Module 1 · Lesson 1

Probability: The problem of points: how Pascal and Fermat put a number on chance

You'll solve the 17th-century puzzle that launched probability theory: how to divide the stakes of a game that stops before anyone has won. You'll learn to count outcomes, compute expected values, and see why chance evens out over many trials.

What you will be able to do

Calculate simple probabilities and expected values, and explain what the law of large numbers guarantees.

0% of this lesson · 9 items · 1h 3m total · 48m without the optional problem set

Contents of this lesson9 items

ReadingHow Should You Split the Pot of an Unfinished Game?3 min
ReadingFrom Dice to Axioms: How Probability Theory Took Shape4 min
ReadingSample Spaces and Counting Equally Likely Outcomes4 min
ReadingIndependence, Conditional Probability, and Expected Value4 min
ReadingThe Law of Large Numbers4 min
Guided ReadingWorked Example: Solving the Problem of Points by Listing Futures7 min
Guided ConversationIs This Game Fair?12 min
Problem Set · optionalChance, Counting, and Expected Value15 min
Knowledge CheckThe problem of points: how Pascal and Fermat put a number on chance10 min

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Reading 3 min Done

How Should You Split the Pot of an Unfinished Game?

Here is a puzzle you can try before reading on. Two players put equal money into a pot and agree that the first to win three rounds takes it all. Each round is a fair contest, so either player is equally likely to win it. The game stops when player A leads 2 rounds to 1, and it can't be finished. How should they split the pot?

Take a minute with it. Most people reach for one of two answers.

The first answer is to split by the score. A has won two rounds and B has won one, so A takes two-thirds and B takes one-third. That sounds fair until you look at what the score means. A doesn't need twice as much luck as B to win. A needs one more round, and B needs two. The score records the past, but the pot belongs to whoever would have won in the future.

The second answer is to split evenly, since nobody has won yet. That ignores A's lead entirely. If the score had been 2 to 0, an even split would feel wrong to almost anyone, and 2 to 1 is a smaller version of the same problem.

So the fair share depends on something that never happened: the rounds that would have been played. That's the hard part. You have to reason exactly about a future that is uncertain by its nature, and then turn that reasoning into a number of coins.

This is the problem of points, and versions of it had circulated for a long time before anyone settled it. In the summer of 1654, Blaise Pascal and Pierre de Fermat took it up in a short exchange of letters. They also discussed a dice question: how many throws of a pair of dice you need before you can expect a double six. Out of that correspondence, as the MacTutor biography of Pascal puts it, came "the foundation for the theory of probability" (O'Connor and Robertson 1996).

Their key move was to stop looking at the rounds already played and start counting the ways the rest of the game could go. If you can list the possible futures, and if those futures are equally likely, you can count how many of them end in a win for A. That count becomes A's share.

Try it before you look anything up. How many more rounds could the game take at most? What are the possible results of those rounds? In how many of them does A win?

What the branch offers is a way to put exact numbers on events that haven't happened yet and may never happen. Probability doesn't tell you who would have won the unfinished game. It tells you what share of the pot each player's position was worth when play stopped, and the same reasoning now helps price insurance and runs inside computer programs.

References

O'Connor, J. J., and E. F. Robertson. 1996. "Blaise Pascal." MacTutor History of Mathematics Archive, University of St Andrews. Last updated December 1996.
- Free: MacTutor: Pascal

Reading 4 min Done

From Dice to Axioms: How Probability Theory Took Shape

Introduction

Probability is now a branch of mathematics with the same footing as geometry or algebra. It didn't start that way. It grew over about four centuries from practical questions about dice and wagers into a theory built on a short list of rules.

This reading traces that growth through five contributors, from a 16th-century study of dice to a 20th-century set of axioms.

Questions from the Gaming Table

The earliest questions came from gambling. A game of chance is a game, such as dice, in which the outcome depends on luck rather than skill. Players wanted to know which bets were favorable and how to divide stakes fairly.

Girolamo Cardano (1501–1576), an Italian physician and mathematician, gambled for years and wrote a book on games of chance that was probably finished by 1563. It wasn't published until 1663, almost a century later. MacTutor describes it as the first study of questions such as dice rolling built on the idea that scientific principles govern how likely an outcome is (O'Connor and Robertson 1998a). Cardano also considered the problem of dividing the stakes of an unfinished game, as did Luca Pacioli and Niccolò Tartaglia (O'Connor and Robertson 1996).

The Letters of 1654

That stakes question became known as the problem of points: how to divide the stakes fairly between players when a game of chance stops before anyone has won. In the summer of 1654, Blaise Pascal and Pierre de Fermat worked on it in a short exchange of letters. They also took up a dice problem that Cardano had studied: how many times you must throw a pair of dice before you can expect a double six. They solved the problem of points for two players, though not for three or more (O'Connor and Robertson 1996).

