# Lottery - FAQ

Given that a lottery has 10 million potential combinations, what are the odds that someone will win with 90% confidence given that 10 million tickets are sold. Clearly it would not be 100% since some tickets would be duplicates. I am less interested in the answer than in the methodology used to solve it.

Scott from New York, New York

Lets try to rephrase the question. Assuming the lottery has 10 million combinations, and all players choose their numbers randomly (allowing for duplicates), how many tickets would the lottery need to sell so that the probability of at least one person winning is 90%? Lets let p be the probability of winning and n be the number of tickets sold. The probability of 1 person losing is 1-p. The probability of all n people losing is (1-p)^{n}. The probability of at least one winner is 1 - (1-p)^{n}. So we need to set this equal to .9 and solve for n.

.9 = 1 - (1-p)^{n}

.1 = (1-p)^{n}

ln(.1) = ln((1-p)^{n})

ln(.1) = n*ln(1-p)

n = ln(.1)/ln(1-p)

n = ln(.1)/ln(.9999999)

n = 23,025,850.

So, the lottery would need to sell 23,025,850 tickets for the probability of at least one winner to be 90%. In case you were wondering, if the lottery sold exactly ten million tickets the probability of at least one winner would be 63.2%, which is very closely approximated as 1-(1/e).

This is my second question to you, this time the subject is state lotteries. I'm sure you've heard of a group of "investors" which used to wait for a jackpot to reach a certain level, at which point they would purchase tickets with every combination of numbers possible. This would assure them a share of the prize. Assuming the cost of a ticket is $1, how high must the jackpot get for this to be a profitable venture?

Ted

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).

I have been playing lotteries and sweepstakes now for two months straight. Will I ever hit a jackpot? And when?

Mavis from Mattapan, U.S.

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.

I have noticed something over the years watching our Quinto lottery drawing here in Washington State. It is a 52 ’card’ game that draws 5. I have noticed that the vast majority of the time, three suits are drawn. From poker numbers you find that the chance of getting just one suit (a flush) is 5148 out of about 2.6 million. What are the odds of 2,3, or all 4 suits coming up?

Kevin from Tacoma, Washington

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.

In your lottery probability chart for the MD lotto game you make no allowance for the probability of a split jackpot. What effect does this possibility have on expected value?

Bob from Falls Church, Virginia

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 am a student from a very poor home please I will like you to help me with the week draws thanks...

Frank from Benin City, Nigeria

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.

Dear wiz: The horserace track that I attend is introducing video lottery machines. Can you tell me anything about them? Are they the same as slots? Any info you can give would be helpful and appreciated.

Mike S.

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.

If I buy two quick-pick lottery tickets what is the probability I get the same number on both cards. Assume a 6/49 lottery.

Anonymous

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'm assuming you're aware, but if not, in the Italian lottery, there is a twice-weekly drawing of 50 out of number 1 through 90 (five numbers from each of 10 cities). For roughly 2 years, the number 53 has not shown up, leading up to a "number 53 frenzy", to the point where people have committed suicide after betting everything they have on what they were *sure* would be a corrective! So I got to thinking - what *are* the probabilities that the 53 would not come up for two years? ()

Andrew from Hollywood

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 0.93333^{208} = 0.000000585665, or 1 in 1,707,460.

Concerning your answer about the Italian lottery (see Sep 11, 2005 column), you showed that the probability of 53 not being picked in two years is 1 in 1,707,460. You should have followed up with the probability that any one (or more) of the 90 numbers would be missed during the same two-year period; I think that’s what the person asking really wanted to know. Also, you could have explained (once again) why 53 is no more likely to be picked in the next lottery than any other number, despite the unlikely past situation.

Anonymous

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.

They say the probability of winning the Powerball lottery is 1 in 146,107,962. In the recent drawing for a $340 million jackpot the local media said the number of tickets sold was 105,000,000. My questions are if you win what is the probability you will have to share the jackpot, and how much does this reduce the expected value?

