14.3: Odds and Ends

In many situations, we translate probabilities into the odds for or against a given outcome. For example, the probability of rolling a two when you roll a fair die is 1/6, and the probability of not getting a two is 5/6. We say that the odds of rolling a two are 1 to 5 and the odds against it are 5 to 1. The odds in favor of a two is the ratio of the number of ways of getting a two (one way) to the number of ways of not getting a two (five ways). And the odds against rolling a two are the five chances that some other side will come up against the one chance that a two will come up.

The relationship between odds and probabilities is a simple and straightforward:

\[Pr(A) = \dfrac{m}{n}\]

if and only if the odds in favor of A are \(m\) to \(n - m\).

From Probabilities to Odds

If the probability of something is 1/36 (as is the probability of rolling box cars), then the odds in favor of it at 1 to 35 and the odds against it are 35 to 1. We convert probabilities to odds with the following rule: if the probability of a given outcome is m/n, then the odds in favor of it are m to n -m and the odds against it are n - m to m.

From Odds to Probabilities

If your friend says that the odds of OU’s beating Texas A&M are 1:5, what does she think the probability of A&M’s winning is? We get the denominator for this probability by adding the two numbers in this ratio, so the number on the bottom is 6. Your friend believes that there is one chance in 6 that OU will win, which translates into a probability of 1/6. And she also believes that the probability of A&M’s winning is 5/6. If the odds in favor of S are m to n, then the probability of S is the first number (m) over the sum of the first and second numbers (m + n).

Fair Bets

Fair bets are based on the odds. If you want to make a fair bet that a two will come up when you roll a fair die, you should bet $1 that you will get a two and your opponent should bet $5 that you won’t. If you both always bet these amounts, then over the long run you will both tend to break even. Gamblers call such a bet an even-up proposition.

By contrast, if you were to bet $1 that you would roll a two and your opponent bets $6 that you won’t, then over the long haul you will come out ahead. And if you bet $1 that you will roll a 2 and your opponent bets $4 that you won’t, then over the long haul you will lose.

Organized gambling usually involves bets that are not even-up. A casino could not pay its operating expenses, much less turn a profit, if it made evenup bets. The house takes a percentage, which means paying winners less than the actual odds would require. The same is true for insurance premiums. It is also true for state lotteries, which in fact offer far worse odds than most casinos. If you gamble in such settings long enough, you are virtually certain to lose more than you win. Of course, if you enjoy gambling enough, you may be willing to accept reasonable losses as the price of getting to gamble.

Example: Roulette

Roulette is a gambling game in which a wheel is spun in one direction and a ball is thrown around the rim into the wheel in the opposite direction A roulette wheel has many compartments, and players bet on which compartment the ball will land in. In the U.S., roulette wheels have thirty-eight compartments. They are numbered from 1 through 36; there is also a thirty-seventh compartment numbered 0 and a thirty-eighth numbered 00.

There are various bets players can place, but here we will focus on the simplest one, where a player bets that the ball will land on one specific number (say 14), from 1 through 36. Although the game can be complex, the following discussion gives the basic points.

Since there are thirty-eight compartments on the wheel, the probability that the ball will land on any given number, say 14, is 1/38; Pr(14) = 1/38. Hence, the true odds against rolling a 14 are 37 to 1. If you played the game over and over, betting at these odds, you would break even. You would win once every thirty-eight times, and the casino (the “house” or “the bank”) would win the other thirty-seven times. But when you did win, they would pay you $37, which would exactly compensate you for the thirty-seven times that you lost $1 (37x $1 = $37). We say that your bet has an expected value $0.0.

But of course, the house does not pay off at the true odds of 37 to 1. Instead, the house odds or betting odds against rolling a 14 are 35 to 1 (the house has the advantage of the 0 and 00). When you lose, this doesn’t make any difference. But when you win you get only $36 (the $35 plus the original $1 that you bet). This is $2 less than you would get if you were paid off at the true odds of 1 to 37. Since the house keeps $2 out of every $38 that would be paid out at the true odds, their percentage is 2/38, or 5.26%. All but one of the bets you can make at roulette costs you 5.26% over the long haul (the remaining bet is even worse, from the player’s point of view).

If you play just a few times, you may well win. Indeed, a few people will win over a reasonably long run. But the basic fact is that your bet on 14 has a negative expected payoff of -5.26%. This means that over the long run you will almost certainly lose at roulette. The odds are against you, and there are no systems or strategies or tricks that can change this basic fact. Simply put, there is absolutely no way you can expect to win at this game. There are a few highly skilled people who make a living playing poker, blackjack, or betting on the horses. But no one can make a living playing casino games like keno, craps, or roulette.

Exercises on Odds and Probabilities

Calculate the odds and probabilities in each of the following cases.

