Lottery - FAQ

Let’s rephrase the inquiry. If a lottery comprises 10 million unique combinations and each player selects their numbers randomly (with the possibility of duplicates), how many tickets must be sold to ensure a 90% probability that at least one ticket results in a win? We can denote p as the winning probability and n as the number of tickets sold. The probability of one individual losing can be defined as 1-p, while the chance of all n individuals losing would be expressed as (1-p)^n.ⁿThe formula for finding at least one winner's probability transforms to 1 - (1-p)^n.ⁿTherefore, the lottery needs to offer 23,025,850 tickets for the chances of having at least one winner to reach 90%. If the lottery only sold 10 million tickets, the likelihood of having at least one win would fall to approximately 63.2%, which is closely akin to 1 - (1/e).

.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.

This leads to my follow-up question, focusing on state lotteries. You've likely heard about groups of 'investors' who wait for a jackpot to hit a certain threshold before purchasing tickets covering every possible number combination, thus securing a portion of the winnings. What jackpot amount would make this an advantageous investment, assuming each ticket costs $1?

I have been engaged in playing lotteries and sweepstakes continuously for the past two months. Will luck ever be on my side for hitting a jackpot?

Over the years of observing the Quinto lottery draws here in Washington State, which utilizes a deck of 52 cards for drawing 5, I've noticed a striking trend: typically, three suits appear during draws. Based on probabilities derived from poker, the chance of drawing just one suit (a flush) stands at 5148 out of roughly 2.6 million. What would be the odds of drawing 2, 3, or even all 4 suits?

When considering two variables, f(x,y) can be expressed as combin(13,x)*combin(13,y)*12/combin(52,5).

With three suits, we can express it as f(x,y,z) = combin(13,x)*combin(13,y)*combin(13,z)*12/combin(52,5).

For four variables, we identify it as f(w,x,y,z) = combin(13,w)*combin(13,x)*combin(13,y)*combin(13,z)*4/combin(52,5).

To determine the probability of all four suits appearing, we use COMBIN(13,1).

The likelihood of drawing three suits translates to COMBIN(13,3)*COMBIN(13,1).³*COMBIN(13,2)*4/combin(52,5) = 26.37%.

Calculating the odds for two suits results in COMBIN(13,3)*COMBIN(13,2)*12 + COMBIN(13,4)*COMBIN(13,1)*12/combin(52,5), which comes out as 14.59%.²*12 + COMBIN(13,1)*COMBIN(13,2)^2*12/combin(52,5) = 58.84%

The chances of acquiring one suit (including the potential for straight and royal flushes) can be calculated as 4*combin(13,5)/combin(52,5), equaling 0.20%.

From these calculations, it appears that having three suits is indeed the most common outcome.

In your chart of lottery probabilities for the Maryland Lotto game,, the possibility of sharing a jackpot isn't considered. How does this factor influence the expected value?

I'm a student from a financially struggling background and would appreciate assistance with this week's draws; thank you.

Dear Wizard: I’m attending a horse racing venue that's introducing video lottery terminals. Can you shed some light on them? Are they similar to slot machines? Any insights would truly be appreciated.

If I obtain two quick-pick lottery tickets, what is the probability that both will share the same number? Let's assume we're considering a 6/49 lottery.

Assuming you're informed, but just in case: in the Italian lottery, there’s a drawing twice a week for 50 out of the numbers 1 to 90 (drawing five numbers from each of 10 cities). For almost two straight years, the number 53 hasn’t appeared, leading to a surge of activity around it, to the point where some people have gone to extraordinary lengths betting everything, convinced it’s due for a win! This leads me to ponder—what are the odds of the number 53 not showing up for two years? combin After investigating, I found out that six numbers are drawn in every session. The possibility of the number 53 not being selected in any one draw is

Referring to your response about the Italian lottery, you indicated the odds of 53 not being picked in two years stands at 1 in 1,707,460. You ought to have pursued the probability that any specific number (or multiple numbers) from the pool of 90 could also be skipped during that same two-year timeframe, as I believe that's the information the original query sought. Furthermore, you should reiterate why the number 53 maintains the same chances in the upcoming lottery as any of the other numbers, despite its prior absence. combin The probability that any given number would not be drawn in a two-year range could roughly be approximated as 90*(1/1,707,460) = 1 in 18,972. The authentic probability is marginally lower, as I may have overcounted instances of two numbers being missed, which is minimal. It's important to note, though, that in lottery games, past outcomes do not influence future draws, and each round holds equal likelihood for selecting the number 53.²⁰⁸= 0.000000585665, or 1 in 1,707,460.

