anonymous
  • anonymous
Five cards are dealt from a standard 52-card deck. What is the probability that we draw a 1 ace, 1 two, 1 three, 1 four, and 1 five (this is one way to get a “straight”)?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
four choices for each up top right? and \[\binom{52}{5}\] for the denominator
anonymous
  • anonymous
Yes, I've got the denom!
anonymous
  • anonymous
The probability of each on their own is 4/52, but I know that changes as you draw each down to 4/51, 4/50, etc...but how do I express that?

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anonymous
  • anonymous
I still don't know what goes in the denominator. I can reliably determine the denominator, but I have no clue what's on top!
anonymous
  • anonymous
the denomator is the number of ways you can choose 5 cards from a set of 52
anonymous
  • anonymous
OH sorry, I meant numerator!!
anonymous
  • anonymous
in other words "52 choose 5" written as \[\binom{52}{5}\] or sometimes \[_{52}C_5\]
anonymous
  • anonymous
I meant I don't know what goes in the numerator! Silly me! Typo.
anonymous
  • anonymous
computed via \[\frac{52\times 51\times 50\times 49\times 48}{5!}\]
anonymous
  • anonymous
I was thinking, yes yes I know the denominator. Why is he/she telling me? I realize my mistake. Numerator is the concern.
anonymous
  • anonymous
ooh numerator
anonymous
  • anonymous
Yeah sorry! Thanks for your patience!
anonymous
  • anonymous
4 choices for each, so \(4\times 4\times 4\times 4\times 4\)
anonymous
  • anonymous
I don't really get where that's coming from.. hmm. Why can it just be 4s?
anonymous
  • anonymous
4 choices of aces right?
anonymous
  • anonymous
Yes, but there's more going on it seems. Why is it not including the reducing total or something like 4/52, 4/51, 4/50, /4/49
anonymous
  • anonymous
that is taken care of in the denominator
anonymous
  • anonymous
It is? Ahh alright...Your answer is right, but I'll have to ask my teacher for clarification about why it's taken care of. It's late and I won't mess with that. Thank you!
anonymous
  • anonymous
think of it this way it is all the "counting principle" if the number of ways to do one thing is n and another m, then there are \(n\times m\) ways to go it together
anonymous
  • anonymous
4 ways to get an ace, 4 ways to get a 2 etc so \(4^5\) ways all together
anonymous
  • anonymous
Okay, I think I understand. The numerator is always the wild card for me
anonymous
  • anonymous
then the number of ways to pick the 5 cards is 52 choose 5, which is again really the counting principle in action
anonymous
  • anonymous
if it was for example 3 spades and two clubs, the answer would be \[\frac{\binom{13}{3}\binom{13}{2}}{\binom{52}{5}}\]
anonymous
  • anonymous
these kind of problems, the question is always "how many ways"
anonymous
  • anonymous
I don't get what you've said. I'm sorry.
anonymous
  • anonymous
I get how they apply in their own situations, but I'm SO bad at generalizing these sorts of rationales
anonymous
  • anonymous
The next question is asking me how ANY straight could be obtained and I don't even know where to start.
anonymous
  • anonymous
oh that is ok we are almost there
anonymous
  • anonymous
Like what goes on top? No idea.
anonymous
  • anonymous
it's no longer just 4s, and it's not comb(13,2) either. So it's a complete effing mystery to my mind.
anonymous
  • anonymous
we got one straight right? one that begins with an ace and ends with a 5
anonymous
  • anonymous
yep!
anonymous
  • anonymous
ok so now we know the probability of that particular straight (that is why that question came first)
anonymous
  • anonymous
now all we have to do is figure out how many different possible straights there are
anonymous
  • anonymous
no idea!
anonymous
  • anonymous
so we count them!
anonymous
  • anonymous
not a poker player i guess huh?
anonymous
  • anonymous
Not in a million years.
anonymous
  • anonymous
ace to 5 2 to 6 3 to 7 4 to 8 5 to 9 6 to 10 7 to jack 8 to queen 9 to king 10 to ace
anonymous
  • anonymous
i count ten
anonymous
  • anonymous
Okay.
anonymous
  • anonymous
btw you can't have for example jack to 2
anonymous
  • anonymous
so we are done
anonymous
  • anonymous
multiply the above answer by 10
anonymous
  • anonymous
oh. So .000394 *10
anonymous
  • anonymous
i.e. \[\frac{10\times 4^5}{\binom{52}{5}}\]
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=%2810*4^5%29%2F%2852+choose+5%29
anonymous
  • anonymous
oh if you had the previous answer then yeah move the decimal one to the left
anonymous
  • anonymous
right
anonymous
  • anonymous
whatever
anonymous
  • anonymous
That's not quite right.
anonymous
  • anonymous
The answer is .00355
anonymous
  • anonymous
i get \[0.00394\]
anonymous
  • anonymous
So did I, from what you told me. The textbook says it's .00355
anonymous
  • anonymous
they are wrong
anonymous
  • anonymous
umm..
anonymous
  • anonymous
ok?
anonymous
  • anonymous
unless...
anonymous
  • anonymous
you are supposed to subtract the number of flushes which i doubt
anonymous
  • anonymous
I have no idea. This is all bullcrap to me. I don't like math.
anonymous
  • anonymous
Let STATA do it, that's what I say.
anonymous
  • anonymous
But I'll put down the answer we worked out and ask later.
anonymous
  • anonymous
Thanks for your help.

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