To show this, we prove a property even more general. Supposing that instead of
a distribution, *D*, we have a multiset of positive real numbers,
*S*, of size *z*=*n*+*m*. The first frog picks, arbitrarily
and uniformly, an element *s* of *S*, removes it from *S*, then
hops a distance of *s*. It then repeats this procedure *n* times for
*n* hops. The second frog then repeats the same procedure on the remaining
*m* elements of *S*. Our stronger claim is that the probability of
winning will remain *n*/(*n*+*m*)=*n*/*z* even in
this case, regardless of *S*.

Once we show that the claim is true in this scenario, showing that it is true
for any *D* is simple: first, draw *z* numbers, *S*_{1},
..., *S*_{z} from *D*, then have the frogs jump these
distances. The probability density of drawing any particular sequence of length
*z* is the same as the probability density of allotting any permutation of
the same sequence. The win probability over all these, together, is, per the
claim, *n*/*z*, regardless of *D*. Taking a weighted mean over
all possible values of *S* that can be drawn from *D*, we necessarily
also get *n*/*z*.

The case *n*=*m*=1, for a multiset *S* containing two distinct
numbers yields a probability of 0.5 by simple symmetry. Translating to *D*
terminology, This handles the case of a distribution *D* where the
probability of a tie is zero.

More interesting is the case *n*=2, *m*=1. We divide this into two
cases. In the simple case, some *S*_{3} is greater or equal to
*S*_{1}+*S*_{2}. In this case, the frog that chose
*S*_{3} wins with probability 1, satisfying a total winning
probability of 2/3, as desired. The remaining case is one where the elements of
*S* can form a triangle. The *n*-frog wins if the angle between its
two hops is larger than the angle between the two hop lengths in the triangle.
Because the sum of a triangle's angles is 180 degrees, the average is 60, so
the win probability is, again 2/3.

The case *n*=1, *m*=2 yields a probability of 1/3, by symmetry.

Suppose, now, we have *z*>3.
Define *P*(*n*;*z*) to be the probability of a frog hopping
*n* strides winning against a frog hopping *z*-*n* strides.
We want to prove *P*(*n*;*z*)=*n*/*z*.
The trick is to look at sequences of steps made by any frog as a single large
step. For example, supposing that we take *S*
and divide it into three
sets of sizes *a*, *b* and *c*. Given any set of angles between
the hops within each set, we can calculate *x*_{a},
*x*_{b} and *x*_{c}, the distances
associated with these hops when viewed as a single large hop.
We do not know, at this point,
the probability that a frog taking the strides of the set of size *a* will
win against a frog taking the rest of the strides, but we do know that if we
take three different frog matches and assign in one match the distances
*x*_{a} and
(*x*_{b},*x*_{c}), in the second match
the distances *x*_{b} and
(*x*_{a},*x*_{c}) and in the third match
the distances *x*_{c} and
(*x*_{a},*x*_{b}), the sum of the three
winning probabilities for the frog making one step to win is 1.

Because this sum is a constant, the fact that we take a complicated weighted
mean over its values when we consider other angles or other values for the
basic step sizes does not matter, and we are sure that
*P*(*a*;*z*)+*P*(*b*;*z*)+*P*(*c*;*z*)=1
for any *a*+*b*+*c*=*z*.
This, combined with the obvious
*P*(*a*;*z*)+*P*(*b*;*z*)=1 for
*a*+*b*=*z*, is enough to prove
*P*(*n*;*z*)=*n*/*z*.

For example, we can show

*P*(2;*z*)=1-*P*(*z*-2;*z*)=*P*(1;*z*)+*P*(1;*z*)=2*P*(1;*z*)

*P*(3;*z*)=1-*P*(*z*-3;*z*)=*P*(1;*z*)+*P*(2;*z*)=3P(1;*z*)

...

*P*(*k*;*z*)=*k*P(1;*z*)

and then combine this with

1=*P*(1;*z*)+*P*(*z*-1;*z*)=*z**P*(1;*z*)

to obtain *P*(1;*z*)=1/*z*, and then
*P*(*n*;*z*)=*n*/*z*, as required.