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# Shuffling a Deck of Cards

### #11

Posted 28 May 2013 - 07:07 PM

### #12

Posted 28 May 2013 - 07:13 PM

### #13

Posted 28 May 2013 - 07:51 PM

I am referring to "riffle shuffles". The random riffle shuffle is modeled by cutting the deck binomially and dropping cards one-by-one from either half of the deck with probability proportional to the current sizes of the deck halves.

How many shuffles must be done, to where every possible card configuration is possible?

I came to know that shuffle by the name “perfect shuffle.” Of course, you should start dropping the cards from the top half, so that the top card does not end up in the same position after the shuffle.

Past prime, actually.

### #14

Posted 28 May 2013 - 08:33 PM

perfect shuffle is when the cards are dropped perfectly alternately. riffle shuffle is where the probability of dropping the next card depends on the size of the remaining undropped deck on that size.

### #15

Posted 28 May 2013 - 10:22 PM

Spoiler for buy that cry tea rear...

on the right track

### #16

Posted 28 May 2013 - 11:20 PM

on this subject: check for a discussion by Persi Diaconis, Professor of Statistics at Stanford. I guess it's a spoiler so do not click if you do not want the answer!

Also if you are intrested, check the discussion on a backgammon site: http://www.bgonline....mes;read=142087

### #17

Posted 07 June 2013 - 05:45 PM

Once you require a "perfect" shuffle, the only *randomness* is the choice of in-shuffle or out-shuffle (where top and bottom cards never move.) Every card moves to a known position, and all you do is calculate whether a particular permutation of the first 52 integers is random.

I'm interested to think what happens when the shuffle itself is random -- anything from "perfect riffle" to cutting the deck at a random card, thus interleaving any number of cards from 1 to 51. A single shuffle could then put the bottom card on top, or vv.* *You could assign a distribution of probabilities governing what happens, but that is all.

*Vidi vici veni.*

### #18

Posted 07 June 2013 - 06:51 PM

then the top half can be partitioned into T segments (keeping the order

of the cards) and, similarly, the bottom half can be partitioned into B

segments. If T=B then there are two ways to interleave these partitions

of cards -- dropping segments of cards alternately from T and B, beginning

with T (the first way) or beginning with B (the second way). If T=B+1,

then there is only one way to interleave the segments -- one must begin

with T. If T+1=B, one must begin with B. Any other relationship between

T and B make them impossible to interleave. We can count the number of

partitions of 26 in all possible arrangements, they are:

#partitions of 26; #of ways to make partitions of this number respecting order

1 1

2 25

3 300

4 2300

5 12650

6 53130

7 177100

8 480700

9 1081575

10 2042975

11 3268760

12 4457400

13 5200300

14 5200300

15 4457400

16 3268760

17 2042975

18 1081575

19 480700

20 177100

21 53130

22 12650

23 2300

24 300

25 25

26 1

We can now count all possible T,B pairs where T and B are at most one

apart. This gives us a grand total of 374,369,872,911,804 possible

riffle shuffles of 52 cards. Considering that 52! is the enormous

80658175170943878571660636856403766975289505440883277824000000000000,

I can't see how to show that there is some number, N, of riffle shuffles

such that one can always produce any particular permutation of the 52!

by applying at most N riffle shuffles to the identity permutation.

### #19

Posted 13 June 2013 - 12:01 AM Best Answer

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