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Subject: Math probabilities
Posted by: Khahan
- [2884979] Fri, Apr 08, 2005, 15:32
Hey guys,
I know we have some math whizzes on here, so I'm wondering if somebody can help me.
I have a standard 52 card deck. I riffle shuffle it X number of times. X doesn't really matter.
How many different orders can the cards come up in?
What is the formula to figure this out?
Isn't it just 52x51x50x49 etc all the way down to 1?
What's the name of the formula?
I just don't know enough about math to know where to begin to look.
Thanks. |
| 1 | Sludge
ID: 54692111
Fri, Apr 08, 2005, 15:41
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If the order matters, and you indicated that it does, the formula you give is correct. It's known as a permutation.
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| 2 | Sludge
ID: 54692111
Fri, Apr 08, 2005, 15:42
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And X does really matter. The commonly accepted number of shuffles required to sufficiently randomize a deck of cards is 7 or more.
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| 3 | Khahan
ID: 2884979
Fri, Apr 08, 2005, 15:43
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No, the order doesn't matter. I simply want to show how many different orders can come up.
Is there an easy way to do this because I want to show it for a 60 card deck vs a 61 card deck.
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| 4 | biliruben
Leader
ID: 589301110
Fri, Apr 08, 2005, 15:48
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If order doesn't matter, your question doesn't make any sense.
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| 5 | Khahan
ID: 2884979
Fri, Apr 08, 2005, 15:51
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Ok, so you shuffle a deck of 60 cards 10 times.
How many possible different ways can the cards be ordered?
Is there a site that you guys know of where I can just plug in a number and it will calculate the permutations?
I just want to know what 60! and 61! are.
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| 6 | FRICK@Work
ID: 220211
Fri, Apr 08, 2005, 15:53
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I think that you can do the 60*59*58 ect calculation on some scientific calculators.
There should be a ! button. If you use this it will do the above calculation for you. IIRC correctly from my wated time in high school calculus the largest number most scientific calculators will do is 69!. I think that is approximately 1E98. Or a really, really large number.
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| 7 | FRICK@Work
ID: 220211
Fri, Apr 08, 2005, 15:54
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Khanan, Excel has a permutation function as well. I have never used it, but it be able to help you.
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| 8 | C.SuperFreak
ID: 243849
Fri, Apr 08, 2005, 15:54
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60! ? try using your computer to do it. I could be incorrect but i believe your windows calculator can do the calculation.
view/scientific
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| 9 | biliruben
Leader
ID: 589301110
Fri, Apr 08, 2005, 15:54
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Oh. Any scientific calculator will do it.
The number is huge, btw.
Start>Accessories>calculator>view>scientific.
Punch is 60 and hit n!
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| 10 | Khahan
ID: 2884979
Fri, Apr 08, 2005, 15:56
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Thanks Frick. Unfortunately, no scientic calculators in this house (go figure, my wife is even a scientist).
Just a basic calculator to do basic accounting.
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| 11 | Khahan
ID: 2884979
Fri, Apr 08, 2005, 15:57
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There we go...thank you all. I didn't know you could change the comp calc into a scientific calc.
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| 12 | FRICK@Work
ID: 220211
Fri, Apr 08, 2005, 15:57
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I just ran them through Excel and assuming you want to use all the cards in the theoretical decks.
60 cards = 8.32E81
61 cards = 5.07E83
That extra card makes a "slight" difference.
Just in case you were wondering a 52 card deck has 8.06E67 possibilities.
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| 13 | GoatLocker
Sustainer
ID: 060151121
Fri, Apr 08, 2005, 15:59
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My scientific based on 61 cards and 10 shuffles came up with an E.
Cliff
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| 14 | GoatLocker
Sustainer
ID: 060151121
Fri, Apr 08, 2005, 16:02
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OK, I take it back.
61 shuffled 10 times comes out to
3.272349153 to the 17th
60 shuffled 10 times comes out to
2.734898472 to the 17th
Cliff
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| 15 | GoatLocker
Sustainer
ID: 060151121
Fri, Apr 08, 2005, 16:11
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OK, unless I'm doing something wrong.
61 cards shuffled 10 times =
565,735,991.3
60 cards shuffled 10 times =
26,958,741.02
Cliff
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| 16 | Khahan
ID: 2884979
Fri, Apr 08, 2005, 16:14
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What does the E81 and E83 mean? Is that the # of places to extend out from that point?
So, if I had a number that was, say: 1.983E3 (i'm trying to keep it simple, just go with the basic concept if possible), it would be: 1,983,000
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| 17 | biliruben
Leader
ID: 589301110
Fri, Apr 08, 2005, 16:25
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Nope. It's the number of places you need to move the decimal point to the right. 1983.
