From Raffle Rules to Net Winnings
In Reading a Distribution from a Context Description, you learned to connect chance outcomes to values of a random variable. A raffle adds an important detail: the prize is not necessarily the amount a player gains. The player pays for a ticket, so the result should include that cost. A ticket that wins a prize worth less than its cost can still produce a net loss, and a ticket that wins nothing loses the ticket cost.
In this tutorial, we will build a probability distribution for one randomly selected raffle ticket. We will treat each ticket as equally likely to be selected, as is appropriate when one ticket is drawn at random from the tickets sold. First translate each raffle outcome into net winnings, and then use the number of tickets that produce each outcome to find its probability.
The subtraction applies to every ticket, not just losing tickets. For example, if a ticket costs $3 and wins a $25 prize, its net winnings are \(25-3=$22\). If it wins nothing, its net winnings are \(0-3=-$3\). The negative sign means the player ends up $3 below the starting amount.
Once you calculate the net outcome for each prize category, assign a probability to each outcome. When all tickets are equally likely, a category with \(k\) tickets out of \(N\) total tickets has probability \(k/N\). If different prize categories produce the same net winnings, combine their ticket counts in the distribution: a probability distribution lists each possible value of \(X\) once.
A Reliable Construction Process
Start by defining \(X\) in context. Then make sure the ticket cost is subtracted from every prize value, including zero for no prize. After that, count how many tickets produce each distinct net amount. Dividing those counts by the total number of tickets gives the probabilities. This applies the distribution-building ideas from Probability Distribution of a Discrete Random Variable and Building a Distribution from a Sample Space to a raffle.
State that \(X\) is the net winnings, in dollars, from one randomly selected ticket.
For each prize value \(G\), calculate \(G-c\). Include \(G=0\) for tickets that win no prize.
Group together all tickets that give the same value of \(X\), even if they came from more than one prize category.
Divide each outcome's ticket count by the total number of tickets. Check that each probability is between 0 and 1 and that the probabilities sum to 1.
A raffle's outcomes should account for every ticket exactly once. A ticket that wins a first prize cannot also be counted as a losing ticket in the same one-drawing model. When the prize categories and no-prize category cover all tickets, their counts should add to the total. As in Checking Whether a Probability Distribution Is Valid, also check the probabilities themselves.
Worked Example: Tickets for a Community Art Raffle
An invented community art raffle sells 200 tickets for $3 each. One ticket wins a framed artwork valued at $100, and four other tickets win small items valued at $25 each. Every remaining ticket wins no prize. Let \(X\) be the net winnings, in dollars, for one randomly selected ticket.
State. Construct the probability distribution of \(X\), including the $3 ticket cost.
Plan. Subtract $3 from each prize value, including $0 for a ticket that wins nothing. Count how many tickets yield each net amount, divide each count by 200, and check that all tickets are represented.
Do. The net winnings are \(100-3=$97\) for the artwork, \(25-3=$22\) for each small-item prize, and \(0-3=-$3\) for a no-prize ticket. There are \(200-1-4=195\) no-prize tickets. The probabilities are:
The distribution is:
| Net winnings, \(x\) (dollars) | Number of tickets | \(P(X=x)\) |
|---|---|---|
| \(-3\) | 195 | 0.975 |
| 22 | 4 | 0.020 |
| 97 | 1 | 0.005 |
The ticket counts add to \(195+4+1=200\). The probabilities add to \(0.975+0.020+0.005=1.000\), and each is between 0 and 1.
Conclude. The model assigns probability 0.975 to net winnings of \(-$3\): a randomly selected ticket has a 0.975 probability of losing the $3 ticket cost. It assigns probability 0.020 to net winnings of $22 and probability 0.005 to net winnings of $97. These are probabilities for one ticket under the stated raffle model.
Reading the Net Amounts Carefully
A net amount of zero means the prize value exactly equals the ticket cost. The player neither gains nor loses money overall, before considering anything outside the raffle. A negative amount means a net loss; a positive amount means the prize value exceeds the ticket cost. These labels refer to net winnings, not to the prize value alone.
The possible values of \(X\) do not have to match the prize values. If every ticket costs $4, for instance, then a $12 prize corresponds to \(X=8\), and no prize corresponds to \(X=-4\). Keeping the variable's definition visible helps prevent accidentally reporting a prize amount where the question asks for net winnings.
The probability of a range or event is found from the net-winnings distribution, just as in Probabilities of Ranges and Inequalities. For example, “net winnings are positive” means \(X>0\), so a break-even ticket with \(X=0\) is not included. “At least break even” means \(X\geq 0\), which does include the zero outcome. Translate the wording into an inequality before adding probabilities.
Worked Example: A Neighborhood Garden Raffle
An invented neighborhood garden raffle sells 120 tickets for $4 each. One ticket wins a planter worth $80, five tickets win gardening kits worth $12 each, and ten tickets win seed packets worth $4 each. The remaining tickets win nothing. Let \(X\) be the net winnings, in dollars, for one randomly selected ticket.
State. Construct the probability distribution and find the probability of breaking even or better.
Plan. Subtract $4 from every prize value and from zero for no prize. Count tickets for each net outcome. For “breaking even or better,” include outcomes with \(X\geq 0\), including the zero-net outcome.
