# elementary but irritating algebra problem

The question itself is very short and sweet, and requires no background.

Find a solution to the following system: $$\left\{\begin{array}{l} a,b,c,d,e,f>0\\\\ a+b+c+d+e+f=1 \\\\\displaystyle\frac{c}{a+c}\geqslant .999 \\\\\displaystyle\frac{d}{b+d}\leqslant .001 \\\\\displaystyle\frac{c}{c+d+e}\leqslant .001 \\\\\displaystyle\frac{d}{c+d+e}\geqslant .999 \end{array}\right.$$ I don't need to find the entire solution set. I literally just need one particular solution. It's basic algebra but it's giving me a headache so maybe someone has some software they can use to find it.

Philosopher Robin Collins wrote a paper where he claims that if $$H$$ and $$K$$ are competing hypotheses and $$E$$ is observed evidence with $$P(E|H)\gg P(E|K)$$, then $$E$$ is "strong evidence" for $$H$$ over $$K$$. This is an ambiguous claim, but my guess is that he means something like the following:

(LP) If $$H\cap K=\emptyset$$ and $$P(E|H)\gg P(E|K)$$, then $$P(H|E)\not\ll P(K|E)$$.

I believe that (LP) is false, but I'm struggling to produce a counterexample. In particular, I need a counterexample where $$P(H\cap K)\neq 1$$, $$.99\leqslant P(E|H),P(K|E)<1$$, and $$0. I then assign variables based on the following Venn diagram:

Finding a solution to the system will disprove (LP).

• Just do $a << c=e << d << b$ such as: $$a=10^{-50}, c=e=10^{-40}, d=10^{-30}, b=10^{-20}, f=1-(a+b+c+d+e)$$ – Michael Jun 11 at 6:29
• This has nothing to do with the algebra, but your interpretation of what you call Collins’s “ambiguous claim” takes a lot for granted. For one thing, only a Bayesian would assent to asking for the probability of a hypothesis conditional on the evidence; frequentists are going to complain that only the data is random, not the hypotheses. The Venn diagram only exaggerates the ontological confusion, making hypotheses and data out to be the same sort of objects. Perhaps you realize all this and just want to solve the toy problem anyway. Just out of curiosity, can you link to the paper? – symplectomorphic Jun 11 at 6:32
• Any thoughts on the answer or on the comments that have been posted, Ben? – Gerry Myerson Jun 12 at 13:12

Let's rewrite things in a more convenient form. $${c\over a+c}\ge.999{\rm\ is\ }c\ge999a$$ $${d\over b+d}\le.001{\rm\ is\ }b\ge999d$$ $${c\over c+d+e}\le.001{\rm\ is\ }{d\over c}+{e\over c}\ge999$$ which will be satisfied if $$d\ge999c$$, $${d\over c+d+e}\ge.999{\rm\ is\ }{c\over d}+{e\over d}\le{1\over999}$$ which will be satisfied if $$d\ge2000c$$ and $$d\ge2000e$$. So let's take $$b=.5$$, $$d=.0005$$, $$c=e=.0000002$$, $$a=.0000000002$$, and $$f=1-a-b-c-d-e$$.