Lottery - FAQ

A factor in the answer is the total number of tickets sold to other players. In the event more than one player hits the jackpot it will have to be shared. Let's call the number of possible combinations n, the total number of other tickets sold t, the rate of return on the small prizes r (in the case of the Big Game r=0.179612), and j be the jackpot size. For this to be a break-even venture j*n/(n+t) + r*n - n=0. This works out to j=(1-r)*(n+t).

The short answer is no, you will never win. The odds of winning the usual 6/49 lottery are 1 in 13,983,816. You would have to play the game ln(.5)/ln(1-1/combin(49,6)) = 9,692,842 times to have a 50/50 chance of winning at least once. Assuming you bought 100 lottery tickets a day, it would take 265.6 years to have a 50% chance of winning. To have a 90% chance of winning, it would take 882.2 years.

Let’s define f(x,y) as the probability of getting x of one suit and y of another. This function is not limited to two terms.

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.

No, I didn’t account for splitting the jackpot. That definitely does depress the value, the more people that play the more it reduces the expected return. I didn’t have enough information about number of players when I wrote that article to properly factor that in.

I take it you mean you want lottery numbers. Sorry but I can’t do any better than you can. However I would recommend you not play at all, especially if you are very poor. There seem to be a lot of former generals and dictators over there trying to wire me 17 million dollars, maybe one of them will give you a scholarship.

Another Mike S., what are the odds? Lots of racetracks permit what is called "class 2" gaming, which must be lottery or bingo based. The way to offer slots under this rule is to have a lottery or bingo game going on behind the scenes and the outcome is displayed in the form of a slot machine win. For example if the lottery game determines that you win 20 times your bet it will display whatever slot machine symbols pay 20. So it is a clever illusion.

The probability of winning correctly picking 6 numbers out of 49 is 1 in combin(49,6) = 1 in 13,983,816. This is also the probability of your two tickets matching.

I did some research and six numbers are chosen each drawing. In any given drawing the probability of the number 53 not making an appearance is combin(89,6)/combin(90,6) = 93.333%. In two years there would be 208 drawings. So the probability of 53 not occurring in a specific two year period would be 0.93333²⁰⁸ = 0.000000585665, or 1 in 1,707,460.

The probability that any number would not be hit in a two-year period could be closely approximated as 90*(1/ 1,707,460) = 1 in 18,972. The actual probability would be slightly less than that because I double counted two numbers being missed, which is very negligible. Of course the past down not matter in the lottery and every drawing has the same probability of picking the 53.

First let’s confirm that probability. The player must match 5 regular numbers from a pool of 55, and one Power Ball from a pool of 42. The probability of winning would be 1 in combin(55,5)*42 = 1 in 146,107,962. So I agree with your probability. I like to use the Poisson distribution for questions such as yours. The mean number of winnings would be 105,000,000/146,107,962 = 0.71865. The general formula for the probability of n winners, with a mean of m, is e^-m*mⁿ/n!. In this case the mean is 0.71865 so the probability of zero is e^-0.71865*0.71865⁰/0! = 0.48741. So the probability of at least one winner is 1-0.48741 = 0.51259. So 0.71865 winners will have to 0.51259 of a jackpot to split up. That is 0.51259/0.71865 = 0.71327 jackpots per winner. So jackpot sharing reduces your expected win to 71.327% of the jackpot amount, or a reduction of 28.673%.

The lottery is always a sucker bet! Briefly, the return of the Powerplay option is 49.28%. The return from a Powerball ticket alone is so much worse that it would be better to buy x/2 tickets with the Powerplay option than x without it. I added details about this option to my lottery section if you want more information.

Let n be your answer. The probability of rolling a 7 or 11 is 8/36. To solve for n:

(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.

According to the Indiana Lottery web site, there is a total of $3,270,000 in prize money given to 325,000 ticket holders. That would make each ticket worth $10.615 each on average, assuming the series sold out. At a cost of $20 each, the return is 50.31%. If only 60,000 tickets were sold then each would be worth $54.50, for a return of 272.50%. The breakeven point is 163,500 tickets sold. If you believe that fewer than that will be sold, then buying tickets becomes a good bet, putting aside tax and the utility of money implications.

I checked the New York and California lottery web sites. Both indicated that if the winner dies before all the payments are made, the rest will be paid to the winner's designated heir or estate.

That is true, but it is not as fishy as you think. According to Understanding Probability: Chance Rules in Everyday Life by H. C. Tijms, the same set of numbers were drawn on June 21, 1995, and December 20, 1986, in bi-weekly drawings. The drawing of December 20, 1986 was the 3,016th drawing. The number of combinations in a 6/49 lottery is combin(49,6) = 13,983,816. The probability the numbers in the second drawing will not match those of the first is (c-1)/c, where c is the number of combinations, or 13,983,816. The probability the third drawing will produce a unique set of numbers is (c-2)/c. So, the probability that every drawing from the 2nd to the 3,016th will produce unique numbers is (c-1)/c × (c-2)/c × ... (c-3015)/c = 0.722413. So, the probability of at least one set of common numbers is 1- 0.722413 = 0.277587, or 27.8%. The following table shows the probability of at least one pair of matching numbers drawn by the number of years, assuming two draws per week.

