%I #7 Mar 12 2024 20:14:24
%S 3,63,1053,16443,250533,28431,56859813,853737003,12811093893,
%T 17472421953,2883131020773,25013333547,648727335888453,
%U 9730949220408843,145964473398624933,128792265384372219,32842036136344638213,3703989419737954191,7389459197057088616293,10076535434903231752353
%N a(n) is the numerator of the ratio of winning probabilities P_A/P_B of winning in a 2player game with a ratio of odds for A and B in a single round of 3:2. To win the game it is necessary to win n rounds in a row.
%C If you play the game with odds of 3:2 with one dice, you can decide what happens to the unused number. In the present sequence, this number is ignored without effect. However, if one defines the occurrence of the 6th number as a draw, which interrupts consecutive wins by both players, then the sequence pair A370825/A370826 results. The addition of a draw increases the odds ratio in favor of the player who has the higher chance of winning a single round. The relative advantage is small with 2 games to be won, i.e., 2/(63/32)1 = 1/63, with 3 rounds 1/27, with 4 rounds 965/16443, but increases with a higher number of consecutive rounds to be won, and reaches asymptotically 1/8.
%F a(n)/A370828(n) = (2/3) * (3/5)^n * ((5/2)^n  1) / (1  (3/5)^n).
%e a(n)/A370828(n) for n = 1..8: 3/2, 63/32, 1053/392, 16443/4352, 250533/46112, 28431/3584, 56859813/4860032, 853737003/49160192.
%o (PARI) a370827(n) = numerator((2/3) * (3/5)^n * ((5/2)^n  1) / (1  (3/5)^n))
%o (Python)
%o from math import gcd
%o def A370827(n): return (a:=3**(n1)*(5**n(1<<n)))//gcd(a,5**n3**n<<n1) # _Chai Wah Wu_, Mar 12 2024
%Y A370828 are the corresponding denominators.
%Y Cf. A370823, A370824, A370825, A370826.
%K nonn,frac,easy
%O 1,1
%A _Hugo Pfoertner_, Mar 09 2024
