However in the truel puzzle the computation is relatively easy; it's determining the best strategy that creates the discussion.

Here it is the computation that is complex, especially when you have to start taking into account the contestant may have been promoted in the pecking order and as a result become the strongest - or seen the ones below him got rid of so he is now the weakest. Given the computation, the overall strategy seems more obvious, though eventually it would depend on factors I haven't given, like the probability each contestant gets an answer wrong.

I did finally get a reasonably elegant, albeit recursive, function for this, in the form f(m,n) to calculate the probability of winning for the nth strongest contestant out of m starters.

An additional difference is that in the truel one may shoot in the air, thus bypassing the need to make an explicit choice as implied in the Weakest Link.

>I did finally get a reasonably elegant, albeit recursive,
>function for this, in the form f(m,n) to calculate the
>probability of winning for the nth strongest contestant out
>of m starters.

Would not you want to share it?

Well, I thought someone might want to figure it out for themselves.

However, since anyone like that could stop reading now....

For n contestants, define
xm = probability the weakest is eliminated in round m, where m=1 indicates the final round, not the first.
ym = probability the strongest is eliminated.
zm = probability someone else is elimnated.

It follows from the Description that zm = (1-xm-ym)/(m-1).

Construct an array S(1..n-1,0..n+1) as follows:
S(m,0) = 0
S(m,1) = xm
S(m,m plus 1) = ym
S(m,n) = zm (1<n<=m)
Other entries are all 0.

The function f(m,n) is then
f(1,n) = S(1,n)
f(m,n) = f(m-1,n-1)*sum(i=0..n-1) S(m,i)
plus f(m-1,n)*sum(i=n plus 1..m plus 1) S(m,i)

With n contestants, there are n-1 rounds, so the initial starting probabilities are given by f(n-1,n)

For 8 contestants A..H with the probabilities originally stated this gives the initial survival rates as

A = 0.076904
B = 0.161087
C = 0.228794
D = 0.2435
E = 0.184062
F = 0.086706
G = 0.018948
H = 0

(PS I originally used square brackets for the array references, but sometimes they got converted to angle brackets, and sometimes they were dropped completely.)

There is a recursive relationship:

Probability p<n,p,q> = a * p<n-1,p,q> + b * p<n-1,p,q+1>

where a is the probability that someone lower in the pecking order was eliminated in the previous round, and b is the probability that someone higher in the pecking order was eliminated in the previous round. a and b are fairly easy to define in terms of n and q.

The initial conditions are p<1,i,j>=1 for i=j, otherwise 0.

This is fairly easy to code into a spreadsheet, and I came up with the same answers.

At first I solved a slightly different problem, with the 3/8 probability in each round being spread between all the contestants. The answer is slightly different, but your best chance was still from being mediocre (i.e. 4th best).

Regards

KL

Neat. Your a and b seem to be essentially the summation expressions in my version: I was a little more explicit in showing one way of calculating them.

I originally built a spreadsheet too. Brilliant invention, spreadsheets.