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Supreme Successive Shuffle (Posted on 2009-12-15) Difficulty: 4 of 5
A deck of M cards is numbered 1 to M and shuffled, and dealt from top to bottom.

Denoting the probability of dealing at least one pair of successive cards in their proper order (that is, a 1 followed by a 2 or, a 2 followed by a 3, and so on) at any position in the deck by s(M), determine s(M) as M → ∞ (The pairs may overlap. For example, for M=5, we have two successive pairs corresponding to 73452.)

As a bonus, what is the expected number of such successive pairs in an M card deck as a function of M?

No Solution Yet Submitted by K Sengupta    
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Solution re: Some thoughts (Spoiler) Finished Comment 9 of 9 |
(In reply to Some thoughts (Spoiler) by Harry)

Your formula for A(m,r) is equivalent to mine.  You also have a nice recursive A(m,0)=A(m) formula.  This is included in the OEIS as

The entry also gives an explicit formula:

a(n) = floor((1/e)*n!*(n+2)+1/2)

As n goes in infinity we can ignore the floor and +1/2
Also, this is offset by 1 so what we really want is

Lim(n->infinity) [(1/e)*(n-1)!*(n+1)]/(n!)
=Lim(n->infinity) [(1/e)*(n+1)]/(n)

  Posted by Jer on 2009-12-18 16:37:57

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