Tetrahedral Trouble

All of the faces must satisfy the triangle inequality: $a + b > c$ (it can't be equal).

Consider an edge $A$ . Without loss of generality, $A$ will create two faces: $A,B,C$ and $A,D,E$ . This means that there is exactly one edge $F$ that doesn't create a face with $A$ (and similarly for each edge that is considered).

Also note that if we swap $B$ and $C$ around, and also swap $D$ and $E$ around, the collection of faces on the right of the diagram don't change (meaning that we can fill in edges that we know later on without the loss of generality).

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Assuming that $A = 17$ , consider the edge with length $23$ . If $17$ was to create a face with $23$ , the third edge of the triangular face cannot be $40$ , $58$ or $65$ as these don't satisfy the triangle inequality so the face would have to be $\{17,23,?\}$ . Alternatively, $17$ and $23$ could be edges that are opposite and don't create a face.

Assuming that $A = 17$ , consider the edge with length $40$ . If $17$ was to create a face with $40$ , the third edge of the triangular face cannot be $23$ , $58$ or $65$ as these don't satisfy the triangle inequality so the face would have to be $\{17,40,?\}$ . Alternatively, $17$ and $40$ could be edges that are opposite and don't create a face.

Now we have $2$ different pairs of options:

Now we must select one option from each row. However, since only one edge can be opposite $17$ we cannot have the purple and orange together (also the $17$ and $?$ can't be on two faces together if blue and red are together). Therefore either the $\color{#3D99F6} \text{blue}$ and $\color{#EC7300} \text{orange}$ statements are true or the $\color{#69047E} \text{purple}$ and $\color{#D61F06} \text{red}$ statements are true.

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Case 1: $\{17,23,?\}$ forms a face, and $17$ is opposite $40$

We can fill in $23$ and $?$ as $B$ and $C$ (proven above that this is without loss of generality) meaning that we get: $A=17$ , $B=23$ , $C=?$ , $D=$ , $E=$ , $F=40$ . Now since we know that $\{B,D,F\}$ forms a face, due to the triangle inequality, $D < 23 + 40$ meaning that $D$ must equal $58$ .

This means that: $A=17$ , $B=23$ , $C=?$ , $D=58$ , $E=65$ , $F=40$ . We can now compare all these numbers with the triangle inequality for all $4$ of the faces that are shown on the diagram. Checking, we see that the ones with known numbers satisfy, and we also get that $25 < ? < 40$ giving us $\boxed{14}$ possible integer values for $?$ .

Case 2: $\{17,40,?\}$ forms a face, and $17$ is opposite $23$

We can fill in $40$ and $?$ as $B$ and $C$ (proven above that this is without loss of generality) meaning that we get: $A=17$ , $B=40$ , $C=?$ , $D=$ , $E=$ , $F=23$ . Now since we know that $\{B,D,F\}$ forms a face, due to the triangle inequality, $D < 40 + 23$ meaning that $D$ must equal $58$ .

This means that: $A=17$ , $B=40$ , $C=?$ , $D=58$ , $E=65$ , $F=23$ . We can now compare all these numbers with the triangle inequality for all $4$ of the faces that are shown on the diagram. Checking, we see that the ones with known numbers satisfy, and we also get that $42 < ? < 57$ giving us another (distinct) $\boxed{14}$ possible integer values for $?$ .

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Binay now knows that there are $14 + 14 = 28$ possibilities for the sixth edge length.

Since Binay won't guess the same thing twice, $P(\text{Not guessing correctly}) = \frac{27}{28} \times \frac{26}{27} \times \frac{25}{26} = \frac{25}{28}$ Therefore, $P(\text{Guessing correctly}) = 1 - \frac{25}{28} = \frac{3}{28} \approx \boxed{0.107}$