A216074 Number of ways in which a four-player Old Maid match lasts for exactly n games until a player has been trolled exactly three times.

16384, 36864, 55296, 61440, 51840, 30240, 10080

Offset

3,1

Comments

[From V. Raman, Dec 12 2012] (Start)

The Old Maid match is played with four players. For every game that is played, one of the four players is trolled at the end. The match ends when a player has been trolled three times, after which he loses the match.

So, to calculate the probability, we represent the four players by the digits 0, 1, 2, 3 in base 4 and then list out all the 18-bit numbers in base 4.

Then a(n) = Number of numbers for which from the left, some digit has occurred three times at the n-th position, i.e. 32 * A220017(n).

For four player match, the maximum number of games needed for a player to be trolled three times is 9. So, we consider with 9-digit base 4 numbers.

Total value of a(i), for i = 3..9 is equal to 4^9 = 262144.

Total value of A220017(i), for i = 3..9 is equal to 4^9/32 = 262144/32 = 8192. (relative chance)

gcd of a(i), for i = 3..9 is equal to 2^5 = 32.

The sequence A220017 gives the relative probability for the match to last exactly for n games. (End)

Links

Table of n, a(n) for n=3..9.

Formula

a(n) = 32 * A220017(n).

Crossrefs

Cf. A220017 (relative probability).

Sequence in context: A069415 A212936 A069275 * A258736 A255666 A220767

Adjacent sequences: A216071 A216072 A216073 * A216075 A216076 A216077

Keyword

nonn,fini,full

Author

V. Raman, Sep 01 2012

Extensions

Edited by V. Raman, Dec 12 2012

Status

approved