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Q&A 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]

0 answers  ·  posted 10mo ago by Chgg Clou‭  ·  edited 10mo ago by Chgg Clou‭

Question lottery
#10: Post edited by user avatar Chgg Clou‭ · 2023-07-23T22:08:11Z (10 months ago)
  • How can players resolve the conflict between a lottery's too negative EV, but passable $\Pr($you win jackpot at least once $|n \text{ plays})$?
  • How to decide whether to buy a lottery with a too negative EV, but passable $\Pr($you win jackpot at least once│n plays)?
  • #### [Daily Keno's too negative Expected Value](https://math.codidact.com/posts/289006) looks scammy.
  • As too many lottery players play the same lottery repeatedly, I consider [$Pr(\text{you win jackpot at least once}|n \text{ plays}) = 1 - (1 - p)^n$](https://math.stackexchange.com/a/4685078).
  • #### But some rational players can sensibly tolerate this fairish $\Pr($ you win jackpot at least once $|n \text{ plays})$.
  • 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 \times 2 plays/day \times 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 $ ightarrow 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)](https://planetcalc.com/7933/) | = 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 $1.25M, particularly when these probabilities cover 6 months.
  • >[Playing the lottery can be worth it, even with negative expected value.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### How can players resolve this strife, as written in this question's title?
  • #### [Daily Keno's too negative Expected Value](https://math.codidact.com/posts/289006) looks scammy.
  • As many play the same lottery repeatedly, I shall consider [$\Pr($ you win jackpot at least once│n plays) $= 1 - (1 - p)^n$](https://math.stackexchange.com/a/4685078).
  • #### 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 $ ightarrow 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)](https://planetcalc.com/7933/) | = 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.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### 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?
#9: Post undeleted by user avatar Chgg Clou‭ · 2023-07-23T22:05:35Z (10 months ago)
#8: Post deleted by user avatar Chgg Clou‭ · 2023-07-23T22:02:34Z (10 months ago)
#7: Post edited by user avatar Chgg Clou‭ · 2023-07-23T06:24:10Z (10 months ago)
  • #### [Daily Keno's too negative Expected Value](https://math.codidact.com/posts/289006) looks scammy.
  • As too many lottery players play the same lottery repeatedly, I consider [$Pr(\text{you win jackpot at least once}|n \text{ plays}) = 1 - (1 - p)^n$](https://math.stackexchange.com/a/4685078).
  • #### But some players tolerate the fairish $\Pr($ you win jackpot at least once $|n \text{ plays})$?
  • 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 rational _Homo Economicus_ can accept these humdrum probabilities, like $1/10K$ or $1/5883$ probability of winning $1.25M, particularly when these probabilities cover 6 months.
  • >[Playing the lottery can be worth it, even with negative expected value.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### How can players resolve this strife, as written in this question's title?
  • #### [Daily Keno's too negative Expected Value](https://math.codidact.com/posts/289006) looks scammy.
  • As too many lottery players play the same lottery repeatedly, I consider [$Pr(\text{you win jackpot at least once}|n \text{ plays}) = 1 - (1 - p)^n$](https://math.stackexchange.com/a/4685078).
  • #### But some rational players can sensibly tolerate this fairish $\Pr($ you win jackpot at least once $|n \text{ plays})$.
  • 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### How can players resolve this strife, as written in this question's title?
#6: Post edited by user avatar Chgg Clou‭ · 2023-07-22T20:54:17Z (10 months ago)
  • How can players resolve the conflict between a lottery's (too) negative EV, but passable $Pr(\text{winning jackpot at least once}|n \text{ plays})$?
  • How can players resolve the conflict between a lottery's too negative EV, but passable $\Pr($you win jackpot at least once $|n \text{ plays})$?
  • #### Daily Keno has too negative Expected Value.
  • I accept that [Daily Keno's EV looks scammy](https://math.codidact.com/posts/289006). As it's unrealistic to presuppose someone playing merely 1 play, I consider [$Pr(\text{winning jackpot at least once}|n \text{ plays}) = 1 - (1 - p)^n$](https://math.stackexchange.com/a/2639469).
