Vtyjkyuk

Problem statement: John and Daisy have devised a system for walking the dog. In a bag they have red marbles and blue marbles. If two marbles of the same color are picked from the bag, then John walks the dog, if it is two different colors than Daisy walks the dog. How many red and blue marbles must there be to give John and Daisy an equal chance of walking the dog.

There is not limit on how many of each marble can be in the bag. so I would love to see some interesting answers. Keep in mind each marble is drawn from the bag without replacement

Let $R$ and $B$ be the number of red and blue marbles respectively in the bag. Then you need $P(John)=frac12=P(Daisy)$.

Taking $P(Daisy)=frac12implies fracC(R,1).C(B,1)C(R+B,2)=frac12$. On simplification it gives $(R-B)^2=(R+B)$.

Few choices are $R=3$ and $B=1$; $R=6$ and $B=3$, etc.

Let's work out the probabilities here.

Say that there are $r$ red marbles, and $b$ blue marbles. Call the total number $n = r + b$. Then, $$P(texttwo reds) = fractext# of ways to draw two redstext# number of ways to draw two marbles.$$ The denominator is simple: $n choose 2$. The numerator is basically the same: $r choose 2$. Thus this works out to be $$P(texttwo reds) = fracr(r - 1)n(n - 1).$$ The formula for blue is the same. Using this, can you work out $P(texttwo of same color)$? Hint: These events are disjoint.

Once you have that probability, what's $P(texttwo of different colors)$? Hint: How is this event related to the previous event?

Once all of this is together, you should have an equation to work with involving constants, $r$, and $b$. (Remember that $n = r + b$.) Can you find $r$ and $b$ that make this equation true?

Let $m$ be the total number of marbles. Now let there be $m-k$ $R$, and $k$ $B$ marbles. First work out the probabilities of picking $BB$, $RR$, $BR$, or $RB$
beginalign*
P(BB)&=frac(m-k)(m-k-1)m(m-1) & P(RR)&=frack(k-1)m(m-1) \
P(BR)&=frac(m-k)km(m-1) & P(RB)&=frack(m-k)m(m-1)
endalign*
We need
beginalign*
P(BR)+P(RB)&=frac2k(m-k)m(m-1)=frac12tag1\
P(RR)+P(BB)&=fracm^2-(2k+1)m+2k^2m(m-1)=frac12tag2
endalign*
On setting the numerators of $(1)$ and $(2)$ equal (since the denominators of both are the same):
beginalign*
2k(m-k)&=m^2-2mk-m+2k^2\
implies m^2-(4k+1)m+4k^2&=0
endalign*
Solve for $m$:
$$m=frac4k+1pmsqrt8k+12tag3$$
Now for $m$ to be an integer $8k+1$ has to be an odd square, say $8k+1=(2u+1)^2$ for some $uinBbb N$. Substituting this term into $(3)$ gives
$$m=frac4k+1pmsqrt(2u+1)^22=frac4k+1pm2upm12$$
Therefore
$$m_+=frac4k+1+2u+12=2k+u+1quadtextandquad m_-=frac4k+1-2u-12=2k-u$$
Now $8k+1=(2u+1)^2$ so $k$ is nonnegative, and $u=frac-1+sqrt8k+12$. So
$$m_+=2k+1+frac-1+sqrt8k+12tag4$$
$$m_-=2k-frac-1+sqrt8k+12tag5$$
All odd squares are of the form $8T_j+1$, where $T_j$ is the $j$ th triangular number ($T_j=frac12j(j+1)$). Therefore, using $m-k$ $R$, and $k$ $B$ for the amount of marbles, and equations $(1)$ and $(2)$ for the probabilities, we have the first $5$ results for the positive root:
$$
beginarrayccccccc
j & k=T_j & m_+ & R_+ & B_+ & P(BR)+P(RB) & P(BB)+P(RR)\hline
1& 1 & 4 & 3 & 1 &tfrac612=tfrac12 & tfrac612=tfrac12\
2& 3 & 9 & 6 & 3 &tfrac3672=tfrac12 & tfrac3672=tfrac12\
3 & 6 & 16 & 10 & 6 & tfrac120240=tfrac12 & tfrac120240=tfrac12\
4 & 10 & 25 & 15 & 10 &tfrac300600=tfrac12 & tfrac300600=tfrac12\
5 & 15 & 36 & 21 & 15 &tfrac6301260=tfrac12 & tfrac6301260=tfrac12
endarray
$$
We see the amount of marbles, $m_+$, has to be a square number $ge4$, in fact $m_+,j=(j+1)^2$; and from $(4)$ find $m_+,j$, $R_+,j$, and $B_+,j$ (subscript to denote sign of root and position) to be, for $jge1$:
beginalign*
m_+,j&=2T_j+j+1=(j+1)^2\
#R_+,j&=T_j+1\
#B_+,j&=T_j
endalign*
For the negative root $m_-$ we have to start from $jge2$, as $j=1$ gives division by zero
$$
beginarrayccccccc
j & k=T_j & m_- & R_- & B_- & P(BR)+P(RB) & P(BB)+P(RR)\hline
1& 1 & 1 & 0 & 1 & -& -\
2& 3 & 4 & 1 & 3 &tfrac12 & tfrac12\
3 & 6 & 9 & 3 & 6 & tfrac12 & tfrac12\
4 & 10 & 16 & 6 & 10 &tfrac12 & tfrac12\
5 & 15 & 25 & 10 & 15 &tfrac12 & tfrac12
endarray
$$
beginalign*
m_-,j&=2T_j-1+j=j^2\
#R_-,j&=T_j-1\
#B_-,j&=T_j
endalign*
As can be seen, for a given square $m$, the parity of the root merely changes the role $R$ or $B$ plays.