Let [math]d[/math] be the clockwise separation between the frogs, modulo 12. Initially [math]d=6[/math].

Each individual frog jump changes [math]d[/math] by either [math]+1[/math] or [math]-1[/math], with equal probability. Thus [math]d[/math] performs a simple random walk on a 12-cycle, stopping when [math]d=0[/math].

For a cycle of [math]n[/math] leaves, the expected hitting time of 0 from distance [math]d[/math] is

[eqn]E_d=d(n-d).[/eqn]

Therefore,

[eqn]E_6=6(12-6)=36.[/eqn]

So the average total number of frog jumps is [math]\boxed{36}[/math].

I had it as the average of 6 and infinity; of which 36 is a valid solution.
May I have partial credit?

>>17042729
These lily pads and frogs are a beautiful clock; thank you.

>>17042729
2 frogs, 1 jump at a time, 12 lilies => (2+1)*12

>>17042747
On a circle of 6-pad separation spaces, 2 frogs square off; ergo
6^2=36

>>17042741
>36
But how did you get from 6 • 6 to 4 • 9?

>>17042741
>>17042744
>>17042753
I’m a natural contrarian, so I’ll bet on not 36

>>17042756
The question is not “what is the answer?” but “what do we expect the answer to be?”.
We expect the answer to be 36.
Not an argument, just a fact. Polls cannot lie. Data simply is.

>>17042759
Still, given all that, I’ll keep my contrarian bet on not 36 being the outcome

>>17042755
>>17042756
Monte Carlo agrees.

Over 10,000,000 simulated games:

[eqn]\text{mean jumps}=36.013804[/eqn]

[eqn]\text{standard error}=0.009167[/eqn]

[eqn]95%\ \text{MC interval}=[35.995837,\ 36.031771].[/eqn]

So the exact prediction [math]36[/math] is comfortably confirmed. The shortest possible game is [math]6[/math] jumps; in this run the longest was [math]484[/math], which is why the single-game distribution has a fairly large spread.

>>17042765
Still, given all that, I’ll keep my contrarian bet on not 36 being the outcome.

>>17042770
It’s average of walks, like how many steos for you from place x to y over 2,000,000 times, it’s not exact each time but it is closer to certain number than any other number

>>17042756
>>17042762
>>17042770
Can this many (YOU)s really be wrong?
Should we not consider both sides of the argument? To be fair to everyone?

>>17042777
Triple-7s. /thread

Excluding the desired end result, you get 3 distinct scenarios: one where the frogs are separated by 6 lily pads, one where they’re separated by 4, and one where they’re separated by 2. You have 3 probabilities to consider throughout: one of the frogs keeping their current gap (2/4), shortening it by 2 (1/4) or widening it by 2 (1/4).
You can set up three different equations for expectation values based on the three scenarios of spacing, which we’ll set up our variables to solve for the expectation value when they’re separated by 6 lily pads (x), 4 lily pads (y), and 2 lily pads (z). Thus, we will have a system of equations as follows:

x = 1 + (1/2)x + (1/4)y + (1/4)y —> x = 2 + y

y = 1 + (1/2)y + (1/4)z + (1/4)x —> y = 2 + (1/2)z + (1/2)x

z = 1 + (1/2)z + (1/4)0 + (1/4) —> z = 2 + (1/2)y

The plus one is added because an even number of turns has to be taken, while the fraction values come from the odds of the impact of the second frog’s random decision. In the equation for x (6 lily pads separation), shortening and widening the gap are the same thing.

Note that in equation for z (2 lily pads), a (1/4) term is multiplied by 0 because after the distance is shortened by 2 here, the frogs are now on the same lily pad.

So, if you solve for x, which is the expected value for number of rounds remaining when the frogs are separated by 6 lily pads, you end up with 18. Multiply that by 2 for total turns, and you get 36.

>>17042776
>>17042777
>>17042788
Still, given all that, I’ll keep my contrarian bet on not 36 being the outcome; even if you round to the nearest even integer. I’m just a retard, I guess.

>>17042799
Cute. Can present in generating function form too.
>>17042807
I don’t think you have the capital to cover a “Not 36” spread, and the House is low on credit.

>>17042807
It’s about asymptomatic, each successive “meet” may not be 36, there is very good chance it’s not 36 each time but after a big very number of times it will approach exactly 36

>>17042818
Coincidentally, 36 is also the number of asymptomatic Covid exposures you can survive before you need to re-up on VaxxJuice.

>>17042818
Also standard deviation is huge.

The stopping time has a long wandering tail. Most games finish moderately quickly, but some frogs repeatedly drift apart, circle the pond, and take hundreds of jumps to meet. Those rare long games pull the spread upward.

In fact the exact second moment is

[eqn]\mathbb E[T^2]=2136,[/eqn]

so

[eqn]\operatorname{Var}(T)=2136-36^2=840,[/eqn]

and therefore

[eqn]\operatorname{SD}(T)=\sqrt{840}\approx28.98.[/eqn]

So the SD is not an error in the simulation; it is genuinely almost as large as the mean. For contrast, the absolute quickest outcome, [math]T=6[/math], happens only with probability

[eqn]2\left(\frac12\right)^6=\frac1{32}=3.125%.[/eqn]

After that, the relative random walk can meander around the circle for a surprisingly long time before it finally hits zero.