1. What are the odds against drawing a king of spades from a full deck of playing cards?
2. What are the odds against drawing a king from a full deck?
3. What are the odds against drawing a face card from a full deck?
4. What are the odds against drawing a king if you have already drawn two cards (one a king, the other a six)?
5. You have a bent coin. The odds of flipping a head are 3 to 2. What is the probability of tossing a tail?
6. In Europe, a roulette wheel only has thirty-seven compartments, one through 36, plus 0. They pay off at the same odds as U.S. casinos. How would this change the probabilities and the odds?
7. If the probability of Duke winning the NCAA basketball tournament championship is 0.166 (= 1/6), what are the odds that they will win? What are the odds against their winning? What are the fair bets for and against their winning? Defend your answers.

Sample Problems with Answers

In each case, explain which rules are relevant to the problem. Your analysis of the problem is more important than the exact number you come up with.

1. You are going to roll a single die. What is the probability of rolling a two or an odd number?
   1. You are asked about the probability of a disjunction; What is Pr(T or O)?
   2. Are the two disjuncts incompatible?
   3. Yes. So, we can use the simple disjunction rule (R4).
   4. It says that Pr(T or O) = Pr(T) + Pr(O).
   5. And Pr(T) + Pr(O) = 1/6 + 3/6 = 4/6 (= 2/3).
2. You are going to draw a single card from a full deck. What is the probability of getting either a spade or a three?
   1. You are asked about the probability of a disjunction; What is Pr(S or T)?
   2. Are the two disjuncts incompatible?
   3. No. They overlap because of the three of spades.
   4. So we must use the more complex disjunction rule (R6), in which we subtract out the overlap.
   5. It says that Pr(S or T) = Pr(S) + Pr(T) - Pr(T & S).
   6. Pr(T & S) is just the probability of drawing the three of spades, which is 1/52.
   7. So Pr(S or T) = Pr(S) + Pr(T) - Pr(T & S) = (13/52 + 4/52) - 1/52.
3. You are going to draw two cards from a full deck without replacing the first card. What is the probability of getting exactly one king and exactly one queen (the order doesn’t matter)?
   1. You are asked about Pr(K & Q), where order doesn’t matter.
   2. There are two different ways for this to occur:
      1. King on first draw and queen on second: (K₁ & Q₂)
      2. Queen on first draw and king on second: (Q₁ & K₂)
   3. So we must calculate the probability of a disjunction: What is Pr[(K₁ & Q₂) or (Q₁ & K₂)]?
   4. The two disjuncts are incompatible, so we use the simple disjunction rule (R4).
   5. But each disjunct is itself a conjunction, and the conjuncts of each conjunction are not independent.
   6. First disjunct is: Pr(K₁ & Q₂). The general rule (R8) for conjunctions tells us that Pr(K₁ & Q₂) = Pr(K₁) x Pr(Q₂|K₁), which is 4/52 x 4/51.
   7. Second disjunct is: Pr(Q₁ & K₂). It works the same way: Pr(Q₁ & K₂) = Pr(Q₁) x Pr(K₂ |Q₁), which is also 4/52 x 4/51.
   8. Now add the probabilities for each disjunct: (4/52 x 4/51) + (4/52 x 4/51).

More Complex Problems

In the next module, we will look at a number of real-life applications of probability. We conclude this module with several problems that are more complex than the ones we’ve dealt with thus far.

Probability theory was formalized in the 1650s. The Chevalier de Méré’s was a wealthy Parisian gambler. He had devised a dice game that was making him money. He would bet even money (betting odds of 1:1) that he could roll at least one six in four throws of a die. Eventually people got wise to this game and quit playing it, so he devised a new game in which he bet even money that he could roll at least one double six (a six on each die) in twenty fours rolls of a pair of dice. But over time he lost money with this bet.

Finally, he asked his friend, the philosopher and mathematician Blaise Pascal (1623-1662), why this was so. Pascal (and Pierre de Fermat, with whom he corresponded) worked out the theory of probability and used it to explain why the first game was profitable while the second one was not. Let’s see how to solve the first problem (the second is left as an exercise).

• What is the probability of rolling at least one six in four throws of a die?

Rolling at least one six means rolling a six on the first roll, or the second, or the third, or the fourth. But it is difficult to work the problem in this way because one must subtract out all the relevant overlaps.

It is easiest to approach this by way of its negation. The negation of the statement that you roll at least one six is the statement that you do not roll any sixes. This negation is equivalent to a conjunction: you do not roll a six on the first throw and you do not roll a six on the second throw and you do no roll a six on the third throw and you do not roll a six on the fourth throw. This conjunction has four conjuncts, but that doesn’t really change anything that affects the probabilities. Each conjunct says that you get something other than a six, and so each has a probability of 5/6.

Furthermore, each conjunct is independent of the other three (the die doesn’t remember earlier outcomes). So, we just multiply the probabilities of the four conjuncts to get the probability that the conjunction itself is true: the probability that you don’t get a six on any of the four rolls is 5/6 x 5/6 x 5/6 x 5/6 (= (5/6)4), which turns out to be 625/1296.

This is the probability that you don’t get any sixes. So the probability we originally asked about (getting at least one six) is just one minus this: the probability of getting at least one six is 1 - (625/1296) (which is approximately 671/1296). This is just a bit more than 1/2. This means that the odds of getting at least one six are 671 to 625, so over the long run the house will come out ahead, and you will lose if you keep playing their game