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What Are the Odds?

Compute odds.

Learning Objectives

After completing this section, you should be able to:

  1. Compute odds.
  2. Determine odds from probabilities.
  3. Determine probabilities from odds.

Computing Odds

The ratio of the number of equally likely outcomes in an event \(E\) to the number of equally likely outcomes not in the event \(E'\) is called the odds for (or odds in favor of) the event. The opposite ratio (the number of outcomes not in the event to the number in the event \(E'\) to the number in the event \(E\) is called the odds against the event.

Computing Odds

Try it.

  1. If you roll a fair 6-sided die, what are the odds for rolling a 5 or higher?
  2. If you roll two fair 6-sided dice, what are the odds against rolling a sum of 7?
  3. If you draw a card at random from a standard deck, what are the odds for drawing a \(♡\)?
  4. If you draw 2 cards at random from a standard deck, what are the odds against them both being \(♠\)?
Solution
  1. The sample space for this experiment is {1, 2, 3, 4, 5, 6}. Two of those outcomes are in the event “roll a five or higher,” while four are not. So, the odds for rolling a five or higher are \(2:4=1:2\).
  2. In , we found the sample space for this experiment using the following table ():

    There are 6 outcomes in the event “roll a sum of 7,” and there are 30 outcomes not in the event. So, the odds against rolling a 7 are \(30:6=5:1\).

  3. There are 13 \(♡\) in a standard deck, and \(52-13=39\) others. So, the odds in favor of drawing a \(♡\) are \(13:39=1:3\).
  4. There are \({}_{13}{C}_{2}=78\) ways to draw 2 \(♠\), and \({}_{52}{C}_{2}-78=1,248\) ways to draw 2 cards that are not both \(♠\). So, the odds against drawing 2 \(♠\) are \(1,248:78=16:1\).

Odds as a Ratio of Probabilities

We can also think of odds as a ratio of probabilities. Consider again the instant-win game from the section opener, with 500,000 winning tickets out of 2,000,000 total tickets. If a player buys one ticket, the probability of winning is \(\frac{500,000}{2,000,000}=\frac{1}{4}\), and the probability of losing is \(1-\frac{1}{4}=\frac{3}{4}\). Notice that the ratio of the probability of winning to the probability of losing is \(\frac{1}{4}:\frac{3}{4}=1:3\), which matches the odds in favor of winning.

We can use these formulas to convert probabilities to odds, and vice versa.

Converting Probabilities to Odds

Try it.

Given the following probabilities of an event, find the corresponding odds for and odds against that event.

  1. \(P(E)=\frac{3}{5}\)
  2. \(P(E)=17\%\)
Solution
  1. Using the formula, we have: \[\begin{array}{lll}\text{odds for}\ E & = & P(E):(1-P(E)) \\ & = & \frac{3}{5}:(1-\frac{3}{5}) \\ & = & \frac{3}{5}:\frac{2}{5} \\ & = & 3:2.\end{array}\] (Note that in the last step, we simplified by multiplying both terms in the ratio by 5.)
    Since the odds for \(E\) are \(3:2\), the odds against \(E\) must be \(2:3\).
  2. Again, we’ll use the formula: \[\begin{array}{lll}\text{odds for}\ E & = & P(E):(1-P(E)) \\ & = & 0.17:(1-0.17) \\ & = & 0.17:0.83 \\ & \approx & 1:4.88.\end{array}\] (In the last step, we simplified by dividing both terms in the ratio by 0.17.)
    It follows that the odds against \(E\) are approximately \(4.88:1\).

Now, let’s convert odds to probabilities. Let’s say the odds for an event are \(A:B\). Then, using the formula above, we have \(A:B=P(E):(1-P(E))\). Converting to fractions and solving for \(P(E)\), we get:

\[\begin{array}{lll}\frac{A}{B} & = & \frac{P(E)}{1-P(E)} \\ A(1-P(E)) & = & B\times P(E) \\ A-A\times P(E) & = & B\times P(E) \\ A & = & A\times P(E)+B\times P(E) \\ A & = & (A+B)\times P(E) \\ \frac{A}{A+B} & = & P(E).\end{array}\]

Let’s put this result in a formula we can use.

Converting Odds to Probabilities

Try it.

Find \(P(E)\) if \(E\):

  1. The odds of \(E\) are \(2:1\) in favor
  2. The odds of \(E\) are \(6:1\) against
Solution
  1. Using the formula we just found, we have \(P(E)=\frac{2}{2+1}=\frac{2}{3}\).
  2. If the odds against are \(6:1\), then the odds for are \(1:6\). Thus, using the formula, \(P(E)=\frac{1}{1+6}=\frac{1}{7}\).

