The Powerball jackpot just broke 200 million. Is the ticket worth what we pay for it? Let's calculate the expected return of a ticket. Let $\mathcal{P}(n,m,q)$ be the probability of matching $q$ out of $m$ balls where each ball is selected from $[n]=\{1,2,\ldots,n\}$ (without replacement). This value is easily calculated:
$$\mathcal{P}(n,m,q) = \frac{\text{# winning tickets}}{\text{# total tickets}} = \frac{{m \choose q} {n-m \choose m-q}}{ {n \choose m} } $$
For the Powerball lottery we have $n = 59$ and $m=5$ and we have an extra powerball which is a number independently chosen from $[35]$. Let $P(\bullet^{\alpha} \circ^{q})$ be the probability of matching $\alpha \in \{0,1\}$ powerballs and $q \in \{0,1,2,3,4,5\}$ regular balls.
$$ P(\bullet^{\alpha} \circ^{q}) = {35}^{-\alpha} \, \mathcal{P}(59,5,q) $$
To calculate the expected return we need to sum over the 9 ways to win $\omega$, the probability $P(\omega)$ times the winnings $W(\omega)$. Let the jackpot be $X$ million.
$$ \sum_{\omega}P(\omega)W(\omega) \;\;\; = \;\;\; 4P(\bullet) + 4P(\bullet \circ) + 7P(\bullet \circ^2) + 7P(\circ^3) + ... \;\;\; = \;\;\; 0.37 + X/175 $$
The jackpot is currently 208 million, so the expected return is 1.56 dollars, 44 cents less than the 2 dollars you pay for the ticket -- and that's before taxes! In fact, your estimated jackpot winnings need to be 285 million before a ticket is worth buying.
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