be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
be the reflection of 
.
intersects line 
and 
.
.
be a quadrilateral inscribed in circle 
with 
.
and 
over lines 
and 
and 
.
meets 
,
meets 
.
.
such that among any 
positive real numbers 
with 
,
Y by mjkkra, CoordinateBash13, Adventure10, Mango247, and 1 other user
In a small pond there are eleven lily pads in a row labeled 
through 
. A frog is sitting on pad 
. When the frog is on pad 
, 
, it will jump to pad 
with probability 
and to pad 
with probability 
. Each jump is independent of the previous jumps. If the frog reaches pad it will be eaten by a patiently waiting snake. If the frog reaches pad it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?
through 
. A frog is sitting on pad 
. When the frog is on pad 
, 
, it will jump to pad 
with probability 
and to pad 
with probability 
. Each jump is independent of the previous jumps. If the frog reaches pad it will be eaten by a patiently waiting snake. If the frog reaches pad it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?
This post has been edited 2 times. Last edited by LauraZed, Dec 27, 2018, 4:41 PM
Reason: fixed capitalization
Reason: fixed capitalization
.
and we use our intuition about symmetry to find that 
.
?
be the probability from lily pad 
(so 
,
)
Adding them all, we get![\[ 1 = p_{10} = \left(
\frac{1}{\binom90} + \frac{1}{\binom91}
+ \frac{1}{\binom92} + \dots + \frac{1}{\binom99}
\right) p_1. \]](http://latex.artofproblemsolving.com/2/2/6/2268bed3c36a298b8959e4a42661246ae870cc70.png)
Equivalently, one can also add up to just 
to derive the equivalent![\[ \frac12 = p_{5} = \left(
\frac{1}{\binom90} + \frac{1}{\binom91}
+ \frac{1}{\binom92} + \frac{1}{\binom93} + \frac{1}{\binom94}
\right) p_1. \]](http://latex.artofproblemsolving.com/e/1/c/e1c234216c2dc9c5988d35e86d3b4100eb849016.png)
In any case, we have![\[
p_1 =
\frac{1}{2\sum_{k=0}^4 \binom9k ^{-1} } = \frac{63}{146}.
\]](http://latex.artofproblemsolving.com/e/0/a/e0a6ad11d4ffc7e2e85f89b117aa7fe34ef956e9.png)
and similar equations to what you have above?
.
?
denote the probability that starting from the 
Using our recursion on 
and 
again we obtain 
from which we obtain 
.
