Going first, second, or last?

There are the following 10 ways to choose 3 out of 5 sticks of lengths 1, 2, 3, 4, 5:

$123,\ 124,\ 125,\ 134,\ 135,\ 145,\ {\color{#D61F06} 234},\ 235,\ {\color{#D61F06} 245},\ {\color{#D61F06} 345}.$

Three facts to note about this list are as follows:

1. Only three combinations $({\color{#D61F06}234}, {\color{#D61F06}245}, {\color{#D61F06}345})$ can generate a non-degenerate triangle.
2. Any combination of numbers that includes the number "1" (the shortest side) cannot generate a triangle because the difference between the longest and second longest sides will always be at least 1.
3. Without "1" among the three sticks chosen (two by Sue, one by Kelly), "4" is the magic number that always give a triangle no matter what the other two numbers are, i.e. whether the other two are 2-3, 2-5, or 3-5. (For example, 3 is not magical since it wouldn't be able to make a triangle if the other two are 2-5 rather than 4-5.)

Now, let's calculate the probability that Sue and Kelly together can get a non-degenerate triangle, for each of the 3 sequences given:

A. Kelly-Sue-Sue (Kelly going first)

As explained, 4 always works as long as 1 is not one of the other two numbers. However, even without 1, the number 2, 3, or 5 doesn't always work because of the possibility of getting 235. So, 4 is the best choice for Kelly, and we need to calculate the probability that, given Kelly's pick of 4, Sue randomly picks two numbers, not including 1, out of 1, 2, 3, 5: $1 - \frac{{1\choose1} \times {3\choose1}}{4\choose2}=1-\frac{3}{6}=0.5. \qquad \left(\text{or equivalently }\frac{3\choose2}{4\choose2}=0.5\right)$

B. Sue-Kelly-Sue (Kelly going second)

1. Sue first picks 1:
   Kelly doesn't even need to make her pick because there is no chance to get a triangle with 1 included. So, the probability here is $\frac15 \times 0=0.$
   
2. Sue first picks 2, 3, or 5:
   Kelly definitely chooses 4 and hopes that Sue will not pick "1" out of the remaining three numbers which include "1" and two numbers other than Sue's pick (2, 3, or 5) and Kelly's pick (4). The total probability of this happening (for all three of 2, 3, and 5) is $3\times \frac15 \times \frac23=\frac25.$
   
3. Sue first picks 4:
   Kelly chooses one of 2, 3, 5 (anything but 1) and hopes for exactly the same as in 2 just above, the probability of which is $\frac15 \times \frac23=\frac2{15}.$

Therefore, all together, the probability of getting a triangle for the sequence Sue-Kelly-Sue is $0+\frac25+\frac2{15}=\frac8{15}\approx 0.53.$

C. Sue-Sue-Kelly (Kelly going last)

As explained in fact #3 above, if Sue happens not to include "1" in her two random choices, all Kelly needs to do is to either choose 4 if it's not there, or choose anything but 1 if 4 is already there. Therefore, the probability to be calculated in this case is just the probability that Sue's two picks don't include 1 : $1 - \frac{{1\choose1} \times {4\choose1}}{5\choose2}=1-\frac{4}{10}=0.6. \qquad \left(\text{or equivalently }\frac{4\choose2}{5\choose2}=0.6\right)$

From A, B, C, we can conclude that the sequence Sue-Sue-Kelly has the greatest probability of 0.6.

Relevant wiki: Triangle Inequality

The three ways to make a non-degenerate triangle are 234, 245 and 345, all others fail.

Every combination of two sticks that does not contain the 1 can thus be made into a triangle (23, 24, 25, 34, 35, 45).

This means it does not matter which sticks you get as long as it's not stick number 1.

When you choose a stick before the end, the chances of drawing the 1 randomly increases (because there are less options to choose from). For that reason alone, picking last offers the best chance.

If Sue's 2 picks include a 1, Kelly can't make a triangle. If Sue's 2 picks don't include a 1, Kelly can always make a triangle. The sequence of picks giving Sue the greatest chance of not picking a one gives Kelly the greatest chance of making a triangle. If Kelly picks before either of Sue's picks, Kelly will not pick 1, and this will increase the chance of Sue picking 1 (as there are fewer sticks remaining). Therefore Kelly should pick after both of Sue's picks.

Relevant wiki: Geometric Probability - Problem Solving

The only valid combinations are $a < b < c$ where $a > c - b$ and $c < a + b$ . This leaves the choices $\{2,3,4\},\{2,4,5\},\{3,4,5\}$ .

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Analysis of KSS :

Each of Kelly's choices leaves four sticks for Sue; there are six equally likely pairs she can pick.

