Truel again

Logic Level 4

Rivals A, B and C want to settle a quarrel by shooting at each other with guns. The rules of the truel (3 person duel) are:

  • They draw lots to determine who fires first, second and third.
  • Each turn, they can choose to fire at a specific person, or not at all.
  • A kills in 100% of his shots, B kills in 80% of his shots and C kills in 50% of his shots. No one is killed by a stray bullet.
  • The truel continues until only 1 person is left alive

Assuming that everyone chooses their ideal strategy, who has the largest chance of surviving the truel? Is this chance greater than or less than 50%?

A, less than 50% C, greater than 50% C, less than 50% A, greater than 50% B, greater than 50% B, less than 50%

Prince Loomba by Prince Loomba , 21, India

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1 solution

Jakob Andersen
Dec 1, 2017

It goes without saying that when there is only two rivals left alive, they will always decide to shoot at each other. With that cleared up, we can look at the slightly more complicated scenario with them all three still alive.

First, we will consider the problem as if they don't have the option not to shoot. This is to realise who they will shoot, in case they decide to do so.

In this problem, the rivals will shoot against the strongest candidate, since they obviously want to duel the weakest of the two opponents after the first kill.

Following this logic, we can evidently say that:

A will shoot B.

B will shoot A.

C will shoot A.

Now, we give the rivals the option not to shoot at all.

From this we can deduce that C won't shoot at all. This is because he knows that A and B will shoot at each other, thus assuring him a seat among the two remaining rivals AND the first shot when the situation occurs.

As a result of this, if A does not shoot, the turn goes to B, and vice versa. Now, you might argue that neither of them will shoot, since this allows all of the rivals to survive with a 100% chance. However, the problem does not allow this, as it is clearly stated that the game goes on until only 1 is left. In other words, someone will inevitably shoot at some point.

This means that A will shoot B, because he knows that otherwise B will eventually take the shot against him. By this logic it also follows that B will shoot A if he can.

So, if A goes before B (50% chance), B will die, and C will then have a 50% of surviving.

Likewise, If B goes before A (50% chance) and MISSES his bullet (20% chance), B will die, and C will then have a 50% of surviving again.

However, if B goes before A (50% chance) and HITS his bullet (80% chance), C will then have a chance of surviving given by:

1 2 \frac{1}{2} ∑ i = 0 ∞ \sum_{i=0}^\infty 1 1 0 i \frac{1}{10^i} = 5 9 \frac{5}{9}

In the first case, C will have 50% chance of winning. In the second case, C will have more than 50% chance. Conclusively, the answer can only be C, greater than 50%

As mentioned, the exact probability of surviving for C can be calculated as:

1 2 \frac{1}{2} (1+ 1 5 \frac{1}{5} ) 1 2 \frac{1}{2} + 1 2 \frac{1}{2} 4 5 \frac{4}{5} 5 9 \frac{5}{9} = 47 90 \frac{47}{90}

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