What made the letters a turning point was the method. Instead of arguing about what seemed fair, both men calculated what each player's position was worth by reasoning about the outcomes still possible. MacTutor credits the exchange with laying the foundation of probability theory (O'Connor and Robertson 1996), though the questions it answered were older.

The First Book and the First Great Theorem

Christiaan Huygens, a Dutch scientist, visited Paris in 1655 and learned of the Pascal–Fermat work there. Back in the Netherlands he wrote a short treatise, De Ratiociniis in Ludo Aleae (On Reasoning in Games of Chance), dated 1657. MacTutor calls it "the first printed work" on the subject (O'Connor and Robertson 1997).

Jacob Bernoulli, a Swiss mathematician, took the subject much further. His Ars Conjectandi (The Art of Conjecturing) was still incomplete when he died in 1705 and was published in Basel in 1713. It contains his law of large numbers, which shows that over many repetitions the fraction of times an event happens settles toward its probability (O'Connor and Robertson 1998b). That result connected the numbers of the theory to frequencies that anyone can observe.

Axioms in 1933

By the early 20th century, probability was used widely in science, but it lacked the kind of foundation that geometry had. David Hilbert's sixth problem, from his famous list of open problems, asked mathematicians to set out axioms for the sciences in which mathematics plays an important part. In 1933 Andrey Kolmogorov, a Russian mathematician, published a monograph that built probability theory rigorously from a few fundamental axioms, which are basic statements accepted without proof from which everything else is derived. MacTutor compares the approach to "Euclid's treatment of geometry" and credits the monograph with a major part in answering Hilbert's problem for probability (O'Connor and Robertson 1999).

Kolmogorov's axioms say, roughly, that a probability is a number from 0 to 1 that measures how likely an event is, that something certain has probability 1, and that the probabilities of events that can't happen together add up. Nearly everything in a modern probability course follows from rules like these.

Who	When	Contribution
Girolamo Cardano	Book finished by about 1563, published 1663	Early study of dice and chance; considered the division of stakes
Pascal and Fermat	Letters of 1654	Solved the problem of points for two players by calculation
Christiaan Huygens	1657	First printed work on probability
Jacob Bernoulli	Ars Conjectandi, 1713	Law of large numbers
Andrey Kolmogorov	1933	Axioms that put probability on a rigorous footing

Conclusion

Probability theory has no single founder. Its questions came from games of chance, its first solutions from the 1654 letters, its first book from Huygens, its first deep theorem from Bernoulli, and its modern foundations from Kolmogorov's axioms. Those axioms remain the standard starting point, though, as philosophers point out, they don't settle what a probability means.

Key Terms

Game of chance: A game, such as dice, in which the outcome depends on luck rather than skill.
Problem of points: The question of how to divide the stakes fairly between players when a game of chance stops before anyone has won.
Axiom: A basic statement accepted without proof, from which the rest of a theory is derived.
Probability: A number from 0 to 1 that measures how likely an event is, where 0 means impossible and 1 means certain.

References

O'Connor, J. J., and E. F. Robertson. 1996. "Blaise Pascal." MacTutor History of Mathematics Archive, University of St Andrews. Last updated December 1996.
- Free: MacTutor: Pascal
O'Connor, J. J., and E. F. Robertson. 1997. "Christiaan Huygens." MacTutor History of Mathematics Archive, University of St Andrews. Last updated February 1997.
- Free: MacTutor: Huygens
O'Connor, J. J., and E. F. Robertson. 1998a. "Girolamo Cardano." MacTutor History of Mathematics Archive, University of St Andrews. Last updated June 1998.
- Free: MacTutor: Cardano
O'Connor, J. J., and E. F. Robertson. 1998b. "Jacob (Jacques) Bernoulli." MacTutor History of Mathematics Archive, University of St Andrews. Last updated September 1998.
- Free: MacTutor: Jacob Bernoulli
O'Connor, J. J., and E. F. Robertson. 1999. "Andrey Nikolaevich Kolmogorov." MacTutor History of Mathematics Archive, University of St Andrews. Last updated January 1999.
- Free: MacTutor: Kolmogorov

Reading 4 min Done

Sample Spaces and Counting Equally Likely Outcomes

Introduction

When Pascal and Fermat discussed dice in 1654, one of their questions was how many throws of a pair of dice you need before you can expect a double six (O'Connor and Robertson 1996). Questions like that rest on one skill: listing every way an experiment can turn out, then counting.

This reading covers sample spaces, counting equally likely outcomes, a common counting mistake, and the scale that every probability sits on.

