Pull the handle
|
||
| Try
your luck. Pull the handle |
||
Many attempt, but most fail when up against the HOUSE. He that hath, more is given. This innocuous fact of life is KEY to understanding why the gambler must ultimately loose.
If you want to learn why this is so, you need to DO THE MATH.
The MATH:
SETTING: A gambler enters a casino with considerably more resources and places a dollar bet. The game itself may even be fair with 50 - 50 odds. The gambler is determined to continue play until he or the house is in ruined. [1: Bibliography]
|
|
Figure 1. Gambler seeking his fortune. Notice that the gambler begins with fewer resources.
In the course of seeking his fortune the gambler will rack up a sequence of wins and losses. If he starts with d dollars and his losses exceed his wins by d, he is ruined. On the other hand, if his wins exceed losses by the HOUSE'S BANK, he leaves a wealthy man. Each sequence of wins and losses is one of many possible outcomes in the gambler's adventure.
The following illustrates one possible outcome. Notice that the gambler's first wager resulted in a win, however, the gambler ultimately lost all his money so was ruined.
Figure 2. The Universal Set. [5] This is a collection of all possible outcomes of the gambler's adventure. The yellow set represents outcomes leading to the gambler's ruin. The ruin set is divided into two subsets: one subset has outcomes where the first wager was a loss, the other where the first wager was a win.
In set notion:
 |
R is the union of the intersections of R with A and B respectively. |
The set A contains outcomes with the first wager a loss, and B the first wager a win. R is all outcomes leading to the gambler's ruin. Notice that R is divided into two disjoint sets. Thus the probability of Ruin is the sum of probabilities:
 |
The probability of a union of disjoint events is a sum of probabilities. [5] |
This can be written:
 |
Conditional Probability:
The probability of R given that A has occurred. [5] |
 |
Here, the first equation results from multiplying by forms of "1" and the second from the definition of conditional probabilities. [5] The probability of R given that A happened in case of the first term. The probability of R given that B happened in case of the second term. Now, let Rd stand for the probability of ruin in the case where the gambler begins with d dollars. Since the conditional probabilities denote wins and losses, an increase and decrease of d by one, the last equation can be written:
 |
Rd stands for the probability of ruin when the gambler has d dollars. This kind of equation is called a difference equation. [1,2,3] |
Here, p represents the probably of winning and q loosing on a particular wager. Notice that p + q =1 since a win or a loss must occur on a wager. The last equation is called a difference equation. Now techniques found in this branch of mathematics can be used to explain why its likely that the gambler is doomed.
 |
The difference equation. [2,3] |
 |
Assume that the solution has form: Rd = kd |
 |
Divide by k d -1 |
 |
Rearrange |
 |
Quadratic formula |
 |
q = 1- p |
 |
Radicand is a perfect square |
 | |
 |
(1) Two solutions of the difference equation |
 |
General solution. Sums and scalar multiples of solutions are also solutions. |
 |
Assumptions:
|
 |
Two equations to find the numbers M and N. |
 |
Solution (1). Unique solution of the difference equation. Notice that the solution is not valid if p = q, the probability of winning and loosing a wager being equal. [1] The following solves this special case. |
 |
(2) Rewrite (1) when p = q. |
 |
Multiply second solution by d when there is a repeated solution. |
 |
General solution when p = q. |
 |
Apply assumptions again when p = q. |
 |
Solution 2. Case of a perfectly fair wager, when the probability of winning and loosing are equal. [1] |
In summary, Solution 1 and Solution 2 give the probability of a gambler's ruin when he starts with d dollars. Solution 1 is the case where the chances of winning and loosing are different and Solution 2 when they are the same. Examples of these cases would be roulette and tossing a fair coin respectively.
| Example: Suppose a gambler has d = $10 and the house has D = $100. Here C = $110. | |
 |
Roulette probabilities when betting on red. It appears nearly certain, a probability of 0.999983, that the gambler will be ruined before he leaves. [5] |
 |
Tossing a fair coin. Here the gambler is ruined in only about 9 in 10 encounters with the house. EVEN WITH A FAIR BET! THE ADVANTAGE IS DUE TO THE FACT THAT THE HOUSE HAS MORE RESOURCES. |
THE HOUSE HAS A CONSIDERABLE ADVANTAGE IN BREAKING THE GAMBLER IN A SINGLE SITTING JUST DUE TO THE HOUSE'S RELATIVELY LARGE RESOURCES.
IN THE CASE OF A FAIR BET WHEN THE GAMBLER DOES BREAK THE HOUSE, SAY ONE IN TEN TIMES AS ABOVE, HE WINS "BIG". SO IN THE LONG RUN PROBABILITY THEORY DOES PREDICT THE EXPECTED RESULT THAT THE HOUSE AND GAMBLER END UP EVEN WHEN ODDS ARE 50-50.
Mathcorner.com researves
all rights on education materials