You are in a free-for-all shootout involving three people, and by some rotten luck, you are the worst shooter of the group. You, shooter A, hit your target 33% of the time. The other two, B and C, have 50% and 100% accuracy respectively.

At least you get to shoot first, with B going after you, and C going last. Everyone takes turns shooting until all but one are dead.

Who will you fire at in the first round to maximize your odds of survival?

Free for All

Assume A shoots at C first and determine possible

outcomes for A to win;

[1]A misses C, B misses C, C shoots B, A shoots C or dies
.67 x .5 x 1.0 x .33 = .11055

[2]A misses C, B shoots C, Dual between A and B with A

shooting first:
.67 x .5 = .335

The dual between A and B ensues as follows with A

shooting first. A shoots at B with a 33% chance of

success. Assuming A misses(67%), B has a .5 chance of

hitting A, so B has a .5 x .67 = .335 chance of that

happing. There's the .33 chance of them both missing.

This distribution happens infinitely, right? But the

probability comes to 50-50 of A winning this one. That

is, the distribution is .33 to .33 or .33/(.33+.33)

So continuing with scenario [2], A has a .335 x .5 =

.1675 chance of winning.

[3] This time A hits C, so B shoots at A first in the

dual.
A hits C happens 33% of the time.

The dual between A and B with B shooting first has a

different probability if B shoots first. In round 1, B

has a 50% chance of hitting A. If B misses, A has a .33

chance, so this happens .5 * .33 times or .165 times.

The other .335 head into the next round. Again,this

distribution continues infinitely for the miss-miss

remainder of the previous round. Roughly, it comes out

to a 75% chance of B winning, and a 25% chance of A

winning. Forget the rounding.

So back at [3], A has a .33 * .25 = .0825 (roughly)

chance of winning this way.

Add up [1]+[2]+[3]= .11055+.1675+.0825=36% chance of A

winning if he shoots at C first.

If A shoots at B first, the following could happen
[1]A misses B, B misses C, and C shoots B, so A better

shoot C, or die. Same probability as [1] above, so

.11055.

[2]A misses B, B shoots C, A shootout with B with A

shooting first.
.67x.5x.5 = .1675

[3] A shoots B, C shoots A, A has no chance to win, so

that's a .33*0=0

Adding these up, we get .1105x.1675 or .27805

Therefore, A should shoot at C first. Am I right?

Posted by Lawrence on 2003-08-27 18:25:50