Let's see can you work this out.

TheBiz

There is a 3 way duel,
Each person has 1 bullet,
Person 1- has a 90% chance of hitting their target,
Person 2-had a 60% chance,
And person 3- has a 10% chance.

Person 3 has to shoot first, what is his/her best strategy if he/she wishes to survive?

DunnoKiddo

He runs away! very fast! .....always works for me

I have no clue

(where is namenotavailabl when ya need him? ...he would know this!)

twerg_85

They say the answer is to shoot into the ground, miss everyone and hope that punter A kills punter B, and then he gets a chance to kill A himself.
Haven't worked it out meself and not convinced the answer above is correct !

Ah right, they have one bullet each. Then if A goes for B (and not for you), your best bet is to miss everyone. Of course, if punter A doesn't like you .....

piuswal

TheBiz wrote: »

There is a 3 way duel,
Each person has 1 bullet,
Person 1- has a 90% chance of hitting their target,
Person 2-had a 60% chance,
And person 3- has a 10% chance.

Person 3 has to shoot first, what is his/her best strategy if he/she wishes to survive?

Fires at 1, if he hits then number 2 has only a 60% chance of hitting him.
If he misses 1 and 2 fire at each other.

piuswal

TheBiz wrote: »

There is a 3 way duel,
Each person has 1 bullet,
Person 1- has a 90% chance of hitting their target,
Person 2-had a 60% chance,
And person 3- has a 10% chance.

Person 3 has to shoot first, what is his/her best strategy if he/she wishes to survive?

Fires at 1, if he hits then number 2 has only a 60% chance of hitting him.
If he misses 1 and 2 fire at each other.

johnny osbourne

he should aim in the middle of them then grab the slash-hook

Cakerbaker

Fire at 1 and miss, 1 then has a 90% chance of success. 1 then fires at 2. As the OP States that person 2 "had" (not has) a 60% chance of hitting their target, it would appear that 1 successfully shot 2. 1 and 3 have both used their one shot and 2 is dead.

Or maybe "had" was a typo and then person 3 is in trouble!

johnny osbourne

Cakerbaker wrote: »

Fire at 1 and miss, 1 then has a 90% chance of success. 1 then fires at 2. As the OP States that person 2 "had" (not has) a 60% chance of hitting their target, it would appear that 1 successfully shot 2. 1 and 3 have both used their one shot and 2 is dead.

Or maybe "had" was a typo and then person 3 is in trouble!

the bookies would love you

TheBiz

'Had' was a typo oops!
But I seen this on an old Qi repeat and the correct idea is to purposely miss.

It's based on 'Nashs equilibrium' (A beautiful mind)
In game theory, the Nash equilibrium is a solution concept of a non-cooperative game involving two or more players, in which each player is assumed to know the equilibrium strategies of the other players, and no player has anything to gain by changing only their own strategy.

The objective was never to kill someone only to survive.

johnny osbourne

TheBiz wrote: »

'Had' was a typo oops!
But I seen this on an old Qi repeat and the correct idea is to purposely miss.

It's based on 'Nashs equilibrium' (A beautiful mind)
In game theory, the Nash equilibrium is a solution concept of a non-cooperative game involving two or more players, in which each player is assumed to know the equilibrium strategies of the other players, and no player has anything to gain by changing only their own strategy.

The objective was never to kill someone only to survive.

so i was right

TheBiz

johnny osbourne wrote: »

so i was right

Minus the slash hook