Game Theory (Part 8)

Game Theory (Part 8)

John Baez

Last time we learned some rules for calculating probabilities. But we need a few more rules to get very far.

For example:

We say a coin is fair if it has probability 1/2 of landing heads up and probability 1/2 of landing tails up. What is the probability that if we flip two fair coins, both will land heads up?

Since each coin could land heads up or tails up, there are 4 outcomes to consider here:

$$(H,H), (H,T),$$ $$(T,H), (T,T)$$

It seems plausible that each should be equally likely. If so, each has probability 1/4. So then the answer to our question would be 1/4.

But this is plausible only because we're assuming that what one coin does doesn't affect that the other one does! In other words, we're assuming the two coin flips are 'independent'.

If the coins were connected in some sneaky way, maybe each time one landed heads up, the other would land tails up. Then the answer to our question would be zero. Of course this seems silly. But it's good to be very clear about this issue... because sometimes one outcome does affect another!

For example, suppose there's a 5% probability of rain each day in the winter in Riverside. What's the probability that it rains two days in a row? Remember that 5% is 0.05. So, you might guess the answer is

$$0.05 \times 0.05 = 0.0025$$

But this is wrong, because if it rains one day, that increases the probability that it will rain the next day. In other words, these outcomes aren't independent.

But if two outcomes are independent, there's an easy way to figure out the probability that they both happen: just multiply their probabilities! For example, if the chance that it will rain today in Riverside is 5% and the chance that it will rain tomorrow in Singapore is 60%, the chance that both these things will happen is

$$0.05 \times 0.6 = 0.03$$

or 3%, if these outcomes are independent. I could try to persuade that this is a good rule, and maybe I will... but for now let's just state it in a general way.

Independence

So, let's make a precise definition out of all this! Suppose we have two sets of outcomes, $$X$$ and $$Y.$$ Remember that $$X \times Y$$, the Cartesian product of the sets $$X$$ and $$Y$$, is the set of all ordered pairs $$(i,j)$$ where $$i \in X$$ and $$j \in Y$$:

$$X \times Y = \{ (i,j) : \; i \in X, j \in Y \}$$

So, an outcome in $$X \times Y$$ consists of an outcome in $$X$$ and an outcome in $$Y$$. For example, if

$$X = \{ \textrm{rain today}, \textrm{no rain today} \}$$

and

$$Y = \{ \textrm{rain tomorrow}, \textrm{no rain tomorrow} \}$$

then

$$X \times Y = \begin{array}{l} \{ \textrm{(rain today, rain tomorrow)}, \\ \textrm{(no rain today, rain tomorrow)}, \\ \textrm{(rain today, no rain tomorrow}, \\ \textrm{(no rain today, no rain tomorrow)} \} \end{array}$$

Now we can define 'independence'. It's a rule for getting a probability distribution on $$X \times Y$$ from probability distributions on $$X$$ and $$Y$$:

Definition. Suppose $$p$$ is a probability distribution on a set of outcomes $$X,$$ and $$q$$ is a probability distribution on a set of outcomes $$Y.$$ If these outcomes are independent, we use the probability distribution $$r$$ on $$X \times Y$$ given by

$$r_{(i,j)} = p_i q_j$$

People often call this probability distribution $$p \times q$$ instead of $$r$$.

Examples

Example 1. Suppose we have a fair coin. This means we have a set of outcomes

$$X = \{H, T \}$$

and a probability distribution $$p$$ with

$$\displaystyle{ p_H = p_T = \frac{1}{2} }$$

Now suppose we flip it twice. We get a set of four outcomes:

$$X \times X = \{(H,H), (H,T), (T,H), (T,T)\}$$

Suppose the two coin flips are independent. Then we describe the pair of coin flips using the probability measure $$r = p \times p$$ on $$X \times X,$$ with

$$\displaystyle{ r_{(H,H)} = p_H p_H = \frac{1}{4} }$$

$$\displaystyle{ r_{(H,T)} = p_H p_T = \frac{1}{4} }$$

$$\displaystyle{ r_{(T,H)} = p_T p_H = \frac{1}{4} }$$

$$\displaystyle{ r_{(T,T)} = p_T p_T = \frac{1}{4} }$$

So, each of the four outcomes — "heads, heads" and so on — has probability 1/4. This is fairly boring: you should have known this already!

But now we can do a harder example:

Example 2. Suppose we have an unfair coin that has a 60% chance of landing heads up and a 40% chance of landing tails up. Now we have a new probability distribution on $$X,$$ say $$q$$:

$$\displaystyle{ q_H = .6, \quad q_T = .4 }$$

Now say we flip this coin twice. What are the probabilities of the four different outcomes that can happen? Let's assume the two coin flips are independent. This means we should describe the pair of coin flips with a probability measure $$s = q \times q$$ on $$X \times X.$$ This tells us the answer to our question. We can work it out:

$$\displaystyle{ s_{(H,H)} = q_H q_H = 0.6 \times 0.6 = 0.36 }$$

$$\displaystyle{ s_{(H,T)} = q_H q_T = 0.6 \times 0.4 = 0.24 }$$

$$\displaystyle{ s_{(T,H)} = q_T q_H = 0.4 \times 0.6 = 0.24 }$$

$$\displaystyle{ s_{(T,T)} = q_T q_T = 0.4 \times 0.4 = 0.16 }$$

Puzzle 1. In this situation what is the probability that when we flip the coin twice it comes up heads exactly once?

Puzzle 2. In this situation what is the probability that when we flip the coin twice it comes up heads at least once?

For these puzzles you need to use what I told you in the section on 'Probabilities of subsets' near the end of Part 7.

Puzzle 3. Now suppose we have one fair coin and one coin that has a 60% chance of landing heads up. The first one is described by the probability distribution $$p,$$ while the second is described by $$q.$$ How likely is it that the first lands heads up and the second lands tails up? We can answer questions like this if the coin flips are independent. We do this by multiplying $$p$$ and $$q$$ to get a probability measure $$t = p \times q$$ on $$X \times X.$$ Remember the rule for how to do this:

$$t_{(i,j)} = p_i q_j$$

where each of $$i$$ and $$j$$ can be either $$H$$ or $$T.$$

What are these probabilities:

$$\displaystyle{ t_{(H,H)} = ? }$$

$$\displaystyle{ t_{(H,T)} = ? }$$

$$\displaystyle{ t_{(T,H)} = ? }$$

$$\displaystyle{ t_{(T,T)} = ? }$$

Puzzle 4. In this situation what is the probability that exactly one coin lands heads up?

Puzzle 5. In this situation what is the probability that at least one coin lands heads up?

Next time we'll go a lot further...