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anonymous
 one year ago
Five cards are dealt from a standard 52card 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”)?
anonymous
 one year ago
Five cards are dealt from a standard 52card 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”)?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0four choices for each up top right? and \[\binom{52}{5}\] for the denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I've got the denom!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0the denomator is the number of ways you can choose 5 cards from a set of 52

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OH sorry, I meant numerator!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in other words "52 choose 5" written as \[\binom{52}{5}\] or sometimes \[_{52}C_5\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I meant I don't know what goes in the numerator! Silly me! Typo.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0computed via \[\frac{52\times 51\times 50\times 49\times 48}{5!}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was thinking, yes yes I know the denominator. Why is he/she telling me? I realize my mistake. Numerator is the concern.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah sorry! Thanks for your patience!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04 choices for each, so \(4\times 4\times 4\times 4\times 4\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't really get where that's coming from.. hmm. Why can it just be 4s?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04 choices of aces right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0that is taken care of in the denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0think 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
 one year ago
Best ResponseYou've already chosen the best response.04 ways to get an ace, 4 ways to get a 2 etc so \(4^5\) ways all together

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I think I understand. The numerator is always the wild card for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then the number of ways to pick the 5 cards is 52 choose 5, which is again really the counting principle in action

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it was for example 3 spades and two clubs, the answer would be \[\frac{\binom{13}{3}\binom{13}{2}}{\binom{52}{5}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these kind of problems, the question is always "how many ways"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't get what you've said. I'm sorry.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get how they apply in their own situations, but I'm SO bad at generalizing these sorts of rationales

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The next question is asking me how ANY straight could be obtained and I don't even know where to start.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh that is ok we are almost there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like what goes on top? No idea.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's no longer just 4s, and it's not comb(13,2) either. So it's a complete effing mystery to my mind.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we got one straight right? one that begins with an ace and ends with a 5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so now we know the probability of that particular straight (that is why that question came first)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now all we have to do is figure out how many different possible straights there are

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not a poker player i guess huh?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not in a million years.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ace 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
 one year ago
Best ResponseYou've already chosen the best response.0btw you can't have for example jack to 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0multiply the above answer by 10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i.e. \[\frac{10\times 4^5}{\binom{52}{5}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%2810*4^5%29%2F%2852+choose+5%29

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh if you had the previous answer then yeah move the decimal one to the left

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's not quite right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer is .00355

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So did I, from what you told me. The textbook says it's .00355

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you are supposed to subtract the number of flushes which i doubt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have no idea. This is all bullcrap to me. I don't like math.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let STATA do it, that's what I say.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But I'll put down the answer we worked out and ask later.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for your help.
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