Answer: Tuesday Teaser #28
I think there are four possibilities for you as the cyborg shooting first. You can either:
- Shoot at Cyborg B (who has 50% accuracy)
- Shoot at Cyborg C (who has 100% accuracy)
- Shoot into the air at nothing/nobody.
Let’s explore these in turn. If you shoot at Cyborg B, you will kill him one-third of the time, in which case you will be immediately killed by Cyborg C. So your probability of staying alive will be:
(2/3)* P(You stay alive having missed with the first shot)
Note that this is the same probability that you would have if you chose the contrived strategy of shooting at yourself with your first shot. That is, assuming you still shoot with 33% accuracy even when shooting yourself!
If you shoot at Cyborg C, you will shoot him one-third of the time, leaving a one on one duel with Cyborg B in which Cyborg B goes first. So your probability of staying alive will be:
(1/3)* P(You win 1-on-1 duel with Cyborg B, going second) + (2/3)* P(You stay alive having missed with the first shot)
Note that this is clearly better than the chance you have if you shoot at Cyborg B with your first shot.
Finally, assuming that cyborgs don’t understand the concept of honour and chivalry, you could shoot away with your first shot, guaranteeing that you miss. In this case, your probability of staying alive is:
P(You stay alive having missed with the first shot)
So there are two probabilities we need to compute in order to figure out whether strategy 2 is better than strategy 3. Let’s tackle them in order.
To calculate the probability that you stay alive having missed with the first shot, let’s consider the scenarios. Cyborg B will undoubtedly shoot at Cyborg C and will kill him 50% of the time. In this instance, you are left in a one-on-one duel with Cyborg B, but crucially you get to go first. In this duel, you win when the following sequences occur: 1. you hit; 2. you miss, he misses, you hit; 3. you miss, he misses, you miss, he misses, you hit, and so on. This occurs with probability:
Sum (n=0 to infinity) {(1/3) * (2/3)^n * (1/2)^n} = 1/2 (Wolfram Alpha)
If Cyborg B misses Cyborg C, Cyborg C will immediately kill Cyborg B and you will be left in a one-on-one duel with Cyborg C, in which you get to go first. Here you only have one shot to kill him, so your chances are 1/3.
So, summing up, the probability that you stay alive having missed with the first shot is (1/2)*(1/2) + (1/2)*(1/3) = 1/4 + 1/6 = 5/12.
To calculate the probability that you win 1-on-1 duel with Cyborg B, going second, we use the same method as before. In this duel, you win when the following sequences occur: 1. he misses, you hit; 2. he misses, you miss, he misses, you hit, and so on. This occurs with probability:
Sum (n=0 to infinity) {(1/3) * (2/3)^n * (1/2)^(n+1)} = 1/4 (Wolfram Alpha)
So let’s review the three options again. If you choose to shoot at Cyborg B (or at yourself), you will survive with probability (2/3)*(5/12) = 5/18.
If you choose to shoot at Cyborg C, you will survive with probability (1/3)*(1/4) + (2/3)*(5/12) = 13/36.
If you choose to shoot away, you will survive with probability 5/12 (or 15/36).
So, in conclusion, the best strategy is to ensure that you miss with your first shot.
