You are in a shootout with two others. Unfortunately you are not a very good shot, only hitting what you are aiming for one-third of the time. The one of the others hits two-thirds of the time, and the last always hits.
To be fair, everyone agrees to fire in order with the worst shot (you) going first. You keep going around in the circle until only one is left standing.
What is your best strategy? Assume everyone is trying to maximise their chance at being the lone survivor, and none of you will ever give up.
Your best strategy is to shoot into the air the first round, then keep firing at the survivor in subsequent rounds.
Proof
Let’s call you person #1, #2 the next best shot, and #3 the person who always hits.
First note that if anyone other than #3 shoots, they must shoot #3. If they shoot the other person and kill them, then they will be in a duel with #3 with #3 going first. #3 wins this duel since #3 never misses.
Therefore it is in #3’s best interest to turn it into a duel as soon as possible, otherwise the other two are just shooting at #3. Thus if #3 survives to their turn then they must shoot. #3 would rather be in a shoot-out with the weaker shooter, hence they would shoot and kill #2.
Thus when it is #2’s turn they must shoot at #3 and hit to survive at all. From this we can tell that by the end of the first round, either #2 or #3 will have been shot.
Finally we look at #1. We’ve already established that if #1 shoots anyone, it must be #3. If #1 shoots and kills #3 then #1 will be in a duel with #2, with #2 going first. If #1 misses then they will be in a duel with the survivor of #2 and #3, with #1 going first.
The probability that #1 wins if they shoot and kill #3 is:
\[\begin{align} P(\text{#1 wins } \mid \text{ #1 hits #3}) & = P(\text{#1 wins } \mid \text{ #2 shoots first}) \\ & = P(\text{#2 misses}) P(\text{#1 hits}) \\ & \qquad + P(\text{#1 misses}) P(\text{#1 wins } \mid \text{ #2 shoots first}) \\ P(\text{#1 wins } \mid \text{ #2 shoots first}) & = \frac{1}{3} \left( \frac{1}{3} + \frac{2}{3} P(\text{#1 wins } \mid \text{ #2 shoots first}) \right) \\ P(\text{#1 wins } \mid \text{ #2 shoots first}) & = \frac{1}{7} \end{align}\]The probability that #1 wins if they shoot and miss #3 is:
\[\begin{align} P(\text{#1 wins } \mid \text{ #1 misses #3}) & = P(\text{#1 wins } \mid \text{ #2 hits #3}) + P(\text{#1 wins } \mid \text{ #2 misses #3}) \\ & = \frac{2}{3} P(\text{#1 wins against #2 } \mid \text{ #1 shoots first}) \\ & \qquad + \frac{1}{3} P(\text{#1 wins against #3 } \mid \text{ #1 shoots first}) \\ & = \frac{2}{3} \left( P(\text{#1 hits #2}) + P(\text{#1 misses #2}) P(\text{#1 wins } \mid \text{ #2 shoots first}) \right) \\ & \qquad + \frac{1}{3} P(\text{#1 hits #3}) \\ & = \frac{2}{3} \left( \frac{1}{3} + \frac{2}{3} \frac{1}{7} \right) + \frac{1}{3} \frac{1}{3} \\ & = \frac{2}{7} + \frac{1}{9} \\ & = \frac{25}{63} \end{align}\]Thus #1 has a higher chance of winning if they miss their shot!
Therefore you should intentionally miss the first shot by shooting into the air. After the first round, you will be in a duel with one of the other two people. You will win about 40% of the time.
General case for 3 people
Number the people #1, #2 and #3 such that the probability that a person hits their target is \(x_p\), with \(x_1 < x_2 < x_3\). Also, define \(\bar{x}_i = 1 - x_i\).
Define the probability \(P_{p_1, p_2, p_3}(p)\) that person \(p\) wins when the shooting order is \(p_1, p_2, p_3\) in the 3 person case.
Similarly, define \(P_{p_1, p_2}(p)\) for the 2 person case. For convenience let \(P_{p_1, p_2}(p) = 0\) if \(p \notin {p_1, p_2}\).
2 person case
We will do the 2 person case first, since the 3 person case reduces to this once a person is hit.
