Understanding Dice

If you find probability analyses to be challenging, you are not alone. Many people find that their intuition can lead to errors when faced with a complicated scenario involving probabilities.

In this, I may be a typical specimen. Between the ages of 8 and 11, I spent beautiful afternoons, well, inside. Specifically, I gambled - black jack, roulette, but above all poker. Antes were 3 cents, so winning $5 was an amazing day. Usually, I won. I rarely bluffed and I had an experiential feel for the probabilities of different hands.

This intuition worked well for simple problems, but I lacked analytical tools. Courses on probability helped, but I found it very easy to make seemingly small mistakes that could lead to large errors.

It seems that even professional mathematicians are not immune. In 1990, the columnist Marilyn vos Savant published a puzzle called the Monty Hall problem. Even though she gave the right answer, many people, including mathematicians, told her she was wrong. Here is how the puzzle goes. Let’s see how you do.

A game show host, Monty Hall, always follows the same protocol. He shows three doors, D1, D2, and D3, to a contestant. Behind one is a valuable gift and Monty knows which one. The other doors have gags. The contestant chooses a door, say D1. Monty doesn’t open the door, but opens another door D2 that has a gag gift. Monty offers the contestant a choice to switch from D1 to the third door D3. Now, the question is: Should the contestant switch or not?

To analyze the situation, start by listing all configurations that have equal probabilities. Then analyze the choice for the contestant in each case. Here, V means a valuable gift and G means a gag gift.

Open table as spreadsheet

D1	D2	D3
V	G	G
G	V	G
G	G	V

If the contestant chooses the correct door to begin with, he wins by not switching. However, in the two other cases, he wins by switching. That is, in two out of three cases, switching is better. So, the contestant should switch. It’s as easy as that.

The strategy of writing down all the equi-probable configurations works in lots of situations. Let’s try a few more.

A challenger, Alice, throws a pair of fair dice, keeps them under an opaque cup and peeks at them. She tells you the total of the pair is 6. She offers you an even money bet: Tell me the maximum die value and I’ll pay you $100. Otherwise, you pay me $100. Do you take it?

Write down the configurations that have the equal probabilities.

5	1
4	2
3	3
2	4
1	5

As you can see, two out of five times, the maximum value is 5. Two out of five times, the maximum value is 4. One out of five times, the maximum value is 3. The probability of winning is too low for an even money payout. You should not take the bet.

Now she tells you that neither die is 5. She still offers you the same bet. Do you take it now?

At this point there are three equi-probable cases:

4	2
3	3
2	4

So, now you guess 4 and will win two times out of three.

Suppose that the two dice have different colors. One is red and one is green. Suppose now that Alice told you that the green die did not have the value 5. Then what are the equi-probable cases?

Open table as spreadsheet

Green	Red
4	2
3	3
2	4
1	5

Now, if you choose 4, your chances of winning are exactly 1/2.

Here is a harder one:

1. Suppose that the green die is unfair. In fact it has a 1/2 chance of being 3 and a 1/10 chance of being any other value. The red die is fair. Alice rolls an 8. She again offers you an even money bet to choose the maximum die value. Do you take it?

Solution to Understanding Dice

1. Suppose that the green die is unfair. In fact it has a 1/2 chance of being 3 and a 1/10 chance of being any other value. The red die is fair. Alice rolls an 8. She again offers you an even money bet to choose the maximum die value. Do you take it?

Again, create a table of equi-probable situations. In this case, we will start with the equi-probable situations including those that don’t lead to 8.

Open table as spreadsheet

Green	Red
1	1, 2, 3, 4, 5, 6
2	1, 2, 3, 4, 5, 6
3	1, 2, 3, 4, 5, 6
3	1, 2, 3, 4, 5, 6
3	1, 2, 3, 4, 5, 6
3	1, 2, 3, 4, 5, 6
3	1, 2, 3, 4, 5, 6
4	1, 2, 3, 4, 5, 6
5	1, 2, 3, 4, 5, 6
6	1, 2, 3, 4, 5, 6

We represented 3 five times because it is five times as likely as the other green values, which are all equally likely. Now cross out those configurations that are impossible because we know the total is 8. That leaves the following:

Open table as spreadsheet

Green	Red
2	6
3	5
3	5
3	5
3	5
3	5
4	4
5	3
6	2

In six of these nine cases, 5 is the maximum value. So your chances of winning are 6/9 = 2/3.