An interesting question (1 Viewer)

[shimmerz_777]

3 people, Albert, Barry and Chris decide to have a gun duel. Albert has a 100% hit ratio, Barry has an 80% hit ratio, and Chris has a 50% hit ratio. they get one shot per turn, then the next person gets to shoot and so on until one is left standing, the person who shoots first is chosen at random, as well as the order for 2nd and 3rd shot. assuming that they fire logaically at who is the most threatening, who is the most likely to survive?

dont need any mathmatic percentages of probability, just a who and a reason why.

 

[insert-username]

Chris is most likely to survive, followed by Albert, and Barry. In the duel, Albert (100%) would much rather shoot at Barry (80%) than Chris (50%), since Barry is much more likely to kill him than Chris. Likewise, Barry would much rather try and hit Albert first, since if he shoots at Chris instead of Albert, he's signing his own death warrant, as Albert never misses. So Chris will not be targeted by the other two (assuming they fire logically), giving him plenty of time to line up whoever's left out of Albert or Barry (more likely Albert, since Barry misses 1 out of 5 times).


I_F

 

[Templar]

Ah, another variation on the truel in game theory.

The original has A, B and C each at 1/3, 2/3 and almost 1 probability of hitting, and they shoot in order of A, B and C.

In fact, it's best for Chris to delibrately miss Albert. This way he reduces the truel into a duel with whoever's left standing, and this increases his chance of survival.

 

[shimmerz_777]

i loved this question when i read it in a book, its the type where you only just have to think about it and do little else. i found it good that it was ironic that the worst of the three was the most likely to live.