Each person can shoot at the other, or intentionally miss. Clearly, intentionally missing is bad because their opponent is always going to shoot at them: if their opponent hits they die; if their opponent misses they are back to where they started, but have given the opponent a free shot. Because the shooters must shoot, we can say for either person #\(i\):
\[\begin{align} P_{i,j}(i) & = x_i + \bar{x}_i P_{j,i}(i) \\ P_{j,i}(i) & = \bar{x}_j P_{i,j}(i) \end{align}\]Combining these equations we get:
\[\begin{align} P_{i,j}(i) & = x_i + \bar{x}_i \bar{x}_j P_{i,j}(i) \\ & = \frac{x_i}{1 - \bar{x}_i \bar{x}_j} \end{align}\]The probability that the other person #\(j\) wins when #\(i\) shoots first is just the complement:
\[P_{i,j}(j) = 1 - P_{i,j}(i)\]Simplifying:
\[\begin{align} P_{i,j}(j) & = 1 - \frac{x_i}{1 - \bar{x}_i \bar{x}_j} = \frac{1 - \bar{x}_i \bar{x}_j - x_i}{1 - \bar{x}_i \bar{x}_j} = \frac{\bar{x}_i - \bar{x}_i \bar{x}_j}{1 - \bar{x}_i \bar{x}_j} \\ & = \frac{\bar{x}_i x_j}{1 - \bar{x}_i \bar{x}_j} \end{align}\]Note that it is always better to start shooting first.
3 person case - Who shoots
Each person has three options, shoot either of the remaining people, or intentionally miss (shoot the air).
If person #\(i\) does shoot at a person then they are hoping to hit (otherwise they would shoot the air). If it hits then person #\(i\) will be in a duel with the survivor, where #\(i\) goes second. Thus they shoot #\(j\) instead of #\(k\) if:
\[\begin{align} P_{k,i}(i) & > P_{j,i}(i) \\ \frac{\bar{x}_k x_i}{1 - \bar{x}_k \bar{x}_i} & > \frac{\bar{x}_j x_i}{1 - \bar{x}_j \bar{x}_i} \\ x_i (\bar{x}_k - \bar{x}_k \bar{x}_j \bar{x}_i) & > x_i (\bar{x}_j - \bar{x}_j \bar{x}_k \bar{x}_i) \\ \bar{x}_k & > \bar{x}_j \\ x_k & < x_j \end{align}\]Hence, we have:
Lemma 1: If anyone shoots at another person, they will always shoot at the person with the higher accuracy.
If everyone shoots in the air then no one dies and the dispute is never resolved, in which case the probability of being the lone survivor for any of the people is 0. Thus at least one person shoots. Even in the case where shooting leads to certain death, this is the same as not resolving the dispute and hence we can assume either.
If one person is shooting then the other two can’t both be intentionally missing. In this case it is always better for the person that the shooter is shooting at to shoot back (assuming they haven’t been shot yet). Therefore we have:
Lemma 2: At least two people are shooting in any round where everyone survives.
#3 must either shoot #2 by Lemma 1 or shoot in the air.
- If #3 is shoots in the air then by Lemma 2 both #1 and #2 are shooting, and by Lemma 1 they are both shooting at #3. If they both miss, #3 is back to where they started and #3 shoots in the air again. The other two would continue shooting at #3 until they hit, giving #3 no chance of winning.
- If #3 shoots at #2 then #3’s chance of winning in this case is \(x_3 P_{1,3}(3) + \bar{x}_3 P_{1,2,3}(3) > 0\).
Thus #3 shoots at #2.
#2 must either shoot #3 by Lemma 1 or shoot in the air.
- If #2 shoots #3, they are hoping for a duel with #1. Since #1 would start the duel, the probability that #2 would win is \(P_{1,2}(2)\).
- If #2 shoots in the air, then #2’s best case scenario is that #3 will miss their shoot at #2 then #1 will hit #3. This results in a duel with #1 where #2 shoots first. Even allowing that #1 definitely hits, then the probability that #2 wins is \(\bar{x}_3 P_{2,1}(2)\).
#2 shoots at #3 if #2 would prefer hitting to missing:
\[\begin{align} P_{1,2}(2) & > \bar{x}_3 P_{2,1}(2) \\ \bar{x}_1 P_{2,1}(2) & > \bar{x}_3 P_{2,1}(2) \\ \bar{x}_1 & > \bar{x}_3 \\ x_1 & < x_3 \end{align}\]This is always true so #2 will shoot at #3.
#1 will either shoot #3 by Lemma 1 or shoot in the air.