Mitch F. from Hopkins, MN

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}*m^{n}/n!. In this case the mean is 0.71865 so the probability of zero is e^{-0.71865}*0.71865^{0}/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%.

Powerplay multiplies non-grand prizes from 2x - 5x. The Powerball site has listed testimonials from it’s winner’s that say "powerplay is the only way to go". I’m thinking it’s a sucker’s bet.

John from Morrisville, NC

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.

Wizard, could you please describe the equivalent odds of the California SuperLotto Plus (1 in 41.4 Million), in terms of number of consecutive times of rolling 7 or 11? I heard it somewhere before. Most people cannot comprehend the lottery odds. But, the rolling of dice -- they can relate.

Tim from Belmont, CA

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.

My wife and I bought a $20 raffle ticket for the Indiana lottery. My understanding of this game is that the drawing for the winning prizes (which number 777) will be held on August 16, 2007, regardless of the number of actual tickets sold and with the absolute maximum available number of tickets being 325,000. As of today, only 60,000 tickets have been sold. Would it be a good gamble to buy a few additional tickets? What would our odds of winning a prize be?

David B. from Evansville, IN

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.

Is it true that state lottery jackpot annuities cease to pay if you die before the annuity runs its term? I heard this was true of New York. This seems outrageous to me.

Alex from Montreal

I checked the 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.

I heard that the same numbers were chosen in the German 6/49 lottery on different days. Something seems fishy with that. What are the odds?

Lotte from Hamburg

That is true, but it is not as fishy as you think. According to 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.

### Probability of MatchingNumbers in 6/49 Lottery

Years | Probability |

5 | 0.009640 |

10 | 0.038115 |

15 | 0.083800 |

20 | 0.144158 |

25 | 0.215822 |

30 | 0.295459 |

35 | 0.379225 |

40 | 0.463590 |

45 | 0.545437 |

50 | 0.622090 |

55 | 0.691985 |

60 | 0.753800 |

65 | 0.807008 |

70 | 0.851638 |

75 | 0.888086 |

80 | 0.917254 |

85 | 0.940000 |

90 | 0.957334 |

95 | 0.970225 |

100 | 0.971954 |

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

What are the odds of the Proline parlay bets offered by the ?

Anonymous

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%

Do you think the probability of a split jackpot should be factored into the calculation of the expected value of lottery tickets? If so, what is that probability?

rdw4potus

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.

### Split Jackpots in Powerball and Mega Millions

Item | Powerball | Mega Millions |

Probability of winning jackpot | 1 in 195,249,054 | 1 in 175,711,536 |

Average jackpot offered | $73,569,853 | $65,792,976 |

Average sales per draw | $23,051,548 | $25,933,833 |

Average expected winners per draw | 0.118 | 0.148 |

Average probability of a split jackpot per draw | 0.74% | 1.29% |

Loss in return due to shared jackpots (unadjusted) | 4.01% | 6.59% |

Loss in return due to shared jackpots (adjusted) | 1.41% | 2.31% |

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 .

I heard there is a lottery game in Massachusetts that is vulnerable to advantage play. Do you know anything about it?

Anonymous

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%.

### Ca$h WinFall Drawing â€” July 18, 2011

Catch | Pays | Combinations | Probability | Return |
---|---|---|---|---|

6 | 2392699.1 | 1 | 0.00000011 | 0.12772207 |

5 | 19507 | 240 | 0.00002562 | 0.24990768 |

4 | 802 | 11700 | 0.00124909 | 0.50088509 |

3 | 26 | 197600 | 0.02109574 | 0.27424465 |

2 | 0.24 | 1370850 | 0.14635171 | 0.01756221 |

1 | 0 | 3948048 | 0.42149293 | 0.00000000 |

0 | 0 | 3838380 | 0.40978479 | 0.00000000 |

Total | 9366819 | 1.00000000 | 1.17032170 |

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, 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 .

What is your opinion of the lawsuit against the Quebec Lottery over allegedly non-random tickets?

