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How to intuit P(win the same lottery twice) $= p^{2}$ vs. P(win the same lottery twice | you won the lottery once) $= p$?

+1
−2

Each lottery draw is independent, with probability $0 < p ≪ 1$. $W_i$ denotes the event of winning the $i^{th}$ draw of the same lottery. I seek intuition, not asking about computations.

$\Pr(\color{green}{W_1} \cap \color{crimson}{W_2}) = p^{2}$

"the probability of winning the lottery multiplied by winning the lottery the second time", because "the two events are independent. In other words, winning the lottery once doesn't some how increase or decrease your chances of winning it a second time."

$\Pr(\color{crimson}{W_2} | {\color{Green}{W_1}}) = \dfrac{\color{crimson}{\Pr(W_2)} \cap \color{Green}{\Pr(W_1)}}{\color{Green}{\Pr(W_1)}} = \dfrac{\color{crimson}{\Pr(W_2)} \times \color{Green}{\Pr(W_1)}}{\color{Green}{\Pr(W_1)}} = \dfrac{\color{red}{p} \times \color{limegreen}{p}}{\color{limegreen}p} = p.$

“If someone already wins the lottery, then the chance that the person wins the lottery a second time will be exactly the same as the probability they win the lottery if they had not previously won the lottery before,” Harvard statistics professor Dr. Mark Glickman tells CNBC Make It.

If you've already won the lottery in week one, however, the odds of winning the next week will be unaffected by the outcome and so remain the same as for any other individual event.

My misgivings

$\Pr(\color{green}{W_1} \cap \color{crimson}{W_2}) \neq \Pr(\color{crimson}{W_2} | \color{green}{W_1})$ feels contradictory. Intuitively, why doesn't $\Pr(\color{crimson}{W_2} | \color{green}{W_1}) = p^2$?

Doesn’t a player's "odds for winning the next time are the same as if they'd never played before" contradict $\Pr(\color{green}{W_1} \cap \color{crimson}{W_2})= p^{2}$. How can I reconcile these 2 probabilities intuitively?

$\Pr(\color{crimson}{W_2} | \color{green}{W_1}) = p$ feels deceitful. Given that you won the lottery once, $\Pr(\color{crimson}{W_2} | \color{green}{W_1}) = p$ implies that winning again is the same probabilistically as your first win! But this is wrong! $\Pr(\color{crimson}{W_2} | \color{green}{W_1}) = p$ OUGHT, but wholly fails to, disclose that winning any lottery twice is way less probable than winning it once.

I ask not about $\Pr($you win a lottery at least twice) $= 1 - \Pr($you never win) $- \Pr($you win the lottery once).

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I may be misreading your post, but it seems like there is some confusion between these two things:

  1. Winning the lottery twice
  2. Winning the lottery a second time

All of the remarks about the probability being the same are relevant to the second situation. Your calculations, however, address the first situation.

Perhaps it will be easier to avoid confusion if we use specifics. Let’s say that this is a lottery that will only ever be run twice, namely next Monday and Tuesday. Let’s say that both draws have a thousand entrants—in fact, the same thousand—and that all of them have an equal chance of winning ($p=\frac{1}{1000}$) in each draw. And let’s say that you, Chgg Clou, win Monday’s lottery.

Now, situation 2 is this: What is the probability that you win Tuesday’s lottery? It’s $\frac{1}{1000}$. The fact that you won Monday’s lottery doesn’t affect that (barring any rules forbidding a double win, or using your winnings to bribe the operators, or…).

Situation 1, meanwhile, is this: What is the probability that you win both Monday’s and Tuesday’s lotteries? Before Monday, we would have simply said that you (and everyone else) had a $\frac{1}{1\,000\,000}$ chance of doing so. But now?

Intuition’s a funny thing, but to me it seems very intuitive that this is improved by the fact that you won on Monday. You’re the only person who can possibly do it! Everyone else’s chances dropped to zero the moment Monday’s result came up. Your chances got better, and theirs got worse.


More broadly, you can compare it to a sport or other contest. (For the best analogy, think of one like volleyball where you need to be the first to reach a certain score, rather than simply ending the game with a higher score. But if we’re after intuitive acceptance, that may not matter.)

Does it make intuitive sense that your chances of winning are higher when you’re in the lead? Even assuming your chances of scoring at any given point are always the same (no psychological advantage to being in the lead, no pressure not to blow it), being in front makes you more likely to win.

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Many thanks. As you wrote "I may be misreading your post, but it seems like there is some confusion b... (2 comments)
+3
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It feels contradictory for P(you win the same lottery twice) $\neq$ P(you win the same lottery twice|you won the lottery once).

Would you expect P(you win the lottery exactly zero times) = P(you win the lottery exactly zero times | you won the lottery once)?

Intuitively, why aren't these two probabilities equal?

There are at least two ways of looking at it.

  1. P(you win the same lottery twice|you won the lottery once) is a simpler case than most conditional probabilities. It can be straightforwardly rephrased as P(you win the lottery for a second time).

  2. Conditional probability always excludes some possibilities. The possibility that you never win the lottery must be taken into account when calculating P(you win the same lottery twice) but it must be ruled out when calculating P(you win the same lottery twice|you won the lottery once).

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$\color{sienna}{1.}$ I cannot answer your first question, because "P(you win the lottery exactly zero... (2 comments)

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