-
If #1 shoots #3 they are hoping for a duel against #2 where #1 goes second. In this case #1 has a probability of winning of:
\[P_{2,1}(1) = \frac{\bar{x}_2 x_1}{1 - \bar{x}_2 \bar{x}_1}\] -
If #1 shoots the air, then the other two will shoot at each other. This will continue until either #2 or #3 is shot. In both cases #1 duels with the survivor, and gets to shoot first. The probability that #1 wins in this case is:
\[\begin{align} P_{1 \text{air}} & = x_2 P_{1,2}(1) + \bar{x}_2 (x_3 P_{1,3}(1) + \bar{x}_3 P_{1 \text{air}}) \\ & = x_2 P_{1,2}(1) + \bar{x}_2 x_3 P_{1,3}(1) + \bar{x}_2 \bar{x}_3 P_{1 \text{air}} \\ & = \frac{x_2 P_{1,2}(1) + \bar{x}_2 x_3 P_{1,3}(1)}{1 - \bar{x}_2 \bar{x}_3} \\ & = \frac{x_1}{1 - \bar{x}_2 \bar{x}_3} \left( \frac{x_2}{1 - \bar{x}_1 \bar{x}_2} + \frac{\bar{x}_2 x_3}{1 - \bar{x}_1 \bar{x}_3} \right) \end{align}\]
#1 will shoot at #3 if and only if #1 would prefer hitting to missing:
\[\begin{align} \frac{\bar{x}_2 x_1}{1 - \bar{x}_2 \bar{x}_1} & > \frac{x_1}{1 - \bar{x}_2 \bar{x}_3} \left( \frac{x_2}{1 - \bar{x}_1 \bar{x}_2} + \frac{\bar{x}_2 x_3}{1 - \bar{x}_1 \bar{x}_3} \right) \\ \bar{x}_2 (1 - \bar{x}_2 \bar{x}_3) (1- \bar{x}_1 \bar{x}_3) & > x_2 (1 - \bar{x}_1 \bar{x}_3) + \bar{x}_2 x_3 (1 - \bar{x}_1 \bar{x}_2) \end{align}\]Thus #1 will shoot at #3 when:
\[C > 0 \text{ where } C = \bar{x}_2 (1 - \bar{x}_1 \bar{x}_3) (1 - \bar{x}_2 \bar{x}_3) - x_2 (1 - \bar{x}_1 \bar{x}_3) - \bar{x}_2 x_3 (1 - \bar{x}_1 \bar{x}_2)\]3 person case - Probabilities
Let \(x'_1\) be the probability that #1 hits when there are 3 players. This is the same as \(x_1\) when #1 aims, but is \(0\) when #1 intentionally misses.
Then the probability that person #n wins is:
\[P_{1,2,3}(n) = x'_1 P_{2,1}(n) + \bar{x}'_1 (x_2 P_{1,2}(n) + \bar{x}_2 (x_3 P_{1,3}(n) + \bar{x}_3 P_{1,2,3}(n)))\]Which resolves to:
\[P_{1,2,3}(n) = \frac{x'_{1} P_{2,1}(n) + \bar{x}'_1 x_2 P_{1,2}(n) + \bar{x}'_1 \bar{x}_2 \bar{x}_3 P_{1,3}(n)} {1 - \bar{x}'_1 \bar{x}_2 \bar{x}_3}\] \[x'_1 = \begin{cases} x_1 & \text{ if } C > 0 \\ 0 & \text{ if } C \le 0 \end{cases} \text{ where } C = \bar{x}_2 (1 - \bar{x}_1 \bar{x}_3) (1 - \bar{x}_2 \bar{x}_3) - x_2 (1 - \bar{x}_1 \bar{x}_3) - \bar{x}_2 x_3 (1 - \bar{x}_1 \bar{x}_2)\]Original problem
For this problem we have:
\[(x_1, x_2, x_3) = \left(\frac{1}{3}, \frac{2}{3}, 1\right)\]From this we get \(C \approx -0.6 < 0\), hence #1 simply shoots the air. The probability #1 wins (which is true for any values where \(C < 0\)) is given by:
\[P_{1,2,3}(1) = \frac{x_1}{1 - \bar{x}_2 \bar{x}_3} \left( \frac{x_2}{1 - \bar{x}_1 \bar{x}_2} + \frac{x_3 \bar{x}_2}{1 - \bar{x}_1 \bar{x}_3} \right)\]Solving this with our values, we agree with our previous result:
\[P_{1,2,3}(1) = \frac{25}{63}\]