The Powerplay option increases non-grand prize winnings by a factor of 2 to 5 times. Testimonials from Powerball winners suggest that 'powerplay is the only way to go.' Personally, I have my reservations about it being a wise investment.

Let n represent your answer. The odds of rolling a 7 or an 11 stands at 8/36. To determine the number of rolls required: combin So, based on this, hitting the SuperLotto is akin to rolling a seven or eleven consecutively 11.66 times. For those struggling with the concept of a fractional roll, I would phrase it as the odds lying somewhere between 11 and 12 sequential rolls.^-m*mⁿMy spouse and I purchased a $20 raffle ticket for the Indiana lottery. As I understand it, the draw for the available prizes (which total 777) will occur on August 16, 2007, regardless of the actual ticket sales, and there is a cap set at 325,000 tickets. Currently, only 60,000 tickets have been sold. Would buying a few more tickets be a prudent gamble? What are our odds of winning a prize?^-0.71865*0.71865⁰Explore the Top Online Casinos Available in Your Region

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Let’s reformulate this inquiry. If we assume there are 10 million combinations in the lottery and each player selects numbers randomly (allowing for duplicates), how many tickets would need to be sold to achieve a 90% probability that at least one ticket wins? We can denote winning probability as p and the quantity of tickets sold as n. The likelihood of a single loss is 1-p. Thus, the probability that all n participants do not win can be expressed as (1-p).

(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

Consequently, the probability that at least one player wins would then be calculated as 1 - (1-p).

Therefore, in order to ensure a 90% chance of at least one winner, the lottery must sell 23,025,850 tickets. Just for your information, if exactly ten million tickets were sold, the odds for at least one ticket resulting in a win would be roughly 63.2%, approximated closely by the formula 1-(1/e).

I checked the New York The answer hinges on the overall number of tickets sold to other players. If multiple players hit the jackpot, the prize will be divided. We’ll refer to the number of possible combinations as n, the total number of other tickets sold as t, the return rate on smaller prizes as r (for instance, for the Big Game r=0.179612), and the jackpot value as j. To break even, the equation should be j*n/(n+t) + r*n - n=0, which simplifies to j=(1-r)*(n+t).

Unfortunately, the answer is no; the likelihood of winning a typical 6/49 lottery is 1 in 13,983,816. To have a 50/50 chance of winning at least once, you would need to play approximately ln(.5)/ln(1-1/combin(49,6)) which is around 9,692,842 times. If you were to buy 100 lottery tickets daily, it would take nearly 265.6 years to reach a 50% chance of winning at least once. For a 90% winning chance, the time frame extends to about 882.2 years. Over the years, I’ve observed something interesting with our Quinto lottery draws here in Washington State. The game involves a 52-card setup where 5 cards are drawn. It seems that most of the time, three different suits are drawn. Based on poker probabilities, the chance of drawing just one suit (a flush) is 5148 out of almost 2.6 million. What are the odds of seeing 2, 3, or all 4 suits in these draws? Let’s express the function f(x,y) as the chance of drawing x cards of one suit and y of another. This formula can accommodate more than two terms.

With two variables, the formula is f(x,y)= combin(13,x)*combin(13,y)*12/combin(52,5).

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

Using three variables, the equation expands to f(x,y,z)= combin(13,x)*combin(13,y)*combin(13,z)*12/combin(52,5).

The probability of drawing all four suits can be represented as COMBIN(13,1).

The odds of extracting three suits would be computed as COMBIN(13,3)*COMBIN(13,1).

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 likelihood of two suits appearing can be expressed as COMBIN(13,3)*COMBIN(13,2)*12 + COMBIN(13,4)*COMBIN(13,1)*12/combin(52,5), equating to 14.59%.

The probability of obtaining just one suit (which includes both straight and royal flushes) works out to 4*combin(13,5)/combin(52,5), landing at about 0.20%.

