Is that a string triangle in your pocket or are you just happy to see me?

You have three coins in your pocket, each of which can be a penny, nickel, dime or quarter with equal probability. You might have three different coins, three of the same coin or two coins that are the same and one that is different.

Each of these coins can buy you a string whose length in centimeters equals the value of the coin in cents, i.e., the penny buys 1 cm of string, the nickel buys 5 cm of string, etc. After purchasing your three lengths of string, what is the probability that they can be the side lengths of a triangle?

First let's determine which combinations of $1,$ $5,$ $10,$ and $25$ can be combined to form a triangle. We can eliminate any combination of three distinct lengths, since none of the combinations satisfy the triangle inequality \begin{align*} 1^2 + 5^2 &\lt \min \{ 10^2, 25^2 \} \\ 1^2 + 10^2 &\lt 25^2 \\ 5^2 + 10^2 &\lt 25^2\end{align*}

If we are focusing now on any selection of two number from $1,$ $5,$ $10,$ and $25,$ then we can only have $3$ possible outcomes that result in a triangle. Namely, let $m$ and $n$ be distinct selections from $\{1, 5, 10, 25\}.$ Then we can only have a triangle if we select two copies of $\max\{m,n\}$ and one selection of $\min\{m,n\},$ since \begin{align*} 2 \cdot 1^2 &\lt \min \{5^2, 10^2, 25^2\}\\ 2 \cdot 5^2 &\lt \min \{10^2, 25^2\}\\ 2 \cdot 10^2 &\lt 25^2\end{align*} There are $\binom{3}{2} = 3$ ways to choose two of the maximum of $m$ and $n$ and $\binom{4}{2} = 6$ ways to choose two distinct coins, so there are $18$ total combinations of triangles with two distinct sidelengths.

Adding in the $4$ possible ways of obtaining equilateral triangles, we get a total of $22$ coin combinations that yield triangles out of a possible space of $4^3 = 64.$ So you only have an $\frac{11}{32} = 34.375\%$ chance of generating a string triangle. I suppose the flip side is that, you probably are happy to see me then?