Ask the Wizard: Lottery - FAQ

With two arguments f(x,y)= combin(13,x)*combin(13,y)*12/combin(52,5).

With three arguments f(x,y,z)= combin(13,x)*combin(13,y)*combin(13,z)*12/combin(52,5).

With four arguments f(w,x,y,z)=combin(13,w)*combin(13,x)*combin(13,y)*combin(13,z)*4/combin(52,5).

The probability of all four suits is COMBIN(13,1)³*COMBIN(13,2)*4/combin(52,5) = 26.37%.

The probability of three suits is COMBIN(13,3)*COMBIN(13,1)²*12 + COMBIN(13,1)*COMBIN(13,2)^2*12/combin(52,5) = 58.84%

The probability of two suits is COMBIN(13,3)*COMBIN(13,2)*12 + COMBIN(13,4)*COMBIN(13,1)*12/combin(52,5) = 14.59%

The probability of one suit (including straight and royal flushes ) is 4*combin(13,5)/combin(52,5) = 0.20%.

So three suits are the most frequent outcome.

(8/36)ⁿ = 1/41,400,000

log((8/36)ⁿ) = log(1/41,400,000)

n × log(8/36) = log(1/41,400,000)

n = log(1/41,400,000)/log(8/36)

n = -7.617 / -0.65321

n = 11.6608

So there you go, the probability of hitting the SuperLotto is the same as rolling a seven or eleven 11.66 times in a row. For those who can’t comprehend a partial throw I would rephrase as the probability falls between 11 and 12 consecutive rolls.

In case you were wondering, the number of draws for the probability of a matching draw to first exceed 50% is 4,404.

For example, on the Monday Night football game of November 9, 2009 they have the following choices:

Steelers win by 3.5 or more: pays 1.9 for 1
Broncos win by 3.5 or more: pays 3.25 for 1
Margin of victory 3 or less: 3.65 for 1

The sum of the inverses is (1/1.9)+(1/3.25)+(1/3.65) = 1.107981. The inverse of that number is 1/1.107981=0.902543. So, the expected return is 90.25. For a parlay, take the product of the return of all picks made.

I looked at several events, and the return per event ranged from 75.4% to 90.3% (from the above example). The average was 82.6%. Based on that average, here is the expected return according to the number of picks:

2: 68.2%
3: 56.3%
4: 46.5%
5: 38.4%
6: 31.7%

I do indeed think that is a factor that should be considered, albeit a somewhat minor one, in the decision to purchase a lottery ticket. To answer your question, I used the jackpot amount and sales figures found at lottoreport.com . I looked at the Powerball going back to Jan. 2008, because that is as far back as that website has data. I also looked at Mega Millions going back to June 2005, a point at which there was a change in the rules. The following table summarizes my results.

So the average probability that a jackpot will be split is 0.74% in Powerball and 1.29% in Mega Millions. As the jackpot gets higher, and sales go up, so does the probability of splitting the jackpot. The reason the split jackpot probability is higher in Mega Millions is because the probability of winning is greater and there is more competition from other players.

All things considered, I show 4.01% is lost in return due to jackpot sharing in Powerball and 6.59% is lost in Mega Millions. However, those figures don’t account for taxes, or that jackpots are paid in the form of an annuity. To adjust for that, I assumed the player gets only half of it, either by taking the lump sum option or the loss in value from choosing the annuity. I also assumed that 30% of the rest is lost in taxes, so the winner can expect to receive 35% after both factors. After that adjustment, I show a 1.20% loss in return due to jackpot sharing in the Powerball and 1.98% in Mega Millions.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

You must be referring to the Ca$h WinFall game. I first learned about this from the article A game with a windfall for a knowing few at the boston.com web site.

It is not unusual for the jackpot in progressive lottery games to get so big that the return exceeds over 100%, before considering taxes, annuitized payments, the small odds of winning, and the decreasing utility of money for enormous jackpots. After factoring in these things, the lottery is almost never a good bet.

What makes the Ca$h WinFall game different is that when the jackpot exceeds two million dollars, and nobody hits it, they roll all but $500,000 of the prize money down to the lower prizes. That makes a hefty profit almost ensured for groups with six-figure bankrolls.

Let's take a look at the recent July 18, 2011 drawing as an example. It is a simple 6-46 game in which the player chooses six numbers from 1 to 46 and the lottery does the same. The more your numbers match the lottery's, the more you win. The following table shows the probability and return for each event. The win for catching two numbers is a free ticket, which I show separately as having a value of 26¢. Each ticket costs $2, so the return column is product of the probability and win, divided by ticket price. The lower right cell shows an expected return of 117%, or a player advantage of 17%.

Since I initially wrote my answer, the Massachusetts Lottery has restricted the number of tickets any one store can sell in a day to 2,500, or $5,000 worth, according to the article, Lottery restricts high-level players at boston.com. This would obviously make it more difficult to get down huge amounts of money, but it may be good for smaller-bankrolled players, as it should minimize competition from other players. There is only so much money in the "roll down," so the less competition for it the better. I would certainly at least try to play it if I lived anywhere close to Massachusetts.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

For those not familiar with the story, the Quebec Lottery offers a game titled Extra . The machine randomly picks a 7-digit number, and the player has to match as many digits as possible, in order, from a random draw. The matching digits can be aligned in either direction. The smallest prize is $2 for matching the right-most digit only. The largest prize is $1,000,000 for matching all seven digits.

What the plaintiffs in the lawsuit noticed was that if you purchased ten Extras then for the first and last digits the game picked one of each numeral. In other words, if you looked at the first, or last, position only then you would see all ten numbers from 0 to 9. The plaintiffs claim this gives them only two chances to win and is non-random.

I see their point. Almost all the variance in that game comes from the $1,000,000 jackpot. The standard deviation of ten completely independent random tickets would be 1002.845. The way the Quebec Lottery does it the standard deviation of ten tickets purchased at the same time is almost the same at 1002.833.

In my opinion if the player buys multiple quick pick lottery tickets each ticket should be independent of the others. However, I find the lawsuit of $20 million almost totally frivolous. If I were the judge, I would award the plaintiffs $1.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

The following table shows the three most frequent set of numbers chosen according to the Quebec Lottery . The number of tickets is out of a total of 366,518 sold for the Jan. 30, 2010 lottery. For those who don't recognize the pattern of the third set, those were the "Lost" numbers, which played a significant role on that show.

By extrapolating, if the numbers 7-14-21-28-35-42 were drawn by the Lotto 6/49 game then the jackpot would be split among thousands of players, each receiving only 0.03% of the jackpot.

My advice, if you must play the lottery, is to go with a Quick Pick.