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”)?

- anonymous

- chestercat

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- anonymous

four choices for each up top right? and \[\binom{52}{5}\] for the denominator

- anonymous

Yes, I've got the denom!

- 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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## More answers

- 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

the denomator is the number of ways you can choose 5 cards from a set of 52

- anonymous

OH sorry, I meant numerator!!

- anonymous

in other words "52 choose 5" written as \[\binom{52}{5}\] or sometimes \[_{52}C_5\]

- anonymous

I meant I don't know what goes in the numerator! Silly me! Typo.

- anonymous

computed via \[\frac{52\times 51\times 50\times 49\times 48}{5!}\]

- 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

ooh numerator

- anonymous

Yeah sorry! Thanks for your patience!

- anonymous

4 choices for each, so \(4\times 4\times 4\times 4\times 4\)

- anonymous

I don't really get where that's coming from.. hmm. Why can it just be 4s?

- anonymous

4 choices of aces right?

- 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

that is taken care of in the denominator

- 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

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

4 ways to get an ace, 4 ways to get a 2 etc so \(4^5\) ways all together

- anonymous

Okay, I think I understand. The numerator is always the wild card for me

- 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

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

these kind of problems, the question is always "how many ways"

- anonymous

I don't get what you've said. I'm sorry.

- anonymous

I get how they apply in their own situations, but I'm SO bad at generalizing these sorts of rationales

- anonymous

The next question is asking me how ANY straight could be obtained and I don't even know where to start.

- anonymous

oh that is ok we are almost there

- anonymous

Like what goes on top? No idea.

- 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

we got one straight right? one that begins with an ace and ends with a 5

- anonymous

yep!

- anonymous

ok so now we know the probability of that particular straight (that is why that question came first)

- anonymous

now all we have to do is figure out how many different possible straights there are

- anonymous

no idea!

- anonymous

so we count them!

- anonymous

not a poker player i guess huh?

- anonymous

Not in a million years.

- 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

i count ten

- anonymous

Okay.

- anonymous

btw you can't have for example jack to 2

- anonymous

so we are done

- anonymous

multiply the above answer by 10

- anonymous

oh. So .000394 *10

- anonymous

i.e. \[\frac{10\times 4^5}{\binom{52}{5}}\]

- anonymous

http://www.wolframalpha.com/input/?i=%2810*4^5%29%2F%2852+choose+5%29

- anonymous

oh if you had the previous answer then yeah move the decimal one to the left

- anonymous

right

- anonymous

whatever

- anonymous

That's not quite right.

- anonymous

The answer is .00355

- anonymous

i get \[0.00394\]

- anonymous

So did I, from what you told me. The textbook says it's .00355

- anonymous

they are wrong

- anonymous

umm..

- anonymous

ok?

- anonymous

unless...

- anonymous

you are supposed to subtract the number of flushes
which i doubt

- anonymous

I have no idea. This is all bullcrap to me. I don't like math.

- anonymous

Let STATA do it, that's what I say.

- anonymous

But I'll put down the answer we worked out and ask later.

- anonymous

Thanks for your help.

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