Probability Exercises

Model answers are in red.

Comments are in blue.

1) Problems 1 and 2 are classical probability problems going back centuries.

The Chevalier de Mere bets he can get a "6" in four rolls of a fair die.  If he get a "6" in four throws, you give him a dollar.  If he doesn’t, he gives you a dollar. 

Do you want to play? Yes No 

Explain your answer:

The chance of beating the Chevalier is equal to 625/1296. The Chevalier has a better chance of winning.

A complete answer should explain where the number 625/1296 comes from. It can be found several ways. Any one (or more) of the following explanations will do:

  1. If four dice are rolled (or a single die is rolled 4 times) there are 6*6*6*6 = 1296 ways for the four dice to come up. (The asterisk means multiply.) That's because on each die there are 6 faces. On the other hand, there are only 5*5*5*5 = 625 ways for all four dice to come up with numbers other than "6". All 1296 rolls are equally likely. There are 625 outcomes that favor you, so the chance of you winning is 625/1296.
  2. The chance of not rolling a 6 in one throw is 5/6. Now picture a bazillion people rolling dice. After the first roll, 5/6 of these people have not lost to the Chevalier. After the second roll, 5/6 of the survivors of the first roll have still not lost. That means (5/6) of (5/6) of the original bazillion remain. After the third throw (5/6) of these remain, etc. After four rolls, the proportion remaining is (5/6)*(5/6)*(5/6)*(5/6) = 625/1296.
  3. The probability of not getting a "6" in one roll is 5/6. So, the probability of not getting a "6" in four rolls is (5/6)*(5/6)*(5/6)*(5/6) = 625/1296.

The first way keeps the "experiment-outcome-event" thought-model out in the open. The second way emphasizes thinking about probability in terms of relative frequency. It's a different kind of thought-mdel (but hopefully you see the close relation to the first model). The third approach uses a rule for probability calculations, namely: If two events are independent, then the probability of both ocurring together is equal to the product of their probabilities.

Who would come out better off if we played this bet over and over and over? the Chevalier   Why?

In the long run, he will win more often. In fact, by the Law of Large Numbers, he will win in close to 671/1296 = approximately 51.77% of the time.

If we played 1,000 times, how much would I probably win (or lose)?

The Chevalier would be expected to win about 518 times. He would come out about 18 dollars ahead.


The Chevalier de Mere bets he can get a "double 6" in 24 throws of a pair of fair dice.  If he gets a "double 6" in 24 throws, you give him a dollar.  If he doesn't don’t, he gives you a dollar. 

For your information:

35^24 =11419131242070580387175083160400390625;

36^24 = 22452257707354557240087211123792674816

(35/36)^24 = 0.508596 (approximately)

(The symbol "^" means exponentiation; a^3 means a*a*a.)

Do you want to play? Yes No

Explain your answer:

3) This problem tests your understanding of the terms "experiment", "outcome" and "event".

a)  In problem 1), what is the experiment? Rolling four dice.

b)  What set would be most reasonable to take for the collection of outcomes?

The set of all 4-term sequences of the digits 1, 2, 3, 4, 5, 6. In other words the outcomes are: 1111, 1112, 1113, 1114, 1115, 1116, 1121, 1122, ..., 5656, 5661, 5662, 5663, 5664, 5665, 5666, 6666. (I skipped writing a whole bunch.)

c)  Are there other sets that would be reasonable candidates for the “set of outcomes”?  For example?

Actually, in this case, any other choice would be more clumsy or difficult to use. You could make up a symbol---say "N"---to stand for "any number but 6". Then the outcomes would look like: NNNN, NNN6, NN6N, NN66, N6NN, N6N6, N66N, N666, 6NNN, 6NN6, 6N6N, 6N66, 66NN, 66N6, 666N, 6666. But now these oucomes ARE NOT EQUALLY LIKELY. So it's much harder to think about them.

d)  Consider the set of all 4-term-sequences built out of the numbers 1, 2, 3, 4, 5, 6.  How many elements are there in this set? 1296

e)  How can the set in c) be viewed as the collection of outcomes for the experiment in 1)? Each sequence would tell the way each of the four dice came up. For example, NNN6 would mean that the first three dice got numbers other than 6 and the last got a 6.  Are the outcomes equally likely? No. The chance of NNNN is (5/6)^4. The chance of 6666 is (1/6)^4.

f)  Using the set in c) to represent outcomes, what is the event that problem 1) asks about?  How many outcomes are in this event?

4) This problem concerns another situation where the language of outcome and event is useful, but not obvious. Without explicitly saying so, this problems also introduces the ideas of "conditional probability" and "dependent events" and is related to tests of statistical significnce for a relationship between two boolean variables.

In a certain city ( very similar to East Baton Rouge), 33/100 of all registered voters are black and 67/100 are white.  Also, 3/10 of all registered voters are Republican and 7/10 are Democrat.  Moreover, 29/100 of all registered voters are BOTH white and Republican.  As you can see, almost all Republicans are white.

a)  How would you answer the following questions instinctively? In this city, is a Republican more likely white or black? In this city, is a Democrat more likely white or black? Is a white voter more likely Republican or Democrat?

b)  Draw a two-by-two table with rows labeled “Black” and “White” and columns labeled “Republican” and “Democrat”.  Fill in the squares with the percentages in each of the 4 categories.

row total
column total

c)  Use this to answer the three questions above.  Were your instincts correct?  If someone’s instincts on this were wrong, do you think it is a sign of prejudice?

d)  Viewing “choosing a voter at random” as an experiment, explain why the four boxes in the table from b) may be viewed as outcomes.  Taking this point of view, is “choosing a Democrat” an event or an outcome?

5) Though this is problem involves using tree diagrams to calculate probabilities, like problem 3, it is mostly about understanding and using the language of outcomes and events.

a)  You have a jar containing 3 red, 2 green, and one white marble.  You choose three marbles without replacing any into the jar.  Explain why it would be reasonable to think of (RRR, RRG, RGR, GRR, RRW, RWR, WRR, RGG, RGW, RWG, GRG, GRW, WRG, GGR, GWR, WGR, GGW, GWG, WGG) as the set of outcomes.

b)  Draw a tree diagram that shows how these outcomes can be achieved by three successive draws.  Label the branches with the appropriate probabilities, and use this to determine the probability of each of the outcomes.

c)  Are the outcomes equally likely?  Explain.

d)  What is the probability of each of the events:1) getting no red marbles, 2) choosing a green marble on the last draw, 3) leaving a marble of each color in the jar.

e)  Discuss variants of this problem.  For example, suppose I replace marbles?  Or, suppose I replace all marbles drawn with red ones.  What if I replace red marbles with green ones, greens with whites and whites with reds?