Problem 1:
  R red and B black cards are given. Pick a random pair.
  Find R, B that the probability of picking the same color is 1/2.

Problem 2:
  R red and B black cards are given. Pick a random pair.
  Find R, B that the probability of picking red cards only is 1/2.

Problem 3:
  Find integral solutions for the hypergeometric distribution
  1/2  =  C(R,r)*C(B,b) / C(R+B,r+b).






















Solution 1:
  Prob(same color) = (C(R,2)+C(B,2))/C(R+B,2)

      2*(C(R,2)+C(B,2)) = C(R+B,2)
  <=> (R - B)^2 = R + B

  So let t = R - B. (Wlog R >= B). Then we have
  R - B  =  t
  R + B  =  t^2
  R      =  C(t+1, 2)
  B      =  C(t,2)
  
  The smallest solutions (R,B) are (3,1), (6,3), (10, 6), (15, 10), ...
  As Prob(same color) + Prob(different color) = 1 we have
  1/2  =  (C(R,2)+C(B,2))/C(R+B,2)  =  C(R,1)*C(B,1))/C(R+B,2)
  and is therefore a special case of problem 3.
 

Solution 2:
  Prob(red only) = C(R,2)/C(R+B,2)

  Solve 2*C(R,2) = C(R+B,2) <=> 2 R (R-1) = (R+B)(R+B-1).
  Let p = 2(R+B)-1 and q = 2R-1 we get the Pell equation p^2 = 2 q^2 - 1.
  Recurrence of the solutions for p^2 = 2 q^2 +- 1 
  (p(n+1),q(n+1)) = 2(p(n),q(n)) + (p(n+1),q(n+1)) 
  with (p0,q0) = (1,0) and (p1,q1) = (1,1).
  Sequences: 
    <p> = <1,1,3,7,17,41,99,239,577,1393,...>, 
    <q> = <0,1,2,5,12,29,70,169,408, 985,...>.
  Formulars:
     p(n) = ((1+sqrt(2))^n + (1-sqrt(2))^n)/2,
     q(n) = ((1+sqrt(2))^n - (1-sqrt(2))^n)/(2sqrt(2)).
  For n odd we get solutions with the minus sign. As p and q are both odd
  we always get a solution of the original problem.
  Solutions: (R,B) = (3,1), (15, 6), (85, 35), (493, 204), ...


Appendix:
  Find Binomial Coefficients with 2*C(n,k) = C(m,k).
  These come in pairs as we have the identity:
  C(r+s,s) / C(r+s+t,s)  =  C(r+t,t) / C(r+s+t,t).
  So 1/2  =  C(n,1) / C(2n,1)  =  C(2n-1,n) / C(2n,n)
  which we call the trivial solutions.


References:

Martin Gardner;
   Wheels, Life, and Other Mathematical Amusements
   Freeman (1983) New York
   (german: Martin Gardner's Mathematische Denkspiele, Hugendubel)
   Chap 5: Nontransitive Dice and Other Probability Paradoxes
      n-card monte: R red and B black cards are given pick a pair
      prob(same color) = (C(R,2)+C(B,2))/C(R+B,2)

Karl Josef Jacquemain;
   Denkspiele f\"ur Tennisspieler,
   it 1727, Insel Verlag, 1995
   Problem 11: Linksh\"anderduell
     Find R, L with 2*C(L,2) = C(R+L,2).

Ivan Morris;
   99 neunmalkluge Denkspiele,
   dtv 1243, Deutscher Taschenbuch Verlag, 1977
   Problem 75: Gaststudenten
     Find R, L with 2*C(L,5) = C(R+L,5).


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