$\begin{array}{llr} \hline \text{Kelly} & \text{Sue's winning picks} & \text{probability} \\ \hline 1 & \text{(none)} & 0 \\ 2 & 34, 45 & 1/3 \\ 3 & 24, 45 & 1/3 \\ 4 & 23, 25, 35 & 1/2 \\ 5 & 24, 34 & 1/3 \\ \hline\end{array}$ Thus if Kelly picks first, her best odds are with stick #4, giving probability of winning $1/2$ .

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Analysis of SKS :

Consider the possibilities of the first draw, and the following choices for Kelly. We can immediately rule out $K = 1$ . After Kelly's choice, Sue will have three remaining sticks to pick from, each with probability $1/3$ .

$\begin{array}{lllr} \hline \text{Sue} & \text{Kelly} & \text{Sue's winning picks} & \text{probability} \\ \hline 1 & \text{any} & \text{none} & 0 \\ \hline 2 & 3 & 4 & 1/3 \\ & 4 & 3,5 & \boxed{2/3} \\ & 5 & 4 & 1/3 \\ \hline 3 & 2 & 4 & 1/3 \\ & 4 & 2,5 & \boxed{2/3} \\ & 5 & 4 & 1/3 \\ \hline 4 & 2 & 3,5 & \boxed{2/3} \\ & 3 & 5 & 1/3 \\ & 5 & 2,3 & \boxed{2/3} \\ \hline 5 & 2 & 4 & 1/3 \\ & 3 & 4 & 1/3 \\ & 4 & 2,3 & \boxed{2/3} \\ \hline \end{array}$

Thus, for $S = 1$ there is no chance of winning; for $S \geq 2$ , Kelly can always choose a stick to make the remaining chance of winning $2/3$ . The combined probability is $\tfrac15\times 0 + \tfrac45\times\tfrac23 = \tfrac8{15}.$

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Analysis of SSK :

There are 10 possible combinations for Sue to draw initially. Four of these contain length 1, with no chance of winning. For the remaining six (23, 24, 25, 34, 35, 45), Kelly can always choose a stick to complete the triangle. Thus the combined probability is $6/10 = 3/5$ .

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Conclusion: the probabilities compare as $\frac 3 5 > \frac 8{15} > \frac 12 \ \ \ \ \ \therefore\ \ \ \ \ \ \boxed{SSK} > SKS > KSS.$

234, 245, and 345 are the only options, which can be found by quickly using the triangle inequality. The important observation is that 1 is in no triangle. In fact, after listing all possibilities, only one degenerate triangle out of seven does not contain a 1: 235. It is clear that minimizing the chance of a 1 getting chosen will maximize the probability of a triangle. Intuitively, having Sue pick last is the worst option since there are less "safe" sticks to choose from. But I'll quickly do the math anyway:

The probability Sue picks a 1 will always be the probability she picks 1 on her first turn + the probability she doesn't pick 1 on her first turn but picks 1 on her second turn.

If Kelly chooses first, then Sue has a $\frac{1}{4}+\frac{3}{4}*\frac{1}{3}=\frac{1}{2}$ probability of picking 1.

If Kelly chooses second, then Sue has a $\frac{1}{5}+\frac{4}{5}*\frac{1}{3}=\frac{7}{15}$ probability of picking 1.

If Kelly chooses last, then Sue has a $\frac{1}{5}+\frac{4}{5}*\frac{1}{4}=\frac{2}{5}$ probability of picking 1.

Therefore, Kelly will have a greater chance of creating a triangle if she picks last.

Claim: The probability of Kelly making a triangle is the same as the probability of Sue NOT selecting the stick of length 1. (See below for the reasoning.) If this claim is true, the best thing Kelly can do is to delay her choice and leave as many non-1 sticks in the pile as possible during Sue's selections. So later is better. $P(KSS) = 1 \times \frac{3}{4} \times \frac{2}{3} = \frac{1}{2} = \frac{15}{30}$ $P(SKS) = \frac{4}{5} \times 1 \times \frac{2}{3} = \frac{8}{15} = \frac{16}{30}$ $P(SSK) = \frac{4}{5} \times \frac{3}{4} \times 1 = \frac{3}{5} = \frac{18}{30}$ Reasoning behind the claim:

• If Sue randomly selects 1, Kelly will never be able to form a triangle, no matter when she selects.

• If Sue randomly avoids 1, Kelly will always be able to form a triangle, no matter when she selects. Kelly just has to avoid the number 1 and the combination 2-3-5.
  Both of these are always doable for Kelly, no matter when she selects.
  

4 must be one of the sticks chosen. Since Sue is choosing randomly, the probability of the pick sequence creating a non-degenerate triangle is maximized when Sue chooses a 4. Thus, Sue should make both picks before Kelly.