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 A220017 Relative probability for a four-player Old Maid match to last for exactly n games until a player has been trolled exactly three times. 1
 512, 1152, 1728, 1920, 1620, 945, 315 (list; graph; refs; listen; history; text; internal format)
 OFFSET 3,1 COMMENTS 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 A216074(n) = Number of numbers for which from the left, some digit has occurred three times at the n-th position, i.e. 32 * a(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 A216074(i), for i = 3..9 is equal to 4^9 = 262144. (true chance) Total value of a(i), for i = 3..9 is equal to 4^9/32 = 262144/32 = 8192. gcd of A216074(i), for i = 3..9 is equal to 2^5 = 32. The sequence A216074 gives the true number of ways for the match to last exactly for n games. LINKS Table of n, a(n) for n=3..9. FORMULA a(n) = A216074(n)/32. CROSSREFS Cf. A216074 (true probability). Sequence in context: A043423 A341886 A045058 * A234879 A202454 A230543 Adjacent sequences: A220014 A220015 A220016 * A220018 A220019 A220020 KEYWORD nonn,fini,full AUTHOR V. Raman, Dec 12 2012 STATUS approved

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Last modified April 23 09:22 EDT 2024. Contains 371905 sequences. (Running on oeis4.)