Choosing lottery numbers strategy (based on heuristics such as choose as many odd numbers as even numbers)

Question

So I got this question in brilliant.org's probability fundamentals:

Every week you play a lottery game that requires choosing 4 numbers between 1 and 80. In this particular game, you win "outright" and do not have to potentially share the prize with others. Your friends give you some advice for the next week on which numbers to pick.

Which advice is the most reasonable?

1. You should choose two even and two odd numbers.
2. You should not pick 4 consecutive numbers (numbers in a row such as 3, 4, 5, 6).
3. You should pick some low numbers, high numbers, and numbers in the middle.
4. It doesn't matter which four numbers you pick. It's just luck.

brilliant.org says the correct answer is advice #4. I do agree that any particular combination of four numbers is just as likely as any other combination; however, combinations having certain characteristics are more likely than combinations not having those characteristics. For example, a combination having 2 even and 2 odd is about six times as likely as a combination having all even or all odd (the math for this is below), so advice #1. is beneficial. Is brilliant.org wrong or am I not thinking about this correctly? Is the question too vague?

I did the math the for advice #1.:

1. You should choose two even and two odd numbers.

The probability that the winning sequence has two even and two odd numbers is:

$P(2,2) = \binom{4}{2} * \frac{40*39*40*39}{80*79*78*77} \approx 0.385$

whereas the probability that the winning sequence has three even and one odd is:

$P(3,1) = P(1,3) = \binom{4}{3} * \frac{40*39*38*40}{80*79*78*77} \approx 0.250$

Finally, the probability that the winning sequence has four even and zero odd is:

$P(4,0) = P(0,4) = \binom{4}{4} * \frac{40*39*38*437}{80*79*78*77} \approx 0.058$

Given that $P(2,2)$ is greater than $P(3,1)$, $P(1,3)$, $P(4,0)$, and $P(0,4)$, wouldn't we be better off following the advice of option #1. than not following it?

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Likewise, if I do the math on advice #2. and #3., I'm very sure that the math would show that the advice is beneficial. It should be possible to follow all three pieces of advice (#1., #2., and #3.) as well.

I'm trying to reconcile this with brilliant.org's correct answer that basically says the advice (#1., #2., and #3.) is irrelevant.

Thanks!

Answer 1

You're almost going about it backwards. Let's say there are five people: Alfonso follows advice #1: He picks two evens and two odds randomly. (We're assuming order doesn't matter here.) Barbara follows advice #2: She picks randomly but makes sure to not pick $4$ consecutive numbers. Cory follows advice #3: he picks one "high number" (say, $>60$), one "low number" (say, $\leq 20$), and one "in the middle" (in the interval $(20,60]$), plus one randomly. Denise follows advice #4 and just goes random. And just for fun, Ed decides that he's always going to pick $(1,2,3,4)$ because that's his phone passcode and he doesn't want to forget it.

As you calculated, Alfonso has a $\binom 42\frac{40\cdot39\cdot40\cdot39}{80\cdot79\cdot78\cdot77}\approx38.5\%$ chance of being "right," i.e. the winning combination has 2 evens and 2 odds. But that's not his probability of winning. No, because Alfonso picked just one of $\frac{40\cdot39\cdot40\cdot39}{2!2!}$ possible "two even, two odd" tickets, and only one of those possible tickets is a winner. Therefore, Alfonso's probability of winning is $\frac{2!2!}{40\cdot39\cdot40\cdot39}\cdot\binom 42\frac{40\cdot39\cdot40\cdot39}{80\cdot79\cdot78\cdot77}=\frac{4!76!}{80!}$ shot at winning.

Let's move on to Barbara. What are the odds that Advice #2 is correct? Well, the only way that two consecutive numbers can win is if the winning combo is $(1,2,3,4),(2,3,4,5),\dots,(77,78,79,80)$. That's only $77$ possible tickets, out of the $\frac{80!}{4!76!}$ available. So that means that the total chance that advice $2$ is correct is $\frac{\frac{80!}{4!76!}-77}{\frac{80!}{4!76!}}\approx99.995\%$. Pretty good, right? So Barbara is in the money? Well, not quite. You see, Barbara has just one ticket, out of the possible $\frac{80!}{4!76!}-77$ possible "non-consecutive" tickets, so her shot is really $\frac1{\frac{80!}{4!76!}-77}\cdot\frac{\frac{80!}{4!76!}-77}{\frac{80!}{4!76!}}=\frac{4!76!}{80!}$, the same as Alfonso.

Hmm. What about the rest of them? Can you do the math out and show that Cory, Denise, and Ed all have the same chances to win? Or can you convince yourself why the probabilities "cancel out" this way?

Answer 2

It's more likely for the lottery numbers to be two-odd-two-even; however, there are more two-odd-two-even selections to choose from, for the exact same reason. Not only do you have to hope the lottery ends up in your chosen category, but you also have to hope that the lottery numbers hit your exact choice in the chosen category. When that category is large (two-odd-two-even), the first event is more likely but the second event is less likely; when that category is small (all numbers less than 10), the first event is less likely and the second event is more likely.

Another analogy: I have a two-story warehouse full of $N$ candy bars, and one of the candy bars (uniformly randomly determined) contains a golden ticket. You get one guess to point to a candy bar and hope that's the one with the golden ticket. You could just look at the shelf next to you and point to one of its candy bars. Does it benefit you to follow the advice "the second story is bigger than the shelf, so go point to a candy bar on the second story instead"? No—you still have a $1$ in $N$ chance of finding the golden ticket.