can you please help me with this question please Find the probability of the following five-card poker hands from a 52-card deck. In poker, aces are either high or low. Two pair (2 cards of one value, 2 of another value) A. 20592/433115 B. 22464/433115 C.3432/433115 D. 5148/433115
we can work through this i am sure, but it is probably better to just google this all stuff about poker on the web
I did am like 3 hours trying to figure it out
ok maybe it would be easier if we figured out what the probability of getting a pair of 2 and a pair of 3 it would be \[\frac{\binom{4}{2}\binom{4}{2}\binom{44}{1}}{\binom{52}{5}}\] i think
reasoning 2 out of the 4 twos 2 out of the 4 threes and one out of the remaining cards that are neither 2 nor 3
but how do I get the answer?
is it more or less clear how i got the answer above?
we are not done, but almost
i computed it for one particular pair, a pair of twos and a pair of threes there are 13 face values in the deck, and you are picking two of them so we have to multiply by \(\binom{13}{2}\) to get the final answer it should be \[\frac{\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{44}{1}}{\binom{52}{5}}\]
are you asking how to compute that number?
2*13= 26
am confuse
\(\binom{13}{2}=\frac{13\times 12}{2}=78\)
\(\binom{44}{1}=44\) and \[\binom{4}{2}=\frac{4\times 3}{2}=6\]
so your numerator is \[78\times 6\times 6\times 44=123552\]
denominator is \[\binom{52}{5}=2598960\]
final answer is \[\frac{123552}{259862}\] or whatever you get when you reduce that fraction
it's not close to any of the answers giving in A B C or D
i am going to stick with that answer especially since that is what wiki says it should be
wiki answer says frequency is \(123552\) and the denominator has to be \(\binom{52}{5}=2598960\)
Yea but am suppose to choose one of the above
here is the reduces fraction you see the percent is what wiki says it should be http://www.wolframalpha.com/input/?i=123552%2F2598960
do I have to sing up to see how I can work the problem?
i'll just guess
thanks for your help
Am smarter than you I chose A and I got it correct yea meeee
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