I want to calculate the expected value of a ticket in a lottery game which gives players a probability $p$ of winning a jackpot prize of $j$ dollars. The total number of tickets in play is $t$.
If every winning ticket gets the full prize amount, the expected value for a ticket is given by $jp$. However, if winners must evenly split the prize in case of multiple winners, then the expected value depends on the number of winners $W$.
The expected number of winners is $tp$. The probability that the number of winners $W$ is $w = 0, 1, 2, \dotsc$, follows a Poisson distribution with the expected number of winners as its parameter:
$$P(W=w) \sim Pois(tp) = \frac{tp^we^{-tp}}{w!}$$
I don't know how to get from there to calculating an accurate expected value for the ticket as a function of the number of tickets in play.
In reading online, I've found two different methods each used by several sources. If I'm following them correctly, then they give different results. My question is 1) which one is correct? 2) what is the error in reasoning (or in my understanding/implementation) in the incorrect method?
Method 1: Number of Winners
The first method calculates the probability that the number of winners $W$ will be $w = 0, 1, \dotsc, t$, given that there is at least one winner:
$$P(W=w | W>0) = \frac{P(W>0|W=w)P(W=w)}{P(W>0)}$$
Where,
- $P(W>0|W=w)$ is $\left\{
\begin{array}{lr}
0 & : w = 0\\
1 & : w > 0
\end{array}
\right.$ - $P(W=w)$ is the probability of $w$ winners: $\frac{tp^we^{-tp}}{w!}$
- $P(W>0)$ is the probability of more than one winner: $1 – P(W=0)$
So the expected value of the ticket is given by:
$$p\sum_{w=1}^{t} \frac{j}{w}\frac{P(W=w)}{1-P(W=0)}$$
For a numerical example, we'll tabulate the first few values of $P(W=w)$ for a lottery with a 1/34,220 chance of winning \$100,000 jackpot, with 6,000 tickets in play, so $p = 1/34,220; j = 100,000; \text{and } t = 6,000$
$$\begin{array}{c|c|c|c|c|}
\text{Winners} & \text{Probability} & \text{Conditional Probability} & \text{Share} & \text{Contribution } \\
w & P(W=w) & P(W=w|W>0) & j/w & (j/w)P(W=w|W>0) \\ \hline
0 & 0.839 & 0 & \text{\$0} & \text{\$0} \\ \hline
1 & 0.147 & 0.913 & \text{\$100,000} & \text{\$91,300} \\ \hline
2 & 0.013 & 0.081 & \text{\$50,000} & \text{\$4,050} \\ \hline
\end{array}$$
Summing the contribution column and multiplying by $p$ gives an expected value of $2.79.
Online resources which use Method 1
- "Powerball Odds" by Durango Bill – see the section titled "Sample Calculation to Find the Expected Shared Jackpot Amount When a Large Number of Tickets are in Play"
- "I Am A Statistician and I Buy Lottery Tickets" by DC Woods.
- "Is it Ever Worth it to play Mega Millions?" by David Torbert
Method 2: Number of Other Winners
The second method calculates the probability that the number of total winners $W$ is $w = 0, 1, \dotsc, t$, given that our ticket is a winner:
$$P(W=w|Winner) = \frac{P(Winner|W=w)P(W=w)}{P(Winner)}$$
Where,
- $P(Winner)$ is the probability that our ticket is a winner: $p$
- $P(Winner|W=w)$ is the probability that our ticket is a winner given $w$ winning tickets: $w/t$
- $P(W=w)$ is the probability of $w$ winners: $\frac{tp^we^{-tp}}{w!}$
Plugging those figures in shows that $P(W=w|Winner)$ reduces to $P(W=w-1)$:
$$\frac{w}{t}\frac{P(W=w)}{p} = \frac{tp^{w-1}e^{-tp}}{(w-1)!} = P(W=w-1)$$
So the expected value is given by:
$$p\sum_{w=1}^{t}\frac{j}{w}\frac{tp^{w-1}e^{-tp}}{(w-1)!}$$
Using the same lottery numbers as above, the first few values of $w$ are given in the following table.
$$\begin{array}{c|c|c|c|c|}
\text{Winners} & \text{Probability} & \text{Conditional Probability} & \text{Share} & \text{Contribution } \\
w & P(W=w) & P(W=w|Winner) & j/w & (j/w)P(W=w|Winner) \\ \hline
0 & 0.839 & 0 & \text{n/a} & \text{\$0} \\ \hline
1 & 0.147 & 0.839 & \text{\$100,000} & \text{\$83,900} \\ \hline
2 & 0.013 & 0.147 & \text{\$50,000} & \text{\$7,350} \\ \hline
\end{array}$$
Summing the contribution column and multiplying by $p$ gives an expected value of $2.67.
Online Resources Which Use Method 2
- "Mega Millions and Powerball Odds: Can You Ever Expect a Ticket to be Profitable?" by Jeremy Elson. See especially his "Computing the Expected Jackpot: The Gory Details".
- The accepted answer for the math.stackoverflow question, "What's the expected value of a lottery ticket?", gives a nice formula which is equivalent to Method 2: $\dfrac j t (1-(1-p)^t)$
- Mark Adler's answer to the math.stackexchange question "Is Mega Millions Positive Expected Value?"
Clearly the expected payout for the example lottery above cannot be both \$2.79 and \$2.67, but I'm having a difficult time reasoning my way to the correct method. Any hints will be appreciated!
Best Answer
Proof that Methods 2 and 3 are equivalent:
Method 3 has a very intuitive appeal as the correct approach when interpreted as someone purchasing $all$ $t$ tickets. Now we'll show that Method 2 gives the same result if we use the binomial for the distribution of winning tickets instead of the Poisson.
The expected value for Method 2 is: $$p\sum_{w=1}^t\frac jw {t-1 \choose w-1}p^{w-1}(1-p)^{t-w}$$ $$=j \sum_{w=1}^t \frac {(t-1)!}{w!(t-w)!}p^{w}(1-p)^{t-w}$$ $$=\frac jt \sum_{w=1}^t {t \choose w}p^{w}(1-p)^{t-w}$$ $$=\frac jt (1-(1-p)^t ) $$
which is also the result for Method 3.
If we had started with the Poisson distribution as an approximation to the binomial, then the expected value for Method 2 reduces to $$ \frac jt(1-e^{-tp}) $$
The Poisson will be a good approximation to the binomial when $t$ is large, $p$ is small and $tp$ is moderate, which holds here.