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| 18 | biliruben
Leader
ID: 589301110
Fri, Apr 08, 2005, 16:26
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=1.983 x 10^3
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| 19 | C1-NRB
ID: 17348117
Fri, Apr 08, 2005, 16:42
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Not to pick nits with vastly superior mathematical minds, but how broadly (or narrowly) are you defining "shuffle"?
Assuming a clean 50/50 cut, is "one" random card from the left half (L) moving to a random point in the right half (R) a "real" shuffle?
For example, when my preschool offspring shuffles a deck of 52 cards, the top three cards 1L, 2L, 3L move into position 1 (the top) of R and cards 1R-26R are now between L3 and L26. After a 50/50 cut, L is now L1-L3, R1-23 and R is R4-26, L24-26.
Would anyone say these cards had been "shuffled"?
How many cards must be inserted from L to R for a shuffle to be a true shuffle?
Semantics...
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| 20 | JeffG
Leader
ID: 1584348
Fri, Apr 08, 2005, 17:00
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A few things....
52x51x50x49x48....x3x2x1 is notated 52! and called FACTORIAL not permiatation. That is the number of different outcomes of a random deck of 52 cards.
I remember reading somewhere that a perfectly ordered deck needs to be shuffled 7 times to become random. A shuffle consists of a cut of the cards and the mixing of the two half-decks together. Overhand shuffles take longer.
So shuffling 10 times instead of 7 cannot make it more random.
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| 21 | Ender
ID: 11033120
Tue, Apr 12, 2005, 22:22
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JeffG, 52P52 = 52! so it is a permutation, it is just the special case where you are permuting all 52 cards.
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| 22 | FRICK@Work
ID: 220211
Wed, Apr 13, 2005, 12:05
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Ender was my math right on the difference between a 60 and 61 card deck?
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| 23 | Sludge
ID: 54692111
Wed, Apr 13, 2005, 15:38
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52x51x50x49x48....x3x2x1 is notated 52! and called FACTORIAL not permiatation. That is the number of different outcomes of a random deck of 52 cards.
Yes, yes. Semantics. The formula is known as the factorial (btw, does anyone know why 0! = 1?), but a particular ordering of the 52 cards is a permutation. So n! is the number of permutations of n items.
And since we're on a nitpick fest, the "number of different outcomes of a random deck of 52 cards" IS a factorial, NOT a permutation.
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| 24 | Sludge
ID: 54692111
Wed, Apr 13, 2005, 15:48
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OK, unless I'm doing something wrong.
61 cards shuffled 10 times =
565,735,991.3
60 cards shuffled 10 times =
26,958,741.02
Since nobody else has corrected Goat, I'll be the one. Your terminology is incorrect (and this isn't just semantics). In a permutation nPr = n!/(n-r)! ("n items permute r" as you'll often hear it said), r is not the number of "shuffles". It is the number of items to be drawn from the larger group of n items. In other words, your computations for 61P10 and 60P10 represent the number of possible orderings of 10 cards drawn from decks of 61 and 60, respectively.
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| 25 | Aris
ID: 166571316
Wed, Apr 13, 2005, 17:51
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Sludge 23:
0!=1 so that formulas such as N!/((N-k)!K!) work for K=0.
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| 26 | Sludge
ID: 54692111
Wed, Apr 13, 2005, 18:10
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Correct.
Although a better argument (fitting the theme of this thread) would be to discuss permutations, not combinations. The general formula for the number of ways to permute r items from n is nPr = n!/(n-r)!. How many ways are there to permute n items from n (that is, how many ways are there to arrange all n items)? Well, n!. Plug that into the formula, and you get nPn = n!/(n-n)! = n!/0!. So 0! must be equal to 1 for it to work.
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| 27 | TacoJohn
ID: 53311411
Thu, Apr 14, 2005, 13:18
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If you're wondering what difference a 61st card makes over a 60 card deck in the number of possible orders it's pretty simple.
The 61 card deck has exactly 61 times as many possibilities as the 60 card deck.
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| 28 | Ref
Donor
ID: 539581218
Fri, Apr 15, 2005, 02:35
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Now I remember why I took my required stats class pass/fail in college!!! ;)
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| 29 | Khahan
ID: 2884979
Mon, Apr 18, 2005, 13:40
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Ref...and now I remember why I wanted to be a communications major in college: It required the least amount of math.
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| 30 | PBJ 08
ID: 31542292
Sun, Jun 29, 2008, 03:42
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For riffle outshuffles, is there a formula for working out how many shuffles it takes to get 52 cards back into its original order?
Ofcourse, assuming that they are all perfect riffles.
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