Probability puzzle about duels

I get 128/315=40.6% chance. I haven't double checked my math but it meets the smell test. I also think A has a better chance if he pulls a Hamilton and misses intentionally on the first shot, but again, I haven't done all the math.

'A' intentionally missing the first shot is equivalent to 'B' being the first to shoot. It makes 'A' the last to shoot.

I don't believe this helps 'A'. Being the last to shoot can't be a good thing.

There is a 75% probability that B will kill C leaving A a 4 out of 7 chance of winning.

Nope...

(assuming B goes first)

After B kills C, B then gets to shoot at A (without A getting a shot).

This means A has a 75% chance of being shot before having a chance to shoot.

Ooops, it looks like I misunderstood the rules... sorry. The phrasing confused me.

No problem. Here is what I calculated.

To solve, you first need to know the probability of C beating A in a one on one matchup and A beating B in a one on one matchup.

The probability of C beating A (Pca)

Pca = 50% + (50%)(50%)*Pca
Pca = 2/3, so if B is killed and C is next to shoot, C has a 2/3 chance of winning.

The probability of A beating B (Pab)

Pab = 50% + (50%)(25%) Pab
Pab = 4/7, so if C is killed and A in next to shoot, A has a 4/7 chance of winning.

Now we can solve the original problem.

Pa = the sum of...
50% (A kills B, now it is 1x1 between C and A) * (1-Pca) +
50% (A misses) * 75% (B kills C, now it is 1x1 between A and B) * Pab +
50% (A misses) * 25% (B misses) * 50% (C misses) * Pa (everyone missed, start over)

Pa = 50%(1/3) + (50%)(75%)*4/7 + (50%)(25%)(50%)* Pa
Pa = 0.3809 + .0625 Pa
Pa = .4062

So if A intentionally misses B as long as C is alive, it looks like this:

Pa' = 75% (B kills C)* Pab + 25% [ 50% (C misses A) * Pa'
Pa' = 3/4 * 4/7 + 1/8 Pa'
7/8 Pa' = 3/7
Pa' = 24/49 = 49.0%

It looks like A should take the high road and miss that first shot. If all three duelists are mathematicians, they will all intentionally miss over and over until they run out of ammo and go home.

You might be confusing A "winning" (i.e. being the only survivor) from A not dying one round.

The possibilities assuming that A intentionally misses the first shot.

(B kills C) (A kills B) => 3/8 [A is the only survivor]
(B kills C) (A misses B) => 3/8 [A and B both survive and by the rules continue]
(B misses C) (C kills A) => 1/8 [and A is dead]
(B misses C) (C misses A) => 1/8 [a push, they all survive and nothing changes]

There is only a 3/8 chance of an outright win for A after 3 shots.

If A and B are left... B get's a 3/4 chance of killing A each round... and the first shot.

I do agree that A should skip the first shot. However I would rather be B (if I have a choice).

I don't think so. I am assuming that they have motivation to participate in the duel anyway (there must be some payoff). I think that B has the expected value to take the shot each time, and to keep shooting until she wins. This puts the other two mathematicians into play.

So after one round:
3/8 A wins
1/8 A loses
3/8 A is alive, but B is in control and has a 6/7 chance of winning it from there
1/8 a push.

So A's overall chance of winning is [3/8 + (1/7)(3/8) ] / (7/8) = 49.0%

If B kills C, A is in the driver's seat and will win 4 of 7 times, so B does not want to kill C if A is still in the game.

Being a programmer, I'll bet you could code up a Monte Carlo in 5 minutes and run a million trials if you want to test that result.

Maybe A is a really off the mark shooter and knocks off C before B even gets a chance. Can A now fire again and then do in B?

So what is B's overall chance of winning (assuming A misses the first shot)?

Ok, completed the MC program. The first result, everyone shooting to kill, 1 million trials.

A: 405705 wins
B: 228638
C: 365657

This is A's strategy if the most important thing is to kill B.

If A intentionally misses and everyone else is playing for keeps:

A: 489940 wins
B: 489732
C: 20328

This is A's strategy to preserve his own skin and kill C, but B can absolutely kill this strategy. If A intentionally misses, then B intentionally misses then A is facing death every turn at C's hand, so if B pursues an intentional miss strategy, then A must start shooting.

If A and B intentionally miss while C is playing straight, you get:

A: 0
B: 857112
C: 142888

If A is playing straight and B is missing intentionally you get:

A: 221797
B: 286429
C: 491774

But a lot of this is due to A going first. What if A is not playing to miss, but just misses the first shot. After all, a large number of B's losses are due to being killed with the first bullet.

If B is playing straight and A misses the first shot, you get:
A: 479400
B: 456605
C: 317483

But if B contains his happiness over surviving and then follows a miss by A with an on purpose miss, you get:

A: 111575
B: 570942
C: 317483

So if B survives the first shot whether A is ducking or not, he should miss on purpose.

Cool Engineer.... I was working on a graph.

That is wonderfully counter-intuitive.

Now we put it all together!

A up to shoot. If A kills B, then A is facing a showdown with C but C is shooting first and A will lose 2 out of 3 times. If A doesn't shoot B, his probability of survival rises, so A chooses to miss.

B up to shoot. If B kills C, then B is facing a showdown with A but A is shooting first and B will lose 4 out of 7 times, not bad, but not favored. If B doesn't shoot C, his probability of survival rises, so B chooses to miss.

C up to shoot. If C kills A, then C is facing a showdown with B but B is shooting first and C will lose 6 out of 7 times. No way is C killing A.

So no one kills anyone and everyone gets drinks afterward.

I confirmed your Monte Carlo results (I wrote a little groovy script and got the same results). With three people, you should expect odds of 1 in 3. Anything better than that is a win (in my book).

Edited: (I originally said B should shoot to kill... but I am wrong again.

If B passes and A passes... C will be the only one shooting and A will die.
If B passes and A shoots.... B has a 57% chances of winning.

So B wins by passing no matter what A does. So B should pass unless C is also passing (unlikely since C is screwed anyway).

By the way, I am wondering if there is anyone else here nerdy enough to care about this.

Yes, went off the deep end on this one, but I enjoy these types of problems.

Another one I worked on recently:

Two teams are flipping coins with the first team trying to get two straight heads and the other trying to get a heads followed by a tails. If neither team has their goal, they do the next flip. If both teams meet their goals at the same time, they start again from scratch. They continue to flip until someone wins. What is the probability of team A winning?

My intuition says 50%. The order or the teams doesnt matter, and each team has exactly one side of the coin that wins per flip.

Am I missing the rules again?

ohhhh... I see.