Reading a Payoff Table
A payoff table connects possible outcomes to the amounts gained or lost when they occur. To find the expected payoff, use the probability of each outcome as its weight. Outcomes that are more probable contribute more to the weighted average; outcomes with negative payoffs contribute negative amounts.
Before calculating, clarify what the payoff measures. It might be a prize, a net gain after costs, or a reimbursement. A negative payoff means a loss on the scale being used. If a table gives a net payoff, do not subtract the cost again; if it gives a gross payout, do not silently treat it as a net amount.
The outcomes must form a complete, non-overlapping set of possibilities: each play or case belongs to exactly one row. The table must also assign a definite payoff to each row. If an outcome has several possible payoffs, the table needs more detail before this calculation can be made.
A useful way to read a row is: “This payoff occurs with this probability.” Multiply the payoff by that probability, then add the products for all rows. The calculation uses the probabilities, not the number of rows. In particular, two rows with the same payoff are still two outcomes, but their probabilities can be combined when considering that payoff.
Worked Example: A Prize Game With Gains and Losses
A hypothetical game has four mutually exclusive outcomes. Its table reports net payoff in dollars, so any costs have already been accounted for. Find the expected payoff per play.
| Outcome | Probability | Net payoff |
|---|---|---|
| No prize | 0.70 | −$2 |
| Small prize | 0.20 | $0 |
| Medium prize | 0.08 | $8 |
| Large prize | 0.02 | $48 |
First check that the outcomes cover the possible results and that the probabilities sum to one: \(0.70+0.20+0.08+0.02=1.00\). Then multiply each net payoff by its probability and add:
The expected net payoff is $0.20 per play. This is a probability-weighted average, not a promise that a particular play will earn 20 cents. A single play has one of the four listed payoffs; $0.20 is not among them.
When Different Outcomes Have the Same Payoff
A table can distinguish outcomes that lead to the same payoff. For example, two different colors might each win the same prize, or two different administrative outcomes might each result in the same loss. In the row-by-row calculation, include every outcome with its own probability and payoff. Alternatively, first add the probabilities of outcomes with identical payoffs, then use the combined probability for that payoff.
This combining step is valid because the outcomes are mutually exclusive. If either of two distinct outcomes leads to the same payoff, the probability of receiving that payoff is the sum of their probabilities. The payoff should not be counted twice using that combined probability; do one calculation per distinct payoff after combining.
Worked Example: Combining Outcomes With the Same Prize
A hypothetical spinner has four possible colors. Landing on yellow or blue awards $12; red awards $0; green results in a net loss of $6. Find the expected net payoff.
| Spinner outcome | Probability | Net payoff |
|---|---|---|
| Yellow | 0.25 | $12 |
| Red | 0.30 | $0 |
| Blue | 0.25 | $12 |
| Green | 0.20 | −$6 |
The probabilities sum to \(0.25+0.30+0.25+0.20=1.00\). Keeping all outcome rows, the expected payoff is
The two $12 outcomes can instead be combined: their total probability is \(0.25+0.25=0.50\). The distinct payoffs are then $12 with probability 0.50, $0 with probability 0.30, and −$6 with probability 0.20. Using that shorter table gives
Both calculations agree. The expected net payoff is $4.80 per spin. It is not one of the possible payoffs, and it does not say that each spin earns $4.80.
Why Combining Rows Preserves the Expectation
The arithmetic behind combining repeated payoffs is simple. Suppose a table has outcomes \(O_i\) with probabilities \(p_i\) and payoffs \(v_i\). Now group outcomes that share a payoff. Let the distinct payoffs be \(w_1,\ldots,w_m\), and let \(I_j\) be the set of row indices whose payoff equals \(w_j\). The combined probability for payoff \(w_j\) is \(q_j=\sum_{i\in I_j}p_i\). The original row-by-row sum can be regrouped as follows:
The first equality just groups the original terms according to payoff. Within group \(j\), every \(v_i\) equals \(w_j\), so \(w_j\) can be factored out of the sum. The last expression is the expected payoff computed from the grouped table. Thus grouping repeated payoffs changes the layout, not the result.
Worked Example: Grouping Reimbursement Outcomes
Consider a hypothetical reimbursement arrangement with four mutually exclusive outcomes for a single service episode. The table gives the net payoff in dollars. Find the expected payoff and show the effect of combining outcomes with the same payoff.
| Outcome | Probability | Net payoff |
|---|---|---|
| Full approval | 0.55 | $120 |
| Partial approval | 0.25 | $60 |
| Denial without appeal | 0.15 | −$40 |
| Denial after appeal | 0.05 | −$40 |
The probabilities sum to \(0.55+0.25+0.15+0.05=1.00\). Calculate from the four rows:
The two denial outcomes both have a payoff of −$40. Their combined probability is \(0.15+0.05=0.20\). The grouped calculation is
The expected net payoff is $73 per service episode under the stated probabilities and payoff table. Since $73 is not one of the possible payoffs, it should be understood as a weighted average, not as the reimbursement that must occur for an individual episode.
Checks That Prevent Common Errors
A careful calculation starts with the table, not with the multiplication. Use these checks before interpreting the result:
- Check the probability total. The table’s mutually exclusive, exhaustive outcomes should have probabilities summing to 1. A total other than 1 may indicate missing outcomes, overlapping rows, or a transcription error.
- Use each row’s own probability. Do not give every row equal weight unless the table actually assigns equal probabilities.
- Keep the payoff sign. A loss of $6 is a payoff of −$6, so its contribution is probability multiplied by a negative number.
- Track the payoff definition and units. Do not subtract a cost twice if the listed amount is already net, and state the expected value in the same units as the payoff.
- Combine probabilities only for matching payoffs. Different outcomes can be grouped for this calculation when their payoffs are identical; their payoff values are not added together.
A frequent mistake is to average the payoff entries without using the probabilities. For example, in the spinner example, averaging the four row values would give \((12+0+12-6)/4=4.50\) dollars. That calculation treats all four colors as equally likely, contradicting the table. The correct calculation uses the stated probabilities and gives $4.80. Another mistake is to count a repeated payoff only once while keeping just one of its row probabilities; that leaves out the probability of the other outcome with that payoff.
Check Your Understanding
Use the probabilities and payoff definitions given in each question.
- A table lists three exhaustive outcomes with probabilities 0.50, 0.30, and 0.20 and payoffs of $4, $0, and −$5. Calculate the expected payoff.
- Two mutually exclusive outcomes each lead to a payoff of $10, with probabilities 0.15 and 0.25. What is the combined probability of that payoff?
- Why is it incorrect to average the payoff entries in a table when the outcomes have unequal probabilities?
- A payoff table’s listed probabilities sum to 0.92. Name one check you should make before calculating its expected payoff.
- Can an expected payoff be different from every payoff listed in the table? Explain briefly.