Question

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”)?

Answer 1

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

Answer 2

Yes, I've got the denom!

Answer 3

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?

Answer 4

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!

Answer 5

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

Answer 6

OH sorry, I meant numerator!!

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

I was thinking, yes yes I know the denominator. Why is he/she telling me? I realize my mistake. Numerator is the concern.

Answer 11

ooh numerator

Answer 12

Yeah sorry! Thanks for your patience!

Answer 13

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

Answer 14

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

Answer 15

4 choices of aces right?

Answer 16

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

Answer 17

that is taken care of in the denominator

Answer 18

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!

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

then the number of ways to pick the 5 cards is 52 choose 5, which is again really the counting principle in action

Answer 23

if it was for example 3 spades and two clubs, the answer would be \[\frac{\binom{13}{3}\binom{13}{2}}{\binom{52}{5}}\]

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

oh that is ok we are almost there

Answer 29

Like what goes on top? No idea.

Answer 30

it's no longer just 4s, and it's not comb(13,2) either. So it's a complete effing mystery to my mind.

Answer 31

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

Answer 32

yep!

Answer 33

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

Answer 34

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

Answer 35

no idea!

Answer 36

so we count them!

Answer 37

not a poker player i guess huh?

Answer 38

Not in a million years.

Answer 39

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

Answer 40

i count ten

Answer 41

Okay.

Answer 42

btw you can't have for example jack to 2

Answer 43

so we are done

Answer 44

multiply the above answer by 10

Answer 45

oh. So .000394 *10

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

right

Answer 50

whatever

Answer 51

That's not quite right.

Answer 52

The answer is .00355

Answer 53

i get \[0.00394\]

Answer 54

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

Answer 55

they are wrong

Answer 56

umm..

Answer 57

ok?

Answer 58

unless...

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

Thanks for your help.