Can We Hit the Lotto Again?

[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. [img]/images/graemlins/tongue.gif[/img]

[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