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Q&A

Without trial and error, how can I effortlessly deduce all $n, k_i ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?

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With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k_i ∈ ℕ[]() ∋ \dbinom n { k_1, k_2, \ldots , k_i} = c$ ? For example below, $i = 1, \color{limegreen}{c = 4,072,530}$. Rule out trial and error!

Context

You get two sets of six numbers from 1-45 per $1 play.

Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.

But as I prefer to pick unpopular "numbers to reduce the number of ways you split the prize", I loathe that

2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated numbers from the class of one to 45.

I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied

I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers $k ≤ 20,$ $n$ to satisfy $\dbinom n k =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires $k ≤ 10,$ because research proves that players dislike picking over $10$ integers, which they find inconvenient. Remember, many players buy physical paper tickets.

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x-post https://math.stackexchange.com/questions/4770854/effortlessly-without-trial-and-error-how-can-... (1 comment)
Sockpuppet accounts (2 comments)

1 answer

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I prefer to pick integers ≥32 to boost my probability of winning

That's silly.

All combinations of numbers have the same chance of winning.


<Soapbox>

You've asked a bunch of questions here that indicate you are trying to find a scheme to win the lottery. There isn't one. On average, you're going to lose. The purpose of a lottery is not to redistribute wealth, but to be a source of income to the government. Since they are making money from it, on average, the lottery players must be losing.

The lottery is basically a voluntary tax. Actually it's a stupidity tax, or a tax on those who didn't pay attention in high school math class.

These lotteries are big enough that they can hire smart people who clearly know lots more math than you do. These folks probably have a degree in actuarial math. You're not going to out-math them. They make sure there aren't loopholes that let a player gain an advantage over all the other players. That's their job.

Then there are the lottery marketing people who know that humans are bad at statistics when looking at situations intuitively, and even worse when getting emotionally involved. Lotteries are therefore promoted on emotional grounds. Targeting dumb people helps spending the marketing dollars more effectively. Looks like its working.

Normally I don't say this because I like other people paying my taxes for me, but since this is the math site: "Don't be an idiot by playing the lottery.".

</Soapbox>

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2 comment threads

All numbers have the same chance of winning, but the "rationale" is that less popular numbers will sp... (1 comment)
Some people play the lotto for entertainment value. (1 comment)

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