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"Maximized Asymptotic Return System " The MARS Equation
A Guide to Succesful Money Managment
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R=1.20 | R=0.80 | |
After 1 play | $100 * (1.21) = $120.00 | $100 * (0.81) = $80.00 |
After 2 plays | $100 * (1.22) = $144.00 | $100 * (0.82) = $64.00 |
After 3 plays | $100 * (1.23) = $172.80 | $100 * (0.83) = $51.20 |
After 10 plays | $100 * (1.210) = $619.17 | $100 * (0.810) = $10.74 |
After 20 plays | $100 * (1.220) = $3833.76 | $100 * (0.820) = $1.15 |
   [1.2 * 1.2 * 1.2 * ...(½ million times)] * [.8 * .8 * .8 * ...(½ million times)]Notice that the actual order of the wins and losses has no has no effect on the overall return. The average return per flip is the one millionth root of the above overall return.
= [1.2500,000] * [.8500,000]
   [1.2500,000 * .8500,000](1/1,000,000)Which comes to 0.98.
= [1.2.5] * [.8.5]
Therefore, the long term player should expect his bankroll to shrink at an average rate or 2% per play. He should expect to lose money on this seemingly fair game. The longer he plays, the closer he will become to experiencing the asymptotic return rate of 0.98.
Now, let us generalize the above specific example:
Let W represent the return rate if you win.Now, following the above example:
Let p represent your probability of winning.
Let L represent the return rate if you lose.
Let q represent your probability of losing.
Let B represent return rate for any unbet money (bank interest)
Let X represent what portion of your money is bet on each play.
Let R represent your asymptotic return rate.
Assume p + q = 1 (no ties, you either win or lose)
R=[{return on a win}p] * [{return on a loss}q]This equation gives the asymptotic return rate "R" that results from using money management strategy "X" in playing a game under conditions {W,p,L,q,B}.
R=[{(1-X)B + WX}p] * [{(1-X)B + LX}q]
(Equation #1)
Maximum R occurs when X = [L + (W-L)p - B] / [(W+L) - WL/B - B]
(Equation #2)
This is the MARS equation
Imagine a casino that has coin flipping game; paying double or nothing; you call heads or tails.Example #2 ~ A Favorable Game:
For this game:This proves the optimal way to play a fair game is not to bet. You should only bet when the game is in your favour.
W=2     p=.5      B=1      L=0      q=.5
Using the MARS equation,
Optimal X = [0 + (2-0).5 - 1] / [(2+0) - (2*0)/1 - 1]
                 = 0
Imagine you study the action of the above game and you discover a way to correctly predict the outcome 60% of the time. Here is a favorable situation! But how should you play it? If you bet 100% of you money on every play, you will go broke on the first loss. If you bet only pennies, you will be wasting time.Example #3 ~ A Favorable Game:For this game:Thus you should bet 20% of your current bankroll on every play. Assuming you start with $100 bankroll, how quickly will your money grow?
W=2     p=.6      B=1      L=0      q=.4
Using the MARS equation,
Optimal X = [0 + (2-0).6 - 1] / [(2+0) - (2*0)/1 - 1]
                 = .20
Using Equation #1:Thus you can expect your initial $100 bankroll to grow at an average rate or 2.03% per play. If you stayed for 100 plays, always betting 20% of your current bankroll, your theoretical outcome would be $100 * (1.023100) which comes to $749. The longer you play, the more certain you can be of achieving this therotical outcome.
R = [{(1-.2)1 + 2*.2}.6] * [{(1-.2)1 + 0*.2}.4]
    = 1.0203
Imagine example #1 is modified so the casino pays triple or nothing.Example #4 ~ A Lottery:For this game:Thus you should bet 25% of your current bankroll on every play. How quickly will your money grow?
W=3     p=.5      B=1      L=0      q=.5
Using the MARS equation,
Optimal X = [0 + (3-0).5 - 1] / [(3+0) - (3*0)/1 - 1]
                 = .25
Using Equation #1:Thus you can expect your money to grow at an average compound rate of 6.07% per play.
