[ExI] who should white shoot?

[The Avantguardian]

--- spike <spike66 at att.net> wrote:

> The point of the exercise is in the math.  I found it quite astounding to
> pose a question that involves only thirds, and end up with an answer (still
> unverified) that involves sevenths and eighths.  Sevenths!  Isn't that an
> oddball number to pop up?

Congratulations, Spike, you have rediscovered Bayes Law. The good reverend
would be pleased. :-) 
 
> Do let us restate the problem without Mr. Black, and just with two guys who
> want to have a duel in the form of taking turns firing at each other until
> one perishes.  White hits his target a third of the time, Brown two thirds
> of the time.  I calculate that if Brown fires first, his chances of survival
> are 7 in 8, whereas if White fires first, White's chances are 3 in 7.
> 
> Am I the only one that finds that result so astounding as to be difficult to
> believe?  I cannot find the error in my calcs.  Anyone?

I have been meaning to get back to you on that, Spike. I got essentially the
same results as you except your 7/8 should be 6/7. And the probability of White
winning if Brown goes first is 1/7 not 1/8.

Here is how I did it:

W:= Mr. White, R:= Mr. Brown, N:= # of rounds.

P(W hits)= 1/3, P(W misses)= 2/3, P(R hits)= 2/3, P(R misses)= 1/3,
P(both miss)= (2/3)*(1/3)= 2/9

The probability that the duel will last N rounds is determined by the chance
they both miss N-1 times

P(N)=(2/9)^(N-1)
 
P(W wins in,N)= P(W hits|N)*P(N)= (1/3)*(2/9)^(N-1)

P(W loses in N)= P(R hits|W misses)*P(W misses|N)*P(N)= (2/3)*(2/3)*(2/9)^(N-1)

P(W wins)=             P(wins in N)
          -------------------------------------
             P(W wins in N) +P(W loses in N)

P(W wins)=(1/3)*(2/9)^(N-1)/[(1/3)*(2/9)^(N-1)+(2/3)^2*(2/9)^(N-1)]

Notice how the indefinite number of possible draw rounds, (2/9)^(N-1),
conveniently cancels out of the numerator and denominator leaving:

P(W wins)=(1/3)/[(1/3)+(2/3)^2]=(3/9)/(7/9)= 3/7

Note that we could added up all the probabilities of him winning in the 1st
round, 2nd round, 3rd round ... up to infinity and the draw rounds would still
cancel out. Isn't Bayes Law great? 

If Brown goes first same logic applies and the draws cancel out leaving:

P(W wins)=(1/3)^2/[(1/3)^2+(2/3)]=(1/9)/[(1/9)+(6/9)]= 1/7

Thus you are right, it is best for White to let someone else take the first
kill so that he may get the first shot in the two person subduel; going first
gives White three times the chances of winning against Brown and is his only
hope against Black.


Stuart LaForge

"A portion of mankind take pride in their vices and pursue their purpose; many more waver between doing what is right and complying with what is wrong." - Horace