Listing Everything That Can Happen

An outcome is one possible result of an experiment, such as rolling a die or tossing a coin. The sample space is the complete list of all possible outcomes. Writing it out is the first step in almost every probability calculation.

Here are three sample spaces, from small to larger:

Experiment	Sample space	Number of outcomes
Roll one die	1, 2, 3, 4, 5, 6	6
Toss two coins	HH, HT, TH, TT	4
Roll two dice	Every pair, from (1, 1) to (6, 6)	36

For two coins, HT and TH count as separate outcomes. One means the first coin landed heads and the second tails, and the other means the reverse. For two dice, think of one die as red and one as blue. The red die has 6 possible faces, and for each of them the blue die has 6, so there are $6 \times 6 = 36$ pairs.

An event is a set of outcomes you care about. "The die shows an even number" is the event made of the outcomes 2, 4, and 6.

Counting When Outcomes Are Equally Likely

Outcomes are equally likely when none of them is more likely to happen than any other, as with the faces of a fair die. When that's true, a probability is a count:

P(\text{event}) = \frac{\text{number of outcomes in the event}}{\text{total number of outcomes}}

Here $P(\text{event})$ is read "the probability of the event." For an even number on one die, the event holds 3 of the 6 outcomes, so the probability is $3/6 = 1/2$ .

Now try two dice. Which pairs give a sum of 7? There are six: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). So the chance of rolling a 7 is $6/36$ , which is $1/6$ . A double six, the pair Pascal and Fermat asked about, is a single outcome, so its chance on one throw is $1/36$ .

The Trap: Outcomes That Aren't Equally Likely

The formula works only when the outcomes you count are equally likely. A tempting mistake is to count the wrong things.

The possible sums of two dice run from 2 to 12, which is eleven values. You might conclude that each sum has a chance of $1/11$ . That's wrong, because the sums aren't equally likely. Only one pair, (1, 1), gives a sum of 2, but six pairs give a 7. The table shows how many of the 36 equally likely pairs produce each sum.

Sum	2	3	4	5	6	7	8	9	10	11	12
Number of pairs	1	2	3	4	5	6	5	4	3	2	1

So 7 is the most likely sum, with probability $6/36 = 1/6$ , while 2 and 12 each have probability $1/36$ . The fix is always the same. Go back to outcomes that really are equally likely, here the 36 pairs, and count those.

The same trap appears with two coins. "Two heads," "one head," and "no heads" are three results, but they aren't equally likely. One head can happen two ways, HT or TH, so its chance is $2/4 = 1/2$ , while two heads has a chance of $1/4$ .

The Scale from 0 to 1

A probability is a number from 0 to 1 that measures how likely an event is, where 0 means impossible and 1 means certain. Rolling a 13 with two dice has probability $0/36 = 0$ . Rolling a sum between 2 and 12 has probability $36/36 = 1$ . Everything else falls in between, and you can write it as a fraction, a decimal, or a percentage: $1/6$ is about 0.167, or about 16.7 percent.

That scale gives you a quick check. If a calculation produces a probability below 0 or above 1, something has gone wrong in the counting.

Conclusion

Counting is the oldest method in probability and still one of the most useful. It depends on two steps: list a sample space of equally likely outcomes, then count the outcomes in your event. Most errors in simple problems come from skipping the first step and counting results that aren't equally likely.

Key Terms

Outcome: One possible result of an experiment, such as one face of a rolled die.
Sample space: The complete list of all possible outcomes of an experiment.
Event: A set of outcomes you're interested in, such as rolling an even number.
Equally likely: Describes outcomes when none of them is more likely to happen than any other.
Probability: A number from 0 to 1 that measures how likely an event is, where 0 means impossible and 1 means certain.

References

O'Connor, J. J., and E. F. Robertson. 1996. "Blaise Pascal." MacTutor History of Mathematics Archive, University of St Andrews. Last updated December 1996.
- Free: MacTutor: Pascal

Reading 4 min Done

Independence, Conditional Probability, and Expected Value

Introduction

Gamblers in the 16th and 17th centuries asked questions that a single count can't answer. What is the chance of at least one six in several rolls? What is a bet worth before it's played? Cardano, and later Pascal and Fermat, worked on dice questions of this kind (O'Connor and Robertson 1996).

This reading covers three tools for such questions: independence, conditional probability, and expected value.

Independent Events Multiply

Two events are independent events when one happening doesn't change the chance of the other. A coin has no memory, so heads on one toss doesn't affect the next toss. Separate rolls of a die are independent in the same way.

For independent events, you find the chance that both happen by multiplying their probabilities. The chance of two heads in two tosses is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ , which matches the count of one outcome, HH, out of four.