R = [{(1-.25)1 + 3*.25}.5] * [{(1-.25)1 + 0*.25}.5]
    = 1.0607
Imagine you are invited to buy a lottery tickets and your chance of winning is one in a million. Tickets cost $1 each. Because of previous unclaimed prizes, the winner of this lottery will be paid $5 million. This is clearly an attractive situation; but what should you pay for a ticket?Example #5 ~ A Horse Race:For this game:Thus if your entire net worth is $200,000, you should pay $200,000 * .0000008 = 16¢ for a single ticket (See comment section below).
W=500,000     p=1/1,000,000      B=1      L=0      q=999,999/1,000,000
Using the MARS equation,
Optimal X = [0 + (5,000,000-0).000001 - 1] / [(5,000,000+0) - (5,000,000*0)/1 - 1]
                 = .0000008
An experienced handicapper estimates that a horse has a 40% chance of winning a race. The horse has odds of 3 to 1 (one dollar bet will produce a $3 profit).Example #6 ~ The Stock Market:For this game:Thus the handicapper should wager 20% of his bankroll on this horse.
W=(odds +1) = 4     p=.4      B=1      L=0      q=.6
Using the MARS equation,
Optimal X = [0 + (4-0).4 - 1] / [(4+0) - (4*0)/1 - 1]
                 = .20
The MARS equation is so very useful at the racetrack, we will restate the equation in a special handicapper's form:
Wager in % of bankroll = {[(Odds +1) * chance of winning -1] / Odds} * 100%
Imagine you are considering buying stock of a small mining exploration company. If the company finds ore, the stock will be tripled a year from now. But if no ore is found, the stock will be at one third of its current value. The chance of finding ore is 50%. You can make 10% annual interest on your money by simply leaving it in the bank.Example #7 ~ A Business Investment:For this game:Thus you should invest 43% of your portfolio in the stock and leave 57% in the bank.
W=3     p=.5      B=1.10      L=.33      q=.5
Using the MARS equation,
Optimal X = [.33 + (3-.33).5 - 1.10] / [(3+.33) - (3*.33)/1.10 - 1.10]
                 = .43
Now suppose the bank raises its interest rates to 15%. How should you react?
B increases from 1.10 to 1.15Thus you should sell some of your stock, so that your bank account is increased from 57% of your portfolio to 61%.
Optimal X = [.33 + (3-.33).5 - 1.15] / [(3+.33) - (3*.33)/1.15 - 1.15]
                 = .39
Imagine you are thinking about starting a small business. What is the optimal size of bussiness to start? In a good year, the company will make a profit of 35% on the current investment; while in a bad year the company will lose 20% of the current investment. Six years out of ten will be good years and four years will be bad. You have no way of predicting whether a given year will be good or bad. You only know there is a 60% chance of having a good year. You can make 10% annual interest on your money by simply leaving money in the bank.Example #8 ~ A Business Investment:For this game:Thus you should invest 44% of your money into the business and leave a cash reserve of 56% in the bank collecting interest. This cash reserve will tide you over in case of a string of bad years.
W=1.35     p=.6      B=1.10      L=.80      q=.4
Using the MARS equation,
Optimal X = [.80 + (1.35-.80).6 - 1.10] / [(1.35+.80) - (1.35*.80)/1.10 - 1.10]
                 = .44
Suppose example # 7 is modified so there is a 70% chance of a good year.For this game:With X greater than 100%, MARS is telling you to invest all of your money, plus 25% additional money borrowed at 10% interest rate. Your business should have one part debt to four parts equity. Your optimal debt / equity ratio is .25.
W=1.35     p=.7      B=1.10      L=.80      q=.3
Using the MARS equation,
Optimal X = [.80 + (1.35-.80).7 - 1.10] / [(1.35+.80) - (1.35*.80)/1.10 - 1.10]
                 = 1.25
"Never bet a portion of your bankroll that exceeds you chance of winning"