Table beneath shows that Pr(Keno's jackpot) > Pr(Grand's jackpot) > Pr(Lotto 6/49's jackpot). $\color{magenta}{\text{This Probability inequality tips you to buy Keno!}}$
But the Expected Value of 1
6/49 play $= \dfrac{$5E6}{13,983,816} - \$3 = -\$2.64$.
Grand play $= \dfrac{$7E6}{13,348,188} - \$3 = -\$2.48$.
Keno play $= \dfrac{$2.5E6}{2,147,181} - \$10 = -\$8.84$.
Hence, E(Keno) < E(6/49) < E(Grand) < 0. $\color{red}{\text{This Expected Value inequality tips you to buy Grand!}}$ Plainly, the outcomes from the $\color{magenta}{\text{Probability}}$ and $\color{red}{\text{Expected Value}}$ inequalities clash!
Intuitively, why does Keno have the highest probability, but Grand highest Expected Value? How do I intuit this contradictory information? Please unravel this cognitive dissonance that's freezing my brain!
Presuppose I can live snugly on any one of these jackpots, and fancy winning whichever one. How ought I construe the above computations? How ought I decide which lottery to buy?
Data for 3 lotteries from the Ontario Lottery and Gaming Corporation
| LOTTO 6/49 | DAILY GRAND | DAILY KENO | |
|---|---|---|---|
| Jackpot | \$5 million | \$2.5 million | |
| Odds of winning jackpot | $\dfrac1{13,983,816}$ | $\dfrac1{13,348,188}$ | $\dfrac1{2,147,181}$ |
| Matching numbers required to win jackpot | 6/6 | 5/5 + Grand Number | 10/20 on a \$10 bet |
| Number pool | 49 | 49 + Grand Number (7) | 70 |
| Number of plays one person can buy | 10 | 5 | 2 |
| Price per play | \$3 | \$3 | \$10 |
Such tradeoffs between higher probability of winning vs higher expected gain are found everywhere -- consider this toy example of a game that costs \$1 to play: you can win \$2 with 99% probability, or win \$980 with 1% probability. The expected values are 0.98 and 9.8 respectively, but people may prefer to play the first game because it's almost guaranteed to earn something. Compare this to investing in 1 index fund, vs hand-picking individual stocks that you think will do better. Something you may want to compute (if possible) is the variance, where higher variance is generally more risky.
If you're only buying 1 ticket then I would maximize P(any win) because expected value is not terribly relevant for a one-off event. Since you are allowed multiple "plays" per ticket, your odds will become 10/13983816, 5/13348188, and 2/2147181, meaning P(Keno) > P(649) > P(Grand). Personally, I would not play any of these because they're all negative expected value.