The probability of breaking the bank

What is the likelihood of a gambler winning all the cash in a casino? A better question is: what is the likelihood of a gambler losing all the money he or she brings into the casino?

There had been casinos going out of business for sure. But those bankruptcies were the results of business failures, e.g. not attracting enough customers to come into the casino, and were not for the results of gamblers breaking the bank. The business model of the casino is based on a built-in advantage called the house edge, which in mathematical term is the the casino’s average profit from a player’s bet. The house edge differs for different games but is always positive (e.g. the house edge for the game of roulette is 5.26%).

As long as a casino has a steady stream of customers willing to plunk down cash to play various gambling games, the casino will always win. For example, for every $100 bet in the game of roulette, the house wins on average $5.26. This post discusses a problem called gambler’s ruin that will further highlight a gambler’s dim prospect against the house.

The problem of gambler’s ruin features two players A and B, betting on the outcome of a coin. At the start of the game, there are n units of wealth between the two players, with player A owning i units and player B owning n-i units. In each play of the game, a coin is tossed. If the result of the coin toss is a head, player A collect 1 unit from player B. If the result of the coin toss is a tail, player A pays player B 1 unit. What is the probability that player A ends up with all the n units of wealth? What is the probability that player B ends up with all the n units of wealth? Probability of winning is winning for one player is the probability of losing for the other player. The following further clarifies the probabilities being discussed.

If the total wealth n is a large number and player A owns a small proportion of that, then player A is like the gambler and player B is like the house. Then the the probability of player A owning all n units of initial wealth is the probability of player A breaking the bank and the probability of player B owning all n units of wealth is the probability of ruin for player A.

In playing this game, player A gains 1 unit for each coin toss that results in a head and loses 1 unit for each coin toss that results in a tail. Theoretically player A can go broke (losing all the i units that he starts with) or break the bank (owning the total n units between the two players). Losing it all for player A is certainly plausible especially if there is a long run of tails in the coin tosses. On the other hand, winning everything is plausible for player A too, especially if there is a long run of heads. Which is more likely?

The answer to the problem of gambler’s ruin will produce two formulas that can tell us the likelihood of player A winning, and the probability of player A losing everything.

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Gambler’s Ruin

The answers had been worked out in a recent post in a companion blog on probability. In this post, we discuss the answers and make some remarks on gambling. We first discuss the case for even odds and then the general case.

Even Odds. Suppose that player A and player B bet on tosses of a fair coin. Suppose further that player A owns i units of wealth and player B owns n-i units at the beginning. The long run probability of player A winning all the units is \displaystyle A_i=\frac{i}{n}. The long run probability of player B winning all the units is \displaystyle B_i=\frac{n-i}{n}.

The probability of a player owning all the original combined wealth is the ratio of the initial wealth of that player to the total initial wealth. Putting it in another way, the probability of ruin for a player is the ratio of the initial wealth of the other player to the total initial wealth. For the player with only a tiny percentage of the combined wealth, the probability of ruin is virtually certain. For example, if the house (player B) has 10,000,000 units and you the gambler (player A) has 100 units, then the probability of ruin for you is 0.99999 (99.999% chance). This is based on the assumption that all bets are at even odds, which is not a realistic scenario when playing against the house. So in the uneven odds case, the losing probabilities for the gambler would be worse.

Uneven Odds. Suppose that player A and player B bet on tosses of a coin such that the probability of tossing a head is p with p not 0.5. Let q=1-p. Suppose further that player A owns i units of wealth and player B owns n-i units at the beginning. The long run probability of player A winning all the units is \displaystyle A_i=\frac{1-\biggl(\displaystyle \frac{q}{p} \biggr)^i}{\displaystyle 1-\biggl(\frac{q}{p} \biggr)^n}. The long run probability of player B winning all the units is \displaystyle B_i=\frac{1-\biggl(\displaystyle \frac{p}{q} \biggr)^{n-i}}{\displaystyle 1-\biggl(\frac{p}{q} \biggr)^n}.

