Probabilistic reasoning, oriented

For a single pick, count what matches, count everything possible, and divide. A box of 40 pens with 14 red ones gives a 14 ÷ 40 = 35% chance of red. This works only when every outcome is equally likely. A spinner with one half and two quarter sections has three sections, but landing on the half is a one-in-two chance, not one in three.

Every probability lies between 0 and 1. An answer above 1, or above 100%, means something was added that should have been multiplied, or counted twice. Convert freely between forms to compare: 5/12 is about 41.7%, slightly more than 40%. Odds are a different way of writing the same thing: "7 to 3 against" means 7 losing ways and 3 winning ways out of 10, a probability of 3/10.

When outcomes are easy to miscount, list them. Two dice have 6 × 6 = 36 ordered outcomes, with a 5 then a 6 different from a 6 then a 5. A total of 11 comes from just those two, so its chance is 1/36 + 1/36 = 1/18, and totals near the middle have many more combinations than the extremes. Each extra coin doubles the outcomes, so six flips give 2^6 = 64, and any one exact sequence has a chance of 1/64. For cards, remember 4 suits of 13, with each rank appearing once per suit.

Rule: count equally likely outcomes, divide, and check the answer lies between 0 and 1.

The chance something does not happen is one minus the chance it does. A 45% chance of a delay means a 55% chance of leaving on time. "At least one" is fastest through its complement: in four coin flips, no heads has a chance of 1/16, so at least one head has a chance of 15/16.

For two independent events both happening, multiply. A 30% chance and an independent 90% chance both happen 0.3 × 0.9 = 27% of the time. Adding them to 120% is the classic error, and it gives an impossible number. The same rule explains backups: two independent backups that each fail 20% of the time both fail only 4% of the time, so at least one works 96% of the time.

For "A or B" when both cannot happen at once, add. From 8 red, 5 blue and 7 green counters, one draw is red or blue with a chance of 13/20. When both can happen together, subtract the overlap once: a heart or a 7 from a deck is 13/52 + 4/52 − 1/52 = 4/13, because the 7 of hearts was counted twice.

Two checks catch most errors. Independence means one event does not change the other's chances, so make sure before multiplying. And a combined event can never be more likely than either part alone, however believable the detailed version sounds.

Rule: subtract for "not", multiply for independent "and", add for exclusive "or", and remove any overlap once.

When you draw twice and the first item is not put back, the second draw faces a smaller collection. A jar holds 7 red and 3 blue balls. The first draw is red with a chance of 7/10. If that ball stays out, the jar holds 6 red among 9, so the chance that both draws are red is 7/10 × 6/9 = 42/90, which simplifies to 7/15.

If the first ball goes back, nothing changes between draws, and the draws are independent. Then the chance of two reds is 7/10 × 7/10 = 49/100, a little higher than 7/15, because a red ball is not missing from the second draw. Both answers usually appear among the options, so read carefully whether the item puts the item back.

The common slips are adjusting only the top of the second fraction or only the bottom. Each draw removes one item of its kind and one item from the total, and both changes matter. For three draws, keep going: the third red has 5 left among 8.

This is also why draws without replacement are not independent: what came out first changes what can come out next. Recognizing that tells you the simple product rule needs adjusted fractions.

Rule: after each draw without replacement, reduce both the count of that kind and the total, then multiply.

"Given that" and "of those" shrink the denominator to the known group. Of 80 club members, 30 play chess and 12 of those also play go. Among the chess players, the chance of a go player is 12/30 = 40%, not 12/80 = 15%. In a two-way table, find the row or column for the condition and read only inside it.

A conditional reversed is a different number. Most professional basketball players are tall, but most tall people are not professional basketball players, because the two statements divide by different groups.

Test results are conditionals, and counts make them clear. Picture 10,000 people with a condition affecting 0.5% of them, which is 50 people. A test catching 98% of cases flags 49 of them. If it wrongly flags 4% of the 9,950 healthy people, that is 398 more. Of the 447 positive results, only 49 are real, about 11%. The test's 98% describes how often it catches the condition, not how often a positive result is right.

The pattern is general. False positives come from the large healthy group, so when a condition is rare, even a small error rate can produce more false alarms than real cases. Ignoring how common the condition is, the base rate, is called base-rate neglect, and it is the error these items test.

Rule: divide by the group you are in, and for test results, compare true positives with all positives.

The expected value of a game is each outcome times its probability, added together. A game that pays $30 on heads and costs $10 on tails is worth 0.5 × $30 − 0.5 × $10 = $10 per play on average. Write losses as negative numbers before you add.

When outcomes are not equally likely, weight each by its own chance. A 20% chance to win $50 and an 80% chance to lose $5 give 0.2 × $50 − 0.8 × $5 = $10 − $4 = $6. Averaging the two amounts as if they were equally likely ignores that the loss happens four times as often.

Expected value describes the average over many plays, not any single play. The game above pays $50 or costs $5 each time and never pays exactly $6, but over hundreds of plays the average settles near $6. Use expected value to compare bets, not to predict one result.

Long shots need special care. A $5 ticket with a 1-in-10-million chance of $20 million is worth $20,000,000 ÷ 10,000,000 = $2 on average, so after the ticket price the expected value is −$3. A huge prize can hide a poor bet until you multiply it by its tiny chance.

Rule: weight every outcome by its own probability, subtract what playing costs, and read the result as a long-run average.

After a fair die shows no 6 in ten rolls, the next roll is still a 1/6 chance. Believing an outcome is due is the gambler's fallacy. Real randomness also clumps: in 20 fair coin flips, a run of four or more of the same face happens about three times in four.

Small samples swing more than large ones. A fair coin flipped 10 times comes up at least 70% heads about 17% of the time, while over 1,000 flips that almost never happens. Expect the most extreme figures from the smallest groups. Relatedly, an unusually high result usually contained good luck, so the next result tends to sit closer to average. That drift, regression to the mean, can make an intervention look effective when it did nothing.

Judging likelihood by how easily examples come to mind, the availability heuristic, makes vivid or widely reported events seem more common than their rate supports. Ask for the rate.

Knowledge changes probabilities. With ten doors and one prize, if a host who knows where the prize is opens eight empty doors, your first pick still has a 1/10 chance and the other closed door carries 9/10.

Finally, a sample chosen by its outcome misleads. Studying only the companies still in business hides the ones that failed.

Rule: before trusting a feeling about chance, ask about independence, sample size, luck and who is missing.