How to decide whether to buy a lottery with a too negative EV, but passable $\Pr($you win jackpot at least once│n plays)? [closed]
Closed as off topic by Peter Taylor on Jul 19, 2023 at 07:17
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Daily Keno's too negative Expected Value looks scammy.
As many play the same lottery repeatedly, I shall consider $\Pr($ you win jackpot at least once│n plays) $= 1 - (1 - p)^n$.
But some rational players can sensibly tolerate Keno's fairish $\Pr($ you win jackpot at least once │n plays).
Before COVID, I spent $5K USD on leisure travel. But I hanker to, and can, retire on $1.25M. Then I can travel less, and spend $3650 CAD/year (e.g. $5/play × 2 plays/day × 365 days/year) buying the 10 PICK $5 Daily Keno, _**twice daily.**_ I prefer Daily Keno's teensy chance of winning jackpot, over traveling's $\rightarrow 0^{+}$ chance — assuming that I don't meet a multimillionaire during travel, and marry him!
| n | n/2 = days | $1 - (1 - \dfrac1{2147181})^n$ |
|---|---|---|
| 21 | 10.5 | = 0.00010 ≈ 1/97,599 |
| 215 | 107.5 (= 3 months, 17 days) | = 0.00010 ≈ 1/10k |
| 366 | 183 (= half a year) | = 0.00017 ≈ 1/5883 |
| 730 | 365 (= 1 year) | = 0.00034 ≈ 1/2941 |
| 1460 | 730 (= 2 years) | = 0.00068 ≈ 1/1471 |
| 2150 | 1075 (= 2 years, 11 months) | = 0.0010 ≈ 1/1000 |
Some Homo Economicus can logically accept these humdrum probabilities, like $1/10K$ or $1/5883$ probability of winning $1.25M, particularly when these probabilities cover 6 months.
Playing the lottery can be worth it, even with negative expected value.
How can rational players decide between Keno’s EV and $\Pr($ you win jackpot at least once │n plays), as written in this question's title?
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