Can We Hit the Lotto Again?

[IRV]

The company I work for had an individual win the Florida lottery about 6 years ago. Can it be proven mathematically that our odds are worst now since we already had a winner?

Another related question. What has a more +EV, spending $1 per week on lotto or $5? I can't see your win percentage changing much.

[knsmith85]

[ QUOTE ]
The company I work for had an individual win the Florida lottery about 6 years ago. Can it be proven mathematically that our odds are worst now since we already had a winner?

Another related question. What has a more +EV, spending $1 per week on lotto or $5? I can't see your win percentage changing much.

[/ QUOTE ]

1) There is no change to your company's chances, unless Florida has changed the way they do the lotto. In fact, there is just as much chance that 1-2-3-4-5-6 will be the winning numbers tomorrow as any random combination. It turns out, though, that playing these numbers is unwise solely because there are a handful of people out there that play these every time, and so if it does hit you'll be splitting with a larger number of people than most other number combinations.

2) The lottery is NOT +EV. Play $1/day gives you BETTER EV, but NOT +EV. Doing $1/day has an EV that is negative, but closer to 0 than $5/day.

[other1]

I've always thought buying multiple tickets was pointless. Lets say it's a huge megamillions jackpot.. The odds of winning, if I calculated it correctly, are 1 in 135,145,920. Is 5 in 135,145,920 really any better? I'd say it's a pretty insignificant change in the odds.

[knsmith85]

[ QUOTE ]
I've always thought buying multiple tickets was pointless. Lets say it's a huge megamillions jackpot.. The odds of winning, if I calculated it correctly, are 1 in 135,145,920. Is 5 in 135,145,920 really any better? I'd say it's a pretty insignificant change in the odds.

[/ QUOTE ]

Well, you have a 5 times greater chance of winning. It's just that to get to a significant level, you need like a 10,000 or 100,000 out of 130 million chance in winning.

The key with the lottery is the "Utility Function". Now, I don't even really know what it means, but I saw someone else mention it and from what I can gather it's that even though the lottery is -EV you can justify it because the result of winning is so life-changing that the losses from playing are reasonable. I'm not sure how you'd use this "Utility Function" in an actual mathematical equation, though - can someone enlighten me?

Realize that this is just the opposite of a common situation in poker - you're in a NL tournament where you are by far the best player. On the first hand, you pick up a flush draw, and someone moves all-in and there is a call behind them. You have the pot odds here to make the call, but the risk of going out when you can win anyways is much too great to justify a call. That's why it can be good to enter -EV situations as well as avoid +EV ones.

[shummie]

Your place of employment may have even greater odds of hitting now if this win has encouraged others to play.

- Jason

[niin]

Why would your chances be worse because you won in the past? Do the lottery balls 'remember' you and hate you now because you won 6 years ago, and magically know what numbers you've picked and conspire to not bring up your numbers?

Lottery balls don't have a memory.

Sorry for the dramatic tone, but it makes no sense that you'd have a worse chance now simply because you won previously.

In fact, since you won previously, you might get more people buying in and, thus, might collectively have a better chance of winning (the more entries you buy, the better your microscopic chances are).

[Hack]

Doesn't the lottery become +EV when the jackpot exceeds your odds of winning it? If the jackpot reaches say, 300 million, and the odds of winning it are 200 million to 1, then isn't it +EV to buy a ticket?

[Marm]

That was always my contetnion, but now I have a name for it, The utility function, I like it. I play every now and then, just for S&G's (S**ts and Giggles), but the Reward for winning far far far out weighs the $5 I spend a week on it.

ANd since the odds for the MM jackpot are 135,145,920, and taxes take out 48%, then the jackpot would have to be higher than $200,015,961.60 to make this +EV. And it has hit that high on occasion.

[college kid]

[ QUOTE ]
The company I work for had an individual win the Florida lottery about 6 years ago. Can it be proven mathematically that our odds are worst now since we already had a winner?

Another related question. What has a more +EV, spending $1 per week on lotto or $5? I can't see your win percentage changing much.

[/ QUOTE ]

I haven't read the other posts yet so I might be saying what has already been said.