First, let’s unpack these formulas. Let’s say the coin has a slight bias with the probability of a head being 0.49. Then p = 0.49 (the house edge is 1%) and q = 0.51 with \frac{q}{p}=\frac{0.51}{0.49} = 1.040816327. Let’s say the total initial wealth is 100 units and player A owns 10 units at the beginning. Then A_{10} is the following:

    \displaystyle A_{10}=\frac{1-\biggl(\displaystyle \frac{0.51}{0.49} \biggr)^{10}}{1-\biggl(\displaystyle \frac{0.51}{0.49} \biggr)^{100}}=0.00917265

In the even odds case, player A would have a 10% chance of winning everything (by owning 10% of the total initial wealth). With the house having just a 1% edge, the chance for winning everything is now 0.917%, less than 1% chance!

To calculate the probability of player B winning everything, plug into the ratio \frac{p}{q} instead of \frac{q}{p}. In the numerator, \frac{p}{q} is raised to n-i. In other words, \frac{p}{q} in the numerator is always raised to the initial wealth of the player being calculated. In the example of p = 0.49, plug in \frac{p}{q}=\frac{0.49}{0.51} to obtain B_{10} = 0.99082735. There is more than 99% chance that player B (the house) will own all 100 units.

Note that A_{10}+B_{10} equals 1.0. In fact, A_{i}+B_i is always 1.0. Thus B_{i} can be calculated by 1-A_{i}. In other words, the probability of breaking the bank plus the probability of ruin is always 1.0. When a gambler plays at a casino, there are only two possibilities, either breaking the bank or ruin. There is no third possibility, e.g. the game goes indefinitely with no winner.

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More Calculations

Using Excel, the winning probabilities for player A are calculated for two scenarios, one for total wealth of 100 units and one for total wealth of 10,000 units. The results are in Table 1 and Table 2. Each table uses four values of p, starting with the fair game of p = 0.5. The column for p = 0.49 does not correspond to an actual casino game. It is used to demonstrate what happen when the house has a 1% edge. The last two columns are for the game of craps and the game of roulette.

    Table 1 – Probabilities of Player A Breaking the Bank
    Total Initial Wealth = 100 Units

    \begin{array}{ccccccccc}  \text{Player A} &  \text{ } & \text{ } &  \text{ } & \text{ }  & \text{ } & \text{ } & \text{ } & \text{ } \\    \text{Initial} &  \text{ } & \text{Fair} &  \text{ } & \text{House}  & \text{ } & \text{Craps} & \text{ } & \text{Roulette} \\  \text{Wealth} &  \text{ } & \text{Game} &  \text{ } & \text{Edge 1\%}  & \text{ } & \text{ } & \text{ } & \text{ } \\  \text{i} &  \text{ } & p=0.5 &  \text{ } & p=0.49  & \text{ } & p=0.493 & \text{ } & p=0.474 \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   i=10 &   \text{ }& 0.10 &   \text{ } & 0.0092 & \text{ } & 0.0209  & \text{ } & 0.0001 \\   i=20 &   \text{ }& 0.20 &   \text{ } & 0.0229 & \text{ } & 0.0486  & \text{ } & 0.0002 \\   i=30 &   \text{ }& 0.30 &   \text{ } & 0.0433 & \text{ } & 0.0852  & \text{ } & 0.0007  \\  i=40 &   \text{ }& 0.40 &   \text{ } & 0.0737 & \text{ } & 0.1337  & \text{ } & 0.0019  \\  i=50 &   \text{ }& 0.50 &   \text{ } & 0.1192 & \text{ } & 0.1978  & \text{ } & 0.0055  \\   i=60 &   \text{ }& 0.60 &   \text{ } & 0.1870 & \text{ } & 0.2826  & \text{ } & 0.0155  \\   i=70 &   \text{ }& 0.70 &   \text{ } & 0.2881 & \text{ } & 0.3949  & \text{ } & 0.0440 \\  i=80 &     \text{ }& 0.80 &   \text{ } & 0.4390 & \text{ } & 0.5434  & \text{ } & 0.1247 \\  i=90 &     \text{ }& 0.90 &   \text{ } & 0.6641 & \text{ } & 0.7400  & \text{ } & 0.3531 \\    \end{array}