Condensed: the full section is in OpenStax Contemporary Mathematics.

Formulas

  • For an event \(E\), \[\begin{array}{l}\text{odds for}\ E=n(E):n({E}^{'})=P(E):P({E}^{'})=P(E):(1-P(E)) \\ \text{odds against}\ E=n({E}^{'}):n(E)=P({E}^{'}):P(E)=(1-P(E)):P(E)\end{array}\]
  • If the odds in favor of \(E\) are \(A:B\), then \(P(E)=\frac{A}{A+B}\).

Sada ti Nema kalkulatora koji bi riješio ovaj problem, ali njegovi dijelovi se mogu izračunati. Pokušajte jedan od gore navedenih ili upišite svoj.

Vježba (3)

Prvo isprobajte svaki na papiru. Otkrijte odgovor da provjerite; provjereni se mogu otvoriti u rješavaču za svaki korak.

    1. If you roll a fair 6-sided die, what are the odds for rolling a 5 or higher?
    2. If you roll two fair 6-sided dice, what are the odds against rolling a sum of 7?
    3. If you draw a card at random from a standard deck, what are the odds for drawing a \(♡\)?
    4. If you draw 2 cards at random from a standard deck, what are the odds against them both being \(♠\)?
    Otkrij odgovor
    1. The sample space for this experiment is {1, 2, 3, 4, 5, 6}. Two of those outcomes are in the event “roll a five or higher,” while four are not. So, the odds for rolling a five or higher are \(2:4=1:2\).
    2. In , we found the sample space for this experiment using the following table ():

      There are 6 outcomes in the event “roll a sum of 7,” and there are 30 outcomes not in the event. So, the odds against rolling a 7 are \(30:6=5:1\).

    3. There are 13 \(♡\) in a standard deck, and \(52-13=39\) others. So, the odds in favor of drawing a \(♡\) are \(13:39=1:3\).
    4. There are \({}_{13}{C}_{2}=78\) ways to draw 2 \(♠\), and \({}_{52}{C}_{2}-78=1,248\) ways to draw 2 cards that are not both \(♠\). So, the odds against drawing 2 \(♠\) are \(1,248:78=16:1\).

  2. Given the following probabilities of an event, find the corresponding odds for and odds against that event.

    1. \(P(E)=\frac{3}{5}\)
    2. \(P(E)=17\%\)
    Otkrij odgovor
    1. Using the formula, we have: \[\begin{array}{lll}\text{odds for}\ E & = & P(E):(1-P(E)) \\ & = & \frac{3}{5}:(1-\frac{3}{5}) \\ & = & \frac{3}{5}:\frac{2}{5} \\ & = & 3:2.\end{array}\] (Note that in the last step, we simplified by multiplying both terms in the ratio by 5.)
      Since the odds for \(E\) are \(3:2\), the odds against \(E\) must be \(2:3\).
    2. Again, we’ll use the formula: \[\begin{array}{lll}\text{odds for}\ E & = & P(E):(1-P(E)) \\ & = & 0.17:(1-0.17) \\ & = & 0.17:0.83 \\ & \approx & 1:4.88.\end{array}\] (In the last step, we simplified by dividing both terms in the ratio by 0.17.)
      It follows that the odds against \(E\) are approximately \(4.88:1\).

  3. Find \(P(E)\) if \(E\):

    1. The odds of \(E\) are \(2:1\) in favor
    2. The odds of \(E\) are \(6:1\) against
    Otkrij odgovor
    1. Using the formula we just found, we have \(P(E)=\frac{2}{2+1}=\frac{2}{3}\).
    2. If the odds against are \(6:1\), then the odds for are \(1:6\). Thus, using the formula, \(P(E)=\frac{1}{1+6}=\frac{1}{7}\).
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Kako: What Are the Odds?

  1. Compute odds.
  2. Determine odds from probabilities.
  3. Determine probabilities from odds.
  4. If you roll a fair 6-sided die, what are the odds for rolling a 5 or higher?
  5. If you roll two fair 6-sided dice, what are the odds against rolling a sum of 7?
  6. If you draw a card at random from a standard deck, what are the odds for drawing a
  7. If you draw 2 cards at random from a standard deck, what are the odds against them both being
  8. The sample space for this experiment is {1, 2, 3, 4, 5, 6}. Two of those outcomes are in the event “roll a five or higher,” while four are not. So, the odds for rolling a five or higher are

Pitanja koja ljudi postavljaju

Mean or median, which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree. The gap is itself information.

What does a p-value actually say?

The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

Why divide by n − 1 for the sample variance?

The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

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