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marbles
12-16-2002, 05:57 PM
With this Wednesday's jackpot at approx \$160.0 mil, Powerball is once again approaching record levels. So it's a good time to brush up on our Powerball probabilities.

The rules of the game: five white balls are selected from a numbered set of 53, and one red ball (the Powerball) is selected from a numbered set of 42.

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

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?

3. Here is the payoff schedule:
Match the Powerball (PB) only: \$3
Match PB and one white ball: \$4
Match PB + 2: \$7
Match PB + 3: \$100
Match PB + 4: \$5,000
Match PB + 5: \$160.0 Mil (important: jackpot is chopped in the event of multiple winning tickets, use numbers from #2, assume no more than 3 winners)
Match 3 white, no PB: \$7
Match 4 white, no PB: \$100
Match 5 white, no PB: \$100,000.

What is the Expected payout on one ticket?

marbles
12-16-2002, 06:08 PM
1. approximately 1/120,526,770

2. zero tickets: approx 0.6079
One winner: approx 0.3026
Two winners: approx 0.0753
Three winners: approx 0.0125
More than 3: approx 0.0017

3. Just under \$1.05. Yes, that means a \$1 Powerball ticket is actually a good investment if standard deviation means nothing to you.

PseudoPserious
12-16-2002, 08:14 PM
My folks play the Maine State lottery iff the EV is positive, but they also factor in the rather significant tax they'd have to pay on their winnings.

PP

Webster
12-17-2002, 12:15 AM
So I still have a chance then! LOL Actually isn't lightning striking you a greater chance?

BruceZ
12-17-2002, 12:42 AM
The chance of being killed in a traffic accident on your way to buy a ticket are vastly greater than your chance of winning the jackpot.

lorinda
12-17-2002, 12:51 AM
Bruce, you are assuming that the buying of a ticket and the traffic accident are independant events.

If you are in a store that sells tickets anyway, then you can reduce that edge to almost zero (unless you are killed in a freak supermarket trolley accident)

Lori

BruceZ
12-17-2002, 03:01 AM
Or in the extra time it takes to buy the ticket, you could be shot by the guy robbing the store. Maybe not in England though.

BruceZ
12-17-2002, 08:28 AM
I get that the expected payout per ticket for the grand prize alone is \$1.04, but when the other prizes are added, the expected payout comes to \$1.21.

Using your numbers (which I confirmed to be correct) the grand prize alone would be worth:

160 million*(.6079 + .3026/2 + .0753/3) = 125,488,000

125,488,000/120,526,770 = \$1.04

The EV for getting the power ball + n white balls is:

prize*(1/42)*C(5,n)*C(48,5-n)/C(53,5)

The EV for getting just n white balls with no power ball is:

prize*(41/42)*C(5,n)*C(48,n)/C(53,5)

Here are the EVs with the power ball:
<pre><font class="small">code:</font><hr>
prize ev white balls w/power ball
3 0.042620507 0
4 0.032288263 1
7 0.010045237 2
100 0.009358917 3
5000 0.009956294 4
125482281.6 1.041115444 5
</pre><hr>
Here are the EVs without the power ball:
<pre><font class="small">code:</font><hr>
prize ev white balls w/no power ball
7 0.026860091 3
100 0.008164161 4
100000 0.034017339 5

</pre><hr>
Sum of all EVs = 1.214426

As a sanity check, I also included the 3 cases where we lose a dollar, and subtracted a dollar from all the prizes since we don't get that back when we win. This came out to an EV of 21 cents for buying a ticket, which is consistent with the above.

marbles
12-17-2002, 11:01 AM
We agree on the 17 cents for everything not related to the powerball, so I'll focus on just the jackpot, where I got 87.7 cents vs. your \$1.04:

First, a few symbols for easier notation:
P(me)=Probability I hit the jackpot = 1/120526770
P(me')=Probability I don't hit the jackpot = (1-1/120526770)
P(0)=Probability of zero winners = 0.6078
P(0')=Probability that there are more than zero winners = 1-0.6078=0.3921
P(1)=probability of one winner = 0.3026
P(2)=probability of two winners = 0.075319
P(3)=probability of three winners = 0.0125

So, the probability of me being the only winner, given that there are not no winners = P(me)*P(1)/P(0'). Multiply that result by \$160.0 mil and you get \$0.772

Likewise, the probability of me winning and being one of two = P(me)*P(2)/P(0'). Multiply that result by \$80.0 mil, and you get \$0.096.

Finally, the probability of me winning and being one of three = P(me)*P(3)/P(0'). Multiply result by 160/3 mil, and you get \$0.01.

Sum the three numbers, and you get roughly \$0.87.

Your approach seems to work as well, although it's just a digit off. Your starting formula:

"160 million*(.6079 + .3026/2 + .0753/3) = 125,488,000"

Note that this formula pays out \$160.0 to the winner when there is NO WINNER, and pays out two winners when there is ONE WINNER, etc. Notice that by shifting these numbers to:
\$160 Mil * (0.3026* 0.07532/2* 0.0125/3) = \$41,512,589. If you divide this result by P(0') to avoid double-counting the non-win, you get 41.5M/(0.3921) or \$105,860,993. This number, divided by the \$125,488,000 from your initial result, is proportionate to the difference between our final results of 87 cents/\$1.04.

Hope this makes sense. It's hard to write probability formulas in this format!

BruceZ
12-17-2002, 01:42 PM
I understand what you did because I almost did it myself. Unfortunately, it isn't correct.

Your approach seems to work as well, although it's just a digit off. Your starting formula:
"160 million*(.6079 + .3026/2 + .0753/3) = 125,488,000"
Note that this formula pays out \$160.0 to the winner when there is NO WINNER, and pays out two winners when there is ONE WINNER, etc.

When I win, the probability of there being NO other winner out of the other 59,999,999 tickets is essentially the same as there being no winner out of 60,000,000 tickets or .6079. Similarly, I split the pot when there is ONE other winner which is .3026 of the time that I win, etc.

Here's where you (and almost I) went wrong by subtley misapplying Bayes' theorem:

So, the probability of me being the only winner, given that there are not no winners = P(me)*P(1)/P(0').

P(1)/P(0') is the probability of exactly 1 winner given that there is at least one winner. We don't want that. We want the probability of there being exactly 1 winner given that "ME" won, and that is the same as P(0).

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. This is fallacious for the same reason.

marbles
12-17-2002, 01:54 PM
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
12-17-2002, 02:37 PM
"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
12-17-2002, 02:38 PM
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
12-17-2002, 02:45 PM
"If you consider how much you actually win after taxes, it is still negative EV."
Well, now you're just raining on my parade! /forums/images/icons/wink.gif

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

Homer
12-17-2002, 04:22 PM
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
12-17-2002, 05:49 PM
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
12-17-2002, 07:22 PM
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.

&lt;giggle&gt; That's pretty funny, BruceZ.

Easy E
12-27-2002, 01:10 PM
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? /forums/images/icons/wink.gif

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
12-27-2002, 01:53 PM
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.