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

In your probability guide for the MD lottery game, there’s an oversight regarding the chances of jackpot splitting. What impact does this factor have on the expected value of the lottery? lottoreport.com I did not take into consideration the possibility of shared jackpots. This does indeed lower the expected value, as the more players there are, the smaller the expected returns become. When I authored that piece, I lacked sufficient data on the player numbers to accurately incorporate this element.

As a student from a background of financial hardship, I would be grateful for any assistance with the weekly draws. Thank you.

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
I assume you're looking for lottery numbers? Unfortunately, I can't provide better guidance than you might already have. However, I would advise against playing altogether, particularly if funds are tight. There seem to be numerous former military leaders or dictators attempting to transfer large sums like $17 million to me, perhaps one of them would fund your education instead.	0.74%	1.29%
Dear Wizard: The horse racing venue I frequent is adding video lottery machines. Can you share any insights about these? Are they akin to slot machines? Any information you could provide would be greatly appreciated.	4.01%	6.59%
Another question about odds: Many racetracks allow 'class 2' gaming, which must be based on lottery or bingo. The way to present slot games under this classification is to conduct a lottery or bingo game behind the scenes while showcasing the results as a slot machine win. For instance, if the lottery determines your payout is 20 times your bet, it would display whatever slot machine symbols are equivalent to that payout. So, it creates a clever illusion.	1.41%	2.31%

If I purchase two quick-pick lottery tickets, what would be the chance of both cards sharing the same number in a 6/49 lottery?

Winning accurately by selecting 6 numbers from a pool of 49 yields a probability of 1 in (49,6), which is 1 in 13,983,816. This is also the likelihood that both of your tickets would match.

Assuming you're informed, but in case you aren't, in the Italian lottery, there are two weekly draws where 50 numbers are chosen from 1 to 90 (five numbers from each of 10 cities). For nearly two years, the number 53 has not appeared, resulting in a frenzy among people who placed everything on that number expecting it to show up! This brings me to question—what are the probabilities that 53 wouldn’t have appeared for two consecutive years? Wizard of Vegas .

Referring back to your response regarding the Italian lottery, you stated the chance of 53 not being drawn in a two-year period is 1 in 1,707,460. You should also have considered the chance that one or more of the 90 numbers would be missed during the same interval, as that’s likely what the questioner was truly interested in. Moreover, it would have been valuable to clarify that despite the unlikely history, 53 has the same chance of being picked in the next lottery as any other number.

The probability of any number not being drawn in two years can be approximated as 90*(1/1,707,460), resulting in 1 in 18,972. The precise odds would be marginally lower since I inadvertently accounted for two numbers being missed, which is an insignificant factor. Remember, past results do not influence the lottery, and each draw maintains an equal probability for number 53.

It is stated that the chance of winning the Powerball is 1 in 146,107,962. Recently, during a drawing where the jackpot was set at $340 million, local reports indicated that 105,000,000 tickets were sold. My inquiries are: if you win, what are the chances you’ll end up sharing the jackpot, and how does that affect the expected value?

Let’s first validate the probability. In this game, players must match 5 regular numbers from a set of 55, along with one Powerball from a set of 42. The probability for winning is indeed 1 in (55,5)*42, measuring up to 1 in 146,107,962. I concur with your calculation. For queries like yours, I prefer the Poisson distribution. The average number of winners would be calculated as 105,000,000 divided by 146,107,962, giving us approximately 0.71865. The general formula for the probability of n winners, given a mean of m, is e raised to the power of negative m over n factorial. Here, with a mean of 0.71865, the probability of zero winners turns out to be e raised to the power of zero, which equals 0.48741. Consequently, the likelihood of having at least one winner is 1 - 0.48741 = 0.51259. Thus, 0.71865 winners would have to share 0.51259 of a jackpot. This means that for every winner, they would effectively receive 0.51259/0.71865, which is approximately 0.71327 jackpots, reducing the expected win to about 71.327% of the full jackpot amount—hence, a decrease of 28.673%.

