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Comments on How to decide whether to buy a lottery with a too negative EV, but passable $Pr ($ you win jackpot at least once│n plays)?

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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]

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Closed as off topic by Peter Taylor‭ on Jul 19, 2023 at 07:17

This question is not within the scope of Mathematics.

This question was closed; new answers can no longer be added. Users with the reopen privilege may vote to reopen this question if it has been improved or closed incorrectly.

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 $5 K U S D o n l e i s u r e t r a v e l . B u t I h a n k e r t o, a n d c a n, r e t i r e o n$ 1.25M. Then I can travel less, and spend $3650 C A D / y e a r (e . g .$ 5/play × 2 plays/day × 365 days/year) buying the 10 PICK $5 D a i l y K e n o,_{*} * t w i c e d a i l y . * *_{I} p r e f e r D a i l y K e n o^{'} s t e e n s y c h a n c e o f w i n n i n g j a c k p o t, o v e r t r a v e l i n g^{'} s$ \rightarrow 0^{+}$ chance — assuming that I don't meet a multimillionaire during travel, and marry him!

n	n/2 = days	$1 - (1 - \frac{1}{2147181})^{n}$
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 $1 / 10 K$ 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.

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).

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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Chgg Clou‭

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3 comment threads

x-post https://www.reddit.com/r/maths/comments/157sctg/how_to_decide_whether_to_buy_a_lottery_with_a_... (1 comment)

x-post https://math.stackexchange.com/questions/4741280/how-to-decide-whether-to-buy-a-lottery-with-a... (1 comment)

Off topic (3 comments)

Peter Taylor‭ wrote over 1 year ago

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The question "But why do most answers on lotteries consider the Pr(winning jackpot in 1 play)" might be on-topic as a meta question, but the answer is going to be along the lines of "Because that's what the questions ask about". As to Homo economicus, this is a mathematics site, and while tightly limited questions about the non-linearity of utility can be mathematical the topic itself is more philosophical.

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Comments on How to decide whether to buy a lottery with a too negative EV, but passable Pr(you win jackpot at least once│n 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)? [closed]