Multiplication also answers a classic gambling question: what is the chance of at least one six in four rolls of a die? Counting the ways to get one, two, three, or four sixes is slow. It's faster to find the chance of the opposite, no six at all. Each roll misses six with probability $5/6$ , and the rolls are independent, so

P(\text{no six in four rolls}) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296} \approx 0.48

Then the chance of at least one six is

1 - \left(\frac{5}{6}\right)^4 = \frac{671}{1296} \approx 0.52

That's slightly better than even, about 52 percent.

Not every pair of events is independent. Suppose you draw two cards from a standard deck of 52 without putting the first one back. If the first card is an ace, only 3 aces remain among 51 cards, so the first draw changes the chances for the second. Multiplying the two original chances would then give the wrong answer.

Updating on New Information

Conditional probability is the chance of one event given that another has already happened. It answers questions like this: once you know something, how should the chance change?

Take two dice. Before you look, the chance of a double (both dice the same) is $6/36 = 1/6$ . Now suppose you're told only that the sum is at least 10. That leaves six possible pairs: (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), and (6, 6). Two of them are doubles. So the chance of a double, given a sum of at least 10, is $2/6 = 1/3$ .

The new information shrank the sample space, the list of possible outcomes, and you counted again inside the smaller list. For independent events the information changes nothing. Knowing that the first toss was heads leaves the chance of heads on the second toss at $1/2$ .

What a Bet Is Worth

Expected value is the average result you would get per play if you repeated a game or bet many times. To compute it, multiply each possible payoff by its probability and add up the results.

Here is a bet. You pay $1 to roll a die. If it shows six, you get $5 back. Your net result is a gain of $4 with probability $1/6$ , or a loss of $1 with probability $5/6$ . The expected value is

4 \times \frac{1}{6} + (-1) \times \frac{5}{6} = \frac{4}{6} - \frac{5}{6} = -\frac{1}{6} \approx -0.17

On average, you lose about 17 cents per play. No single play loses 17 cents, since each play either wins $4 or loses $1. The expected value describes the average over many plays. Over 600 plays, for example, you would expect to be down about $600 \times \frac{1}{6} = 100$ dollars, though your actual total would land somewhat above or below that.

A fair game is one whose expected value is zero, so neither side gains on average. This bet would be fair if a six paid $6 back, because then the net gain of $5 times $1/6$ would exactly balance the loss of $1 times $5/6$ .

Tool	Question it answers	How to compute
Independence	Do these events affect each other?	If not, multiply their probabilities
Conditional probability	How does the chance change once I know something?	Count again inside the outcomes that remain possible
Expected value	What is a bet worth on average?	Multiply each payoff by its probability and add

Conclusion

These three tools carry most everyday reasoning about chance. Independence lets you combine events, conditional probability lets you update when you learn something, and expected value tells you what a bet or decision is worth on average. All three still depend on listing outcomes correctly.

Key Terms

Independent events: Events where one happening doesn't change the chance of the other, such as separate tosses of a coin.
Conditional probability: The chance of one event given that another event has already happened.
Expected value: The average result per play over many repetitions, found by multiplying each payoff by its probability and adding.
Fair game: A game whose expected value is zero, so neither side gains on average.

References

O'Connor, J. J., and E. F. Robertson. 1996. "Blaise Pascal." MacTutor History of Mathematics Archive, University of St Andrews. Last updated December 1996.
- Free: MacTutor: Pascal

Reading 4 min Done

The Law of Large Numbers

Introduction

Toss a fair coin ten times and you might get seven heads. Toss it ten thousand times and you'll almost never get seven thousand. Jacob Bernoulli turned that familiar pattern into a theorem.

This reading covers what the law of large numbers says, what it guarantees, what it doesn't, and why it counts as a landmark of the field.

Bernoulli's Result

Jacob Bernoulli, a Swiss mathematician, worked on the result for years. MacTutor dates his law to about 1689, though it reached print only in 1713 in Ars Conjectandi (The Art of Conjecturing), published eight years after his death (O'Connor and Robertson 1998b).

The law concerns relative frequency: the fraction of trials in which an event happens. If you toss a coin 100 times and get 54 heads, the relative frequency of heads is $54/100 = 0.54$ . The law of large numbers says that as an experiment is repeated many times under the same conditions, the relative frequency of an event settles toward its probability. For a fair coin, the fraction of heads settles toward $1/2$ .

What It Guarantees

The law is a statement about the long run, meaning a very large number of repetitions. The more tosses you make, the less room the fraction of heads has to wander from $1/2$ .

