Multiple Weighted Coin Toss Probability 1) Is it possible to weight two coins in such a way that if the two coins are tossed, the three possible outcomes (both heads, both tails, or one head and one tail) all have probability $\frac{1}{3}$? The two coins do not have to be weighted the same as each other.
2) What about three coins? Can you weight them so that the four possible outcomes (all heads, two heads and one tail, one head and two tails, all tails) are equally likely?
3) Can you generalize to an arbitrary number of coins?
My intuition is that it is not possible to weight two coins so that the three possible outcomes all have probability $\frac{1}{3}$. It would then follow that the other two questions have no solution as well. However, I do not know how to go about proving this.
 A: We deal with the general case of $n\ge 2$ coins.
Let the $i$-th coin have probability $p_i$ of landing heads, and $q_i$ of landing tails. (Of course $p_i+q_i=1$, but that will turn out to be of no importance!)  
Then the probability of $k$ heads and $n-k$ tails is the coefficient of $x^k$ in the product
$$(p_1x+q_1)(p_2x+q_2)(p_3x+q_3)\cdots(p_nx+q_n).$$
If the probabilities are to be all equal, this product must be identically equal to
$$\frac{1}{n+1}(x^n+x^{n-1}+x^{n-2}+\cdots +x+1).$$ 
However, $x^n+x^{n-1}+x^{n-2}+\cdots+x+1=0$ has some non-real roots, and therefore cannot be expressed as a product of $n$ linear factors with real coefficients. 
Remark: Though we did not use the term, this was a generating function argument. Note how smoothly a generating function argument can sidestep potentially messy algebra. 
A: Let $p,q$ denote the respective bias of the first and second coin:
$$
\mathbb{P}\{C_1=\mathsf{Heads}\} = p,\qquad \mathbb{P}\{C_2=\mathsf{Heads}\} = q
$$
Then, as long as $p,q\in[0,1]$ satisfy $p+q=1$ and $pq=1/3$, one can easily show the three conditions are satisfied (this is necessary):
\begin{align}
\mathbb{P}\{C_1=\mathsf{Heads}, C_2=\mathsf{Heads} \} = pq &= \frac{1}{3} \\
\mathbb{P}\{C_1=\mathsf{Tails}, C_2=\mathsf{Tails} \} (1-p)(1-q) &= \frac{1}{3} \\
\mathbb{P}\{C_1\neq C_2 \} = p(1-q) +q(1-p) &= \frac{1}{3} \\
\end{align}
Sadly, this is not possible. Show why.
Hint: $p+q=1$ and $pq=1/3$ imply $1=(p+q)^2 = p^2+q^2 + \frac{2}{3}$; and therefore $$\frac{1}{3} = p^2+q^2 = p^2 + (1-p)^2 = 1+2p^2-2p.$$ But the equation $$ p^2-p+\frac{1}{3} = 0$$
has no real root.