1. No, there is no effect on your chances of winning because someone you know, work with, love, once knew, met on the street, saw in a supporting role in a movie, or read a novel by won.

2. The lotto has a high "house" edge and it is therefore the better EV play to play $1 a week than $5, if your only choice was one of the two. Obviously, not playing has the highest expectation, unless of course you know how to legally rig the lotto or get extra information--in which case you might have an edge by playing. But that is shady and most everything you can possible do would not be advantage play, it would be cheating--which is definately -EV with the high risk involved and teh fact that you don't keep the money if you are caught.

[niin]

[ QUOTE ]
Doesn't the lottery become +EV when the jackpot exceeds your odds of winning it? If the jackpot reaches say, 300 million, and the odds of winning it are 200 million to 1, then isn't it +EV to buy a ticket?

[/ QUOTE ]

Only if you're the only one to win... if multiple people win, then it's no longer a positive bet.

[knsmith85]

[ QUOTE ]
That was always my contetnion, but now I have a name for it, The utility function, I like it. I play every now and then, just for S&G's (S**ts and Giggles), but the Reward for winning far far far out weighs the $5 I spend a week on it.

ANd since the odds for the MM jackpot are 135,145,920, and taxes take out 48%, then the jackpot would have to be higher than $200,015,961.60 to make this +EV. And it has hit that high on occasion.

[/ QUOTE ]

I think this is actually pretty easy to represent in an equation.

Without the Utility Function (U), the EV equation is:

EV = -x + J*(x/p)

Where p = probability, J = jackpot value, and x = bet amount. If we take x to be 1, when the Jackpot exceeds the reciprical of the probability, you have a +EV. Since this doesn't really happen (because of multiple winning and more notably taxes), you always have a -EV.

However, things change when we consider U. Say your net income is $36.5k per year, or $100 per day (we'll say there are no taxes whatsoever on typical income). Of that $100/day, a loss of 1% of it is what we'll say the cutoff is for significance. So we have part of the utility function... that which we take from our willingness to lose a small amount of money.

Now, we need the part of U that comes from the enormous signifiance in winning a certain amount. For this exercise, we'll call that amount $1 million dollars.

So U is applicable when x <= $1.

So we can call U = J/1000000 as long as x <= $1.

Simply adding this to our original EV equation, we get:

EV = -x + J(x/p) + U, or EV = -x + J(x/p) + J/1000000

This value now becomes positive when J and p are in a certain ratio.

I think I probability messed this up somewhere, so if someone could please correct me, I'd appreciate it.

Thanks,
Kyle

[Marm]

Check that and correct that. I did the 48% math wrong, but you get my point.

[Bataglin]

Lol... buy ALL the lottery tickets and see what happens...

[IRV]

My thoughts exactly but we still have employees who say this often.

It's like playing 1,2,3,4,5,6 in the lotto. Same odds, but would you really play these numbers?????

[TomCollins]

1,2,3,4,5,6 will hit just as often as any other 6 number combination.

However, since jackpots are split, you are better off choosing numbers people are unlikely to play.

[knsmith85]

[ QUOTE ]
1,2,3,4,5,6 will hit just as often as any other 6 number combination.

However, since jackpots are split, you are better off choosing numbers people are unlikely to play.

[/ QUOTE ]

Thanks for covering this topic, I was concerned that it hadn't been discussed already. /images/graemlins/tongue.gif

[TomCollins]

Since he still didn't seem to get it, it seemed neccesary.

But thanks for the informative post.

[college kid]

Well, if the numbers are one in about seven million, and we can figure the average number of players and the chance of more than one person hitting, then we can come up with a figure at which point it would be +EV to get a seven million dollar loan and do just that. Unfortunately, I don't think any loan places would take "but it's positive expectation!!!"

[MortalWombatDotCom]

[ QUOTE ]
I think this is actually pretty easy to represent in an equation.

Without the Utility Function (U), the EV equation is:

EV = -x + J*(x/p)

Where p = probability, J = jackpot value, and x = bet amount. If we take x to be 1, when the Jackpot exceeds the reciprical of the probability, you have a +EV. Since this doesn't really happen (because of multiple winning and more notably taxes), you always have a -EV.