The numbers show how fast the settling happens. With 10 tosses of a fair coin, the fraction of heads lands between 0.3 and 0.7 about 89 percent of the time, so values like 0.3 or 0.7 are common. With 10,000 tosses, the fraction lands between 0.49 and 0.51 about 96 percent of the time.

Bernoulli's version goes further than "it settles eventually." It says that for any margin you choose, however small, you can make the chance of landing within that margin as close to certain as you like by making enough trials.

What It Doesn't Guarantee

The law says nothing about the next toss. Suppose a fair coin lands tails five times in a row. Many people feel that heads is now "due." It isn't. The tosses are independent, so the chance of heads on the next one is still $1/2$ . Believing otherwise is called the gambler's fallacy: the mistaken belief that after a run of one result, the other result becomes more likely, as if chance must catch up.

How does the fraction settle, then, if nothing pushes it back? The answer is dilution rather than correction. Five extra tails in the first ten tosses is a large share of ten. After 10,000 tosses, those same five tails are a tiny share, swamped by everything that came after. The count of heads minus tails can stay far from zero, and often grows, even as the fraction approaches $1/2$ .

The law says	The law doesn't say
The fraction of heads approaches $1/2$ over many tosses	Heads becomes more likely after a run of tails
Large samples very likely land close to the probability	Small samples land close to the probability
The fraction's wandering shrinks as trials grow	The difference between heads and tails shrinks to zero

Why It Matters

Before Bernoulli, a probability was a number from a calculation, such as $1/6$ for a six. The law connects that number to something you can observe: how often the event happens over many trials. It's the bridge between the theory and the world.

The bridge carries real weight. A casino offers games whose expected value per play slightly favors the house. On any one play the house may lose, but across many plays the house's average gain per play settles toward that expected value. An insurer works the same way across many policies. Neither can predict one result, but both can plan on the average of many.

Conclusion

The law of large numbers guarantees that long-run frequencies approach probabilities, and nothing more. It doesn't make short runs balance out, and it doesn't make a coin owe you anything. It does make probability something you can test against observation.

Key Terms

Relative frequency: The fraction of trials in which an event happens, such as 54 heads in 100 tosses, or 0.54.
Law of large numbers: The result that as an experiment is repeated many times, the relative frequency of an event settles toward its probability.
Long run: A very large number of repetitions of an experiment.
Gambler's fallacy: The mistaken belief that after a run of one result, the other result becomes more likely, as if chance must catch up.

References

O'Connor, J. J., and E. F. Robertson. 1998b. "Jacob (Jacques) Bernoulli." MacTutor History of Mathematics Archive, University of St Andrews. Last updated September 1998.
- Free: MacTutor: Jacob Bernoulli

Guided Reading 7 min Done

Worked Example: Solving the Problem of Points by Listing Futures

Introduction

Two players stop a game before anyone has won. How should they divide the pot? This question, the problem of points, is the puzzle Blaise Pascal and Pierre de Fermat solved in their letters of 1654 (O'Connor and Robertson 1996).

This worked example solves one case step by step, using the method of listing possible futures that runs through Fermat's side of the correspondence. It then checks the answer a second way, close to the route Pascal took.

The Problem

Two players, A and B, each put the same amount into a pot. They agree that the first to win three rounds takes the whole pot. Each round is a fair contest: A and B are equally likely to win it, and no round affects any other. The game stops with A leading 2 rounds to 1, and it can't be resumed.

You're asked what share of the pot each player should take. "Fair" here has a definite meaning. Each player should receive the share of the pot that matches their chance of winning, had the game been played out.

So the real question is a probability question. What is A's chance of winning from a score of 2 to 1?

Working It Through

Step 1: Count how many rounds remain at most

A needs one more win. B needs two. Ask how long the game could last. If A wins the next round, the game ends at once. If B wins the next round, the score is 2 to 2, and one more round decides it. So at most two more rounds would settle the game.

This step matters because it sets the size of the list you need to make. You never need to think about a third round, because one player or the other will have three wins by then.

Step 2: List every way two more rounds could go

Write each future as a pair of letters, showing who wins the first extra round and then who wins the second. There are four possibilities:

AA: A wins both
AB: A wins the first, B wins the second
BA: B wins the first, A wins the second
BB: B wins both

Each round is a 50–50 contest, and the rounds don't affect each other. That makes the four futures equally likely: each has a chance of $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ . These four futures form the sample space, the complete list of possible outcomes.

Step 3: Mark the winner in each future

Go through the list and ask who reaches three wins first.

Future	Who wins the game	Why
AA	A	A's first extra win makes 3
AB	A	A's first extra win makes 3
BA	A	B ties at 2 to 2, then A wins
BB	B	B wins twice and reaches 3

A wins in three of the four futures. B wins only in BB.

