# Computing the Expected Jackpot: The Gory Details

Computing the expected value of the jackpot given the number of tickets in play is a little tricky (well, for me, anyway). I describe two separate methods here, both of which agree. I'm grateful to Mark Eichenlaub for pointing out an error in my original formulation.

Each method will make the same assumptions:

• All tickets are independent.
• There are $n$ tickets in play.
• Each has a probability $p$ of winning the jackpot.
• There's a discount $d$ applied to all winnings to account for net present value and taxes.
• When we illustrate with specific numbers, we use the MegaMillions drawing from March of 2012, where $jackpot=640,000,000$, $n=651,915,940$, $p=1/175,711,536$ and $d=0.63\times0.59=0.3717$.
• We are computing just the jackpot's contribution to a ticket's EV.

Mark's method is the most straightforward. Each ticket's EV (not counting non-jackpot prizes) is the expectation of the jackpot to be paid to all tickets divided by number of tickets. The expected jackpot, in turn, is the jackpot times the probability there's at least one winner, which we get from the Poisson distribution with $\lambda = n p$: $$\text{EV Per Ticket} = \frac{\text{Expected Total Payout}}{n} = \frac{d \times jackpot \times (1 - \text{PMF}(\text{Winners}=0))}{n}$$ The PMF of $\text{Pois}(np)$ at 0 is 0.0245, so this comes out to $0.3560. ### Method 2 This method is a lot more cumbersome, but also a little more informative. Just seeing the ticket's EV isn't as illustrative as knowing which ties are the most likely and how much of the jackpot you win in each case. Here, our goal is to compute the expected jackpot payout to a winner. That is, on the condition that I've won, how many other people have won, too? We can compute the probability that 1, 2, 3... total players have won, the expected split in each case, and the sum over all those cases to get the expected size of a jackpot to a winner. First, assuming I've won, what's the probability there are $W$ total winners? In other words, what is $P(W=1,2,3...|\text{I won})$? We can convert this using Bayes to $$P(W=w|\text{I won}) = \frac{P(\text{I won}|W=w) \times P(W=w)}{P(\text{I won})}$$ Now, we compute each of the three terms: • $P(\text{I won}|W=w)$ is $w/n$: the chance I've won given that $w$ out of the $n$ tickets in play won • $P(W=w)$ is the PMF of $\text{Pois}(np)$ at $w$ • $P(\text{I won}) = p$ As a sanity check, I asked Mathematica to verify $$\sum_{w=1}^{\infty}P(W=w|\text{I won})=1$$ Now we can build a table of each case: the number of total winners $w$, our share of the post-tax jackpot $\frac{jackpot\times d}{w}$, and the probability of this case happening:  Winners(Including You) Probability Jackpot Share($millions) Jackpot SharePost-Tax NPV Contribution toExpected Jackpot($millions) 1 0.0244739 640. 237.89 5.82 2 0.0908017 320. 118.94 10.8 3 0.168444 213.33 79.3 13.36 4 0.208317 160. 59.47 12.39 5 0.193222 128. 47.58 9.19 6 0.143377 106.67 39.65 5.68 7 0.0886581 91.43 33.98 3.01 8 0.0469907 80. 29.74 1.4 9 0.0217928 71.11 26.43 0.58 10 0.00898383 64. 23.79 0.21 11 0.00333314 58.18 21.63 0.07 12 0.00112422 53.33 19.82 0.02 Total 62.55 Summing over all cases gives the expected jackpot payout to a winner as$62.55 million. Multiplying that by $p$ gives the jackpot's contribution to the expected value of a ticket as \$0.3560 -- the same as Method 1.

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