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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.
1

%I #15 Dec 14 2012 11:49:59

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

%N 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.

%C [From _V. Raman_, Dec 12 2012] (Start)

%C 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.

%C 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.

%C 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).

%C 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.

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

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

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

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

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

%Y Cf. A220017 (relative probability).

%K nonn,fini,full

%O 3,1

%A _V. Raman_, Sep 01 2012

%E Edited by _V. Raman_, Dec 12 2012