Step 4: Turn the count into shares

Because the four futures are equally likely, A's chance of winning is the number of futures A wins divided by the total:

P(\text{A wins}) = \frac{3}{4}

B's chance is $1/4$ . So A should take three-quarters of the pot and B one-quarter, a split of 3 to 1.

In money terms, if each player staked 32 coins, the pot is 64 coins. A takes $\frac{3}{4} \times 64 = 48$ coins and B takes 16. Pascal's letter of July 29, 1654, works through this same position, with each player having staked 32 pistoles (gold coins) in a game "in three throws," meaning first to three wins, and gives the leader 48 pistoles and the other player 16 (Fermat and Pascal [1654] 1929, Pascal to Fermat, July 29, 1654).

Step 5: Notice the trick that made the counting work

Look again at AA and AB. In both, the game would really end after the first extra round, because A would already have three wins. The second round would never be played. So why list it?

Listing it is what makes the outcomes equally likely. If you listed only the futures that really happen, you would get three: "A wins at once," "B then A," and "B then B." Those three aren't equally likely. "A wins at once" has a chance of $1/2$ , while each of the others has a chance of $1/4$ . Counting them as one each would give A two chances in three, which is wrong.

By pretending that both rounds are always played, you split "A wins at once" into AA and AB, each with chance $1/4$ . Now every item on the list has the same chance, and counting works. The extra round changes nothing about who wins. It only makes the list fair to count.

The two men discussed this assumption in their letters. Fermat, taking up Pascal's own word, called the extension of the game "this fiction" and wrote that it serves "only to make the rule easy" and to make all the chances equal (Fermat and Pascal [1654] 1929, Fermat to Pascal, September 25, 1654).

Key Considerations

Checking the answer another way. You can reach the same split without imagining any extra rounds. Look at the very next round. If A wins it, A takes the whole pot. If B wins it, the score is 2 to 2, and the players are now even, so each deserves half. A is sure of at least half the pot either way. The other half depends on a single fair round, so A should get half of that half as well. A's share is $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ , matching the count. This is close to the way Pascal reasoned in his letter of July 29, 1654. He started from what is certain and split the uncertain part evenly, rather than listing combinations (Fermat and Pascal [1654] 1929, Pascal to Fermat, July 29, 1654). He called Fermat's method of combinations "very just" for two players, although the two men argued over how it extends to three players (Fermat and Pascal [1654] 1929, Pascal to Fermat, August 24, 1654).

The common mistake: splitting by the score. The most natural wrong answer splits the pot in proportion to rounds already won: 2 to 1, giving A two-thirds and B one-third. It's a mistake because the score describes the past, and the pot goes to whoever would win in the future. What matters is how many wins each player still needs: one for A, two for B. A score of 2 to 1 in a race to three is worth 3 to 1, not 2 to 1. The same score in a longer race, say first to ten, would be worth much less of an edge, which shows that the past score alone can't be the measure.

Two conditions behind the method. The listing method relies on the rounds being fair and independent. If A were the stronger player, the four futures wouldn't be equally likely, and you'd need to weight each one by its probability instead of counting.

Summary

The fair split depends on each player's chance of winning from the current position, and you can find that chance by listing the equally likely ways the game could finish.

Future Winner
AA A
AB A
BA A
BB B
A wins in 3 of 4 equally likely futures, so A's chance is 3/4. Split the pot 3 to 1: three-quarters to A, one-quarter to B.

Future	Winner
AA	A
AB	A
BA	A
BB	B

The list includes the second round even when the first would end the game, so that all four futures are equally likely.
The answer checks out by a second route: A is sure of half the pot and has an even chance at the other half, giving $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ .
Splitting by the score so far (2 to 1) gives A only two-thirds and undervalues A's position.

References

Fermat, Pierre de, and Blaise Pascal. (1654) 1929. "Fermat and Pascal on Probability." Translated by Vera Sanford. In A Source Book in Mathematics, edited by David Eugene Smith. New York: McGraw-Hill.
- Free: University of York: Fermat and Pascal on Probability (PDF)
O'Connor, J. J., and E. F. Robertson. 1996. "Blaise Pascal." MacTutor History of Mathematics Archive, University of St Andrews. Last updated December 1996.
- Free: MacTutor: Pascal

Guided Conversation 12 min Done

Is This Game Fair?

In this conversation, you'll design a small coin or dice game of your own, with a price to play and a prize. Then you'll work out your chance of winning and what the game is worth on average, with hints when you want them. You'll leave knowing whether your game is fair and what would happen to a player over many plays.

You'll have this conversation with an AI assistant, using your own account. Choose a button to open a new chat with the prompt already filled in, then press send to start. If the chat opens empty, copy the prompt and paste it in.

