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

Question

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).

Answer 1

Intuitively, why aren't these two probabilities equal?

Let $W_i$ denote the event of winning on the $i$th lottery, for the sake of brevity.

$P(W_1 \text{ and } W_2)$ measures how likely it is to win two lotteries, from the very start. You have not won any lotteries and know nothing yet about your outcomes, just the probability of winning a single lottery.

For $P(W_2 \mid W_1)$, you are given that you won the first lottery. You may think of it as knowing, at this moment in time, you have already won it (or will win it if you'd rather frame this as being at the start). You only have one more lottery in front of you to participate in as a result (at least, only one more that you could possibly lose), and hence the probability of winning the second lottery is just the probability of winning that lottery (because you have already won one).

Answer 2

It feels contradictory that P(you win the same lottery twice) ≠ P(you win the same lottery twice|you won the lottery once). Intuitively, why aren't these two probabilities equal?

Foremost, they describe two distinct things. Namely they are measured under different conditions.

The joint probability measures your expectation that you will win both when you have no knowledge of the lottery outcomes.

The conditional probability measures your expectation that you will win the second should the first be won.

Answer 3

It feels contradictory that $\Pr(\color{green}{W_1} \cap \color{crimson}{W_2}) \neq \Pr(\color{crimson}{W_2} | \color{green}{W_1})$. Intuitively, why aren't these two probabilities equal?

I feel that $\Pr(\color{crimson}{W_2} | \color{green}{W_1})$ OUGHT $= p^{2}$ because $\Pr(\color{crimson}{W_2} | \color{green}{W_1})$ OUGHT capture, but wholly fails to disclose, that $\Pr(\color{green}{W_1} \cap \color{crimson}{W_2}) ≪ \Pr(\color{crimson}{W_2} | \color{green}{W_1})$.

$\def\Wone{\color{green}{W_1}} \def\Wtwo{\color{crimson}{W_2}} $(Revision 8 of the question)

I am confused, do you expect $\Pr(\Wone \cap \Wtwo) \ll \Pr(\Wtwo \mid \Wone)$, but at the same time feel contradictory that $\Pr(\Wone \cap \Wtwo) \neq \Pr(\Wtwo \mid \Wone)$?

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I am missing how your intuition relates $\Pr(\Wtwo\mid\Wone)$ and $\Pr(\Wone\cap \Wtwo)$. Their ratio ($\ne 1$) exactly discloses how (un)likely $\Wone$ is:

$$\Pr(\Wone\cap\Wtwo) = \Pr(\Wone)\Pr(\Wtwo\mid \Wone) \ll \Pr(\Wtwo \mid \Wone)$$

As $\Pr(\Wone) = p \ll 1$, this value also discloses that $\Pr(\Wone\cap\Wtwo) \ll \Pr(\Wtwo \mid \Wone)$.

Intuitively to me, the probability of achieving both $\Wone$ and $\Wtwo$ is the product of

• the probability of $\Wone$, and
• the probability of $\Wtwo$ given that one already achieved $\Wone$.

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How do you see the meanings of $\Pr(\Wtwo\mid\Wone)$ and $\Pr(\Wone\cap \Wtwo)$ alternatively, and see a contradiction?