I'll trying posting this question again without all of the superfluous
data I gave last time.

There is a casino gambling game where you will win 49.5% of the time,
and lose 50.5%.  The bet is even money, so when you win you get $1 for
every dollar bet.  If you start with $1, and bet exactly $1 per hand
until you either reach your goal or go broke, what are the odds of
reaching the following goals?

$3
$4
$5
$6
$7
$8
$9
$10
$11

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Clarification of Question by awstick-ga on 25 Sep 2006 22:26 PDT

@stan:
The problem with solving it like that though, is that the extra 1%
chance of losing is the whole point.  If it were a 50/50 game then the
odds of any goal would just be 1 divided by the goal amount.  What I'm
looking for is how much that 1% house edge affects the odds of
reaching the goal.

I've tried calculating using the actual numbers in the manner you
described, but the process seems to break down for me after the $3
goal.  When I tried to calculate the odds of reaching $4, I figured
that since the two points at which I stop playing are when I'm down $1
or up $3.  I started to calculate the odds of losing, since it should
be easier and I would only need to calculate one or the other, but it
didn't seem to work out.  Every losing scenario will have one more
loss than wins, and since losing $1 ends the trial, every loss before
the last one will have to be accompanied by a one win.  So the total
chance of losing should be the sum of the sequence: .505 x (.505
x.495)^n, where n goes from 0 to infinity.  But this process only
worked for the process of getting from $1 to $3, and can't be applied
to any other goals because the answer would always be the same no
matter what the goal was.  This is where I got stuck and need someone
with more knowledge of maths than I have to step in.  There must be a
way to solve this without having to map out every single possible
situation in each trial by hand.

Here's a start:
Assume probability of winning one bet is 1/2 (For a small number of
bets, that is close enough). Assume w means win and l means lose.
With a start of 1, you could get to 3 by ww, wlww, wlwlww, etc.
This works out to (1/2)^2 + (1/2)^4 + (1/2)^6
This works out to .25 + .0625 + .015
This works out to .33
You can work out the other targets in the same manner.

I have to get some sleep, but I'll get back to this tomorrow. The
reason that these probabilities are low is not because of the house
505/495 advantage, but rather the fact that you have only a dollar,
whereas the house has infinite capital (You might consider starting
with more capital). Of course, the house 505/495 advantage becomes
important when you have a long string of bets.
I'm sure that I can continue with this.

OK, I think I have figured out how to solve this as requested now:

Here are the odds:
$1: .4950
$2: .3267
$3: .2426
$4: .1921
$5: .1584
$6: .1344
$7: .1164
$8: .1024
$9: .0912
$10:.0820
$11:.0743

To calculate these, I (you guessed it) made an Excel spreadsheet. 
Here is a sample:
	 $-   	 $1 	 $2 	 $3 	 $4 	 $5 	 $6 
1	0.505	0.495					
2	0.755	0.000	0.245				
3	0.755	0.124	0.000	0.121			
4	0.817	0.000	0.123	0.000	0.060		
5	0.817	0.062	0.000	0.091	0.000	0.030	
6	0.849	0.000	0.077	0.000	0.060	0.000	0.015
7	0.849	0.039	0.000	0.068	0.000	0.037	0.000
8	0.868	0.000	0.054	0.000	0.053	0.000	0.022
9	0.868	0.027	0.000	0.053	0.000	0.037	0.000
10	0.882	0.000	0.040	0.000	0.045	0.000	0.025

The chart reads... Across the top, you have the dollar amount
obtained.  Along the side, you have the number of gambles.  The body
of the chart displays the odds of acheiving the given dollar amount
after the given number of gambles.  Here are the formulas:
After 1 gamble, we know that $0 is .505 and $1 is .495
Along the $0 column, the odds are always the $0 odds above it + .505 x
the $1 odds above it.  So after 2 gambles, $0 odds = .505 + .505 x
.495 (the trend continues below).
Along the $1 column (after the first gamble), the odds are always .505
x the $2 odds above it.  (notice that you can never move from $0 to
$1, that is why this column has a different formula than the rest of
the positive values).
For every other column, the formula is: 
For dollar amount X:
odds above for 'X-1' * .495 + odds above for 'X+1' * .505
(the reason is that you have a .495 chance of moving from the dollar
amount below and a .505 chance of moving from the dollar amount above.

To find the odds of hitting any specific dollar amount, you must then
eliminate all the columns to its right (because once you reach $5, you
won't keep playing to try to reach $5).  Then sum the entire column to
find the odds of reaching your desired amount (notice that the rows
must be infinite to get a precise value... my values are based on a
maximum of 200 gambles).

And finally, here is the expected value for each dollar amount you
attempt to acheive:
$1: $.495
$2: $.6534
$3: $.7277
$4: $.7683
$5: $.7922
$6: $.8066
$7: $.8150
$8: $.8194
$9: $.8211
$10:$.8205
$11:$.8175

Notice that this maxes out when attempting to acheive $9 with an
expected gain of $.8211 (assuming this is a free $1 you're playing
with)... also notice that in my comment to your other question, the
expected value can reach .9268 by using the method that I suggested
there.  You will have better success if you gamble all the money every
time (as you suggested over there, the bonus encourages you to be
risky).

Let me know if I'm off my rocker or you need any clarification.

Thanks a lot for your help.  The numbers were so much better than I
expected.  When applying them to my situation I found that the
expected weekly profit is still increasing significantly after winning
11 hands.  Using your numbers for winning $10 and $11 I get this:

(.0820 x $1000) - (.9180 x $67) = $20.49
(.0743 x $1100) - (.9257 x $67) = $19.71

So the odds of winning, times the profit, minus the odds of losing
times the loss, is equal to the expected profit for one single
deposit.  However, since I can claim this bonus as many times as I
want until I cash out, my total profit would be the EV for one
deposit, divided by my chance of success.  This makes my profit for a
10 unit win $250 (20.49 / .082), and my profit for an 11 unit win $265
($19.71 / .0743).

I think the way I described of flat betting $100 per hand is the best
way I could play this.  I know that doubling the bet after every win
would be better, but the casino has a maximum bet size of $100 per
hand, so I could never bet more than that.

I'm still trying to digest the information you gave me about the
method you used to solve it, so I could apply it myself in future. 
But what you have given me should be sufficient.  I calculated my
expected profit for $1 to $11, and looking at the numbers I think the
breakeven point would be around $18 with profit of roughly $320.  Even
if my profit is going to be a bit higher by continuing after winning
$1100, I don't think I would have the guts to risk doing it.  I'll
probably just try and get to the $1000 winning point, as it would be
incredibly disappointing to risk losing all of that for a few extra
dollars in EV.