Run this conversation in whichever assistant you already use:

Open in Claude Open in ChatGPT

Claude desktop app

To use another LLM, simply copy and paste the prompt into its chat window.

Show the full prompt (it lists misreadings to watch for, so skip it if you would rather come to the conversation fresh)

Guided Conversation: Is This Game Fair? (about 12 minutes)

Note to the learner: press send to start. Everything below is facilitator guidance for the AI. It lists misconceptions to watch for, so skip it if you'd rather come to the conversation fresh.

Please facilitate a coached problem session with me. I'm an adult who is comfortable with basic high-school algebra but may be rusty or uneasy with math, exploring what probability is and whether to study it further. I'm studying simple probabilities and expected values. Follow this guidance for the whole conversation.

GOAL
Help me calculate the chance of winning and the expected value for a simple game I design, and decide whether it's fair.

HOW TO RUN THE CONVERSATION
- Ask one question at a time, then wait for my reply. Keep each of your turns under about 120 words.
- Don't lecture. Explain a point only when I need it to continue, then return to my game.
- Be curious, collegial, and patient. Define terms briefly on first use. Welcome disagreement when I give a reason.
- Coach, don't solve. Ask for my approach before any hint. Give one hint at a time, and never state an answer before I've attempted that step.
- Write math in plain text, such as 1/6 or (5/6)^4.
- Plain conversation only: don't search the web or create files or documents.
- Aim for about 12 minutes, mostly on topics 2 and 3. If my replies are brief, offer one concrete game: pay 1 dollar, toss two coins, get 3 dollars back if both land heads. If I seem uncertain, shorten the conversation to 5-7 minutes. Always reach the final topic.
- Start now. Open with one or two warm sentences: this is a conversation, not a quiz; a clear method matters more than a fast answer; I can ask you to clarify anything. Then ask me to describe a simple game with coins or dice, a cost to play, and a prize.

TOPICS, IN ORDER
1. Design the game. Help me pin down what is tossed or rolled (no more than two coins or two dice), what counts as a win, the cost, and the prize. Follow up: ask me for my net gain on a win and net loss on a loss.
2. Outcomes and the chance of winning. Ask me to list the sample space, the complete list of possible outcomes, and to check that the outcomes are equally likely. Then ask me to count the winning outcomes and find the probability. Follow up: is it between 0 and 1?
3. Expected value and fairness. Ask how I'd find the average result per play. Let me try, then guide me to multiply each net payoff by its probability and add. Ask whether the game is fair (expected value 0) and, if not, what prize would make it fair.
4. Closing. Ask what would happen to a player's total winnings over 1,000 plays, and why. Help me say it in one sentence with a number. This is something I can take into practice problems on probability and expected value.

CORRECT METHOD AND ANSWER
- Method: list equally likely outcomes; P(win) = winning outcomes / total outcomes. Use net payoffs (prize minus cost for a win; minus the cost for a loss). Expected value = net gain x P(win) + net loss x P(lose). Fair means expected value 0. Over n plays the expected total is about n x expected value, and actual totals vary around it.
- Two coins: HH, HT, TH, TT, each 1/4. Two dice: 36 equally likely pairs; a sum of 7 comes from 6 pairs, so P(7) = 1/6. Independent events multiply: P(two heads) = 1/2 x 1/2 = 1/4.
- Reference game: pay 1 dollar, toss two coins, get 3 dollars back on two heads. P(win) = 1/4. Expected value = 2 x 1/4 - 1 x 3/4 = -1/4, a loss of 25 cents per play on average. Paying back 4 dollars would make it fair. Over 1,000 plays, a player should expect to lose about 250 dollars.
- Favorable over total works only for equally likely outcomes. Expected value is a long-run average, not the result of one play. Past results don't change future independent results.

MISCONCEPTIONS TO CORRECT GENTLY
When one appears, name the accurate version briefly, then return to my game.
- "After five tails, heads is due": each toss is independent, so heads stays at 1/2.
- "A positive expected value guarantees a win": it's a long-run average; a single play or a short run can lose.
- "All sums of two dice are equally likely": 7 is the most likely sum (6 of 36 pairs); 2 and 12 come from 1 pair each.
- Using the prize instead of the net gain: subtract the cost to play.

LIMITS
- Don't encourage gambling or suggest real bets, and don't discuss real casinos, lotteries, or betting strategies.
- Don't compute my answers for me.
- Keep to coins and dice, counting, multiplying, and expected value.

TO FINISH
After I state what happens over 1,000 plays, close in one short turn:
- Affirm one specific thing I worked out, in my own words where possible.
- Suggest one or two next steps that fit how the conversation went. Possible steps: change the prize to make my game exactly fair; try a two-dice version; find the chance of at least one six in four rolls; review why sums of two dice aren't equally likely.
- Restate my result on its own line, labeled "My result", so I can copy it.