    Table 2 – Probabilities of Player A Breaking the Bank
    Total Initial Wealth = 10,000 Units

    \begin{array}{ccccccccc}  \text{Player A} &  \text{ } & \text{ } &  \text{ } & \text{ }  & \text{ } & \text{ } & \text{ } & \text{ } \\    \text{Initial} &  \text{ } & \text{Fair} &  \text{ } & \text{House}  & \text{ } & \text{Craps} & \text{ } & \text{Roulette} \\  \text{Wealth} &  \text{ } & \text{Game} &  \text{ } & \text{Edge 1\%}  & \text{ } & \text{ } & \text{ } & \text{ } \\  \text{i} &  \text{ } & p=0.5 &  \text{ } & p=0.49  & \text{ } & p=0.493 & \text{ } & p=0.474 \\    \text{ } &   \text{ } & \text{ } & \text{ } & \text{ } &  \\   i=100 &   \text{ }& 0.010 &   \text{ } & 0.0000 & \text{ } & 0.0000  & \text{ } & 0.0000 \\   i=500 &   \text{ }& 0.050 &   \text{ } & 0.0000 & \text{ } & 0.0000  & \text{ } & 0.0000 \\   i=1000 &   \text{ }& 0.100 &   \text{ } & 0.0000 & \text{ } & 0.0000  & \text{ } & 0.0000  \\  i=5000 &   \text{ }& 0.500 &   \text{ } & 0.0000 & \text{ } & 0.0000  & \text{ } & 0.0000  \\  i=9800 &   \text{ }& 0.980 &   \text{ } & 0.0003 & \text{ } & 0.0037  & \text{ } & 0.0000  \\   i=9850 &   \text{ }& 0.985 &   \text{ } & 0.0025 & \text{ } & 0.0150  & \text{ } & 0.0000  \\   i=9900 &   \text{ }& 0.990 &   \text{ } & 0.0183 & \text{ } & 0.0608  & \text{ } & 0.0000 \\  i=9950 &     \text{ }& 0.995 &   \text{ } & 0.1353 & \text{ } & 0.2466  & \text{ } & 0.0055 \\  i=9990 &     \text{ }& 0.999 &   \text{ } & 0.6703 & \text{ } & 0.7558  & \text{ } & 0.3531 \\    \end{array}

Table 1 describes the gambling situations where the two players have a combined initial wealth of 100 units. It is not a realistic scenario for the situation of gambler vs. house as the total wealth is quite moderate. Perhaps it best describes the interpersonal gambling situations where the initial wealth positions of the two players are moderate in size. It is clear that the values of p are critical to the eventual success or ruin for player A. Let’s look at the row for i = 50 (both players in equal initial position). For a fair game, there is equal chance for breaking the bank and ruin. As the game becomes more unfair, the chance for breaking the bank is less than 1% (roulette).

In Table 1, the same pattern for i = 50 appears for the other initial positions as well. As p drops below 0.5 (as the game becomes more unfair to player A), the less likely it is for player A to win and the more likely it is for player B (the player who has the edge) to win.

Table 2, with the total wealth of 10,000 units, is designed to more realistically reflect the gambling situation of a typical gambler versus the house. With the higher combined wealth, the effect of p is more pronounced. For example, in the top half of Table 2, there is virtually no chance for player A to win at any of the unfair games, i.e. there is a virtually certain chance for ruin for the gambler. Even when player A has the same initial wealth as the casino, there is still no chance to win at the unfair games.

The bottom half of Table 2 shows that when player A has substantially more wealth than the casino, player A begins to have a positive chance of winning (but the positive probability is still very small). Of course the gambler having more wealth than the casino is far from a realistic scenario.

The formulas for A_i and B_i only provide the long run probability for player A (the gambler) to break the bank and the long run probability of ruin for player A, respectively. They give no information about the number of plays of the game in order to reach the eventual success or ruin.

For example, in table 2, if the initial wealth of player A is 100, the probability of winning at the game of roulette is zero. This means that eventually player A will lose the 100 units. But the results in Table 2 do not tell us on average how many plays of the game have to happen before eventual ruin. If the gambler is lucky, he or she may be able to rack up some winnings for a period of time before getting to the point of ruin. One thing is clear. When the game is unfair (to the gambler), there is a virtually certain chance that the gambler will lose all the money that he or she bring into the casino. This is essentially the story told by the formulas discussed here.

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\copyright 2017 – Dan Ma

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3 Responses to The probability of breaking the bank

  1. Pingback: The gambler’s ruin problem | A Blog on Probability and Statistics

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  3. Pingback: Five probability problems to help us think better | Math in the Spotlight

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