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I'm sure the integral isn't that hard to evaluate but after setting it up, I decided I'd spare myself that agony and used my calculator. It spit out an exact answer of (2*ln(2)-1)/12 or about 0.0322. Now, that's the probability that just one side is concave. It can be easily seen that no more than one region of a quadrilateral that doesn't self-intersect (like this one) can be concave. As such, we simply take this probability and multiply it by four to find the probability that any of the four sides is concave. Subtract this value from 1 to obtain (4 - 2*ln(2))/3 ≈ 0.8712. Hey, what can I say? I tried. Edit 2: I forgot to mention that my prediction of P(one region concave) = 0.0322 does not agree with my Excel spreadsheet. That's showing a probability of something like 0.0228 + 0.0002. I have no idea if that would help anyone figure out the real solution.
The integral of the integral of the integral of the integral of the integral of the integral of 1 dy_I dx_I dy_IV dx_IV dy_II dx_II as y_I ranges from 0 to y_II - (y_II + y_IV)/(x_II + x_IV)*(x_II + x_I) as x_I ranges from 0 to y_II*(x_II + x_IV)/(y_II + y_IV) - x_II as y_IV ranges from 0 to y_II*x_IV/x_II as x_IV ranges from 0 to 1 as y_II ranges from 0 to 1 as x_II ranges from 0 to 1
Plugging in the asymptotic formula C(2n,n)~4n/sqrt(pi*n) gives the approximation 1/2 + (sqrt(pi*n) - 1)/(4n) for large n, which is another way to show that the limit is infinity. The long explanation for this is as follows: The best strategy is: for each card, choose the color which has more cards remaining; since you make a choice at each stage, the aim is to maximize the expected number of cards guessed correctly for each card alone. When there are an equal number of red and black cards, it doesn't matter what you choose; the long-term success probability will not be affected. For simplification purposes, I will assume that in this case you choose red and black with equal probability (1/2 each). It turns out that there is a combinatorial approach to this problem. Model the deck as a path of up and right steps starting from (0,0) and ending at (n,n) (such as shown in https://i.stack.imgur.com/vGlBj.gif for n=1 to 3) where you go through the deck and go up if the card is red and right if the card is black. Then there is a surprising relation between the expected number of correct guesses (and thus the long-term success probability) and the number of times this path contacts the line y=x (henceforth called "the diagonal"). Following the strategy above, consider each time the path goes between contact points on the diagonal, say, from (a,a) to (b,b). At (a,a), the random choice of red or black adds 1/2 to the expected number of correct guesses so far. After this card, you choose black (if it was red) or red (if it was black) until the path reaches (b,b); this adds b-a correct guesses. Thus, from (a,a) to (b,b), the expected number of correct guesses so far is increased by b-a + 1/2. Summing over the whole path gives as the expected number of correct guesses n + E/2, where E is the average number of contacts with the diagonal, other than at (n,n), over all such paths from (0,0) to (n,n). So the long-term success probability is (n + E/2)/(2n) = 1/2 + E/(4n). The problem then becomes one of calculating E. Generating functions A Dyck path is an up-right path from (0,0) to (n,n) that never goes below the diagonal. The generating function for Dyck paths is D(x)=(1-sqrt(1-4x))/(2x). I will define a "strict Dyck path" as a Dyck path which never contacts the diagonal except at (0,0) and (n,n). A strict Dyck path is just a Dyck path from (0,1) to (n-1,n), with an additional up-step added to the beginning and an additional right-step at the end. I also allow the strict Dyck path to acquire two orientations: one where it is above the diagonal and one where it is reflected so as to be below the diagonal. The generating function for strict Dyck paths on two orientations is 2xD(x) = 1-sqrt(1-4x). Thus an unrestricted up-right path from (0,0) to (n,n) can be viewed as a sequence of k≥1 strict Dyck paths on two orientations, and the number of contacts is k. Normally, we take the generating function 1/[1-(1-sqrt(1-4x))] = 1/sqrt(1-4x), which indeed is the generating function for the sequence C(2n,n). On the other hand, we can take the bivariate generating function F(x,z)= 1 / [1 - (1-sqrt(1-4x))z], which in addition counts the number of the diagonal contacts other than (n,n) in the variable z. To calculate the average number of contacts E, we first take the sum total of contacts, which we do by taking the derivative of F(x,z) in z and then setting z=1. Since d/dz F(x,z) = (1-sqrt(1-4x))/[1 - (1-sqrt(1-4x))z]2, this gives (1-sqrt(1-4x))/(1-4x) = 1/(1-4x) - 1/sqrt(1-4x). The coefficient of xn of this generating function is 4n - C(2n,n). Thus the average number of contacts E is 4n/C(2n,n) - 1. Substituting into 1/2 + E/(4n) gives the formula 1/2 + (4n/C(2n,n) - 1)/(4n), as above. Math is full of stuff like that. Possible reasoning based on heuristics for why there could be infinitely many Mersenne primes (and finitely many Fermat primes) is given here: http://math.stackexchange.com/questions/838678/intuitively-what-separates-mersenne-primes-from-fermat-primes . Of course not a valid proof of anything, but one of the reasons that people use heuristics is because we know so little about some things in math. I'll answer this later if no one else does, but I have some questions. For (1), does it matter what the fourth hand is? Can it be anything, including the last King-Queen suited? For (2), do we ignore the fourth hand completely, as in, we can assume that it has been folded and plays no further part in the deal? Also, as a question of interest, in the deal that you mentioned (three players with King-Queen suited, 10, J, A and two others on the table), do you remember if all three players have nut hands? That is, from their point of view, they all had a hand which from their point of view could never lose no matter what anybody else had? If so, that would have been very interesting. The usual result being that all three players go all-in and a huge amount of excitement is generated, only to end in a split pot anyway. :D
n=10 0.616887 n=100 0.541867 n=1000 0.513764 n=10000 0.504406 n=100000 0.501399 n=1000000 0.500443