Is there a math to winning the lottery?

Lottery mathematics is used to calculate probabilities of winning or losing a lottery game. It is based primarily on combinatorics, particularly the twelvefold way and combinations without replacement.

en.wikipedia.org - Lottery mathematics - Wikipedia

Are odds always 50 50?

The sum of all possible outcomes is always 1 (or 100 percent) because it is certain that one of the possible outcomes will happen. This means that...

Last updated Mar 27, 2019

Can you tell when slot machine will hit?

You cannot tell when a modern slot machine will hit because the outcome of each spin is random. No matter how many times a machine has spun, and no...

Last updated Mar 1, 2021

Make Your Bookie Cry And Pull The Hair By Generating $23,187

Promotion

Probabilities of winning or losing a lottery game

Choosing 6 from 49 [ edit ]

In a typical 6/49 game, each player chooses six distinct numbers from a range of 1-49. If the six numbers on a ticket match the numbers drawn by the lottery, the ticket holder is a jackpot winner—regardless of the order of the numbers. The probability of this happening is 1 in 13,983,816. The chance of winning can be demonstrated as follows: The first number drawn has a 1 in 49 chance of matching. When the draw comes to the second number, there are now only 48 balls left in the bag, because the balls are drawn without replacement. So there is now a 1 in 48 chance of predicting this number. Thus for each of the 49 ways of choosing the first number there are 48 different ways of choosing the second. This means that the probability of correctly predicting 2 numbers drawn from 49 in the correct order is calculated as 1 in 49 × 48. On drawing the third number there are only 47 ways of choosing the number; but of course we could have arrived at this point in any of 49 × 48 ways, so the chances of correctly predicting 3 numbers drawn from 49, again in the correct order, is 1 in 49 × 48 × 47. This continues until the sixth number has been drawn, giving the final calculation, 49 × 48 × 47 × 46 × 45 × 44, which can also be written as 49 ! ( 49 − 6 ) ! {displaystyle {49! over (49-6)!}} or 49 factorial divided by 43 factorial. This works out to 10,068,347,520, which is much bigger than the ~14 million stated above. However; the order of the 6 numbers is not significant. That is, if a ticket has the numbers 1, 2, 3, 4, 5, and 6, it wins as long as all the numbers 1 through 6 are drawn, no matter what order they come out in. Accordingly, given any set of 6 numbers, there are 6 × 5 × 4 × 3 × 2 × 1 = 6! or 720 orders in which they could be drawn. Dividing 10,068,347,520 by 720 gives 13,983,816, also written as 49 ! 6 ! ∗ ( 49 − 6 ) ! {displaystyle {49! over 6!*(49-6)!}} , or more generally as ( n k ) = n ! k ! ( n − k ) ! {displaystyle {n choose k}={n! over k!(n-k)!}} binomial coefficient and multinomial coefficient. This function is called the combination function. For the rest of this article, we will use the notation ( n k ) {displaystyle {n choose k}} . "Combination" means the group of numbers selected, irrespective of the order in which they are drawn. An alternative method of calculating the odds is to note that the probability of the first ball corresponding to one of the six chosen is 6/49; the probability of the second ball corresponding to one of the remaining five chosen is 5/48; and so on. This yields a final formula of ( n k ) = ( 49 6 ) = 49 6 ∗ 48 5 ∗ 47 4 ∗ 46 3 ∗ 45 2 ∗ 44 1 {displaystyle {n choose k}={49 choose 6}={49 over 6}*{48 over 5}*{47 over 4}*{46 over 3}*{45 over 2}*{44 over 1}} The range of possible combinations for a given lottery can be referred to as the "number space". "Coverage" is the percentage of a lottery's number space that is in play for a given drawing. Odds of getting other possibilities in choosing 6 from 49 [ edit ] One must divide the number of combinations producing the given result by the total number of possible combinations (for example, ( 49 6 ) = 13 , 983 , 816 {displaystyle {49 choose 6}=13,983,816} ). The numerator equates to the number of ways to select the winning numbers multiplied by the number of ways to select the losing numbers. For a score of n (for example, if 3 choices match three of the 6 balls drawn, then n = 3), ( 6 n ) {displaystyle {6 choose n}} describes the odds of selecting n winning numbers from the 6 winning numbers. This means that there are 6 - n losing numbers, which are chosen from the 43 losing numbers in ( 43 6 − n ) {displaystyle {43 choose 6-n}} ways. The total number of combinations giving that result is, as stated above, the first number multiplied by the second. The expression is therefore ( 6 n ) ( 43 6 − n ) ( 49 6 ) {displaystyle {6 choose n}{43 choose 6-n} over {49 choose 6}} .

