Probability that 6 Peas will Sink

Date: 1/23/96 at 14:36:11
From: Anonymous
Subject: probability

We need to know the probability that all six of these durn peas 
will be floating on top of the water before they are all six sunk 
down to the bottom of the jug.

Conditions:

1 jug of magic math water and six special peas.

Start:  four peas floating and two peas resting on the bottom.

Each second, peas can move:

0, 1, or 2 peas can go top to bottom with equal probability for 
each case.  0, 1, or 2 peas can go bottom to top with equal probability 
for each case

If there is exactly one pea in either location,
0, or 1 pea moves with equal probability

When all peas are in either location, then game is done.

Date: 6/1/96 at 17:7:10
From: Doctor Gary
Subject: Re: probability

Let's call the probability of all the peas going to the top before 
they all go to the bottom "t" and the probability of all the peas going 
to the bottom before they all go to the top "b".  Eventually, one or the 
other will happen, so t + b  =  1.   

Since all we care about is which happens first, and not how long it 
will take to happen, we can safely ignore those so-called "moves" in 
which no peas change location.   

We can calculate t as a sum:

There are only four possible non-null first moves, each with 
probability .25.  

1. If the first non-null move is 2 from bottom to top, the game is 
over. This means that t will be AT LEAST .25.

2. If the first non-null move is 1 from top to bottom, we have 3 peas 
on top and 3 on the bottom. If moves are independent of one another, 
the chances are now even that all the peas will be on the top before 
they are all on the bottom, and vice versa. Add another .125 to our 
calculation of t.

3. If the first non-null move is 2 from top to bottom, then there will 
be 4 peas on the bottom and 2 on the top.  FROM THIS POINT ON, 
the probability that the peas will all be on the bottom before they are 
on the top is the same as it was at the beginning of our game that they 
would all be on the top before they were all on the bottom, namely t.  
The probability that they will all be on the top before they are all on 
the bottom is b, which is equal to (1-t).   This means that our 
computation of t will include a term of .25(1-t).

4. If the first non-null move is 1 from bottom to top, then there are 
only two possibilitites for the SECOND non-null move, each with 
probability .5. The second such move must either end the game with all 
the peas on the top (add .125 to our calculation of t) or restore the 
original situation with 4 on top and 2 on the bottom (add .125t to our 
calculation of t).

t can now be expressed as the sum of:

 .250 + .125 + .250(1-t) + 
.125 + .125t

which equals:

  .750 - .125t

 Now we can solve the equation:

 t  =  .750 - .125t

 8t  =  6 - t

 9t  =  6

 t  =  2/3

-Doctor Gary,  The Math Forum