Two dice: totals are not equally likely
Totals from 2 to 12 are eleven outcomes, so each has chance 1/11.
There are 36 equally likely ordered rolls. A total of 7 arises six ways; 2 and 12 arise one way each.
The mistake is counting totals rather than rolls. Treat the dice as distinguishable — one red, one white — and each of the 6 × 6 = 36 pairs is equally likely.
| Total | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
Dice games travelled widely in Asia: Japan's sugoroku board games and India's moksha patam (the ancestor of snakes and ladders) both rely on rolled moves. Anyone designing such a board is, knowingly or not, shaping these frequencies.
The birthday problem
Problem
How many people must be in a room before there is a better than 50% chance that two share a birthday? (Ignore 29 February and assume all 365 days are equally likely.)
About 183 — half of 365.
23 people. The chance is about 50.7%.
Show the worked solution
Step 1 — count the complement. It is easier to find the chance that all birthdays differ.
Step 2 — build it person by person. The second person avoids the first with chance 364/365, the third avoids both with 363/365, and so on.
Step 3 — multiply. For 23 people: (365 × 364 × … × 343) ÷ 36523 ≈ 0.493. So the chance of at least one match is about 1 − 0.493 = 0.507.
Why intuition fails. We imagine comparing ourselves with others (22 comparisons). But every pair counts: 23 people form C(23, 2) = 253 pairs.
The gambler's fallacy
After five heads in a row, many people feel tails is "due". For a fair coin the next toss is still 1/2. The coin has no memory. What is true is that over many tosses the proportion of heads tends toward 1/2 — not because tails catches up, but because a few early streaks are diluted by thousands of later tosses.
Test the assumption by counting: of the 64 equally likely sequences of six tosses, exactly 32 end in heads and 32 in tails — including among those that begin with five heads (HHHHHH and HHHHHT, one each).
Three doors
Problem
A prize is behind one of three doors. You pick one. The host, who knows where the prize is, opens a different door showing nothing and offers a switch. Should you switch?
Show the worked solution
Your first pick is right with chance 1/3. The host's action never changes that: he can always open an empty door. So "stay" wins 1/3 of the time and "switch" wins the remaining 2/3. Listing cases makes it concrete: if the prize is behind door A and you pick A, switching loses; if you pick B or C, the host must open the other empty door, and switching wins. Two of three cases favour switching.
Limit test: if the host opened a door at random and it happened to be empty, switching would win only 1/2. The host's knowledge is part of the problem.
The I Ching: early enumeration and uneven odds
The Yijing (I Ching, "Book of Changes") from ancient China organises 64 hexagrams, each a stack of six lines that are either broken or unbroken. That is a complete listing of 26 = 64 possibilities. In 1703 Gottfried Leibniz noted its connection to binary numbers after correspondence with the Jesuit Joachim Bouvet.
The traditional yarrow-stalk method of generating lines is a fine probability exercise. Its four outcomes are not equally likely:
| Line | Yarrow stalks | Three coins |
|---|---|---|
| Old yin (changing) | 1/16 | 1/8 |
| Young yang | 5/16 | 3/8 |
| Young yin | 7/16 | 3/8 |
| Old yang (changing) | 3/16 | 1/8 |
The assumption that "both methods are equivalent" fails on counting. Each gives yin and yang overall with chance 1/2, but the changing lines differ. Two procedures with the same headline result can still distribute chance differently.
Source context
The modern theory of probability is usually dated to the 1654 letters between Pascal and Fermat. Counting traditions in China and India, such as the arithmetic triangle, came earlier but served different purposes.