Do. The net amounts are \(80-4=$76\), \(12-4=$8\), \(4-4=$0\), and \(0-4=-$4\). The number of no-prize tickets is \(120-1-5-10=104\). The distribution is:
| Net winnings, \(x\) (dollars) | Number of tickets | \(P(X=x)\) |
|---|---|---|
| \(-4\) | 104 | \(104/120=0.8667\) |
| 0 | 10 | \(10/120=0.0833\) |
| 8 | 5 | \(5/120=0.0417\) |
| 76 | 1 | \(1/120=0.0083\) |
The counts total \(104+10+5+1=120\). Using the unrounded fractions, the probabilities total \((104+10+5+1)/120=120/120=1\). For breaking even or better, add the probabilities for \(X=0\), \(X=8\), and \(X=76\):
The same result comes from adding the rounded table entries: \(0.0833+0.0417+0.0083=0.1333\). The probability is rounded to four decimal places.
Conclude. According to the model, a randomly selected ticket has probability 0.1333 of breaking even or better. The ten seed-packet tickets count as breaking even because their $4 prize value equals the $4 ticket cost; they do not count as a positive net win.
Combining Categories and Finding Event Probabilities
Sometimes two listed prize categories produce the same net winnings. This happens when the prize values are the same, and it can also happen if the raffle lists separate ways to receive a prize of equal value. In the distribution, put that net amount in one row and add the counts. For instance, if three tickets in one category and two in another each give net winnings of $6, then \(P(X=6)=(3+2)/N\), not two separate entries for \(X=6\).
After the table is complete, questions about chance refer to the random variable's net amount. To find \(P(X>0)\), add probabilities for positive net outcomes. To find \(P(X<0)\), add probabilities for negative net outcomes. For \(P(X\geq 0)\), include a break-even value if one appears. As with other discrete distributions, include exactly the values allowed by the inequality and avoid counting any ticket more than once.
The model describes the chance for one randomly selected ticket under the raffle rules. Its probabilities do not promise what will happen to a particular player. The interpretation in Interpreting a Probability Distribution in Context still applies: the probability describes the model's chance for the event, and long-run interpretations refer to comparable repetitions of the chance process.
Worked Example: A Greenhouse Fundraiser Raffle
An invented greenhouse fundraiser sells 500 tickets for $2 each. One ticket wins a garden bench worth $250, nine tickets win tool sets worth $20 each, and 40 tickets win seed trays worth $2 each. All other tickets win nothing. Let \(X\) be the net winnings, in dollars, for one randomly selected ticket. Find the distribution and the probability of a positive net win.
State. Construct the distribution of \(X\) and calculate \(P(X>0)\).
Plan. Subtract the $2 cost from every prize value and from zero for no prize. Divide each category's ticket count by 500. To find \(P(X>0)\), add the probabilities of outcomes strictly greater than zero; a break-even outcome is excluded.
Do. The net amounts are \(250-2=$248\), \(20-2=$18\), \(2-2=$0\), and \(0-2=-$2\). There are \(500-1-9-40=450\) tickets that win nothing. Thus:
| Net winnings, \(x\) (dollars) | Number of tickets | \(P(X=x)\) |
|---|---|---|
| \(-2\) | 450 | \(450/500=0.900\) |
| 0 | 40 | \(40/500=0.080\) |
| 18 | 9 | \(9/500=0.018\) |
| 248 | 1 | \(1/500=0.002\) |
The counts sum to \(450+40+9+1=500\), and the probabilities sum to \(0.900+0.080+0.018+0.002=1.000\). For a strictly positive net win, include the tool-set and bench outcomes but not the seed-tray outcome:
Conclude. The model assigns probability 0.020 to positive net winnings for one randomly selected ticket. The 40 seed-tray tickets break even, so they are not included in \(P(X>0)\), even though they win a prize.
Common Mistakes and AP Exam Tips
- Listing prize values instead of net winnings. If the question defines \(X\) as net winnings, subtract the ticket cost from the prize. A $20 prize on a $2 ticket gives \(X=$18\), not \(X=$20\).
- Forgetting the cost on a losing ticket. A no-prize ticket does not have net winnings of $0 when it cost money. Its net amount is \(0-c=-c\).
- Calling a break-even result a positive win. If the prize equals the ticket cost, \(X=0\). This is included in “at least break even” but not in “positive net winnings.”
- Using the prize count as the total. Probabilities use the total number of tickets, including losing tickets, as the denominator. Check that the ticket counts for all outcomes add to that total.
- Listing the same net value more than once. A distribution pairs each distinct value of \(X\) with one probability. Combine ticket counts whenever outcomes give the same net amount.
- Interpreting the distribution as a guarantee. A probability describes the raffle model's chance for a ticket, not what must happen to a specific player or a fixed group of players.
A full-credit response clearly defines \(X\), shows the subtraction of the ticket cost, and gives the probabilities based on the total number of tickets. It checks that every ticket is accounted for and that the probabilities form a valid distribution. When asked for an event probability, it identifies which net values satisfy the wording before adding them and interprets the result in context.
Key Takeaway
To build a raffle distribution for net winnings, subtract the ticket cost from every prize value, including zero for no prize. Then group tickets by distinct net amount and divide each group count by the total number of tickets. The resulting distribution distinguishes losses, break-even outcomes, and positive net wins.
Check Your Understanding
For each raffle, define net winnings as prize value minus ticket cost. Show how you account for every ticket.
- A raffle has 100 tickets costing $2 each. One ticket wins $50, four tickets win $5 each, and the rest win nothing. Construct the distribution of net winnings.
- A ticket costs $6. What is the net amount for a prize worth $6? Is that outcome a positive net win?
- A raffle sells 200 tickets. Three tickets win prizes worth $20, and seven tickets win prizes worth $2. Each ticket costs $2; all others win nothing. Find the probability of positive net winnings.
- Two different prize categories each produce net winnings of $8. One category has two tickets and the other has five. How should the distribution represent \(P(X=8)\), if there are 100 tickets in total?
- Explain why a no-prize ticket in a raffle with a $4 ticket cost has net winnings of \(-$4\), not $0.