>>17042817
>>17042818
Still, given all that, I’ll keep my contrarian bet on not 36 being the outcome. My bet includes the fact that you’ll never get a 36 before you go belly up under my steamroller

>>17042837
I don’t think you understand the problem, it’s not about each time, there is only a 2.56% chance to land on 36, that’s why the problem asks for the average, with a huge sd 36 will only show up as average after millions of times

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>>17042729
>jumps randomly
>with equal probability

>>17042840
I’m pretty sure I understand the problem better than you lol.

>>17042837
I just so happen to have a very good tip from a very reliable friend of ours who knows all about these types of things that these two particular frogs do not quite get along with each other very well and so do not exactly randomly hop from pad to pad with equivalent probabilities.
If you get my drift.
>>17042843
Yes.

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>>17042843
>>17042729

>>17042876
it's not random unless the probability coefficient is also randomized every step

>>17042729
Very nice thread OP thank you for making this image.

>>17042746
Concur.

>>17042729
I scrolled down a bit and saw no obviously explicit Linear Algebra answer, which was my immediate first thought, so I guess I'll do it. I'd imagine someone who has more probability background would think of something else which I'll read later

There are 4 states: 6 apart, 4 apart, 2 apart, and 0 apart. After a transition/turn/hop, there are probabilities to move between them. This would suggest we label the states as (1,0,0,0)^T, ..., (0,0,0,1)^T, respectively, and use a matrix whose columns describe the probabilities of transitioning for each state. Notice the last column is all 0s cause we dgaf about what happens after they touch each other (lewd).
All we care about is when it hits the 0 state (0,0,0,1)^T, so we can single it out using a dot product on (0,0,0,1).

The equation is simply

[math] \displaystyle
E[N] = \sum_n n\cdot P(n) =
\sum_{n=0}^\infty \ (0 \ 0 \ 0 \ n)
\left( \begin{matrix}
.5 & .25 & 0 & 0 \\
.5 & .5 & .25 & 0 \\
0 & .25 & .5 & 0 \\
0 & 0 & .25 & 0 \\
\end{matrix} \right)^n
\left( \begin{matrix}
1\\ 0\\ 0\\ 0\\
\end{matrix} \right)
[/math]

Now just do the factoring, do the matrix multiplication, and do the sum. If you can't understand the math, this is stuff I learned in High school, and AI can teach HS stuff well so just ask them about it or ask here for me or someone else to answer. I just don't wanna make a long post.
Using sympy to do the matrix math for me, I get the sum to be

[math] \displaystyle
E[N] =
(-1,\ -\tfrac{1}{3},\ 8e_3,\ \tfrac{4}{3}e_4 ) \cdot
\sum_{n=0}^\infty n
\ (e_1,\ e_2^n,\ e_3^n,\ e_4^n)
[/math]

where the eigenvalues [math] e_i [/math] are [math] (0, \tfrac{1}{2}, \tfrac{2-\sqrt{3}}{4}, \tfrac{2+\sqrt{3}}{4} ) [/math]. Given that all the eigenvalues are < 1 and [math] \displaystyle \sum_{n=0}^\infty Ana^n = \tfrac{Aa}{(1-a)^2}[/math] (again, HS shit so either look it up or ask separately), we get that it is exactly (again, sympy)
[math] \boxed{ E[N] = 18 \text{ jumps} } [/math]

If you want std dev, then
[math] \displaystyle
E[N^2] =
(-1,\ -\tfrac{1}{3},\ 8e_3,\ \tfrac{4}{3}e_4 ) \cdot
\sum_{n=0}^\infty n^2
\ (e_1,\ e_2^n,\ e_3^n,\ e_4^n)
= \sum_{i,n} A_in^2 e_i^n = \sum_i \tfrac{A_i(e_i+e_i^2)}{(1-e_i)^3}
[/math]
which gives that the std.dev is

[math] \boxed{ \sqrt{\text{Var}[N]} = \sqrt{210} \approx 14.5 \text{ jumps} } [/math]

The method should be right (unless yall point something out to me), but if the numbers are wrong, I'm not really checking my work so meh

>>17042974
it might actually be 8e_4 and 4/3e_3 but wtv

>>17042974
18 jumps seems really high

>>17042974
Oooooh, so it looks like everyone is taking it take each frog INDIVIDUALLY counts as a jump. Whereas, I count after BOTH frogs jump.

Ok yeah, so everyone is answering 36 cause it's double counting mine. It's not like they can land on each other on an odd number of jumps given the not-my definition. So my answer is basically matching what the others are saying in a sense. I mean, you can change the matrix so now that it isn't 4 dimensional but 7 dimensional, but uh, nah. And again, it ain't like you can ever get an odd number state anyway.

>>17042977
huh? the others are saying 36 tho

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Per the wording of the problem, the second frog always jumps in the same direction as the first, so they never end up on the same pad.

Y'all dumbfucks need to learn to read more carefully.

Who the fuck calls a lily pad a water lily leaf? What fucking language was this translated from?

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Troll version