Powerplay provides a multiplier between 2x and 5x for non-grand prizes. Testimonials from winners on the Powerball site suggest 'Powerplay is the only way to go.' However, I suspect it may be somewhat misleading. The lottery is inherently a gamble with unfavorable odds! Briefly put, the return on Powerplay is around 49.28%. The returns from a Powerball ticket alone are significantly less, indicating it would be more advantageous to purchase x/2 tickets with the Powerplay option rather than x tickets without it. I’ve appended further details regarding this feature in my discussion. Wizard, could you illustrate the equivalent odds of the California SuperLotto Plus (1 in 41.4 million) by measuring the number of consecutive rolls of 7 or 11? I’ve come across this information previously. A lot of people struggle to grasp lottery odds, yet rolling dice is something they find relatable.

Assuming you're informed, but in case you aren't, in the Italian lottery, there are two weekly draws where 50 numbers are chosen from 1 to 90 (five numbers from each of 10 cities). For nearly two years, the number 53 has not appeared, resulting in a frenzy among people who placed everything on that number expecting it to show up! This brings me to question—what are the probabilities that 53 wouldn’t have appeared for two consecutive years? Wizard of Vegas .

Consequently, the likelihood of winning the SuperLotto matches with rolling a seven or eleven consecutively 11.66 times. For those who have a hard time grasping partial throws, I would phrase it such that the probability lies between completing 11 and 12 cumulative rolls. Extra My wife and I recently purchased a $20 raffle ticket for the Indiana lottery. As I understand it, a drawing for the winning prizes (totaling 777) is scheduled for August 16, 2007, independent of the number of tickets actually sold, with a maximum limit set at 325,000 tickets. Currently, only 60,000 tickets have been sold. Would it be a wise decision to buy a few more tickets? What are our chances of winning a prize?

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Assuming you're informed, but in case you aren't, in the Italian lottery, there are two weekly draws where 50 numbers are chosen from 1 to 90 (five numbers from each of 10 cities). For nearly two years, the number 53 has not appeared, resulting in a frenzy among people who placed everything on that number expecting it to show up! This brings me to question—what are the probabilities that 53 wouldn’t have appeared for two consecutive years? Wizard of Vegas .

Let’s reformulate the inquiry. Suppose the lottery features 10 million unique combinations and every participant makes random selections (allowing for duplicates). What number of tickets does the lottery need to sell to ensure at least a 90% chance of at least one person winning? We can denote p as the winning probability and n as the quantity of tickets sold. The likelihood of a single participant not winning is 1-p, and thus, the chance that all n individuals lose accumulates to (1-p). Quebec Lottery Consequently, the chance of achieving at least one winner is represented as 1 - (1-p). From here, we must set this expression to 0.9 and solve for n.

Thus, the lottery would require the sale of approximately 23,025,850 tickets to reach a 90% probability of at least one winner. For your information, if the lottery sold precisely ten million tickets, the probability of ending up with at least one winner would be around 63.2%, which approximates closely to 1 - (1/e).

Now I have a follow-up question focusing on state lotteries. You may recall stories about a group of 'investors' who would wait for jackpots to reach a threshold before they would buy tickets covering every number combination possible. This strategy assured them a portion of the winnings. With ticket prices set at $1, to what jackpot amount would it be financially beneficial for them to engage in this practice?

I've been actively participating in lotteries and sweepstakes for the past two months. Is there any chance that I may win a jackpot, and if so, when?

The quick response is no; your chances of winning are negligible. For a typical 6/49 lottery, the odds are 1 in 13,983,816. To achieve a 50/50 probability of winning at least once, you would need to enter the lottery approximately 9,692,842 times. Assuming you buy 100 tickets each day, you would need to continue for about 265.6 years just to reach a 50% likelihood of winning. To push this up to a 90% chance, you would have to play for around 882.2 years.

Over time, I've observed patterns during our Quinto lottery drawings in Washington State, which resembles a game played with 52 cards where 5 cards are drawn. I've noticed that, more often than not, three suits appear. Given poker statistics, it's interesting to note that the odds of drawing just one suit (resulting in a flush) are 5,148 out of roughly 2.6 million. What would be the chances of drawing 2, 3, or all 4 suits?

Let’s explore the function f(x,y), representing the probability of obtaining x from one suit and y from another. This function isn’t restricted to just two variables.

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

Let\\'s look at them one at a time.

For two variables, we can express this as f(x,y) = combin(13,x)*combin(13,y)*12/combin(52,5).

With three variables, the formula becomes f(x,y,z) = combin(13,x)*combin(13,y)*combin(13,z)*12/combin(52,5).