This can be written in a general form for all lotteries as:

( K B ) ( N − K K − B ) ( N K ) {displaystyle {K choose B}{N-K choose K-B} over {N choose K}} where N {displaystyle N} is the number of balls in lottery, K {displaystyle K} is the number of balls in a single ticket, and B {displaystyle B} is the number of matching balls for a winning ticket.

The generalisation of this formula is called the hypergeometric distribution.

This gives the following results:

Score Calculation Exact Probability Approximate Decimal Probability Approximate 1/Probability 0 ( 6 0 ) ( 43 6 ) ( 49 6 ) {displaystyle {6 choose 0}{43 choose 6} over {49 choose 6}} 435,461/998,844 0.436 2.2938 1 ( 6 1 ) ( 43 5 ) ( 49 6 ) {displaystyle {6 choose 1}{43 choose 5} over {49 choose 6}} 68,757/166,474 0.413 2.4212 2 ( 6 2 ) ( 43 4 ) ( 49 6 ) {displaystyle {6 choose 2}{43 choose 4} over {49 choose 6}} 44,075/332,948 0.132 7.5541 3 ( 6 3 ) ( 43 3 ) ( 49 6 ) {displaystyle {6 choose 3}{43 choose 3} over {49 choose 6}} 8,815/499,422 0.0177 56.66 4 ( 6 4 ) ( 43 2 ) ( 49 6 ) {displaystyle {6 choose 4}{43 choose 2} over {49 choose 6}} 645/665,896 0.000969 1,032.4 5 ( 6 5 ) ( 43 1 ) ( 49 6 ) {displaystyle {6 choose 5}{43 choose 1} over {49 choose 6}} 43/2,330,636 0.0000184 54,200.8 6 ( 6 6 ) ( 43 0 ) ( 49 6 ) {displaystyle {6 choose 6}{43 choose 0} over {49 choose 6}} 1/13,983,816 0.0000000715 13,983,816

When a bonus number is included, the adjusted odds are:[1]

What are the 4 types of pulse?

There are seven types of pulse. Temporal: It is felt in the head. Carotid: It is felt in the neck. Branchial: It is felt in the elbow. Femoral: It...

Last updated Jun 10, 2022

What happens if you get caught being a bookie?

Charged as a felony, California bookmaking carries the following potential penalties: Felony (formal) probation; Sixteen (16) months, two (2) years...

Last updated Nov 10, 2020

Promotion

Score Calculation Exact Probability Approximate Decimal Probability Approximate 1/Probability 5, bonus not won ( 6 5 ) ( ( 43 − 1 ) 1 ) ( 49 6 ) {displaystyle {6 choose 5}{(43-1) choose 1} over {49 choose 6}} 0.0000180208 55,491.33 5, bonus won ( 6 5 ) ( ( 43 − 1 ) ( 1 − 1 ) ) ( 49 6 ) {displaystyle {6 choose 5}{(43-1) choose (1-1)} over {49 choose 6}} 0.0000004291 2,330,636

Ensuring to win the jackpot [ edit ]

There is only one known way to ensure winning the jackpot. That is to buy at least one lottery ticket for every possible number combination. For example, one has to buy 13,983,816 different tickets to ensure to win the jackpot in a 6/49 game. Lottery organizations have laws, rules and safeguards in place to prevent gamblers from executing such an operation. Further, just winning the jackpot by buying every possible combination does not guarantee to break even or make a profit. If p {displaystyle p} is the probability to win; c t {displaystyle c_{t}} the cost of a ticket; c l {displaystyle c_{l}} the cost for obtaining a ticket (e.g. including the logistics); c f {displaystyle c_{f}} one time costs for the operation (such as setting up and conducting the operation); then the jackpot m j {displaystyle m_{j}} should contain at least m j ≥ c f + c t + c l p {displaystyle m_{j}geq c_{f}+{frac {c_{t}+c_{l}}{p}}}

to have a chance to at least break even.

