What are the odds that the frog jumps into the pond?

Let f(x) = the odds that the frog lands in the pond in 1 jump, starting from position x. As many other answerers have made abundantly clear, f(0) = 1/2. In fact, for 0 <= x < 1, f(x) = 1/2

Let g(x) = the odds that the frog "eventually lands" in the pond, starting at position x. We can map g(x) recursively.

g(x) = 1/2 + Integral ( z = x to 1, dz/2 * g(z) )

dz/2 represents the probability that the frog will jump to position z. g(z) is the probability that the frog will make it from there, so by Integrating over the interval (x, 1), we account for all possible leap combinations that the frog can use to reach the pond.

Let G(x) be the anti-derivative of g(x), with initial condition G(1) = 0. By the Fundamental Theorem of Calculus:

g(x) = 1/2 + 1/2 * (G(1) - G(x)) = 1/2 * (1 - G(x))

Now we have a simple differential equation. (For the sake of convention/habit, let y = G(x))

y' = 1/2 * (1 - y), -2y' = y - 1, y + 2y' = 1

After doing some Diff EQ work, I got:

y = G(x) = 1 - e^(1/2 - x/2)

g(x) = G'(x) = 1/2 * e^(1/2 - x/2)

Therefore g(0) = 1/2 * sqrt(e) ≈ 82.436%

Relatively good odds, I must say.

@jimbot: This number seems to agree with your simulations, so I'm pretty sure my math is correct

-- EDIT --

I should note that the function g(x) ONLY works for x in [0, 1)

Once x < 0, the odds of reaching the pond in a single jump change - changing the definition of g(x).

I actually perceived a green frog with warts, although frogs do not have them, it was running from something. Maybe in an animal's natural fight or flight mode, it jumped into the air and into the rippling water. The "plopping" sound became silence after the frog was beneath it's blanket. I'd like to know if this means anything now.

The odds are clearly better than 1/2, because if on first jump the frog lands to the left of the pond, he would have second chance on second attempt to jump.

The odds f(x) depend on initial position of frog x, for example f(-0) = 1/2. We are asked to find f(x=-1).

1/2, the frog has 1/2 a chanse touch land before and 1/2 an chanse to

land behind the edge of the pond.

Half of the points are on land, half are in the water.

0 ---- land ---- 1 ---- pond ---- 2

The length of the pond is 1 unit.

The reach of the frog's jump is 2 units

Answer:

P(frog lands in pond) = ½

50%, assuming that the frog can jump only in one direction directly toward the pond (in the shortest direction).

If this is not the case you will have to use areas of circles to help with this. What level math is this for?

OK I see, sorry for the confusion.

My gut tells me that it should be 5/6, but I can't prove it.

Millions of scripted frogs tell me that it's between 0.824 and 0.825, I'd love to see the math behind it though.

Yes Torquest, your math does look correct, nicely done.

The theoretical probability is 50% in the land or 50% in the water. In the experimental probability it depends of the ability of the frog to jump.