  • #### But some players can tolerate the fairish $Pr(\text{winning jackpot at least once}|n \text{ plays})$.
  • 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 rational _Homo Economicus_ can accept these humdrum probabilities, like $1/10K$ or $1/5883$ probability of winning $1.25M, particularly when these probabilities cover 6 months.
  • >[Playing the lottery can be worth it, even with negative expected value.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### How can players resolve this strife, as written in this question's title?
  • #### [Daily Keno's too negative Expected Value](https://math.codidact.com/posts/289006) looks scammy.
  • As too many lottery players play the same lottery repeatedly, I consider [$Pr(\text{you win jackpot at least once}|n \text{ plays}) = 1 - (1 - p)^n$](https://math.stackexchange.com/a/4685078).
  • #### But some players tolerate the fairish $\Pr($ you win jackpot at least once $|n \text{ plays})$?
  • 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 rational _Homo Economicus_ can 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.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### How can players resolve this strife, as written in this question's title?
#5: Post edited by user avatar Chgg Clou‭ · 2023-07-22T20:46:24Z (10 months ago)
  • How do I resolve the conflict between a lottery's (very) negative Expected Value, but passable Pr(winning jackpot at least once|n plays)?
  • How can players resolve the conflict between a lottery's (too) negative EV, but passable $Pr(\text{winning jackpot at least once}|n \text{ plays})$?
  • #### Daily Keno has too negative Expected Value.
  • I accept that [Daily Keno's EV looks scammy](https://math.codidact.com/posts/289006). As it's unrealistic to presuppose someone playing merely 1 play, I consider [$Pr(\text{winning jackpot at least once}|n \text{ plays}) = 1 - (1 - p)^n$](https://math.stackexchange.com/a/2639469).
  • #### But some players can tolerate the fairish $Pr(\text{winning jackpot at least once}|n \text{ plays})$.
  • 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 rational _Homo Economicus_ can 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.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### Then, how can players resolve this strife between a too negative EV, and a passable $Pr(\text{winning jackpot at least once}|n \text{ plays})$ that they can brook?
  • #### Daily Keno has too negative Expected Value.
  • I accept that [Daily Keno's EV looks scammy](https://math.codidact.com/posts/289006). As it's unrealistic to presuppose someone playing merely 1 play, I consider [$Pr(\text{winning jackpot at least once}|n \text{ plays}) = 1 - (1 - p)^n$](https://math.stackexchange.com/a/2639469).
  • #### But some players can tolerate the fairish $Pr(\text{winning jackpot at least once}|n \text{ plays})$.
  • 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 rational _Homo Economicus_ can 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.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### How can players resolve this strife, as written in this question's title?
#4: Post edited by user avatar Chgg Clou‭ · 2023-07-22T20:45:07Z (10 months ago)
  • #### Daily Keno has too negative Expected Value.
  • I accept that lotteries are scammy, as outlined [here](https://math.codidact.com/posts/289006/289007#answer-289007). It's unrealistic to presuppose someone playing merely 1 play! Thus I consider Pr(winning jackpot at least once|n plays) = [$1 - (1 - p)^n$](https://math.stackexchange.com/a/2639469).
  • #### But some players can tolerate the fairish Pr(winning 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 rational _Homo Economicus_ can accept these humdrum $Pr($winning jackpot at least once$|$n plays) , like $1/10K$ or $1/5883$ probability of winning $1.25M, particularly when these probabilities cover 6 months.
  • >[Playing the lottery can be worth it, even with negative expected value.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### Then, how can players resolve this strife between a too negative EV, and a passable Pr that they can brook?
  • #### Daily Keno has too negative Expected Value.
  • I accept that [Daily Keno's EV looks scammy](https://math.codidact.com/posts/289006). As it's unrealistic to presuppose someone playing merely 1 play, I consider [$Pr(\text{winning jackpot at least once}|n \text{ plays}) = 1 - (1 - p)^n$](https://math.stackexchange.com/a/2639469).
