#1
given a jack, a king, a queen, an ace, and a ten, how many unique hands can you make? a hand must have at least one card and obviously no more than five. order matters.

examples:

Q
J A
A J
J K Q T
J Q K T
Q K T A J

are all unique hands. how many are there?

if you a programmer, you might be able to solve this by brute force. later questions on this topic will be impossible to answer by brute force.

side note: by the end of this thread i will demonstrate a method for calculating the number of unique shuffles in a 52 card deck and give the result.
i don't know why i feel so dry
#5
The benzo withdrawal's getting worse.
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#6
I could do this if I had a calculator but I don't know where mine is

The first question on my stats final two years ago was almost exactly like this though
cat
#7
Quote by guitarxo
I could do this if I had a calculator but I don't know where mine is

The first question on my stats final two years ago was almost exactly like this though
Poor excuse.
If you're on a computer and can't do maths because you don't have a calculator... (or a smartphone for that matter)

EDIT:
I don't know the answer I'm going to work through a guess...

5 one card possibilities
+
20 two card possibilities (each card can be combined with one of four others. 5*4 =20)
+
60 three card possibilities (each of the above two card possibilities can have one of three remaining cards added to get a three card possibility 20*3 = 60)
+
120 four card possibilities (each of the above 60 three card possibilities can have one of two remaining cards added to it to give cour card possibilities 60*2=120
+
120 five card possibilities each of the four card possibilities can have one possible remaining fifth card added to it 120

so 120+120+60+20+5 = 325?


I'm sure I've missed something...
Si
#9
using calculators on a computer sux though
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#12
Quote by 20Tigers
Poor excuse.
If you're on a computer and can't do maths because you don't have a calculator... (or a smartphone for that matter)

Well I know the principles of how to do it so I really don't care what the final answer is
cat
#13
skimed through it, didnt understand. imma stick to war or go fish.
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This thread topic is gold. I've been on this website for 8 years and I've never come up with anything like this. So yeah. Great job TS[457undead].
#14
Quote by guitarxo
Well I know the principles of how to do it so I really don't care what the final answer is



---------------
I edited my previous post and had a crack. Not sure if it's right though.
Si
#15
20T has done it. nice work.

that was the setup for the better question: given n unique cards, how many unique hands can you create with the same rules as before?

you will wind up with a tricky formula. there is a satisfying approximation involving the constant e.
i don't know why i feel so dry
#18
h(1) = number of unique 1 card hands
h(2) = number of unique 2 card hands
h(3) = number of unique 3 card hands
...
h(n) = number of unique hands with n cards

T = Total number of combinations

T=h(1)+h(2)+h(3)...+h(n)

h(1) = n

h(2) = h(1)*(n-1) = n(n-1)

h(3) = h(2)*(n-2) = n(n-1)(n-2)
...

h(n) = h(n-1)*(n-(n-1)) = n!

T = n + n(n-1) + n(n-1)(n-2)...n!

Yeah I'm giving up, at least for now. I'm pretty sure that if I sat at it long enough I would figure it out but it's been a looong time since I've done mathematics and I know this would take me a while because there's a lot I'd have to remember before I could get there. Maybe I'll come back to it when I have more time.

The number of ways a deck can be shuffled is 52!. It's not the same as the above scenario obviously because you will always have 52 cards. So that's just a factorial of 52. (i.e. 80,658,175,170,943,900,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000)
Si
#19
is it n! (factorial) sir?

EDIT: 20Tigers ninja'd me
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#20
nah it's not n!

if n = 6 then n! is 6*5*4*3*2*1

the answer to his question would be

6*5*4*3*2*1 + 6*5*4*3*2* + 6*5*4*3 + 6*5*4 + 6*5*4 + 6*5 + 6

so n! + n! + n!/2 + (n!/2)/3 ... etc I'm not quite sure how to write that...
Si
#21
I have a problem that was posed years ago on another site, and no one there ever answered it correctly:

"What are the odds that a shuffled pack of cards will contain at least two same-suit cards in sequence (eg 4D-3D, 10S-JS, KC-AC)? It seems to happen every time but the odds are obviously not 100%. Is there a formula?"
#22
Quote by 20Tigers
nah it's not n!

if n = 6 then n! is 6*5*4*3*2*1

the answer to his question would be

6*5*4*3*2*1 + 6*5*4*3*2* + 6*5*4*3 + 6*5*4 + 6*5*4 + 6*5 + 6

so n! + n! + n!/2 + (n!/2)/3 ... etc I'm not quite sure how to write that...


sum from i=0 to n-1

(n!/i!)

But I don't know how to write it without summation notation.
#23
this is like high school math that i've long forgotten because i never use
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#24
Upon observing the card you collapse the wave-function and cause the whole problem to = 1
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#25
Quote by Jehannum
I have a problem that was posed years ago on another site, and no one there ever answered it correctly:


52/52 x 39/51 x 38/50 x 37/49...

right?
#27
let me expand on what you found 20T.

first, let me made a small modification to this:
Quote by TheChaz
sum from i=0 to n-1

(n!/i!)

But I don't know how to write it without summation notation.

i'm going to substitute i for k as k is a cooler letter. next, i'm going to use the proper notation in an image:

where T(n) is the number of possible hands, stealing 20T's notation there. pulling the n! out of the summation and observing the well known infinite sum for e we have:

but as 20T points out, n! is the number ways to order the cards using every card. this result shows us that if you allow yourself to neglect cards in your hand, you cannot get any more than 2.71829... more permutations then if you decide to use every card. bit unintuitive, isn't it?
i don't know why i feel so dry
Last edited by Eastwinn at Jun 20, 2014,
#28
i thought this would be about cards against humanity
*-)
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#29
Quote by lolmnt
i thought this would be about cards against humanity


apples much funnier imo
i don't know why i feel so dry
#30
Quote by Eastwinn
let me expand on what you found 20T.

first, let me made a small modification to this:

i'm going to substitute i for k as k is a cooler letter. next, i'm going to use the proper notation in an image:


See I can't remember what all those symbols mean and would have to do a lot of revision to have been able to get there. Even when you write it out and I can see the formula and only vaguely remember bits of what it's telling me, but not nearly enough for it to actually have any meaning to me.
Si
#32
Quote by Eastwinn
let me expand on what you found 20T.

first, let me made a small modification to this:

i'm going to substitute i for k as k is a cooler letter. next, i'm going to use the proper notation in an image:

where T(n) is the number of possible hands, stealing 20T's notation there. pulling the n! out of the summation and observing the well known infinite sum for e we have:

but as 20T points out, n! is the number ways to order the cards using every card. this result shows us that if you allow yourself to neglect cards in your hand, you cannot get any more than 2.71829... more permutations then if you decide to use every card. bit unintuitive, isn't it?

That is a whole lot more explanatory than what I thought.
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