## anonymous one year ago 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”)?

1. anonymous

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

2. anonymous

Yes, I've got the denom!

3. 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?

4. 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!

5. anonymous

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

6. anonymous

OH sorry, I meant numerator!!

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

ooh numerator

12. anonymous

Yeah sorry! Thanks for your patience!

13. anonymous

4 choices for each, so $$4\times 4\times 4\times 4\times 4$$

14. anonymous

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

15. anonymous

4 choices of aces right?

16. 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

17. anonymous

that is taken care of in the denominator

18. 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!

19. 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

20. anonymous

4 ways to get an ace, 4 ways to get a 2 etc so $$4^5$$ ways all together

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

oh that is ok we are almost there

29. anonymous

Like what goes on top? No idea.

30. 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.

31. anonymous

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

32. anonymous

yep!

33. anonymous

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

34. anonymous

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

35. anonymous

no idea!

36. anonymous

so we count them!

37. anonymous

not a poker player i guess huh?

38. anonymous

Not in a million years.

39. 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

40. anonymous

i count ten

41. anonymous

Okay.

42. anonymous

btw you can't have for example jack to 2

43. anonymous

so we are done

44. anonymous

multiply the above answer by 10

45. anonymous

oh. So .000394 *10

46. anonymous

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

47. anonymous
48. anonymous

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

49. anonymous

right

50. anonymous

whatever

51. anonymous

That's not quite right.

52. anonymous

53. anonymous

i get $0.00394$

54. anonymous

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

55. anonymous

they are wrong

56. anonymous

umm..

57. anonymous

ok?

58. anonymous

unless...

59. anonymous

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

60. anonymous

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

61. anonymous

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

62. anonymous

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

63. anonymous

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