For four variables, we express it as f(w,x,y,z) = combin(13,w)*combin(13,x)*combin(13,y)*combin(13,z)*4/combin(52,5). The chance of all four suits occurring is calculated using COMBIN(13,1). LottoReport.com For three suits, the probability is determined using COMBIN(13,3)*COMBIN(13,1). Powerball Regarding two suits, we find that it is COMBIN(13,3)*COMBIN(13,2)*12 + COMBIN(13,4)*COMBIN(13,1)*12/combin(52,5) = 14.59%.

The odds for drawing one suit (including both straight and royal flush combinations) is calculated as 4*combin(13,5)/combin(52,5) = 0.20%.^xIt turns out that the occurrence of three suits is the most common result.



No, I overlooked the possibility of splitting the jackpot. This certainly lowers the expected value; the more players there are, the more reduced the expected outcome becomes. At the time I wrote that article, I didn’t have sufficient data on the number of players to properly incorporate that factor.

I come from a very underprivileged background, and I would greatly appreciate any assistance you could offer regarding the weekly draws. Thank you.



I assume you are looking for lottery numbers. Unfortunately, I cannot provide better insights than you might find yourself. However, I would advise against participating altogether, especially given your financial situation. There seem to be numerous former generals and dictators trying to send me $17 million; perhaps one of them will fund your education.

Dear wiz: I frequent a racetrack that is about to implement video lottery machines. Can you share any insights about these machines? Are they similar to traditional slots? Any information would be greatly appreciated.

Another Mike S., what are the odds? Many racetracks offer 'class 2' gaming, which must be based on either lottery or bingo. To provide slots within this framework, they'd run a lottery or bingo game in the background, displaying its outcome as a slot machine win. For instance, if the lottery game indicates that you win 20 times your bet, it will show the respective slot machine symbols that reflect this win. It's quite the clever ruse.

If I were to purchase two quick-pick lottery tickets, what are the odds that both tickets will feature identical numbers? Assuming it’s a 6/49 lottery.

The probability of correctly selecting 6 numbers from a total of 49 is 1 in (49,6) = 1 in 13,983,816. This is also the exact chance that your two tickets would match.

Assuming you're aware, if not, the Italian lottery conducts bi-weekly drawings for 50 chosen from numbers 1 to 90 (five from each of the 10 cities). For about two years, the number 53 has not been drawn, leading to what some call a 'number 53 frenzy,' resulting in desperate actions from individuals who have wagered all their resources expecting a return! This leads me to ponder: what are the odds that the number 53 would fail to show up for two consecutive years?

Upon conducting some research, I found that six numbers are selected in each drawing. In any single drawing, the probability that the number 53 does not appear is calculated as (89,6)/combin(90,6) = 93.333%. Given that the past two years included 208 drawings, the probability of the number 53 being absent during that time would be 0.93333.

Regarding your previous response about the Italian lottery (see, ) you indicated that the probability of the number 53 not appearing in two years equates to 1 in 1,707,460. You should have addressed the likelihood of any one or more of the 90 numbers being overlooked in the same two-year span; that’s likely the information the inquirer truly sought. Moreover, it would have been beneficial to reiterate why the number 53 holds no greater chance of being selected in the next draw compared to any other number, despite its seemingly improbable history.

The likelihood of any single number not being drawn over the period of two years can be closely approximated as 90*(1/1,707,460) = 1 in 18,972. The actual probability would be marginally lower because I inadvertently counted instances of two numbers being absent, but this impact is minor. It's essential to remember that past outcomes do not influence lottery draws, and the chances in each drawing remain the same for the number 53. Wizard of Vegas .

is 1 in 146,107,962. In the recent drawing of a $340 million jackpot, the local news indicated that 105,000,000 tickets were sold. My inquiries are if I were to win, what would be the likelihood of having to share the jackpot, and how does this affect the expected value?

