Quadratic Game Show

Stupid solution 1:

Choose $a=b=c=0$ . The "quadratic" has (infinitely many) real solutions, no matter how the host "rearranges" the three zeroes. Thus the probability is $1$ , giving $x = \boxed{100}$ .

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Stupid solution 2:

Note that the probability of winning is always an integer divided by $3$ . To see this, note that there are $6$ equally likely possibilities, so the denominator must be $6$ . Also, since $ax^2+bx+c$ and $cx^2+bx+a$ always both have real roots or both have complex roots (the roots are reciprocals of each other), the numerator must be even; whenever we have a successful case, we have another (the one paired, swapping $a$ and $c$ ). Thus the probability is an integer divided by $3$ . So $\frac{x}{100}$ must be an integer divided by $3$ , and also a probability, so it can only be $0$ or $1$ . Finally, $x \neq 0$ , since we can at least choose $1,2,3$ , which when put properly as $x^2 + 3x + 2$ , gives real solutions. So the probability is $1$ , giving $x = \boxed{100}$ .

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Actual solution, assuming $a,b,c$ are nonzero:

Consider the selections $1,1,-2$ . There are three cases:

• $a=1, b=1, c=-2$ , giving $x^2+x-2 = 0$ . This has real solutions $x = 1, -2$ .
• $a=1, b=-2, c=1$ , giving $x^2-2x+1 = 0$ . This has real solution $x = 1$ .
• $a=-2, b=1, c=1$ , giving $-2x^2+x+1 = 0$ . This has real solutions $x = 1, -\frac{1}{2}$ .

Thus in all cases, the contestant wins, giving $x = \boxed{100}$ for a certain win.

To see how to obtain that triple:

A quadratic $ax^2+bx+c$ has real roots iff $b^2 \ge 4ac$ . But with real coefficients, the left hand side is non-negative. Thus if we pick two positive numbers and one negative number, whenever the negative number falls to $a$ or $c$ , we win. Thus the only possibility we can lose is if $b$ is negative. But if so, just make $b$ really large and $a,c$ really small to ensure $b^2 \ge 4ac$ still applies.

The answer is very simple. The contestant can choose a, b, c in unlimiyed ways in the above questions.

The quadratic will have real roots if b^2 -4ac > 0. The contestant can choose any values of a, b, c but keep the sign of either a or c negative each time.