Paradoxes from Probability

We have seen how attention to mutual exclusivity and independence in the probability calculus can correct our judgments about chance phenomena in surprising ways that seem paradoxical, initially. There are similar corrections deriving from the properties of expectations or expected values in the probability calculus. We shall see two of them here.

The first is the popular game of "chuck-a-luck" or "crown and anchor." A natural supposition that the game favors the player is refuted by computing the expectation.

The second is the extent to which a gambler is harmed by a slightly negative expectation. What is presumed to be a benign effect produces what is called in the probability literature, the "gambler's ruin."

Chuck-a-Luck, Crown and Anchor

The Game

When gambling games are simple, it is fairly easy to see whether they favor the gambler or not. Computing the expected value is likely unnecessary for most of us to see which game is fair and which is not. Things are not always so simple. Then computing the expected value can give useful information.

A game is known as chuck-a-luck in the US and related places; and as "crown and anchor" in England and related places. It is played with three dice. In the US they are numbered one to six. In England they have the four card suits, a crown and an anchor on their sides.

Soldiers playing chuck-a-luck in Harper's Weekly, 1865.
https://commons.wikimedia.org/wiki/File:Harper%27s_weekly_(1865)_(14784937143).jpg

A player is invited to bet on one of the sides for a fee of $1. In the US game, the player may choose any number. For concreteness, let us say that is it a six

. If one of the dice shows a six, the player is paid a net of $1. If two

show, the player is paid that twice: a net of $2. If all three dice show a a six

, the player is paid this $1 amount ten times: a net of $10.

Here is a version of chuck-a-luck from Hoyle's celebrated (1904) book of rules for games. This version is less favorable to the player. It only requires a payment of $3 if all three dice show the player's chosen number:

"There is a one in six chance of a six

on just one die. Since there are three dice, I get three opportunities of a one in six chance. That is, there is a one in two chance of a win. So I have even odds: an equal chance of winning $1 and of losing $1. But wait... when a six

comes up on one die, there are still two others. And if any of those also show a six

, I win there too. If all three, I get ten times as much! Thus the game favors me. I get three ways of winning to the house's one."

The Expectation

Needless to say, this sort of reasoning is just what the house wants the player to think. In practice, the game is slightly favorable to the house. To see it, we need to pay attention to the outcome space. Its elements are outcomes of the form:

We then need to compute the probabilities and then the expected value in this outcome space. Here is how the computation looks:

This calculation shows already where the informal reasoning of the player has failed. The probability of just one six

is merely 75/216 = 0.347, that is, roughly a third. An even odds bet on just this outcome is clearly unfavorable.

P(one six

only)
= P(six

on the first die
and not on the second and third)
+ P(six

on the second die
and not on the the first and third)
+ P(six

on the third die
                        and not on the the first and second)
=(1/6)x(5/6)x(5/6)
     + (5/6)x(1/6)x(5/6)
         + (1/6)x(5/6)x(5/6)
= 3 x (1/6)x(5/6)x(5/6) = 75/216

Expected value
= P(one six

only) x $1
+ P(two sixes

only) x $2
+ P(three sixes

) x $10 - P(no sixes

) x $1
=(75x$1 + 15x$2 + 1x$10 - 125x$1) / 216
= -$10/216
= -$0.0463

The calculation makes it easy to see which alteration would in the payouts would make the game fair. That is, if the player were paid a net of $20 on a roll of three sixes, then the expected value would rise to $0 and game would be fair.

That is, the expected value is a net loss of -$0.0463. What this means is that in repeated playing of the game, the player loses an average of $0.0463 on each play.

This average loss is small enough that it is hard to notice behind the normal variations in wins and losses. If a player makes 100 plays, the player will on average only have lost $4.63. For the player, that is just the fee paid for the opportunity of 100 plays and the enticing chance of a big win. For the house, however, this small loss by the players, is a gain for the house. Over the longer term, these gains build and keep the game profitable for the house.

The Gambler's Ruin

In a fair game, neither the gambler nor the house has any advantage. This fairness entails that the expected return to either on any play is zero. For if one--the house, for example--has a positive expectation, then the other--the gambler--must have a negative expectation. The house would then be favored only because the gambler is disfavored.

This means that gamblers should seek fair games in which their expectation on each play is zero. Of course if a gambler can find a game in which the gambler has a positive expectation, that would be the even better choice. However in practice this option is not offered in casinos unless by mistake. The realistic choice is between games, all of which have some negative expectation. The better choice is the game with the less negative expectation.

The casino at Monte Carlo
https://commons.wikimedia.org/wiki/File:The_Casino,_Monte_Carlo_-_panoramio_cropped.jpg

It is easy to imagine that a small negative expectation will not have much damaging effect. That this is mistaken is the principal conclusion to be drawn from the "gambler's ruin" problem.

A Fair Game

To calibrate what we anticipate should happen, we first consider a simple fair game. The game play has two outcomes--red and black--each with probability 1/2. We could imagine that the outcomes arise from tossing a coin with red on one side and black on the other. The game pays a net of +$1 on red and -$1 on black.