The above theoretical "chance to break-even" point is slightly offset by the sum ∑ i m i {displaystyle sum _{i}{}m_{i}} of the minor wins also included in all the lottery tickets: m j ≥ c f + c t + c l p − ∑ i m i {displaystyle m_{j}geq c_{f}+{frac {c_{t}+c_{l}}{p}}-sum _{i}{}m_{i}} Still, even if the above relation is satisfied, it does not guarantee to break even. The payout depends on the number of winning tickets for all the prizes n x {displaystyle n_{x}} , resulting in the relation m j n j ≥ c f + c t + c l p − ∑ i m i n i {displaystyle {frac {m_{j}}{n_{j}}}geq c_{f}+{frac {c_{t}+c_{l}}{p}}-sum _{i}{}{frac {m_{i}}{n_{i}}}} In probably the only known successful operations[2] the threshold to execute an operation was set at three times the cost of the tickets alone for unknown reasons m j ≥ 3 × c t p {displaystyle m_{j}geq 3 imes {frac {c_{t}}{p}}}

I.e.

n j p c t ( c f + c t + c l p − ∑ i m i n i ) ≪ 3 {displaystyle {frac {n_{j}p}{c_{t}}}left(c_{f}+{frac {c_{t}+c_{l}}{p}}-sum _{i}{}{frac {m_{i}}{n_{i}}} ight)ll 3} This does, however, not eliminate all risks to make no profit. The success of the operations still depended on a bit of luck. In addition, in one operation the logistics failed and not all combinations could be obtained. This added the risk of not even winning the jackpot at all.

Powerballs and bonus balls [ edit ]

Many lotteries have a Powerball (or "bonus ball"). If the powerball is drawn from a pool of numbers different from the main lottery, the odds are multiplied by the number of powerballs. For example, in the 6 from 49 lottery, given 10 powerball numbers, then the odds of getting a score of 3 and the powerball would be 1 in 56.66 × 10, or 566.6 (the probability would be divided by 10, to give an exact value of 8815 4994220 { extstyle {frac {8815}{4994220}}} ). Another example of such a game is Mega Millions, albeit with different jackpot odds. Where more than 1 powerball is drawn from a separate pool of balls to the main lottery (for example, in the EuroMillions game), the odds of the different possible powerball matching scores are calculated using the method shown in the "other scores" section above (in other words, the powerballs are like a mini-lottery in their own right), and then multiplied by the odds of achieving the required main-lottery score. If the powerball is drawn from the same pool of numbers as the main lottery, then, for a given target score, the number of winning combinations includes the powerball. For games based on the Canadian lottery (such as the lottery of the United Kingdom), after the 6 main balls are drawn, an extra ball is drawn from the same pool of balls, and this becomes the powerball (or "bonus ball"). An extra prize is given for matching 5 balls and the bonus ball. As described in the "other scores" section above, the number of ways one can obtain a score of 5 from a single ticket is ( 6 5 ) ( 43 1 ) = 258 { extstyle {6 choose 5}{43 choose 1}=258} . Since the number of remaining balls is 43, and the ticket has 1 unmatched number remaining, 1/43 of these 258 combinations will match the next ball drawn (the powerball), leaving 258/43 = 6 ways of achieving it. Therefore, the odds of getting a score of 5 and the powerball are 6 ( 49 6 ) = 1 2 , 330 , 636 { extstyle {6 over {49 choose 6}}={1 over 2,330,636}} . Of the 258 combinations that match 5 of the main 6 balls, in 42/43 of them the remaining number will not match the powerball, giving odds of 258 ⋅ 42 43 ( 49 6 ) = 3 166 , 474 ≈ 1.802 × 10 − 5 { extstyle {{258cdot {frac {42}{43}}} over {49 choose 6}}={frac {3}{166,474}}approx 1.802 imes 10^{-5}} for obtaining a score of 5 without matching the powerball. Using the same principle, the odds of getting a score of 2 and the powerball are ( 6 2 ) ( 43 4 ) = 1 , 851 , 150 { extstyle {6 choose 2}{43 choose 4}=1,!851,!150} for the score of 2 multiplied by the probability of one of the remaining four numbers matching the bonus ball, which is 4/43. Since 1 , 851 , 150 ⋅ 4 43 = 172 , 200 { extstyle 1,851,150cdot {frac {4}{43}}=172,!200} , the probability of obtaining the score of 2 and the bonus ball is 172 , 200 ( 49 6 ) = 1025 83237 = 1.231 % { extstyle {frac {172,200}{49 choose 6}}={frac {1025}{83237}}=1.231\%} , approximate decimal odds of 1 in 81.2. The general formula for B {displaystyle B} matching balls in a N {displaystyle N} choose K {displaystyle K} lottery with one bonus ball from the N {displaystyle N} pool of balls is:

