# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a265919 Showing 1-1 of 1 %I A265919 #21 Nov 29 2018 16:00:25 %S A265919 0,1,3,17,71,301,1275,5257,21711,88997,363395,1480385,6014423, %T A265919 24393245,98781323,399502841,1614022751,6514800277,26275725843, %U A265919 105904696369,426598453863,1717507247885,6911604624923,27802402448233,111796372691439,449398016848261 %N A265919 Numerator of the probability that Bob wins the following game: Alice and Bob take turn (Alice starts first) to gain 1 or 2 chips randomly and independently with 1/2 chance, and the first player that collects at least n chips is the winner. %C A265919 The formula is proved by _Wing Hong Tony Wong_ and _Jiao Xu_, and then proved by Taoye Zhang and Ju Zhou independently. %C A265919 The probability that Bob wins the game is a(n)/4^(n-1). %H A265919 Michael De Vlieger, Table of n, a(n) for n = 1..1662 %H A265919 Tony W. H. Wong, Jiao Xu, A Probabilistic Take-Away Game, J. Int. Seq., Vol. 21 (2018), Article 18.6.3. %F A265919 a(n) = (4^n - Sum_(k=1..n) 4^(n-k)*(binomial(k, n-k)+binomial(k-1, n-k))^2)/8. %t A265919 Table[(4^n - Sum[4^(n - k)*(Binomial[k, n - k] + Binomial[k - 1, n - k])^2, {k, n}])/8, {n, 100}] %Y A265919 A265920 provides the integer sequence of the numerator of the probability that Alice wins the game. The corresponding terms of these two sequences add up to 4^n. %K A265919 nonn %O A265919 1,3 %A A265919 _Wing Hong Tony Wong_, _Jiao Xu_, Dec 18 2015 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE