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As you probably know, a 6/49 lottery game is a game where there will be 6 numbers drawn, ranging from 1-49 with no duplicates allowed. Today I'm not looking for winning chances though.

If I were looking for the average sum of all 6 winning numbers, I'd assume the result to look much like a Gaussian distribution with the center point right in the middle at 24 per number, because all numbers are equally likely to be drawn, so big numbers cancel out small numbers and long-term the average will be pretty much exactly in the middle. So the average sum I'd expect would be pretty much around 6*24=144. Is that right?

Now I'm wondering though, how would you determine the average sum of the 3 smallest winning numbers? By that I mean the lottery goes as normal, drawing 6 random winning numbers from 1-49, but we only want to sum up the 3 smallest of the winning numbers. So if the winning numbers are 1 4 13 25 46 49, then our sum would be 1+4+13=18. What would the average for this "little sum" be, and how do you get there?

Kenny Wong
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user2015253
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2 Answers2

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Think of it like this:

There is a round table with $50$ seats. Randomly we choose $7$ seats. Then the seat chosen as first is labeled as as number $50$ and starting at that seat clockwise we label the other seats with numbers $1,2,\dots,49$ (so $1$ and $49$ are the numbers of the seats bordering the first chosen seat).

The numbers that correspond with the other $6$ seats are the lottery numbers.

If $X_1,\dots,X_6$ are the chosen numbers with $X_1<X_2<\dots<X_6$ then by symmetry the $7$ variables $X_1, X_2-X_1,X_3-X_2,X_4-X_3,X_5-X_4,X_6-X_5, 50-X_6$ have equal distribution and their summation is $50$.

Then the expectation of each of them is $\frac{50}7$ and on base of that we find:$$\mathbb EX_i=i\times\frac{50}7$$

Then $\mathbb E(X_1+X_2+X_3)=(1+2+3)\times\frac{50}7=\frac67\times50=\frac{300}7$.

This is the average of sum of the $3$ smallest numbers.

drhab
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You're right about the overall sum, except that the middle is $(1+49)/2 = 25$, not $24$. So the expectation of the overall sum is $25 \times 6 = 150$.

The name for the $i$th smallest value when drawing $n$ values from the same distribution is an "order statistic". However these are usually computed with replacement, and you're working without replacement - that is, you can't draw the same ball more than once. For the key formula in this case, see Mean and variance of the order statistics of a discrete uniform sample without replacement . The expectation of the $i$th smallest ball when choosing $n$ balls from $k$ is $E[X_{(i)}] = i(n+1)/(k+1)$. You have $n = 6, k = 49$ so this is $50i/7$.

By linearity of expectation, then, $$E[X_{(1)} + X_{(2)} + X_{(3)} = E[X_{(1)}] +E[X_{(2)}] + E[X_{(3)}] = 50(1+2+3)/7 = 300/7 \approx 42.86.$$

Michael Lugo
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