Anonymous

For those not familiar with the story, the Quebec Lottery offers a game titled . 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 .

What are the most common sets of numbers players pick for lottery tickets?

Anonymous

The following table shows the three most frequent set of numbers chosen according to the . 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.

### Quebec Lottery Most Frequent Picks

Numbers | Number of Sales in Quebec | Frequency |
---|---|---|

7-14-21-28-35-42 | 824 | 1 in 444.8 |

1-2-3-4-5-6 | 424 | 1 in 864.4 |

4-8-15-16-23-43 | 377 | 1 in 972.2 |

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.

Do you agree with the article titled at Business Insider?

Asswhoopermcdaddy

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.

### Mega Millions -- $636 Million Jackpot

Catch | Mega Ball | Pays | Combinations | Probability | Return |
---|---|---|---|---|---|

5 | Yes | $636,000,000 | 1 | 3.86E-09 | 2.456634 |

5 | No | $1,000,000 | 14 | 5.41E-08 | 0.054077 |

4 | Yes | $5,000 | 350 | 1.35E-06 | 0.006760 |

4 | No | $500 | 4,900 | 0.000019 | 0.009463 |

3 | Yes | $50 | 24,150 | 0.000093 | 0.004664 |

3 | No | $5 | 338,100 | 0.001306 | 0.006530 |

2 | Yes | $5 | 547,400 | 0.002114 | 0.010572 |

1 | Yes | $2 | 4,584,475 | 0.017708 | 0.035416 |

0 | Yes | $1 | 12,103,014 | 0.046749 | 0.046749 |

Loser | $0 | 241,288,446 | 0.932008 | 0.000000 | |

Total | 258,890,850 | 1.000000 | 2.630865 |

That shows a $1 ticket will get back $2.630864. After deducting the dollar the ticket cost, the expected profit is $1.630864. Business Insider gets $1.632029. A difference of 0.001164, but no big deal.

However, there are three things that depress the value significantly:

- The lump sum penalty.
- Taxes.
- Jackpot sharing.

Let's look at them one at a time.

Large lottery progressive jackpots are usually paid out in the form of an annuity of about 30 years, including Mega Millions. If the winner wants the money all at once, which most do, he must take a significant reduction. This is only fair because a dollar today is worth more than a dollar in the future. In the case of the December 17, 2013 drawing, the total prize money was reduced to $347.6 million, or 54.65% of the advertised jackpot.

Next, let's look at taxes. The highest federal marginal income tax rate is 39.6%. State taxes range from 0% to 12.3% (), so let's just say 6% on average. After taking out 45.6% in taxes, we're left with $189.1 million.

Now comes the trickiest part -- jackpot sharing. It should be noted that starting with the October 22, 2013 drawing Mega Millions changed the rules to a 75-15 format, where they draw five numbers from 1 to 75, and then one from a separate pool of 1 to 15. This reduced the odds of winning to 1 in 258,890,850, evidently in an effort to have bigger jackpots. Looking at just the 17 drawings since then, using jackpot and sales data from , I find there is an exponential relationship between jackpot size and demand. I found the same thing for the Powerball lottery, by the way. Using exponential regression, my formula for total tickets sold (in millions) is 12.422 × exp(0.0052 × j), where j is the jackpot size (in millions). For example, for a $636 million jackpot, the expected sales would be 12.422 * exp(0.0052*636) = 339.2 (million). Actual sales were $337 million, so pretty close.

Based on the actual ticket sales of 336,545,306, we can expect 336,545,306/258,890,850 = 1.300 winners. The pertinent question is if you win how many other people can you expect to share the money with? This is easily answered looking at the Poisson distribution. Given a mean of 1.3 winners, the probability of exactly x winners is exp(1.3)×1.3

^{x}/fact(x). The following table shows the probability of 0 to 10 other winners, your share of the jackpot in each case, and expected share, assuming that you win.

