Odds
A chance can be told as a share of everything that might happen, or as the ways it happens set against the ways it does not. How do the two descriptions fit together, and how do you move from one to the other?
Where you are 14 in Basic probability
Why this exists
A probability answers the question “what share of all the outcomes make the event happen?” It is a part compared with the whole. That is not the only natural way to talk about chance. People often compare the two sides directly: the ways the event can happen against the ways it cannot. “Three ways it happens for every five ways it does not” is a complete description of a chance, and it never mentions the total of eight.
That second description is called odds. It is a part-to-part ratio, where a probability is a part-to-whole fraction, and the two carry exactly the same information. The point of this lesson is to see why, and to move between them without losing anything on the way. Sports commentary and games of chance often state chances as odds, and some later statistics (the odds ratio in medicine, logistic models) is built on odds rather than on probabilities, so the conversion is worth knowing well.
Definition
Take an experiment and an event E. Every outcome either makes E happen or makes its complement \text{not } E happen, never both.
When the outcomes are equally likely, count them: say s outcomes make E happen and f outcomes do not.
- The odds in favor of E are s : f, read “s to f”. The ways it happens come first.
- The odds against E are f : s. The ways it does not happen come first.
So the odds against are the odds in favor written in the other order.
When the outcomes are not counted, or are not equally likely, the same comparison is made with the probabilities themselves: the odds in favor of E are
and the odds against are P(\text{not } E) : P(E). For equally likely outcomes the two versions agree, because P(E) = \tfrac{s}{n} and P(\text{not } E) = \tfrac{f}{n}, and dividing both parts of a ratio by the same n does not change it.
Odds are a ratio, so any two equivalent ratios state the same odds: 6 : 10, 3 : 5 and 1.5 : 2.5 are one set of odds. They are usually given in lowest whole numbers. Odds of 1 : 1 are called even odds.
Intuition
A bag holds eight marbles, three blue and five yellow, and one is drawn without looking. Lay the eight marbles out in a row.
The odds in favor of blue compare the two brackets on top, 3 : 5. The probability of blue compares the shaded bracket with the bracket underneath, \tfrac{3}{8}. Both are read off the same row, so each can be rebuilt from the other: from the odds, add the two parts to recover the whole; from the probability, take the whole away from the part to recover the other part.
Notice what \tfrac{3}{5} is here. It is the number of blue marbles for each yellow one, the quotient of the two parts. It is not the probability of blue, because the five yellow marbles are not all the marbles. Odds stop being confusing the moment you ask of each number: is this a part, or the whole?
Results
If the odds in favor of E are a : b, with a and b not both zero, then
If P(E) = p, then the odds in favor of E are p : (1 - p) and the odds against are (1 - p) : p. When p = \tfrac{m}{n}, these are m : (n - m) and (n - m) : m.
The odds are even, 1 : 1, exactly when P(E) = \tfrac{1}{2}. The first number of the odds in favor is the larger exactly when P(E) > \tfrac{1}{2}.
Why it is true
Suppose the odds in favor of E are a : b. By definition this means P(E) : P(\text{not } E) is the same ratio as a : b, so the two probabilities are the same multiple of a and b: there is a number k with
The complement rule says P(E) + P(\text{not } E) = 1. So ka + kb = 1, which is k(a + b) = 1 by the distributive property, and k = \tfrac{1}{a + b}. Putting k back gives P(E) = \tfrac{a}{a + b} and P(\text{not } E) = \tfrac{b}{a + b}.
The other direction is shorter. If P(E) = p, the complement rule gives P(\text{not } E) = 1 - p, and the odds in favor are P(E) : P(\text{not } E) = p : (1 - p) by definition. With p = \tfrac{m}{n}, the ratio \tfrac{m}{n} : \tfrac{n - m}{n} has both parts multiplied by n to give m : (n - m), and multiplying both parts of a ratio by the same nonzero number gives an equivalent ratio.
For even odds: p : (1 - p) is 1 : 1 exactly when p = 1 - p, that is when p = \tfrac{1}{2}. And p > 1 - p exactly when 2p > 1, that is when p > \tfrac{1}{2}.
Worked examples
Blue marbles: the three numbers from one row
The bag above holds three blue marbles and five yellow ones, and one is drawn at random. Find the probability of blue, the odds in favor of blue and the odds against blue.
