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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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Recursive two-way formula. | Comment 3 of 9 |
C(M,N)=the number of was of having N successive pairs from an M card deck.

C(1,0)=1, C(1,N≠0)=0
C(M,N) = C(M-1,N-1) + (M-N-1)*C(M-1,N) + N*C(M-1,N+1)

This is an easy way of building successive rows on a chart especially if someone feels like making a program to find, say s(100)
(Hint, hint)

The triply recursive nature makes finding an explicit formula for C very difficult.

  Posted by Jer on 2009-12-16 18:55:43
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