### Expected Jackpot Share Given 1.3 Other Winners on Average

Other winners | Probability | Jackpot share | Expected share |
---|---|---|---|

10 | 0.000001 | 0.090909 | 0.000000 |

9 | 0.000008 | 0.100000 | 0.000001 |

8 | 0.000055 | 0.111111 | 0.000006 |

7 | 0.000339 | 0.125000 | 0.000042 |

6 | 0.001827 | 0.142857 | 0.000261 |

5 | 0.008431 | 0.166667 | 0.001405 |

4 | 0.032429 | 0.200000 | 0.006486 |

3 | 0.099786 | 0.250000 | 0.024946 |

2 | 0.230283 | 0.333333 | 0.076761 |

1 | 0.354295 | 0.500000 | 0.177148 |

0 | 0.272545 | 1.000000 | 0.272545 |

Total | 1.000000 | 0.559602 |

The bottom right cell shows that you can expect to keep 55.96% of the money, and the other 44.04% will go to those darned other winners you'll have to share with.

Now our $636 million jackpot is down to $189.1 × 55.96% = $105.8 million. Let's see what the return table looks like with that figure as the jackpot.

### Mega Millions -- $105.8 Million Jackpot

Catch | Mega Ball | Pays | Combinations | Probability | Return |
---|---|---|---|---|---|

5 | Yes | $105,800,000 | 1 | 3.86E-09 | 0.408666 |

5 | No | $1,000,000 | 14 | 5.41E-08 | 0.054077 |

4 | Yes | $5,000 | 350 | 1.35E-06 | 0.006760 |

4 | No | $500 | 4,900 | 0.000019 | 0.009463 |

3 | Yes | $50 | 24,150 | 0.000093 | 0.004664 |

3 | No | $5 | 338,100 | 0.001306 | 0.006530 |

2 | Yes | $5 | 547,400 | 0.002114 | 0.010572 |

1 | Yes | $2 | 4,584,475 | 0.017708 | 0.035416 |

0 | Yes | $1 | 12,103,014 | 0.046749 | 0.046749 |

Loser | $0 | 241,288,446 | 0.932008 | 0.000000 | |

Total | 258,890,850 | 1.000000 | 0.582898 |

The bottom right cell shows that an expected return of 58.29%. In other words your $1 investment can expect to get back about 58 cents, for an expected loss, or house edge, of about 42%. Does that sound like the math saying you should buy a ticket?

According to the article, "

*So, as long as there are fewer than 730 million tickets sold, a fairly likely situation right now, the expected value of a ticket should be positive, and so you should consider buying a Mega Millions ticket today.*"

There were a lot fewer than 730 million in sales and it still was a terrible value. However, in all fairness, the article went on to say the following:

*"Bear in mind that there are many caveats to this analysis. Taxes will likely hurt your expected winnings pretty severely — the Feds will take about 40%, and your home state will claim anywhere from 0% to around 13%.*

A lot of people have been buying tickets, and as discussed above, this will greatly increase the odds of a tie, and the reduced payout that goes with it."— Business Insider

A lot of people have been buying tickets, and as discussed above, this will greatly increase the odds of a tie, and the reduced payout that goes with it."

Those are some pretty big caveats! They shouldn't just be mentioned in passing at the bottom but factored into the analysis to begin with.

Not that you asked, but I find the math says you should

**never**play Mega Millions. Given the exponential demand for tickets, based on jackpot size, I find the optimal time to play is a jackpot of $545 million. With jackpots larger than that you'll have to share it with too many other winners. At that jackpot size, the player can expect a return of 60.2%, or a loss of 39.8%. That is as good as it gets.

In closing, no, I do not agree with Business Insider for tricking readers with a sensational title and not doing a proper analysis of taxes and jackpot sharing.

This question is asked and discussed in my forum at .

Is the "Jackpot Only" option in the Mega Millions lottery a good value?

Anonymous

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.

Wiz, I know you're a wet blanket when it comes to the lottery, but can you suggest an easier way to turn $2 into a $1,000,000?

Anonymous

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.