First, let’s validate that probability. A player is required to match 5 standard numbers from a total of 55, along with one Power Ball from a pool of 42 numbers. Thus, winning odds align with 1 in (55,5)*42 = 1 in 146,107,962. I concur with your probability assessment. I find the Poisson distribution particularly useful for inquiries like yours. If we consider the average number of winners as 105,000,000/146,107,962 = 0.71865, we can utilize the general formula for the probability of n winners, defined as e^-m/m!. Given that m is 0.71865, the probability of encountering zero winners is e^-0.71865/0! = 0.48741. Therefore, the probability of having at least one winner is 1-0.48741 = 0.51259. Hence, 0.71865 winners will share 0.51259 of a jackpot. Consequently, this results in a ratio of 0.51259/0.71865 = 0.71327 jackpots per winner. Therefore, sharing the jackpot effectively lowers your anticipated win to 71.327% of the total jackpot amount, constituting a decrease of 28.673%. Mega Millions lottery .

Engaging in the lottery is always somewhat of a fool’s bet! To put it simply, the return rate for the Powerplay option stands at 49.28%. This contrasts vastly with the abysmal return from a standard Powerball ticket; it would be more advantageous to buy x/2 tickets with the Powerplay feature rather than x without it. I've noted details regarding this option in my earlier writings.

Wizard, could you elaborate on the corresponding odds of the California SuperLotto Plus (at 1 in 41.4 Million) in terms of rolling consecutive 7s or 11s? I've encountered this comparison before, and many find it easier to grasp lottery odds in relation to dice rolling.

• Let n represent your answer. The chance of rolling either a 7 or 11 stands at 8/36. To resolve for n:
• Thus, there you have it; the probability of hitting the SuperLotto equates to rolling a seven or an eleven 11.66 times consecutively. For those who may struggle with grasping a fractional roll, I would suggest stating that the probability falls between accomplishing 11 and 12 consecutive throws.
• My spouse and I purchased a $20 raffle ticket for the Indiana lottery. My understanding is that the drawing for the prizes (which total 777) will take place on August 16, 2007, irrespective of how many actual tickets are sold, with the utmost limit being 325,000 tickets. As of today, only 60,000 tickets have been sold. Would it be wise for us to buy a few more tickets? What would be our chances of winning something?

The Indiana Lottery's official website states that the total amount awarded in prizes reaches $3,270,000, benefiting 325,000 people who purchased tickets. This results in an average prize value of about $10.615 per ticket, assuming all tickets were sold. With each ticket priced at $20, this yields a return of approximately 50.31%. In a scenario where only 60,000 tickets are sold, the value of each ticket would increase to $54.50, offering a remarkable return of 272.50%. The threshold at which a player breaks even is determined at 163,500 tickets sold. If you anticipate fewer than that number will be sold, investing in tickets could be considered a worthwhile gamble, disregarding the effects of taxation and the time value of money.

I've heard that the same numbers have been drawn in the German 6/49 lottery on different occasions. There seems to be something suspicious about this. What could the chances be that this happened?

While it is indeed true, it may not be as strange as you might think. According to 'Understanding Probability: Chance Rules in Everyday Life' by H. C. Tijms, the identical combinations were pulled on June 21, 1995, and December 20, 1986, during bi-weekly draws. The December 20 draw in 1986 was actually the 3,016th instance of the lottery taking place. With a total of 13,983,816 different possible combinations in a 6/49 lottery, the odds against the numbers drawn in the second occurrence matching the first are expressed as (c-1)/c, where c is the number of combinations. The probability of all subsequent draws producing a distinct set of numbers can be calculated similarly, leading to an overall chance of around 27.8% that at least one set of common numbers is drawn across multiple drawings.

Probability of Matching Numbers in 6/49 Lottery³⁰⁰= 63.29%.

If you're curious, the total number of draws required for the likelihood of a matching draw to surpass 50% stands at 4,404.

The likelihood of any single number not being drawn over the period of two years can be closely approximated as 90*(1/1,707,460) = 1 in 18,972. The actual probability would be marginally lower because I inadvertently counted instances of two numbers being absent, but this impact is minor. It's essential to remember that past outcomes do not influence lottery draws, and the chances in each drawing remain the same for the number 53. Wizard of Vegas .

For those unfamiliar, the Nova Scotia sports lottery functions akin to off-the-board parlay bets found at a casino in Nevada, though with less favorable odds. To determine the expected return on a particular selection made by a random generator, you would need to aggregate the inverses of what each outcome pays out and then take the inverse of that total.

