Lottery - better try to hit numbers-group that appears often, than hope for a "one-in-thousand-years" combination?

Question

Please correct my reasoning:

There is a 5 out of 36 lottery (5 unique numbers out of pool of 36 numbers ranging [1,2,…,36]).

Strategy 1: just play any random combination. Like 12345, or better 11111. Chances of such combination to appear are so low that I need to live thousands of years for it to appear. Maybe in my lifetime, but probably not.

Strategy 2: I know that 198492 (5-number) combinations out of all 376992 (that is 53% !) have some property (say arithmetic mean in range x..y). So I would 100% hit into that group many times in my lifetime! It is better to try to hit something that appears often, rather than some rarely-appearing ghost?

Answer 1

The probability that the winning combination has this property is $53\%$, but you don't get to choose all the combinations from this selection of combinations when you buy your ticket, you only get to choose one. The probability of any individual 5 digit combination is equal - you still have to find a way of choosing one of the combinations with this property. You are not accounting for this, so this is where your reasoning fails.

With strategy $1$, the combination $1,2,3,4,5$ is just as likely as something like $3,11,19,22,34$, but the fact that all the numbers in this second combination seem to follow no pattern is probably why it intuitively may seem more likely to occur. That is because the probability that a "random-looking" combination will win is much greater than the probability that the winning numbers will be consecutive. This is true. But which random-looking combination will win? That's just luck.

Answer 2

The first answer is absolutely correct, and clear, I just wanted to add some more intuition.

Suppose (instead of a 5 out of 36 lottery), you are facing an $n$ out of 2 lottery, where the two numbers are 0 and 1.

This is equivalent to the case where you are asked to guess a sequence of $n$ coin flips, each flip being heads or tails with equal probability. Then, you know that the number of 1's that appear is almost certainly going to be $n/2$ (for large $n$), but this information does not help you "choose" the right sequence of flips. It is still the case that any sequence is equally likely to occur, it just so happens that the majority of random 0-1 sequences have $n/2$ 1's.

Intuitively: approximately half of all possible coin flip sequences have $\approx n/2$ heads, so this set is very dense in the solution space, however, given that it is so dense, the number of choices amongst these sequences is large. If the lottery operator guaranteed you that "tonight's coin flips seuence will have $\approx n/2 $ heads!", your probability of winning just increased two fold. Without that information, you would have had the same luck just buying two tickets :).