Hmm, strategy seems clear: guess the majority color remaining, if any. So the answer can be expressed in terms of all possible shufflings, the N = 52!/(26!)^2 balanced bit strings of length 52, since for each one we know the expected payoff with this strategy (a messy sum): E = Sum_{x balanced bit string length 52} (1/N) * Sum_{1 <= k <=52} etc. —Dan Allan Wechsler wrote: ----- I know how to analyze this, but I haven't tried to actually do the work. But I'm going to make a wild-ass stab based on an intuition. Is it $32.87? On Wed, Dec 26, 2018 at 11:50 AM Thane Plambeck <tplambeck@gmail.com> wrote:
the usual 52-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. the player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong. what is a fair entry price to play? ----=-