On each play, the gambler either wins or loses $1. It follows from the zero expectation that on average, with each play, the gambler gains or loses nothing.

The gambler approaches the game with an initial stake of $50 and plans to keep playing until the gambler's fortune has risen to $100 ("win"), that is, until it has doubled; or the gambler has lost the entire $50 stake, which is the "ruin" of the "gambler's ruin." The overall dynamic, as play proceeds, is for the gambler's fortune to rise and fall.

Here is a simulation of the changing fortunes of ten gamblers, each of whom play for 1,000 bets:

While each gambler's fortune will move either up or down, there is no preferred direction. Overall the gambler's fortunes are as likely to rise and fall.

This lack of a preferred direction follows from the symmetry of the game. If we switched the rules so that the gambler were to lose $1 on a red outcome and gain $1 on a black outcome, then the dynamics would remain the same. Thus the dynamics can show no preference for either of the directions. This reversed game is the one that the house is actually playing.

Because of the symmetry of the game, the chance of winning or of ruin are the same. If we assume that the gambler keeps playing until either happens, then their probabilities must sum to unity and be one half for each:

The gambler starts with a fortune of $50. As play proceeds the gambler's fortune will rise and fall. It will meandering around the average of $50, but gradually drift farther and farther away from it, until it has drifted to $100 (win) or $0 (ruin). In the simulation above only two of the gamblers have drifted far enough to halt play. One drifted to a win and the other to ruin.

We can ask how far the fortunes drift as play proceeds. The drift motion is a classic of stochastic processes, the so-called "random walk." For such processes the distance drifted on average follows a simple rule:

For those who want it, the more precise result is this. The variance of payoff in each individual play is σ². Then the total variance of N plays is Nσ². The precise measure of the spread is given by the standard deviation, which is the square root of the variance, √(Nσ²) = √(N)σ.

We are interested in how many plays are needed for a win or for ruin to arise. There is no single number since some games will end early and other late. However we can determine an expectation value for the number of games that leads to termination by inverting this drift estimate. In the inversion, the number of plays is proportional to (average drift)².

More specifically, we seek how many plays are needed for the drift to extend to $50, either in a gain or a loss, for that marks the end of play. The inverted result tells us that the number of plays for termination is proportional to 50². A more precise analysis (given in the linked pdf) shows that the result is not just proportionality but equality:

The chart above only extends to 1,000 plays, which is well short of this expected duration. If it were to be extended, more of games would terminate by the expected duration of 2,500 plays. We would likely have to extend well beyond this value of 2,500 in order to find all the games terminating.

Roulette: An Unfair Game

There is something close to this fair red/black game in the game of roulette, as it is offered traditionally in casinos. In the American version, there is a wheel with slots numbered 1 to 36 and two slots numbered 0 and 00. The slots numbered 1 to 36 are divided into 18 red and 18 black slots. Gamblers can bet with equal odds on red or black; or on a low number 1-18, or a high number 19-36; or an even number or an odd number. A bet on any of these loses, however, if either the 0 or 00 arises.

Since each number arises with the same probability and there are 36+2=38 of them, a bet on red has the following probabilities and payoffs:

Since 18/38 = 9/19 = 0.474 < 1/2, the game is unfair. The expectation on a bet of $1 on either red or black is

Expectation (red) = Expectation (black)
= $1 x 18/38 - $1 x 20/38 = -$(2/38) = -$(1/19) = -$0.0526

This is a small, negative expectation. It is merely about 5% of the $1 bet. One might imagine that a such small negative expectation would not alter the dynamics of play. That is not so. The chart below shows a simulation of ten gamblers repeatedly making $1 bets on red at roulette. As before, they start with a stake of $50 and continue to play until their fortune has risen to $100 (win); or it has fallen to zero (ruin):

Over the same 1,000 plays simulated for the fair game above, the dynamics are quite different. The fortunes of the gamblers do drift, but they all drift in a downward direction.

The symmetry of the fair game is lost. The gamblers' fortunes all drift downward towards ruin. If we reverse the payoffs above, we have how the house experiences the game. If there is a black or 0 or 00, the house wins $1, with probability 20/38; and if there is a red, the house loses, $1 with probability 18/38. The house has a positive expectation in each play of +$0.0526. The house's fortune, if charted, would show an upward drift.

The probability of ruin in the fair game is 1/2. The probability of ruin in this unfair roulette game is so close to 1 that the difference is indiscernible in ordinary play. That is, any gambler playing this game with this strategy will almost certainly face ruin. A precise calculation of the probability of ruin is given in the linked pdf. The formula uses the ratio of the probabilities of a loss and a win in a single play:

Probability of ruin
= [ (10/9)¹⁰⁰ - (10/9)⁵⁰ ] / [ (10/9)¹⁰⁰ - 1 ]
= [37,649 - 194] / [37,649 - 1 ] = 0.9949

The means that on average, if 200 gamblers play with this strategy, 199 of them will be ruined and only 1 will win. This dismal fate is reflected in the chart above. Over the 1,000 plays simulated, 8 of the 10 gamblers were ruined. The remaining two played on. One was already close to ruin and second has less than the starting stake.