However, things change when we consider U. Say your net income is $36.5k per year, or $100 per day (we'll say there are no taxes whatsoever on typical income). Of that $100/day, a loss of 1% of it is what we'll say the cutoff is for significance. So we have part of the utility function... that which we take from our willingness to lose a small amount of money.

Now, we need the part of U that comes from the enormous signifiance in winning a certain amount. For this exercise, we'll call that amount $1 million dollars.

So U is applicable when x <= $1.

So we can call U = J/1000000 as long as x <= $1.

Simply adding this to our original EV equation, we get:

EV = -x + J(x/p) + U, or EV = -x + J(x/p) + J/1000000

This value now becomes positive when J and p are in a certain ratio.

I think I probability messed this up somewhere, so if someone could please correct me, I'd appreciate it.

Thanks,
Kyle

[/ QUOTE ]

Well, i think this is wrong on many levels.

First of all, you are trying to measure utility in dollars. The utility function maps outcomes (which consist partially but not entirely of the dollar values assigned to those outcomes) to some metric of utility. You can do a better or worse job of this depending on your assumptions, but since we are assuming that utility is in fact non-linear, you are stuck with doing a pretty bad job of it, at best.

Even if you could measure utility in dollars, it still wouldn't make sense to say U(-x) = anything, where x is an amount of money. That is like saying that spending a dollar on a candy bar has the same worth to you as spending a dollar on unnecessary invasive surgery. Spending a dollar on a lottery ticket gives you more pleasure than flushing a dollar down the toilet, so even if some malevolent being rigged the lottery so that you could never win (without your knowledge of course), U(buying a ticket that will lose) is not the same as U(having somebody steal a dollar bill from you).

to calculate your Expected Utility, you must sum, over all possible outcomes, the product of the utility of the outcome times the probability of the outcome (assuming your outcomes are discrete, and even if they are not, you just use the continuous analog of integral for sum, but i digress). so it would look something like

EU = U(buying a losing ticket) * (1-p) +
U(winning all the jackpot, after taxes) * p * (p0) +
U(winning half the jackpot, after taxes) * p * (p1) +
...

where p0 is the probability that nobody else played your numbers, p1 is the probability that only one other person played your numbers, etc.

The correct thing to compare this value EU to is the utility of spending that dollar on something else (this is opportunity cost of spending the dollar on a lottery ticket). So if you have exactly one dollar, you should spend it on the thing with the greatest EU. If you have more than one dollar, you start running into marginal utility. The utility of buying an eating a candy bar is pretty good... you choose to do it sometimes. But if you have a candy bar, the utility of buying a second candy bar and eating that is lower. Now, you can save the dollar until the utility of that second candy bar goes up (say, in a few days), but if you have a million candy bars, the million-and-first candy bar is pretty worthless. So maybe if i had a million dollars and nothing to spend them on except candy bars and lottery tickets, i might buy a lottery ticket... as long as i didn't realize that if i won, i could only use my winnings to buy more lottery tickets or more candy bars...

[shummie]

I like the utility argument, but I have a bit of a counter-argument for people like my father who play like $1 every day.

Why wouldn't you instead take your $356 once a year to a casino and bet it on the double-zero. 356x35 = $12775, and you would hit that once ever 38 years or so. Hell, let it ride... $12775x35 = $447,125... generally a life altering amount. You have a 1 in (38*38) 1444 chance of hitting this... that's better odds than the lotto obviously.

And I'm sure people play a lot more than $1 per day and would have a decent bankroll for some Roulette fun.

I think it's probably true that you can take any number that you deam "life altering" and figure out a Roulette skeem to make that money that give better odds than playing the lotto.

Am I just in love with the wheel and missing something here?

- Jason

[Redd]

Wouldn't we need to facor in the total investment return that one could obtain with their winnings as well? If I could lock $300 million at, say, 5% P.A., that's $15 million a year before any compounding. Over the winner's lifetime (at 5%, the winnings could double about 3 times in ~45 years?) the return would allow the principal to be much smaller (with these fudged numbers, about 1/8th) and still be +EV.