Problem Set 15 minOptional Done

Chance, Counting, and Expected Value

Overview

These five problems give you practice with the core moves of probability: listing outcomes, counting the ones you want, and working out what a bet is worth on average. They start with a single coin and build to the unfinished-game puzzle from 1654. The set is optional, and the work stays with you.

As you go, list or count the outcomes for each problem, compute each probability or expected value, and then compare your work with the answers and methods given.

Problems

Problem 1

What is the probability of heads on one toss of a fair coin?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 1Hide the solution

Answer: 1/2 (or 0.5)

Method:

List the outcomes: heads or tails. That's 2 equally likely outcomes.
Count the favorable outcomes: 1 (heads).
Divide: $1/2$ .

Most common error: Giving the answer as 1 or as 50 instead of 1/2. A probability is a number from 0 to 1. Fifty percent is the same value written as a percentage.

Problem 2

You roll two fair dice. What is the probability that their sum is 7?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 2Hide the solution

Answer: 1/6 (about 0.167)

Method:

List the sample space as ordered pairs, first die then second die. There are $6 \times 6 = 36$ equally likely pairs.
Find the pairs that sum to 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). That's 6 pairs.
Divide: $6/36 = 1/6$ .

Most common error: Treating the eleven possible sums, 2 through 12, as equally likely and answering 1/11. The sums aren't equally likely. Only one pair makes 2, but six pairs make 7.

Problem 3

You roll a fair die four times. What is the probability of getting at least one six? Give your answer as a fraction and as a decimal.

Once you've tried the problem, check your work against the solution.Show the solution to Problem 3Hide the solution

Answer: 671/1296, about 0.518

Method:

Find the chance of the opposite event, no six at all. Each roll misses six with probability $5/6$ .
The rolls are independent, so multiply: $(5/6)^4 = 625/1296$ .
Subtract from 1: $1 - 625/1296 = 671/1296 \approx 0.518$ .

Most common error: Adding $1/6$ four times to get $4/6$ . That method double-counts rolls that contain more than one six, and it breaks down completely for longer runs: with seven rolls it would give a probability above 1.

Problem 4

A game costs $1 to play. You roll one fair die, and if it shows six, you get $5 back. What is the expected value of the game for you, per play?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 4Hide the solution

Answer: about −$0.17 per play (exactly −1/6 of a dollar)

Method:

Work out the net result of each outcome. A six gains you $5 − $1 = $4. Any other face loses your $1.
Find the probabilities: six has probability $1/6$ , and not six has probability $5/6$ .
Multiply each net result by its probability and add: $4 \times \frac{1}{6} + (-1) \times \frac{5}{6} = \frac{4}{6} - \frac{5}{6} = -\frac{1}{6}$ .
Convert: $-1/6$ of a dollar is about −$0.17.

The value is negative because the payout is too small for the chance of winning. A fair game, with expected value zero, would need to pay back $6 for a six.

Most common error: Using the $5 prize as the gain and forgetting the $1 cost. That gives $5/6 - 5/6 = 0$ and makes the game look fair when it isn't.

Problem 5

Two players race to win three rounds. Each round is a fair 50–50 contest. The game stops with the leader ahead 2 rounds to 1. What share of the pot should the leader get?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 5Hide the solution

Answer: 3/4 of the pot

Method:

At most two more rounds would decide the game.
List the four equally likely futures for two rounds, writing L for the leader and T for the trailing player: LL, LT, TL, TT.
The leader reaches three wins in LL, LT, and TL. The trailing player wins only in TT.
The leader's chance is $3/4$ , so the leader should take three-quarters of the pot.

Most common error: Splitting by the score so far, 2 to 1, which gives the leader 2/3. The pot should follow each player's chance of winning from here, not the rounds already won.

Self-Check

When you're done, check that:

Each probability you found lies between 0 and 1
You counted only equally likely outcomes, such as the 36 ordered pairs for two dice
Your expected value in Problem 4 is negative, and you can say why in one sentence
You can explain why splitting by the score gives the wrong answer in Problem 5

Knowledge Check 10 min Done

The problem of points: how Pascal and Fermat put a number on chance

This ungraded knowledge check assesses your understanding of how probability reasons exactly about uncertain events. You'll be asked about the history of probability from Cardano to Kolmogorov, sample spaces and counting, independence and expected value, and the law of large numbers.

Note: Use this to test yourself, review the feedback on any questions you miss, and retry until you feel confident before moving forward.

5 questions · ungraded · retry as often as you like