SolutionThere are 3 + 5 = 8 marbles, each equally likely to be drawn. Three of them are blue, so
The odds in favor put the ways it happens first: three blue against five that are not blue, so the odds in favor of blue are 3 : 5. The odds against put the other side first: 5 : 3.
Check by going back: odds in favor of 3 : 5 give P(\text{blue}) = \tfrac{3}{3 + 5} = \tfrac{3}{8}, which matches. The three descriptions, \tfrac{3}{8}, 3 : 5 and 5 : 3, all say the same thing.
Fifteen raffle tickets: odds that simplify
A box holds fifteen raffle tickets numbered 1 to 15, and one is drawn at random. Find the odds in favor of drawing a number greater than 5, the odds against it, and its probability.
SolutionThe tickets greater than 5 are 6, 7, \dots, 15, which is 15 - 5 = 10 tickets. The other 5 tickets, 1 to 5, do not make the event happen. So the odds in favor are 10 : 5.
Both parts share a factor of 5, and dividing both parts of a ratio by the same number gives an equivalent ratio, so the odds in favor are 2 : 1. The odds against are the reverse, 1 : 2.
The probability is \tfrac{10}{15} = \tfrac{2}{3}. From the simplified odds it is \tfrac{2}{2 + 1} = \tfrac{2}{3} as well: simplifying the odds first does no harm, because \tfrac{a}{a + b} is unchanged when a and b are scaled by the same number. Here the first number of the odds in favor is the larger, matching a probability above \tfrac{1}{2}.
A tennis match: from odds against to a probability
Before a tennis match, a commentator says the odds against one player winning are 7 to 2. A tennis match always has a winner. What probability does the commentator give that this player wins, and that the player loses?
SolutionOdds against put the failures first, so “not winning” stands to “winning” as 7 : 2. Turned round, the odds in favor of winning are 2 : 7. By the result above,
Since a match cannot end in a draw, losing is exactly the complement of winning, and
The two add to 1, as a probability and its complement must. The danger in this example is the order: the 7 belongs to “not winning”, so it goes with the complement, not with the event.
A probability as a decimal
An airline’s records give a probability of 0.35 that a certain evening flight is delayed. State the odds in favor of a delay and the odds against it.
SolutionThe complement rule gives P(\text{not delayed}) = 1 - 0.35 = 0.65. So the odds in favor of a delay are
Multiplying both parts by 100 clears the decimals and gives the equivalent ratio 35 : 65, and both parts are multiples of 5, so dividing by 5 gives 7 : 13. The odds in favor of a delay are 7 : 13 and the odds against are 13 : 7.
Check: \tfrac{7}{7 + 13} = \tfrac{7}{20} = 0.35.
Odds given as “to 1”
Odds are sometimes stated with a second number of 1, such as “odds against are 4.5 to 1”. Find the probability of the event.
SolutionThe odds against are 4.5 : 1. Doubling both parts gives the equivalent whole-number ratio 9 : 2. So the odds in favor are 2 : 9, and
The conversion would have worked without clearing the decimal, since \tfrac{1}{4.5 + 1} = \tfrac{1}{5.5} = \tfrac{2}{11}, but whole numbers make the fraction easier to read. “r to 1” is simply odds scaled so the second part is 1; any odds a : b with b \neq 0 can be written as \tfrac{a}{b} : 1.
Working back from odds to counts
A bag holds only red and blue counters. The odds in favor of drawing red are 4 : 7. When 6 more red counters are added, the odds become even. How many counters of each color were in the bag at first?
SolutionOdds of 4 : 7 say that the red and blue counts are in the ratio 4 : 7, so for some whole number k there are 4k red and 7k blue counters. Adding 6 red counters makes 4k + 6 red, and even odds mean the two parts are equal:
Subtracting 4k from both sides gives 6 = 3k, so k = 2. At first there were 8 red and 14 blue counters, 22 in all.
Check: 8 : 14 simplifies to 4 : 7, and after the addition 14 : 14 is 1 : 1. Notice that the odds alone never fixed the size of the bag: 4 : 7 fits 4 and 7, or 40 and 70. It took the second piece of information to find k.
Combining two events: convert first
E and F are mutually exclusive events. The odds in favor of E are 2 : 3 and the odds in favor of F are 1 : 4. Find the odds in favor of “E or F”.
