Is this lottery calculation correct?

[Ladewig]

The Texas Lottery is now at $70 million. The website says that taking all the money at once will produce a $45.7 million prize. If the federal government takes 35% of that, then $29.7 million is left for the winner.

The chances of winning (from website) are 1 in 25.8 million. The lottery is set up in that same way most state lotteries are in that it is a paramutual game and the grand prize is split evenly among all the winners. Thus, if one were assured of not splitting the jackpot, then the expected value would be positive - an extremely rare event for any state-sponsored game of chance.

So the tricky bit is calculating the odds of splitting the jackpot. The game was changed in 2003, so I looked at all the data after the change date and of the forty pay-outs, only four involved a even split of the money. There were no three-way splits. Can I perform the intermediate step in calculating the expected value in this manner:

((36/40) x 29.7) + ((4/40) x 14.9) = 28.2

If so, wouldn't that result in a positive expected value (or a negative house edge)?

 

[CatOfGrey]

The Texas Lottery is now at $70 million. The website says that taking all the money at once will produce a $45.7 million prize. If the federal government takes 35% of that, then $29.7 million is left for the winner.

The chances of winning (from website) are 1 in 25.8 million.

In calculating 'house odds', then only thing that matters is money in, money out. Since there is more than 25.8 million being distributed, yes, there is a house advantage.

How does this happen? Easy. In the previous lottery, the jackpot was not paid out. In a single paramutual event (such as a horse race) the money would be distributed amongst other bettors. But the lottery isn't a single event, so the jackpot carries forward. So there's an advantage to those playing this particular draw, at the cost of those who participated in the previous draw, who unexpectedly got screwed by a low payout rate.

 

[Jorghnassen]

Technically, you should factor in the amount of tickets purchased, and the more tickets sold, the more likely the event of sharing the jackpot will occur. As larger jackpots increase ticket sales, the house does not actually expect a loss.

 

[Ladewig]

In calculating 'house odds', then only thing that matters is money in, money out. Since there is more than 25.8 million being distributed, yes, there is a house advantage.

How does this happen? Easy. In the previous lottery, the jackpot was not paid out. In a single paramutual event (such as a horse race) the money would be distributed amongst other bettors. But the lottery isn't a single event, so the jackpot carries forward. So there's an advantage to those playing this particular draw, at the cost of those who participated in the previous draw, who unexpectedly got screwed by a low payout rate.

Yes, I understand. I was talking about this drawing by itself.

 

[maddog]

The Texas Lottery is now at $70 million. The website says that taking all the money at once will produce a $45.7 million prize. If the federal government takes 35% of that, then $29.7 million is left for the winner.

The chances of winning (from website) are 1 in 25.8 million. The lottery is set up in that same way most state lotteries are in that it is a paramutual game and the grand prize is split evenly among all the winners. Thus, if one were assured of not splitting the jackpot, then the expected value would be positive - an extremely rare event for any state-sponsored game of chance.

So the tricky bit is calculating the odds of splitting the jackpot. The game was changed in 2003, so I looked at all the data after the change date and of the forty pay-outs, only four involved a even split of the money. There were no three-way splits. Can I perform the intermediate step in calculating the expected value in this manner:

((36/40) x 29.7) + ((4/40) x 14.9) = 28.2

If so, wouldn't that result in a positive expected value (or a negative house edge)?

If the 4/40 is accurately representative of the actual probability (which depends on the number of tickets sold for this particular drawing, presuming all numbers are randomly drawn -- if you select your numbers based on any non-random method, then anybody else's non-random method could change the probability of splitting the pot) - then this is a reasonable step.

If there are 50M tickets sold for *this drawing*, then there is about a 1:1 chance that the pot will be split. If you buy every number combination, based on the positive expected value, you significantly change the probability of the pot being split.

Also, you could include the lesser prizes (like match 3, win $10; match 4, win $250; anything like that) into your expected value calculation, to improve your accuracy.

 

[Ladewig]

Technically, you should factor in the amount of tickets purchased, and the more tickets sold, the more likely the event of sharing the jackpot will occur. As larger jackpots increase ticket sales, the house does not actually expect a loss.

Yes, I thought about that but was unsure how to factor it in.

A unrelated but additional factor is that certain combinations are more popular than other combinations (i.e. because so many people play their birthdays, combinations with lower numbers have higher chances of being duplicates).

ETA: I missed Maddog's response.

 

[Ladewig]

If there are 50M tickets sold for *this drawing*, then there is about a 1:1 chance that the pot will be split. If you buy every number combination, based on the positive expected value, you significantly change the probability of the pot being split.

The current prize is about 3 million more than the previous drawing's prize. If half of the revenue goes directly to the state's coffers, wouldn't that mean 6 million tickets?

Also, you could include the lesser prizes (like match 3, win $10; match 4, win $250; anything like that) into your expected value calculation, to improve your accuracy.

Yes, I was just trying to do thumbnail calculations to see if it was positive of negative.

 

[Jorghnassen]

Yes, I thought about that but was unsure how to factor it in.

That is of course, the hard part of the problem. The state presumably has some data and models for that, but with what you have access to, your "crude" estimate is probably as good as it gets for the available data. You just can't say that because this crude expected return is higher than 1, the house has a negative edge (because the house can sell more tickets than there are possible combinations, and can presumably predict how many tickets will be sold depending on jackpot size with a reasonable degree of accuracy).

A unrelated but additional factor is that certain combinations are more popular than other combinations (i.e. because so many people play their birthdays, combinations with lower numbers have higher chances of being duplicates).

ETA: I missed Maddog's response.

That is another issue. While the draw is random, the tickets are not (their distribution is not uniform). So, if the draw happens to include many popular numbers (e.g. birth dates, the Lost numbers), the chance of sharing the jackpot increases.

 

[maddog]

The current prize is about 3 million more than the previous drawing's prize. If half of the revenue goes directly to the state's coffers, wouldn't that mean 6 million tickets?

Not necessarily - they also need to account for the lesser prizes, payments for advertising, payments to ticket sellers, payments to ticket cashing (lower prizes get cashed at ticket sellers, right?), and other administrative expenses.

But, there probably is *some* direct or nearly-direct ratio between main-prize increase and number of tickets sold. Maybe if you push hard on the lottery system, they might let that ratio / equation be known.

Yes, I was just trying to do thumbnail calculations to see if it was positive of negative.

Understood, just saying...