The simulation charted above also suggests that the expected duration of play is shorter than that of the fair game. In the simulated fair game, only two gamblers of ten had terminated play in 1,000 plays. In this unfair game, in 1,000 plays, 8 of 10 gamblers have been ruined.

The full analysis given in the linked pdf shows that the expected duration of play can be easily approximated in cases like this in which the probability of ruin in near certain. We start with the stake of $50 and allow that, with each play, there is an expected loss of $1/19, as calculated above. The expected duration of play is the number of plays needed for the accumulated losses of $1/19 per play to exhaust the initial stake. That is:

Expected duration of play
= Initial stake / expected loss per play
= 50 / (1/19) = 950

This expected duration of play of 950 is less than half the expected duration of play of the fair game of 2,500. It tells us that the near assured ruin in the unfair game will come more swiftly than any ruin in the fair game.

Cup-board Simulation

A simple mechanical analogy might help clarify the difference between the two cases. Imagine that we have seven small cups in a row on a board. We put a pea in the central cup and will shake the cups and board. The walls between the cups are low so that the pea can bounce between the cups. The walls on the other sides are high so that the pea can only bounce between adjacent cups. There are two, additional destination cups at the left "L" and right "R" ends of the two

Keeping the board horizontal, we shake the board and cups up and down so that the pea jumps around between the cups. There is an equal chance that the pea will jump to the right or to the left. If we keep shaking long enough, the pea's drifting back and forth between the cups will eventually lead it to escape to either the left or right destination cup. From the symmetry of the arrangement, there is an equal probability that the pea ends up in either left or right cup. (This case corresponds to the fair game.)

Now imagine that we incline the board and cups slightly to the right and repeat the shaking. The slight incline means that the pea is slightly more likely to jump to a cup on its right than on its left. Over the course of many shakings, the pea will migrate preferentially to the right and end up in the right destination cup. (This case corresponds to the unfair game.)

While the inclination of the board and cups may be small, in many bounces, it is enough to favor a rightward drift quite strongly. Correspondingly, a small negative expectation in the red bets in roulette is enough to favor quite strongly a downward drift in the gambler's fortune.

How to Gamble, If You Must

The phrase comes from a thorough by very technical treatment of the gambler's ruin problem: Lester E. Dubins, Leonard J. Savage, How to Gamble If You Must: Inequalities for Stochastic Processes. Rev. ed. Mineola, NY: Dover, 2014. Ch. 1.3 provides a brief history of the problem.

"If you must"? It does not take much acquaintance with probability theory to see that casino gambling is a poor idea for someone interested in turning a profit or just breaking even. What the gambler's ruin problem shows is that gambling by the strategy sketched above is not just a poor idea, but a very poor idea. That ruin is assured for this strategy in the unfair game is surely surprising and even paradoxical in the sense of a strikingly unexpected result.

The analysis, however, does suggest an alternative strategy that gamblers can use to mitigate the approaching ruin:

This is hardly a satisfying strategy for someone who actually enjoys time spent at the gambling tables. For the activity is reduced as much as possible to a single bet. But perhaps someone who enjoys time at the gambling table is seeking something more than or other than profit?

That is, if you have a stake of $50, bet it all at once. As before, the bet has a negative expectation of -$50x(1/19) = -$2.63. However the prospect of ruin is greatly reduced. We have

Probability (win $50, fortune to $100) = 18/38 = 0.47
Probability (lose $50, ruin) = 20/38 = 0.53

This strategy has the same result as the single $1 bet strategy: the game ends with the gambler's fortune rising to $100 or falling to $0. However now the probability of ruin is much less. It is just over 1/2.

That this shift in strategy makes such a big difference is something peculiar to the unfair game. In the fair game, the probability of a win or a loss is always 1/2, independently of whether the gambler makes many single $1 bets, bets the entire stake of $50 all at once or has some combination in between. That this is so, follows from the symmetry of the game. As long as the gambler plays so that the only outcomes are a win or ruin, the probabilities will always be 1/2.

Why the Largest Bets Make a Difference

It may seem puzzling that the probability of ruin can be so easily reduced in an unfair game by making the largest bets possible. That it is puzzling may come from the following (flawed) reasoning:

"If I bet all my stake of $50 at once, my expectation is just -$50x(1/19) = -$2.63.

If I divide my stake into 50 single $1 bets, then my expectation in each is -$(1/19). But there are 50 bets, so my overall expectation is the same -50x$(1/19) = -$2.63.

Thus there should be no difference in the effect of the negative expectation in the two cases."

There will be other differences. After a single bet of $50, the gambler's fortune will be either $100 or $0 only. If the gambler places 50 single bets of $1, then the gambler's fortune can be anywhere between $100 and $0, with the extremes of the range very unlikely.

The false assumption is in the second inference "... there are 50 bets..." There are not 50 bets, but very many more in the chosen strategy. We saw above that the expected duration of play is 950 plays. That is, the gambler will actually place 950 bets on average, before ruin, and in each of these incur a negative expectation of -$(1/19).

As before, the accumulated negative expectation is -950x$(1/19) = -$50, which erases the gambler's initial stake.