SolutionThe addition rule for mutually exclusive events works with probabilities, so convert both odds first:
Since E and F cannot happen together,
The complement has probability \tfrac{2}{5}, so the odds in favor of “E or F” are \tfrac{3}{5} : \tfrac{2}{5}, which is 3 : 2 after multiplying both parts by 5.
Adding the odds part by part would give 3 : 7, a probability of \tfrac{3}{10}, which is wrong. The reason is mathematical: the 3 in “2 : 3” counts the outcomes outside E, the 4 in “1 : 4” counts the outcomes outside F, and those two “outside” groups overlap, so their sizes cannot be added. Probabilities all share the same whole, which is why they can be added and odds cannot.
Boundaries
- An impossible event has odds in favor 0 : 1 and odds against 1 : 0. A certain event has odds in favor 1 : 0. These are allowed as ratios, but a ratio with a second part of 0 has no value as a single number \tfrac{a}{b}, so “to 1” form fails for them.
- The ratio 0 : 0 is not odds. The formula P(E) = \tfrac{a}{a + b} needs a + b \neq 0, and both parts of a set of odds are never zero together, because P(E) and P(\text{not } E) add to 1.
- Counts give odds only when the outcomes are equally likely. A bent coin still has one head and one tail, but its odds are not 1 : 1. In that case the odds come from the probabilities, P(E) : P(\text{not } E), not from counting faces.
- Odds are part to part, so the parts must be complements. In a three-way contest (win, draw, lose), the ratio of wins to losses is not the odds in favor of winning; the second part must be everything that is not a win, draws included.
- Odds do not fix the counts. 3 : 5 fits a bag of 8 marbles or a bag of 80. Recovering actual numbers needs the total or some other extra fact.
Where this goes next
Odds turn up wherever a chance is compared with its complement. In conditional probability and Bayes’ rule, new evidence changes the odds of an event in a particularly simple way, by multiplying them, which is one reason statisticians often prefer odds to probabilities. In expected value, a fair game is one where the payouts balance the odds. And because odds are a ratio, everything known about equivalent ratios applies to them directly.
Exercises
From counts to odds
- A bag holds 5 orange and 9 purple tiles, and one is drawn at random. Find the odds in favor of orange and the odds against orange.
- A fair six-sided die is rolled. Find the odds in favor of rolling a 6 and the odds against.
- Ten cards numbered 1 to 10 are shuffled and one is drawn. Find the odds in favor of drawing a prime number, in lowest terms, and the odds against.
- Two fair coins are tossed. Find the odds in favor of getting exactly one head.
- A spinner has ten equal sections: 4 red, 3 blue and 3 white. Find the odds in favor of red and the odds against blue.
From probability to odds
- P(E) = \tfrac{2}{7}. Find the odds in favor of E and the odds against E.
- P(E) = 0.6. Find the odds in favor of E and the odds against E.
- The probability that a train is on time is 45\%. Find the odds in favor of it being on time.
- The probability that a bus arrives on time is \tfrac{5}{6}. Find the odds against it arriving on time.
- P(\text{not } E) = 0.28. Find the odds in favor of E in lowest terms.
From odds to probability
- The odds in favor of E are 3 : 4. Find P(E).
- The odds against E are 11 : 1. Find P(E).
- The odds of an event are even. Find its probability.
- The odds against E are 2.5 to 1. Find P(E) and P(\text{not } E).
- The odds in favor of E are 12 : 18. Find P(E) as a fraction in lowest terms and as a decimal.
Putting odds to work
- A box holds 30 pens, every one black or white. The odds in favor of drawing a black pen are 2 : 3. How many black pens are in the box?
- A jar holds only green and orange beads. The odds against drawing green are 5 : 2, and there are 15 more orange beads than green ones. How many beads of each color are there?
- E and F are mutually exclusive. The odds in favor of E are 1 : 3 and the odds in favor of F are 1 : 5. Find the odds in favor of “E or F”.
- A bag holds 6 red and 10 blue counters. How many red counters must be added so that the odds in favor of drawing red become 3 : 2?
Reasoning
- The odds in favor of an event are 3 : 5. Explain why its probability is not \tfrac{3}{5}.
- Explain why the odds in favor of an event are even exactly when its probability is \tfrac{1}{2}, and why the first number of the odds in favor is the larger exactly when the probability is more than \tfrac{1}{2}.
