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Q: Gambling math question ( No Answer,   4 Comments )
Question  
Subject: Gambling math question
Category: Miscellaneous
Asked by: awstick-ga
List Price: $25.00
Posted: 24 Sep 2006 08:56 PDT
Expires: 25 Sep 2006 12:46 PDT
Question ID: 767997
There is an online casino that has a promotion where you get a 50%
bonus on all of your deposits.  So if you deposit $50, you get a $25
bonus.  The bonus will be awarded as many times as you want, and you
can make a withdrawal at any time, but as soon as you make a
withdrawal you will not be eligible for a bonus for one week.  The
casino also removes all bonuses from your account when you withdraw.

Now there is a game at this casino in which the player will win even
money 49.5% of the time, and lose 50.5%.  The maximum bet size for the
game is $100.  If you were to deposit $67 and get a $33.50 bonus and
bet $100 in this game, then 49.5% of the time you would win $100
($200.50 - $67 deposit - $33.50 bonus) and 50.5% of the time you would
lose $67.  This would make the expected profit from the scenario
$15.67 ($100 x .495 - $67 x .505).  This part of the math is easy, but
it gets much more complicated when we add in more hands.  I can do it
out by hand for a little longer, but with my limited knowledge of
calculus it starts to get confusing pretty quickly.  If we were to
make $100 bets until we won $200 or went broke my math would look
something like this:
Lose first hand: 50.5%
Win first two hands: 24.50% (.495 x .495)
Win, lose, lose: 12.62% (.495 x .505 x .505)
win, lose, win, win: 6.13%
win, lose, win, lose, lose: 3.16%
win, lose, win, lose, win, win: 1.53%
And so on.  

So our chance of winning like this is about 32.67%.  The expected
profit is now (32.67% x $200) - (67.33% x $67) = $20.33.  The expected
profit goes up when you take higher risks.  Since you cannot get a
bonus for a week after making a withdrawal, the goal is to make as
much money as possible each week.  We will always be making $67
deposits, and always be betting $100 per hand on this game where we
win 49.5% of the time.  When we try and turn our $100 into $200 we
profit, and we profit even more when we try and turn it into $300. 
However, there will be a point where trying to make an extra $100 will
actually cause us to lose money.  What I need to know is when this
will occur, and the expected profit for every $100 increment up until
this point.  I'd also be interested in the methodology used to solve
it.

Thanks

Clarification of Question by awstick-ga on 24 Sep 2006 09:10 PDT
I just realized that the question I asked may not really be what I was
looking for.  The profit that I calculated was what you would expect
to make on each individual deposit, but what I am looking for is what
creates the highest profit for the week.   For example, trying to win
one hand has an expected profit of $15.67 per deposit, and since it
would take 2.02 deposits on average to have a winning hand, my profit
for the week would be $31.66.  When trying to get to $300, I will make
$62.23 (1/.3267) x $20.33.  What I'm actually looking for is my
highest weekly profit, which may or may not be the same point at which
it becomes unprofitable on a single bonus.

Clarification of Question by awstick-ga on 25 Sep 2006 10:23 PDT
Sorry, perhaps I should have been clearer in my description.  When I
said the casino removes all bonuses from your account when you
withdraw, what I mean is they will remove the amount of the bonus
credited no matter how much is in your account.  If you have only $67
then the max you can withdraw is $33.50.  If you think of the casino
account balance as a stack of casino chips, the bonus chips would be
at the bottom of the stack, and you would only be betting with them
once you have gone through all of your money.  But this fact certainly
does not make the bonus unprofitable.  The casino is letting you
gamble with some of their money, and the more aggressively you play
the more valuable the bonus is to you.

However, the game has a maximum bet of $100, so you can never bet more
than that on one hand.  If there were no maximum bet, I know that
doubling you bet every time would be the best way to play it.  I'm
pretty sure the best way to play this is to keep betting $100 per
hand, until the level that decreases the total profit for the week. 
But, I'm not really sure how to calculate this without doing it by
hand.  There is probably a way to use calculus to solve this without a
whole lot of work.

I did notice something interesting when I calculated the return for
trying to triple my money, and I'm not sure if it could be a simpler
way to calculate the other samples.  In the one I calculated the total
chance of winning could be expressed as the sum of the odds of winning
every hand (every hand until the goal is reached) plus 1/4th of that
number, plus 1/16th of that number, plus 1/64th of that number, and so
on.  I need to go to class right now but I'll try calculating like
that and doing it out by hand for the trial where I try and quadruple
my money and see if it works.
Answer  
There is no answer at this time.

