Introduction

A common problem related to games of chance is their value: what is the fair deterministic price to charge for participation in a game whose outcome is random. As we know from the Law of Large Numbers, the expected payout of the game is a fair price. Indeed, if the game is played independently @@@n@@@ times, then with probability tending to @@@1@@@, the total payout will be of the order of @@@(1+o(1))n P@@@ with probability @@@1+o(1)@@@.

We will begin through a series of warm-up examples, all can be viewed as paradoxes:

• As we know well, some RVs (are finite but) have infinite expectation. If a game has a payout with infinite expectation, according to the discussion above, the price to change for participation is infinite. St. Petersburg Paradox we introduce in the St. Petersburg Paradox is one such special case.
• Regardless of a fair price (or even the fairness of the game), in most games one can (theoretically) continue playing by increasing the bets, so that the first win will not only wipe all losses, but will also lead to arbitrarily large net gain. We will review this through an explicit example, a continuation of the St. Petersburg Paradox, leading to notions of strategy and resources.
• Speaking of resources, we will return to the example above, but now limiting the resources, and show that no matter how large the resources are, if each round of the game is fair, there is no way to make a net gain.

We will then turn to Doob’s Optional Stopping Theorem which states that under finite resources, strategy doesn’t matter, and the game will remain fair. This is probably why you’re never charged extra for the ability to apply strategy in a casino.

It’s going to be a wild ride, and hope it can substitute the thrill from playing games of chance.

The classic paradox

I’m offering you the following game, which we call Game 1.

Game 1.

• I repeatedly toss a fair coin until it first lands Tails and record the number of tosses, @@@T@@@.
• I pay you @@@2^T@@@.

Nice!

St. Petersburg Paradox

Nothing comes for free. How much are you willing to pay for the game in advance?

As your random payout is @@@2^T@@@, a finite amount, a finite amount, the LLN tells us that a fair price is @@@E[2^T]@@@. Now @@@T\sim \mbox{Geom}(1/2)@@@, and as you can easily verify, @@@E[2^T]=\infty@@@. Oops. This the first paradox, a very famous one, known as St. Petersburg Paradox.

A more general game

Let’s now introduce Game 2.

Game 2.

• You name an amount @@@x@@@.
• I put @@@x@@@ on the table.
• I toss a coin:
  • If lands Heads, we each take what we put back.
  • If lands Tails, you take the entire amount on the table.

What is the fair price for Game 2? Your random payout is either @@@0@@@ or @@@x@@@, each with probability @@@\frac 12@@@, and therefore the fair price is @@@\frac 12 \times 0 + \frac 12 x =\frac{x}{2}@@@. So if you had to pay in advance to play the game, a fair price would be @@@x/2@@@.

Let’s add an assumption:

• You can play Game 2 as much as you want (each time you play it will be called a round), and you have the right to quit after any round (think of this as I being a 24-hour casino and you as a patron).

Relation between the games

How are the two games related? To get Game 1 from Game 2 you should simply

1. Initially set @@@x=2@@@;
2. Double the amount you put each round; and
3. Quit after the first Tails.

In other words, Game 1 is nothing but a specific strategy for you to repeatedly play Game 2: you decide how much to put and how when to quit.

From now on we will assume that you pay for each round you play, and that you don’t really put money on the table, you just name the amount @@@x@@@ and then pay the price for the game @@@x/2@@@.

A second paradox

Recall that the payout for Game @@@1@@@ is @@@2^T@@@ where @@@T@@@ is the number of rounds until the first Tails. Now if you pay for each of the rounds played according to the formula devised above, then you pay @@@2/2@@@ for the first round, @@@4/2@@@ for the second round, … @@@2^T/2@@@ for the @@@T@@@-th round. Altogether, you pay @@@2/2+4/2 \dots + 2^{T}/2 = 2^{T}-1@@@ before you get @@@2^T@@@ from me.

Wait, what? Your net earnings are always @@@2^T - (2^T-1)=1@@@. This is not fair, despite the fact that we made sure that each round was fair. This is a second paradox.

Note that even if I cheated you and used a coin that was heavily Heads-biased without letting you know (and still charging you the price for a fair game), you’d still end up with the guaranteed net earnings of @@@1@@@, because the coin will eventually land Tails. Strategy is very powerful!

The paradox is easy to settle. Remember you have the ability to pick a strategy: you select the amount we’re playing on and when to quit. I can’t do that. The calculation shows that the monetary value of the strategy chosen is @@@1@@@. Other strategies can be even better for you and, in fact, the ability to pick a strategy is priceless. But there’s a catch: let’s see why I won’t charge you for the ability to pick a strategy and why it doesn’t really matter.

A third paradox?

In order to be able to play @@@m@@@ rounds you need to have enough money in your pocket to cover losing all these games. The amount is @@@2/2+\dots + 2^m/2 = 2^m-1@@@. Now if your budget is @@@2^m-1@@@, then your net earnings will be @@@1@@@ if @@@T\le m@@@, and otherwise, you will lose everything, @@@2^m-1@@@. As a result, your expected net earnings are

Zero. No matter how much you started with. Wow! The strategy is irrelevant now. The game is fair again, not matter how big your budget is. This is why I the strategy is not such a great find (at least as long as I don’t go bankrupt before you do). Let’s call it a third paradox? Well, this one too can be settled quite easily. It’s actually a known theorem, called the optional stopping theorem. In our context, its weakest form says that any strategy in which the time to quit is bounded by a constant (any constant) preserves the fairness of the game: the expected net earnings will be zero. Can’t really game the system with finite resources.

Some mathematical theory

Let’s try to put everything into a nice mathematical theory that will generalize Game 2.

I’m running a sequence of games. In game @@@n@@@, you bet an amount @@@X_n\ge 0 @@@ and your net earnings are @@@X_n G_n@@@.

We will assume the following:

1. Games:
2. @@@E[G_n]=0@@@ for all @@@n@@@, that is, each game is fair; and
3. For each @@@n@@@, @@@G_{n+1}@@@ is independent of whatever happened before (this can be slightly relaxed).
4. You can apply a strategy:
5. @@@X_1@@@ is independent of @@@G_1@@@
6. For each @@@n@@@, @@@X_{n+1}@@@ and is a function of @@@X_1,\dots,X_{n}, G_1,\dots, G_{n}@@@. Of course it can also be deterministic.
7. For each @@@n@@@, @@@X_n \ge 0@@@ and @@@E[X_n]< \infty@@@.

The notion of strategy allows you to quit. How do you quit? Bet @@@0@@@ on the next game and all following games, so effectively you’r out. If @@@T@@@ is the round after which you quit, that is:

then since it is part of a strategy, we see that the event @@@\{T=n\}@@@ is a function of @@@X_1,\dots,X_n,G_1,\dots,G_n@@@.

In what follows, we assume that @@@T < \infty@@@ a.s.

Clearly, your net earnings up to quitting are

Let’s also define your accumulated loss:

Here is the amazing result, a special case of Doob’s Optional Stopping Theorem. It states that under a very reasonable assumption, strategy doesn’t matter: the game will remain fair.

• Introduction
• The classic paradox
  • St. Petersburg Paradox
• A more general game
  • Relation between the games
  • A second paradox
  • A third paradox?
• Some mathematical theory