- What are the odds in favor of an impossible event? Why can the odds in favor of a certain event not be written in the form “r to 1”?
- The odds in favor of E are a : b and the odds in favor of F are c : d, all four numbers positive, with \tfrac{a}{b} > \tfrac{c}{d}. Show that P(E) > P(F).
Answers
- In favor 5 : 9, against 9 : 5. Five tiles are orange and nine are not.
- In favor 1 : 5, against 5 : 1. One face is a 6 and five are not.
- In favor 2 : 3, against 3 : 2. The primes are 2, 3, 5, 7: four cards, against six that are not prime, so 4 : 6 = 2 : 3.
- Even odds, 1 : 1. The outcomes HH, HT, TH, TT are equally likely; HT and TH give exactly one head and the other two do not, so 2 : 2 = 1 : 1.
- In favor of red 2 : 3, against blue 7 : 3. Red has 4 sections and 6 are not red, so 4 : 6 = 2 : 3. Blue has 3 sections and 7 are not blue, and odds against put the 7 first.
- In favor 2 : 5, against 5 : 2. With \tfrac{2}{7} the other part is 7 - 2 = 5.
- In favor 3 : 2, against 2 : 3. The odds in favor are 0.6 : 0.4 = 6 : 4 = 3 : 2.
- 9 : 11. The odds in favor are 45 : 55, and dividing both by 5 gives 9 : 11.
- 1 : 5. Not on time has probability \tfrac{1}{6}, so the odds against are \tfrac{1}{6} : \tfrac{5}{6} = 1 : 5.
- 18 : 7. P(E) = 1 - 0.28 = 0.72, so the odds are 72 : 28, and dividing by 4 gives 18 : 7.
- P(E) = \tfrac{3}{7}, from \tfrac{3}{3 + 4}.
- P(E) = \tfrac{1}{12}. The odds in favor are 1 : 11, so P(E) = \tfrac{1}{1 + 11}.
- \tfrac{1}{2}. Even odds are 1 : 1, and \tfrac{1}{1 + 1} = \tfrac{1}{2}.
- P(E) = \tfrac{2}{7} and P(\text{not } E) = \tfrac{5}{7}. Doubling gives odds against 5 : 2, so the odds in favor are 2 : 5.
- P(E) = \tfrac{2}{5} = 0.4. The odds simplify to 2 : 3, and \tfrac{2}{2 + 3} = \tfrac{2}{5}.
- 12 black pens. Black and white are 2k and 3k with 5k = 30, so k = 6 and there are 2 \cdot 6 = 12 black pens (and 18 white).
- 10 green and 25 orange. Odds against green put orange first, so orange is 5k and green is 2k. Then 5k - 2k = 15 gives k = 5. Check: 25 : 10 = 5 : 2.
- 5 : 7. P(E) = \tfrac{1}{4} and P(F) = \tfrac{1}{6}, so P(E \text{ or } F) = \tfrac{3}{12} + \tfrac{2}{12} = \tfrac{5}{12}, and the complement is \tfrac{7}{12}.
- 9 red counters. With x added, (6 + x) : 10 = 3 : 2, so 2(6 + x) = 30 and x = 9. Check: 15 : 10 = 3 : 2.
- The 5 counts only the outcomes where the event fails, not all the outcomes. The whole is 3 + 5 = 8 parts, so the probability is \tfrac{3}{8}; \tfrac{3}{5} is how many successes there are for each failure.
- The odds in favor are p : (1 - p). Two parts are equal exactly when p = 1 - p, which is p = \tfrac{1}{2}. The first part is larger exactly when p > 1 - p, which is 2p > 1, or p > \tfrac{1}{2}.
- An impossible event has odds in favor 0 : 1, since P(E) = 0 and P(\text{not } E) = 1. A certain event has odds in favor 1 : 0; writing them as “r to 1” would need r = \tfrac{1}{0}, and division by zero has no value.
- Write r = \tfrac{a}{b}, so P(E) = \tfrac{a}{a + b} = \tfrac{r}{r + 1} after dividing top and bottom by b, which is 1 - \tfrac{1}{r + 1}. As r grows, r + 1 grows, so \tfrac{1}{r + 1} shrinks and P(E) grows. The same formula gives P(F) = 1 - \tfrac{1}{t + 1} with t = \tfrac{c}{d}. Since r > t, it follows that P(E) > P(F).