Thanks. I'm going back to Omaha now. Buzz |
#6
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Re: probability homework question
[ QUOTE ]
I see another way to read the problem that I had not even considered before. 222A3, 33324, 44435, etc. (but not 44436 or 44425 or 44426 etc.) just exactly one rank above the trips and exactly one rank below the trips. [/ QUOTE ] That was my interpretation. Pococurante's too. |
#7
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Re: probability homework question
guys...thank you so much for the response...bruce...you hit it perfect...dunno why my brain didn't think of that - seems kinda simple when you lay it out.
this forum is the best! thanks again |
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