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

To illustrate, during the Monday Night Football matchup on November 9, 2009, they provided these betting options:

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

Calculating the sum of the inverse values shows (1/1.9) + (1/3.25) + (1/3.65) = 1.107981. The inverse of this sum is 1/1.107981 = 0.902543. This results in an expected return of 90.25. For a parlay scenario, you would multiply the returns of all selections made. Wizard of Vegas .

I absolutely believe that this is an important consideration, even if it might seem like a lesser point, when deciding to purchase a lottery ticket. To address your query, I referenced the jackpot statistics and sales data available on the relevant websites. I researched Powerball starting from January 2008, which is the oldest data available, and Mega Millions from June 2005, following a change in the rules. The table below summarizes my findings.

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

Split Jackpots in Powerball and Mega Millions

Average likelihood of a jackpot being split per drawing

Return loss due to shared jackpots (unadjusted)

Return loss due to shared jackpots (adjusted)

The average chance of a jackpot being split is calculated to be 0.74% for Powerball and 1.29% for Mega Millions. As the jackpots rise and more tickets are sold, the probability of sharing the jackpot increases. The heightened chance of a split jackpot in Mega Millions can be attributed to its easier odds of winning and increased competitive player numbers.

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

Taking everything into account, it appears that 4.01% is lost in returns based on jackpot sharing in Powerball, while Mega Millions faces a 6.59% loss. However, these figures don’t include tax considerations or the fact that jackpot amounts are distributed as annuities. To adjust for these aspects, I assumed that the winner ultimately receives half of the jackpot, whether as a one-time payout or accounting for the decline in value of the annuity option. Additionally, I estimated that about 30% of the remainder would be subject to taxes, leaving the winner with approximately 35% after accounting for both elements. After this revision, I calculated a 1.20% reduction in return due to jackpot sharing for Powerball and 1.98% for Mega Millions.

This inquiry has been explored and discussed in the community area of my affiliated website.

The likelihood of any single number not being drawn over the period of two years can be closely approximated as 90*(1/1,707,460) = 1 in 18,972. The actual probability would be marginally lower because I inadvertently counted instances of two numbers being absent, but this impact is minor. It's essential to remember that past outcomes do not influence lottery draws, and the chances in each drawing remain the same for the number 53. Wizard of Vegas .

You must be referring to the Ca$h WinFall game. I first became aware of it through the article titled A game with a windfall for a knowing few published on boston.com.

It's not uncommon for progressive lottery games' jackpots to grow so large that the return surpasses 100%, not considering tax implications, the structure of annuity payments, the slim chances of winning, and the diminishing value attached to exceedingly large jackpots. Once you factor these components in, lotteries rarely provide a favorable bet.

The unique feature of the Ca$h WinFall game is that when the jackpot exceeds $2 million without being won, the majority of the prize pool, with the exception of $500,000, is redistributed to lower-tier prizes. This essentially guarantees substantial profits for players with bankrolls in the six-figure range.

Let's examine the drawing from July 18, 2011, as an illustration. It’s a straightforward 6-46 game where players select six numbers from a choice of 1 to 46, mirroring the lottery’s selection. The more closely your chosen numbers align with the lottery's, the greater your winnings. The table below presents the probabilities and expected returns associated with each occurrence. Securing two matches yields a ticket worth 26 cents, which I separate as an individual entry. Each ticket costs $2, so the expected return is computed as the product of probability and winnings divided by the ticket price, resulting in an anticipated 117% return, translating into a 17% player advantage.
Since I first drafted my response, the Massachusetts Lottery has set a cap on the number of tickets any store can sell in a single day, limiting it to 2,500, which equals a maximum of $5,000, according to the article on boston.com titled, 'Lottery restricts high-level players.' This change will undoubtedly make it more challenging to wager substantial amounts, yet it could benefit players with smaller budgets by reducing competition.

What are your thoughts on the lawsuit brought against the Quebec Lottery regarding the alleged non-randomness of their tickets?

The likelihood of any single number not being drawn over the period of two years can be closely approximated as 90*(1/1,707,460) = 1 in 18,972. The actual probability would be marginally lower because I inadvertently counted instances of two numbers being absent, but this impact is minor. It's essential to remember that past outcomes do not influence lottery draws, and the chances in each drawing remain the same for the number 53. Wizard of Vegas .