Intuitively, why doesn't picking unpopular integers ($> 31$) lower your probability of winning lotteries?
Let $N \ge 32$ be the largest integer pickable by players. I feel that picking unpopular integers (like N) lowers your probability of winning lotteries, for 2 reasons.
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Winning integers range randomly from 1 to N. But you're transmogrifying this game, by artificially narrowing yourself to $[32, N]$!
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By disregarding the 31 possible winning integers $\in [1, 31]$, you are flouting the lottery’s random distribution of winning integers! Because $[1, N]$ contains more integers than $[32, N]$, picking numbers $\in [1, N]$ proffers more chances to win than picking $\in [32, N]$.
What’s wrong with my intuition? Please correct my intuition. I'm NOT seeking any proof or formality. Please explain at a 16 year old (my son's age) level.
You should also not pick the date - or any date for that matter.
1 answer
Let's say every lottery you buy exactly one ticket and always use the same numbers. Does that lower your odds of winning versus a strategy where you always pick a random set of numbers? No.
Assuming the winning numbers are uniformly distributed, your odds of winning are always directly proportional to the number of (distinct) tickets you buy. Their actual values or how you chose them do not matter.
A bit of probability theory reconciles this with the fact the winning numbers are less likely to be in a subrange. Write $P(A\mid B)$ for the probability that some statement $A$ is true given that $B$ is known to be true. We have $P(A\textrm{ and } B) = P(A\mid B)P(B)$ which states that the probability that $A$ and $B$ hold is equal to the probability that $A$ holds given that $B$ holds times the probability that $B$ holds. This should be intuitively reasonable.
We now have a concrete and an abstract way of illustrating the above. Assume that each number has 100 possible values and we only consider tickets where the first number is 39, say. Then our two statements are $A$="I won." (i.e. the specific ticket I bought, which starts with 39, was chosen), and $B$="The winning ticket started with 39." It should be obvious that $P(B)$ is 1% corresponding to the fact that there is only a 1/100 chance that the winning numbers start with 39. However, if I knew the first number was 39, then any ticket starting with 39 is now 100 times more likely to be a winner, i.e. $P(A \mid B) = 100P(A)$. (This is balanced by the fact that every ticket not starting with 39 now has zero chance of being a winning ticket.) These clearly cancel out giving $P(A\textrm{ and }B) = 100P(A)/100 = P(A)$. It should be clear that this would work for any subset. However much you restrict the possibilities you choose from will be balanced out by having each choice out of those possibilities being more likely given the winning ticket is one of those possibilities.
A more abstract but much simpler proof that $P(A\textrm{ and } B) = P(A)$ in this case is just to note that $A$ implies $B$. That is, "my ticket, which starts with 39, was the winner" implies that "the winning ticket starts with 39". Thus logically $(A\textrm{ and }B)$ is equivalent to just $A$. So we immediately have that they have the same probability, i.e. $P(A\textrm{ and }B) = P(A)$.
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