What is the diet of a jockey?

Training diet for jockeys Jockeys should try to eat three meals per day with foods from each of the food groups; breads and cereals, fruit,...

Last updated Jun 17, 2018

What is the highest NBA score in a game?

186-184 Twenty-two years ago, on Dec. 13, 1983, the Detroit Pistons beat the Denver Nuggets 186-184 in triple overtime. It was, and remains, the...

Last updated Nov 23, 2022

Promotion

K − B N − K ( K B ) ( N − K K − B ) ( N K ) {displaystyle {frac {{frac {K-B}{N-K}}{K choose B}{N-K choose K-B}}{N choose K}}} The general formula for B {displaystyle B} matching balls in a N {displaystyle N} choose K {displaystyle K} lottery with zero bonus ball from the N {displaystyle N} pool of balls is: N − K − K + B N − K ( K B ) ( N − K K − B ) ( N K ) {displaystyle {N-K-K+B over N-K}{K choose B}{N-K choose K-B} over {N choose K}} The general formula for B {displaystyle B} matching balls in a N {displaystyle N} choose K {displaystyle K} lottery with one bonus ball from a separate pool of P {displaystyle P} balls is: 1 P ( K B ) ( N − K K − B ) ( N K ) {displaystyle {1 over P}{K choose B}{N-K choose K-B} over {N choose K}} The general formula for B {displaystyle B} matching balls in a N {displaystyle N} choose K {displaystyle K} lottery with no bonus ball from a separate pool of P {displaystyle P} balls is: P − 1 P ( K B ) ( N − K K − B ) ( N K ) {displaystyle {P-1 over P}{K choose B}{N-K choose K-B} over {N choose K}}

Minimum number of tickets for a match [ edit ]

It is a hard (and often open) problem to calculate the minimum number of tickets one needs to purchase to guarantee that at least one of these tickets matches at least 2 numbers. In the 5-from-90 lotto, the minimum number of tickets that can guarantee a ticket with at least 2 matches is 100.[3]

Information theoretic results [ edit ]

