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• Sugar Land
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August 28, 2019
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When you play a lottery and win a pari-mutuel Jackpot, there's always the possibility that one or more other players will also have won, and you'll be sharing (splitting) the prize money. Can you calculate what the Expected Value is of the Jackpot prize, given you are a winner? Certainly!

We need to assume something first: every ticket bought is a "Quickpick" - so any given set of numbers (a single ticket) can be randomly replicated on any other ticket. Suppose there are "N" such tickets sold (1 of which is yours, matching the winning combination). Suppose the probability of a ticket matching the winning combination is "p". Then, there are a total of "N" possible outcomes of "k" winners *besides* you, all the way from k=0 to N-1. Obviously "k" being more than a few is going to be extremely unlikely, but these terms are still part of the equation.

To explain Expected Value briefly, it is the sum of the product of each outcome by its probability (in other words, a weighted average). In our case the first part of the product is our share of the Jackpot (e.g. if one other winner, we get \$Jackpot/2, if two other winners, we get \$Jackpot/3, and so on. Let's use the symbol "J" for the nominal (advertised) Jackpot - what we would win (before taxes of course) if nobody else has a winning ticket.

EV(Jackpot-\$return | I am a winner) = Sigma_(k=0 to N-1){[J/(k+1)] * C(N-1,k) *  [(1-p)^(N-k-1)] * p^k}

Here "*" means multiply and "^" means to the power of. Sigma means sum each term (in this case, with the index k from 0 to n-1).

Also, C(n,r) = n! / [r! * (n-r)!] , the number of combinations of r items taken from n (where r <= n).

Try the formula yourself on any lottery. Assume a Jackpot and N. The problem you will likely run into with C(n,r) is that for any large N, say 100000000 for example, a math overflow error will result. This is because the factorials are calculated first, and anything above ~70! will be too large of a number to store. The way to get around it is to create function that will calculate the logarithm of C(n,r). Thus, for example, LN(N!) becomes LN(N) + LN(N-1) + LN(N-2) + ... + LN(2) + LN(1). Do this for the numerator and denominator on the right-hand side, and then convert the overall result back to C(n,r) with the EXP() function. For a typical lottery, anything after the ~5th term in the EV summation will probably be small that it can be ignored.

• United States
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This is an interesting topic, nice write up. In the past I have run simulations to find the probability distribution of the number of jackpot winners (under the assumption every ticket is a quick pick to simplify things) varying the parameters of total number of combinations and number of tickets sold.

For example, in a 5/39 game, which is common in many states, there are 575757 combinations (a more manageable number). If 575757 quick picks are sold, the likelihoods of 0, 1, 2, 3, and 4+ jackpot winners are

• 0 winners: ~35%
• 1 winner: ~43%
• 2 winners: ~17%
• 3 winners: ~5%
• 4 or more winners: <1%

So if the jackpot amount is J, then the expected amount each winner would receive, given there is at least one winner, is about 0.53*J.

In a 3-digit daily numbers game, there are 1000 combinations, and if 1000 quick picks are sold then the likelihoods of 0, 1, 2, 3, 4, and 5+ winners are

• 0 winners: ~37.3%
• 1 winner: ~36.0%
• 2 winners: ~18.8%
• 3 winners: ~6.3%
• 4 winners: ~1.3%
• 5+ winners: <1%

These closely resemble Poisson distributions but I think the underlying function is more complicated. Or they may actually be Poissons. I have to think about it a little more. In any case, they could be reasonably approximated by a Poisson with small error.

A non-trivial snag in all this is that in real life only 50%-70% of tickets sold are quick picks, and for people who prefer to pick their own losers, certain combinations are favored.

• Sugar Land
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The underlying distribution is Binomial. Poisson is an approximation of the Binomial Distribution for cases when "p" is small, and "N" is large (both of which should be reasonable for just about any lottery).

By the way, if you want a condensed probability and statistics primer, I recommend "Principles of Statistics" by M.G. Bulmer, who was a professor at Cambridge University in the 1960s. It's a Dover Edition, <\$15. I think the latest edition was published in 1967. It was published in an era when the educational system was not dumbed down to the lowest common denominator like it is today. (Translation: the end of chapter questions are actually challenging.)

• United States
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Quote: Originally posted by Orange71 on Jul 19, 2022

The underlying distribution is Binomial. Poisson is an approximation of the Binomial Distribution for cases when "p" is small, and "N" is large (both of which should be reasonable for just about any lottery).

By the way, if you want a condensed probability and statistics primer, I recommend "Principles of Statistics" by M.G. Bulmer, who was a professor at Cambridge University in the 1960s. It's a Dover Edition, <\$15. I think the latest edition was published in 1967. It was published in an era when the educational system was not dumbed down to the lowest common denominator like it is today. (Translation: the end of chapter questions are actually challenging.)

You're right, I see it now. Poisson is also the wrong one because it has infinite support where as this distribution is finite. Thanks for the book recommendation, I got a shelf full of Dover editions. I can't believe more college courses don't use them just on the price alone, but they are written without much filler and leave the details to the reader, probably not suitable for most college students except math majors and grad students.

• Sugar Land
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Actually, your math doesn't seem to be quite right on the first example. You set N = 1 / p, where N = 575757. OK, no problem. However, let's expand the terms for k = 0 and k = 1 in my equation, the cases where there is no other winner (i.e. besides you/me) and one other winner.

P(k=0) = [(n-1)C(0)] * (1-p)^(N-1) * p^0 = (1) * [(1-p)^(N-1)] = (1-p)^(N-1)

P(k=1) = [(n-1)C(1)] *  [(1-p)^(N-2)] *  p = (n-1) * [(1-p)^(N-2)]  * p  = (p*n - p)  *  [(1-p)^(N-2)]    = (1-p) * [(1-p)^(N-2)] = (1-p)^(N-1)

Therefore, the probability of 0 or 1 additional winner for N = 1/p is the same.

By the way, the editing capability for posts on this site absolutely sucks. I can't even paste an image into the post with the correctly formatted equations.

• Sugar Land
United States
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"N" and "n" are the same. Sorry about that sloppiness. Trying to fight this awful editor was an uphill battle.

• United States
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Quote: Originally posted by Orange71 on Jul 20, 2022

Actually, your math doesn't seem to be quite right on the first example. You set N = 1 / p, where N = 575757. OK, no problem. However, let's expand the terms for k = 0 and k = 1 in my equation, the cases where there is no other winner (i.e. besides you/me) and one other winner.

P(k=0) = [(n-1)C(0)] * (1-p)^(N-1) * p^0 = (1) * [(1-p)^(N-1)] = (1-p)^(N-1)

P(k=1) = [(n-1)C(1)] *  [(1-p)^(N-2)] *  p = (n-1) * [(1-p)^(N-2)]  * p  = (p*n - p)  *  [(1-p)^(N-2)]    = (1-p) * [(1-p)^(N-2)] = (1-p)^(N-1)

Therefore, the probability of 0 or 1 additional winner for N = 1/p is the same.

By the way, the editing capability for posts on this site absolutely sucks. I can't even paste an image into the post with the correctly formatted equations.

You're right. Those were numbers I obtained running simulations. I obviously didn't run enough since the frequencies are off, lol. After gave it some thought I realized the exact distribution really is just binomial and not something else, I couldn't edit the post and then I forgot and moved on to something else. Sorry!

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