Comments  
Subject: Re: Gambling math question
From: maycjay-ga on 25 Sep 2006 02:55 PDT
 
Hello awstick,

Actually it is the most profitable to withdraw the 100$ right after
you deposit the 67$. The simple reason being that this is a game of
negative expected value and the shorter you play it the more money you
will expect to have :)

here is the math:
( to make things simple I decided that after winning the first bet you
take the whole 200$ for the next bet - splitting the it into 100$ bets
makes the expected value even worse)

here the math for each 'level' of the game:

Level 0:
this is the level before any bets
you have the 33$ bonus and that is your expected value: +33$

Level 1:
after the first 100$ bet:
value if you win: +133$ [you made an error here in your calculation]
value if you loose: -67$

total value = 0,495*133 + 0,505*(-67)= 32$ (less than level 0)

Level 2:
If you win the 100$, you make a 200$ bet
Value if you win: 333$
Value if you loose: -67$

Total value = 0,495*0,495* 333 + 0,755 * (-67) = 31,01 (even worse)

Level 3:
If you win the 100$ and 200$ bets, you make a 400$ bet:

Value if you win: 733
Value if you loose: -67

Total value = 0,1213 * 733 + 0,8787 * (-67) = 30,03 (even worse)

... and so on and so on.

Hope this answers you question :)
Subject: Re: Gambling math question
From: jack_of_few_trades-ga on 25 Sep 2006 07:44 PDT
 
Mayc,
I think that this sentence in the question explains away your reasoning:
"The casino also removes all bonuses from your account when you withdraw."

It seems that if the casino gives $33.50 bonus, then it will take away
that $33.50 bonus when the money is withdrawn.  However, if you can
turn that $33.50 into a larger amount then you can withdraw the
excess.

Awstick,
Here is my thought on maximizing your profit:
Only bet the $33.50.  Maycjay did make a good point that the casino is
making money on each bet.  But this is only the case if you bet your
own money... the $33.50 is not your own money so gamble with it.
Example:
After 1 round, you calculated $15.67 profit from betting $100.
If you only bet the $33.5, then it is $33.5 X .495 = $16.58 profit.

So clearly you do not want to bet your own money, only bet the bonus money.

I'll work on maxing your profit and get back to you if I find anything conclusive.
Subject: Re: Gambling math question
From: jack_of_few_trades-ga on 25 Sep 2006 08:26 PDT
 
Level 0:
this is the level before any bets
you have the $33.5 bonus and that is your expected value: $0

Level 1:
after the first $33.5 bet:
Value if you win: $67  **so you could cash out $33
value if you loose: $0

total value = .495*33.5 + .505*0 = $16.58

Now, instead of simply betting $33.5 every time, I suggest betting the
whole amount (above your initial investment) every time... So your bet
should be doubling every time.

Level 2:
After you bet $67
Value if you win: $134 **so you could cash out $100
Value if you loose: $0

Total value = .495*.495* 100 = $24.5


I set this up in an excel spreadsheet with the formula:
.495^x*(33.5*x-33.5) = expected value  **where x is the number of bets
The expected value is maxed out after 6 bets, and your expected profit is $31.05

Do note that in reality you will earn $0 98.52% of the time, and
$2111.5 1.48% of the time.  Also note that the expected value rises
less than $1 from 4 bets to 6 bets, and you will win 4 bets 6% of the
time... so it is probably worth taking the $502.50 6% of the time
rather than the $2111.5 1.48% of the time.  The expected value of 4
bets is $30.17.

Here are the expected values and the % chance of winning for each number of bets:

# bets   %              Expected Value   Total prize
1	0.495	        $16.5825           $33.5
2	0.245025	$24.6250125       $100.5
3	0.121287375	$28.44188944      $234.5
4	0.060037251	$30.16871844      $502.5
5	0.029718439	$30.86259896     $1038.5
6	0.014710627	$31.04677899     $2110.5
7	0.007281761	$30.98025018     $4254.5
8	0.003604471	$30.79119747     $8542.5
9	0.001784213	$30.54305664    $17118.5
10	0.000883186	$30.26721279    $68574.5
Subject: Re: Gambling math question
From: jack_of_few_trades-ga on 25 Sep 2006 10:47 PDT
 
Awstick,
Try this table:
Gambles	Total	Odds	        Withdrawal	Expected Value
0	$100	1	        66.5	        66.5
1	$200	0.495	        166.5	        82.4175
2	$400	0.245025	366.5	        89.8016625
3	$800	0.121287375	766.5	        92.96677294
4	1600	0.060037251	1566.5	        94.0483531
5	3200	0.029718439	3166.5	        94.10343728
6	6400	0.014710627	6366.5	        93.65520892
7	12800	0.007281761	12766.5	        92.96259581
8	25600	0.003604471	25566.5	        92.15371965
9	51200	0.001784213	51166.5	        91.2919536

Formulas:
Total starts at $100 (account balance) and doubles with every win.
Odds = .495^x  (where x is the # of gambles).
Withdrawal = Total - $33.5
Expected Value = Withdrawal * Odds.

Notice that the Withdrawal caps out at 5 gambles; $3200 total, with a
2.97% chance of winning all 5, for a total gain of $94.10 - $67 =
$27.1

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