As a discrete probability space, the probability of any particular lottery outcome is atomic, meaning it is greater than zero. Therefore, the probability of any event is the sum of probabilities of the outcomes of the event. This makes it easy to calculate quantities of interest from information theory. For example, the information content of any event is easy to calculate, by the formula I ⁡ ( E ) := − log ⁡ [ Pr ( E ) ] = − log ⁡ ( P ) . {displaystyle operatorname {I} (E):=-log {left[Pr {left(E ight)} ight]}=-log {left(P ight)}.} In particular, the information content of outcome x {displaystyle x} of discrete random variable X {displaystyle X} is I X ⁡ ( x ) := − log ⁡ [ p X ( x ) ] = log ⁡ ( 1 p X ( x ) ) . {displaystyle operatorname {I} _{X}(x):=-log {left[p_{X}{left(x ight)} ight]}=log {left({frac {1}{p_{X}{left(x ight)}}} ight)}.} For example, winning in the example § Choosing 6 from 49 above is a Bernoulli-distributed random variable X {displaystyle X} with a 1/13,983,816 chance of winning ("success") We write X ∼ B e r n o u l l i ( p ) = B ( 1 , p ) { extstyle Xsim mathrm {Bernoulli} !left(p ight)=mathrm {B} !left(1,p ight)} with p = 1 13 , 983 , 816 { extstyle p={ frac {1}{13,983,816}}} and q = 13 , 983 , 815 13 , 983 , 816 { extstyle q={ frac {13,983,815}{13,983,816}}} . The information content of winning is I X ⁡ ( win ) = − log 2 ⁡ p X ( win ) = − log 2 1 13 , 983 , 816 ≈ 23.73725 {displaystyle operatorname {I} _{X}({ ext{win}})=-log _{2}{p_{X}{({ ext{win}})}}=-log _{2}!{ frac {1}{13,983,816}}approx 23.73725} I X ⁡ ( lose ) = − log 2 ⁡ p X ( lose ) = − log 2 13 , 983 , 815 13 , 983 , 816 ≈ 1.0317 × 10 − 7 shannons . {displaystyle {egin{aligned}operatorname {I} _{X}({ ext{lose}})&=-log _{2}{p_{X}{({ ext{lose}})}}=-log _{2}!{ frac {13,983,815}{13,983,816}}\&approx 1.0317 imes 10^{-7}{ ext{ shannons}}.end{aligned}}} The information entropy of a lottery probability distribution is also easy to calculate as the expected value of the information content. H ( X ) = ∑ x − p X ( x ) log ⁡ p X ( x ) = ∑ x p X ( x ) I X ⁡ ( x ) = d e f E [ I X ⁡ ( x ) ] {displaystyle {egin{alignedat}{2}mathrm {H} (X)&=sum _{x}{-p_{X}{left(x ight)}log {p_{X}{left(x ight)}}} &=sum _{x}{p_{X}{left(x ight)}operatorname {I} _{X}(x)}\&{overset {underset {mathrm {def} }{}}{=}} mathbb {E} {left[operatorname {I} _{X}(x) ight]}end{alignedat}}} Oftentimes the random variable of interest in the lottery is a Bernoulli trial. In this case, the Bernoulli entropy function may be used. Using X {displaystyle X} representing winning the 6-of-49 lottery, the Shannon entropy of 6-of-49 above is H ( X ) = − p log ⁡ ( p ) − q log ⁡ ( q ) = − 1 13 , 983 , 816 log 1 13 , 983 , 816 − 13 , 983 , 815 13 , 983 , 816 log 13 , 983 , 815 13 , 983 , 816 ≈ 1.80065 × 10 − 6 shannons. {displaystyle {egin{aligned}mathrm {H} (X)&=-plog(p)-qlog(q)=-{ frac {1}{13,983,816}}log !{ frac {1}{13,983,816}}-{ frac {13,983,815}{13,983,816}}log !{ frac {13,983,815}{13,983,816}}\&approx 1.80065 imes 10^{-6}{ ext{ shannons.}}end{aligned}}}

en.wikipedia.org - Lottery mathematics - Wikipedia

What is this number 1000000000000000000000000?

Some Very Big, and Very Small Numbers Name The Number Symbol septillion 1,000,000,000,000,000,000,000,000 Y sextillion...

Last updated Sep 1, 2018

Which sport has hardest exercise?

A great deal of strength is also required to progress in boxing. Plus, there's a big chance you'll get punched in the face. To last a whole boxing...

Last updated Oct 1, 2018

Promotion

What is the best strategy for horse racing?

Top 5 Horse Race Betting Strategy That Is Proven Effective The Dutch Betting Strategy. The 80/20 System. The Place Laying System. The Yankee...

Last updated Dec 3, 2021

Promotion

Which country is 1xBet?

1X Corp N.V. (also known as ""1xBet"") is an online gambling company licensed by Curaçao eGaming License. It was founded in 2007 and registered in...

Last updated Mar 16, 2021