Cris, that's pretty convincing, thanks. C(0,0) = 0 C(0,n) = n C(1,1) = 3/2 The simplest nontrivial case seems to be C(1,2). If you make the greedy guess, you have a 2/3 chance of being right and getting 3/2 + 1, and a 1/3 chance of being wrong and getting 2. 2/3 * 5/2 + 1/3 * 2 = 7/3. If you make the perverse guess, you have a 1/3 chance of being right and getting 3, and a 2/3 chance of being wrong and getting 3/2. So the total value is 2; the greedy guess is 1/3 better. On Wed, Dec 26, 2018 at 3:44 PM Cris Moore <moore@santafe.edu> wrote:
I think the key thing is that the state of the cards don’t depend on your bet, so there’s no hard in being greedy - no point in sacrificing for the future.
Cris
On Dec 26, 2018, at 1:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I agree that it seems to make the most sense to be greedy. But I haven't proven it yet! There is something about the logic of this game that keeps tripping me up.
On Dec 26, 2018 3:18 PM, "Dan Asimov" <dasimov@earthlink.net> wrote:
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? ----=-
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