powerball probabilities

[marbles]

Bruce,

I see your point. I hadn't thought of taking my own ticket out of the set like that, and ended up trying to jam a square peg into a round hole...

So I guess this means my Powerball ticket is even MORE valuable than I originally thought. Sweet!

[marbles]

"1. What is the probability of hitting the jackpot, matching all five white balls and the powerball?"
P(winner)=(1/53*1/52*1/51*1/50*1/49*1/42)*5!
The first 5 fractions are the probabilities of hitting each individual ball, but, since the 5 white balls can be selected in any order, you have to multiply by 5!. Take the inverse of this tiny number, and you'll find it to be 120,526,770.

"2. If 60.0 million tickets are sold for this drawing, what is the probability that there is exactly one winning ticket? Two tickets? Three tickets? Zero tickets?"

P(zero winners)=(1-(1/120,526,770))^60,000,000.
Simply the probability of 60,000,000 losers.

P(one winner)=(1/120,526,770)*((1-(1/120,526,770))^59,999,999)*(60,000,000!/1!59,999,999!)
It's simply the p(winner)*p(59,999,999 losers)*(number of ways one unit can be selected out of a set of 60,000,000).

p(two winners)=(1/120,526,770)*((1-1/120,526,770)^59,999,998)*(60,000,000!/(2!)(59,999,998!))
It's simply the p(2 winners)*p(59,999,998 losers)*(number of ways two units can be selected out of a set of 60,000,000).

If you get the idea, try to solve for p(3 winners), or what have you.

Solution for #3 is in the string below.

[BruceZ]

If you consider how much you actually win after taxes, it is still negative EV. On the other hand, if you invest your winnings you can have positive EV over time. That would be true even if jackpot was normal sized.

[marbles]

"If you consider how much you actually win after taxes, it is still negative EV."
Well, now you're just raining on my parade! [img]/forums/images/icons/wink.gif[/img]

You're right, of course... I'd just prefer to think of my $5 to be spent on a "good" investment for now.

[Homer]

1. What is the probability of hitting the jackpot, matching all five white balls and the powerball?

P = 1 / (C(53,5)*C(42,1)) = 1 / (2869685 * 42) = 1/120,526,770

2. If 60.0 million tickets are sold for this drawing, what is the probability that there is exactly one winning ticket? Two tickets? Three tickets? Zero tickets?

P (0 winners) = (120526769/120526770)^60000000 = 60.78%

P (1 winner) = C(60000000,1) * (1/120526770)^1 * (120526769/120526770)^59999999
P (1 winner) = 30.26%

P (2 winners) = C(60000000,2) * (1/120526770)^2 * (120526769/120526770)^59999998
P (2 winners) = 7.53%

P (3 winners) = C(60000000,3) * (1/120526770)^3 * (120526769/120526770)^59999997
P (3 winners) = 1.25%

....too tired to do number 3 right now.

-- Homer

[marbles]

Also, don't forget the "annuity factor," discounting what is usually a 20-25 year annuity to a lump-sum today. Factoring both the taxes and annuity factor together, you're typically looking at about 40% of the actual jackpot if you elect to take it all today.

Kinda screws with my numbers, I know.

[PseudoPserious]

This error reminds me of the joke about how you should always bring a bomb aboard an airplane since then the chance of someone else bringing a bomb is much smaller since the chance of there being 2 bombs is much smaller than the chance of there being just one.

<giggle> That's pretty funny, BruceZ.

[Easy E]

Just considering the jackpot odds alone, and assuming 40% tax loss, the breakeven EV point (assuming no splitting) for the cash option requires an advertised jackpot of $371,996,204...... a threshold which has NEVER been reached in the USA. (For the other national lottery, it's close to $432 million)

But do we really care? If you ended up with a 3-way split and netted "only" $40 million, would you see your Actual Value as positive or negative? [img]/forums/images/icons/wink.gif[/img]

Another way to put it- will buying one less hotdog, or going out for lunch one less time this week, be too much of an EV sacrifice to take a shot at obscene gobs of money? And being able to jump up to the $500/$1000 level without a care, or enter the WSOP with 10 of your friends??

[marbles]

Yes and no, I think. While the typical Powerball player would not consider figuring out the actual odds/EV's, they are all playing the odds to a certain degree. Their barroom-napkin arithmetic is simple: "I'll spend this $1 on a tapper at happy hour, or I'll throw it on a Powerball ticket if the payoff is big enough." While few of them know the precise EV of a ticket, they all have a price. It's all based on what the individual person considers a big enough "gob" of money.

These numbers amazed me:
12/11/02 Powerball drawing, $101M jackpot: 28.5M tix sold
12/25/02 Powerball drawing, $315M jackpot: 171.6M tix sold

I guess my price is around $150.0M, even though that still makes it a -EV proposition. So I guess it doesn't really matter.