Selecting cards to form a fair game Background


  In an old card game we draw 2 cards from a pile of 2 red and 2 black cards without replacement. If the two cards have the same color (for instance red and red) you win. However, if the cards have opposite color you lose. 


This is a rich problem, especially if we instead look at the broader problem of drawing $2$ cards from a pile of $(n,m)$ cards, where we now have $n$ red and $m$ blue cards. After sorting out the information one can show that any solution must satisfy 

$$(n-m)^2 = n + m$$

from which it is not hard to deduce that every solution must be a pair of consecutive triangular numbers.

$$(1,3), \ (3,6), \ (6,10), \ (10,15), \ldots$$

In other words we have

$$T_2(n) = T_2(n-1) + n, \qquad T(n)=0, n\leq 1$$

Which of course also can be expressed as $T_2(n) = n(n+1)/2$. So
$\bigl(T_2(n), T_2(n+1)\bigl)$ forms every solution.

My question is if similar beautiful patterns appear when we increase the number of cards we draw.

Main statement


  Assume we have a pile of $(n,m)$ cards, where $n$ of the cards are red and $m$ are black and we draw $c$ cards from the pile (where $c \leq n + m$). Fix $n$, how do we have to choose $m$ to obtain a fair game? E.g. a game where the probability of drawing a pile of cards of similar colors (red, red ..., red or black, black, ..., black) equals the probability of drawing cards of opposite color (any combination of red and black cards)


For $c = 3$ it seems we have to find integer solutions to

$$n(n-1)(n-2) + m(m-1)(m-2) = 3mn(n+m-2)$$

and this seems really hard. However, it seems $(1,5,3)$ is a solution. After an extensive computer search it seems

$$(1,5), \ (5,20), \ (20,76), \ (76,285), \ (285,1065), \ (1065,3976), \ \ldots$$

Are the first few solutions when drawing three cards. It seems these satisfy
$$
 T_3(u) = 5 T_3(u-1) - 5 T_3(u-2) + T_3(u-3) \ \text{with} \ T_3(1) = 1 \ \text{and} \ T_3(u) = 0 \ \text{if} \ u \leq 0.
$$

EDIT: Seems to boiling down to finding all integer pair such that

$$     \binom{m}{c}\binom{n}{0} \Bigl/\binom{m+n}{c}\Bigr. 
    + \binom{m}{0}\binom{n}{c} \Bigl/\binom{m+n}{c}\Bigr. = \frac{1}{2}, $$
Where again $c \in \mathbb{N}_{\geq 2}$ and $c \leq n &lt; m$. The expression above can be "simplified" to 
$$\prod_{i=0}^{n-1} \frac{m+n-k-i}{m+n-i} + \prod_{i=0}^{m-1} \frac{n+m-k-i}{n+m-i} = \frac{1}{2}$$
and can be quite easily be numerically approximated. However, it does not lead me closer to finding every solution for every $c$.

EDIT 2: While I thought all solutions would be on the form $(a,b)$, $(b,c)$, $(c,d), \ldots$ this does not seem to be the case. 
In particular for $c = 6$ we 
$$T_6(1) = (1,11), \qquad T_6(2) = (2,19)$$
interesting!

Problems

Let $T_c(n)$ be the $n$'th solution when drawing $c$ cards.


Is it true that $T_c(1) = 2c - 1$ for every $c\geq 2$?
Is there a general recurrence relation for $T_c(n)$? Is there a closed expression for $T_c(n)?$
Given a particular $c$ how can we find all pairs $(n,m)$ that form a fair game?
This is a partial answer.


  Is it true that $T_c(1) = 2c - 1$ for every $c\geq 2$?


Yes, it is.

Since$$\begin{align}\frac{1}{2}\binom{m+1}{c}-\binom mc&amp;=\frac 12\cdot\frac{(m+1)!}{(m+1-c)!c!}-\frac{m!}{(m-c)!c!}
\\\\&amp;=\frac{m!}{c!(m+1-c)!}\bigg(\frac{m+1}{2}-(m+1-c)\bigg)
\\\\&amp;=\frac{m!}{c!(m+1-c)!}\cdot\frac{2c-1-m}{2}\end{align}$$
we have$$\frac{1}{2}\binom{m+1}{c}-\binom mc=0\iff m=2c-1$$

So, we can say that


If $m\lt 2c-1$, then $\frac{1}{2}\binom{m+1}{c}\not=\binom mc$. 
If $m=2c-1$, then $\frac{1}{2}\binom{m+1}{c}=\binom mc$.


Therefore, it follows that $(1,2c-1)$ is the first solution for every $c\ge 2$.




  Is there a general recurrence relation for $T_c(n)$? Is there a closed expression for $T_c(n)?$


For $c=3$, as you noticed, it seems that we have
$$T_3(u) = 5 T_3(u-1) - 5 T_3(u-2) + T_3(u-3)\qquad (u\ge 4)$$
$$T_3(1)=1,\qquad T_3(2)=5,\qquad T_3(3)=20$$
Since $x^3-5x^2+5x-1=(x-1)(x-(2-\sqrt 3))(x-(2+\sqrt 3))$, we get
$$T_3(u)=\frac{1}{12}\bigg((3 - \sqrt 3)(2 - \sqrt 3)^u+ (3 + \sqrt 3)(2 + \sqrt 3)^u-6\bigg)$$



For $c=4$, it seems that $(1,7)$ is the only solution.

Let $m+n-2=t\ (\ge 3)$. Then, the equation 
$$\binom m4+\binom n4=\frac 12\binom{m+n}{4}$$
can be written as
$$(2m^2-2mt-4m+2t^2-t+1)^2=3t^4-6t^3+6t^2+1$$So, $t\ (\ge 3)$ has to be an integer such that $3t^4-6t^3+6t^2+1$ is a perfect square. 

Though it seems that $t=6$ is the only such integer, no proof can be obtained. If this is true, then we can say that $(1,7)$ is the only solution.



For $c=5$, it seems that $(1,9)$ is the only solution.

Let $m+n=k\ (\ge 6)$. Then, the equation 
$$\binom m5+\binom n5=\frac 12\binom{m+n}{5}$$
can be written as
$$(2m^2-2mk+k^2-4k+5)^2=\frac{3k^4-28k^3+108k^2-188k+125}{5}$$So, $k\ (\ge 6)$ has to be an integer such that $\frac{3k^4-28k^3+108k^2-188k+125}{5}$ is a perfect square. 

Though it seems that $k=10$ is the only such integer, no proof can be obtained. If this is true, then we can say that $(1,9)$ is the only solution.
"n an old card game we draw 2 cards from a pile of 2 red and 2 black cards without replacement. If the two cards have the same color (for instance red and red) you win. However, if the cards have opposite color you lose. 
This is a rich problem, especially if we instead look at the broader problem of drawing cards from a pile of (n,m)
cards, where we now have n red and m blue cards."

Actually, there is no problem here! What do you want to find?  The probability of winning?

If so drawing two cards, blue or red, leads to four possible results, BB, BR, RB, of RR (B stands for "blue" and R stands for "red").  If there are m blue and n red cards, the probability of "BB" is $\frac{m}{m+ n}\frac{m-1}{m+n-1}$ and the probability of "RR" is $\frac{n}{m+ n}\frac{n-1}{m+n-1}$.  The probability of winning, getting either "BB" or "RR" is the sum of those two.