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

For the benefit of other readers, the Nova Scotia sports lottery is like off the board parlays at a Nevada casino, with worse odds. For the random picker to get the expected return for a given pick, take the sum of the inverse of what each outcome pays. Then, take the inverse of that sum.

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.

No, I don't agree with it. That is a terrible piece of journalism and Business Insider should be embarrassed by it.

To begin, the article was published on Dec. 17, 2013, before the $636 million drawing that evening. Let's look at the math to evaluate the value of a $1 ticket. The following table shows the probability and expected return of all possible outcomes for the $636 million jackpot, before considering such things as the lump sum penalty, taxes, and jackpot sharing. The top three probabilities are in scientific notation because the numbers are so small.

1. The lump sum penalty.
2. Taxes.
3. Jackpot sharing.

If we ignore the effects of taxes, the annuity on the jackpot, and jackpot sharing, then if the jackpot is greater than $224,191,728 you should invoke the "Jackpot Only" option. If we do consider those factors, then you should never invoke but the Megaplier instead.

For more information, please see my page on the Mega Millions lottery.

Yes. The house advantage in most lotto-based lottery games is close to 50%. So, a hypothetical $2 game with a $1,000,000 jackpot, with no smaller prizes, would need to have a probability of winning of 0.5*(2/1000000) = 1 in one million to retain a 50% house edge.

Here is my strategy to turn $2 into $1,000,000 with better odds than that.

• First bet $2 on any number in double-zero roulette. You can find $2 minimums at some Vegas casinos, like the El Cortez and South Point. If you win, you'll be up to $72.
• Next, parlay your $72 onto another single-number bet. If you win, you'll be at $2,592.
• Next, carry that $2,592 into one of the high-end casinos on the Strip, like the Wynn, Venetian, or Bellagio. Bet your $2,592 from roulette on the Banker bet in baccarat. Do this a total of nine times, letting everything ride each time. After your ninth win, you'll be up to $1,056,687. Your ninth bet would be for $541,891, which I'm sure any of these casinos would take if they saw you win it in front of their noses.

The probability of winning a single-number bet in double-zero roulette is 1/38. The probability of winning the Banker bet in baccarat is 50.6825%, not counting ties. So, the probability of two roulette wins and nine Banker wins is (1/38)^2 × 0.506825^9 = 1 in 654,404. Those are better odds than the 1 in a million you would get with a lottery and you'll have a little more than a million dollars too.

First, let's solve for the probability that any given game number has exactly 3 out of 5 days with the same number. That answer is COMBIN(5,3)*(1/80)^2*(79/80)^2 = 0.001523682. This is the number of ways to choose the 3 matching days out of 5 times the probability the second and third day match the first times the probability the other two days don't.

So, the probability there is not a 3 out of 5 day match for any given game number is 1 - 0.001523682 = 0.9984763.

The probability this does not happen for 300 days is 0.9984763³⁰⁰ = 63.29%.

Thus, the probability of the alternative, that there is at least one drawing number with 3 out of 5 days matching the same Bulls Eye number is 36.71%.

This question is asked and discussed in my forum at Wizard of Vegas.

I won't go into all the math, but here are the jackpot points where multiple winners are equally likely as exactly one:

• PowerBall: $975 million
• Mega Millions: $1.65 billion

Not that you asked, but here are the jackpots where the probability of at least one winner is equal to the probability of none (50%).

• PowerBall: $704 million
• Mega Millions: $867 million

This question is raised in discussed in my forum at Wizard of Vegas.

First, let's go over the rules. The bet cost is $2. The game is based on drawing five balls out of 43. Here is the pay table:

• Match 5 = Jackpot
• Math 4 = $200
• Match 3 = $10
• Match 2 = $2

In addition, the player gets what I presume is like a scratch-off card. This has a 1/80 chance of paying $6 and 1/5 chance of paying $2.

The following table shows my analysis of the base game. It shows the value of catching 2 to 4 numbers is $0.287784.

The following table shows my analysis of the Quick Cash instant win feature. The lower right cell shows a value of $0.475

Thus, the value of the non-progressive prizes is $0.287784 + $0.475000 = $0.762784.

Let j equal the value of the breakeven jackpot. Then:

2 = 0.762784 + j × (1/962598)
1.237216 = j × (1/962598)
j = 1.237216 × 962598
j = $1,190,941.95.

The overall rate of return is 0.381392 plus 0.051943 for every $100,000 in the jackpot.

As stated in the question, all this ignores taxes and jackpot sharing.

This question is asked and discussed in my forum at Wizard of Vegas.