  • #### But some players can tolerate the fairish $Pr(\text{winning jackpot at least once}|n \text{ plays})$.
  • 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 rational _Homo Economicus_ can 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.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### Then, how can players resolve this strife between a too negative EV, and a passable $Pr(\text{winning jackpot at least once}|n \text{ plays})$ that they can brook?
#3: Post edited by user avatar Chgg Clou‭ · 2023-07-22T20:40:12Z (10 months ago)
  • Isn't considering Pr(winning jackpot at least once|n plays) more accurate and precise, making lotteries less loony?
  • How do I resolve the conflict between a lottery's (very) negative Expected Value, but passable Pr(winning jackpot at least once|n plays)?
  • I accept that lotteries are scammy, as outlined [here](https://math.codidact.com/posts/289006/289007#answer-289007). But why do most answers on lotteries consider the Pr(winning jackpot in 1 play), rather than Pr(winning jackpot at least once|n plays) = [$1 - (1 - p)^n$](https://math.stackexchange.com/a/2639469)? It's unrealistic to presuppose someone playing merely 1 play! Isn't this latter more accurate and precise? If so, then playing the lottery for entertainment is less kooky as appears?
  • 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 |
  • These Pr(winning jackpot at least once|n plays) are higher than Pr(winning jackpot on 1 play). Doesn't considering $1 - (1 - p)^n$ make lotteries slightly less foolish? Some rational _Homo Economicus_ can find a $1/10K$ or $1/5883$ probability of winning $1.25M appears reasonable, particularly when these probabilities cover 6 months?
  • >[Playing the lottery can be worth it, even with negative expected value.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### Daily Keno has too negative Expected Value.
  • I accept that lotteries are scammy, as outlined [here](https://math.codidact.com/posts/289006/289007#answer-289007). It's unrealistic to presuppose someone playing merely 1 play! Thus I consider Pr(winning jackpot at least once|n plays) = [$1 - (1 - p)^n$](https://math.stackexchange.com/a/2639469).
  • #### But some players can tolerate the fairish Pr(winning 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 \times 2 plays/day \times 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)](https://planetcalc.com/7933/) | = 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 rational _Homo Economicus_ can accept these humdrum $Pr($winning jackpot at least once$|$n plays) , 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.](https://money.stackexchange.com/a/106336)
  • >[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
  • #### Then, how can players resolve this strife between a too negative EV, and a passable Pr that they can brook?
#2: Question closed by user avatar Peter Taylor‭ · 2023-07-19T07:17:00Z (10 months ago)
#1: Initial revision by user avatar Chgg Clou‭ · 2023-07-18T07:43:29Z (10 months ago) 
Isn't considering Pr(winning jackpot at least once|n plays) more accurate and precise, making lotteries less loony?
I accept that lotteries are scammy, as outlined [here](https://math.codidact.com/posts/289006/289007#answer-289007). But why do most answers on lotteries consider the Pr(winning jackpot in 1 play), rather than Pr(winning jackpot at least once|n plays) = [$1 - (1 - p)^n$](https://math.stackexchange.com/a/2639469)? It's unrealistic to presuppose someone playing merely 1 play! Isn't this latter more accurate and precise? If so, then playing the lottery for entertainment is less kooky as appears?

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 \times 2 plays/day  \times 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)](https://planetcalc.com/7933/) | = 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 |

These Pr(winning jackpot at least once|n plays) are higher than Pr(winning jackpot on 1 play). Doesn't considering $1 - (1 - p)^n$ make lotteries slightly less foolish? Some rational _Homo Economicus_ can find a $1/10K$ or $1/5883$ probability of winning $1.25M appears reasonable, particularly when these probabilities cover 6 months? 

>[Playing the lottery can be worth it, even with negative expected value.](https://money.stackexchange.com/a/106336)

>[From a mathematical expected-value standpoint, there is no difference between gambling (e.g. buying a lottery ticket) and investing (e.g. buying a share of stock).](https://money.stackexchange.com/a/63930)
lottery