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| '''Lottery mathematics''' is used here to mean the calculation of the [[probabilities]] in a [[lottery]] game. The lottery game used in the examples below is one in which one selects 6 numbers from 49, and hopes that as many of those 6 as possible match the 6 that are randomly selected from the same pool of 49 numbers in the "draw".
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| ==Calculation explained in choosing 6 from 49==
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| In a typical 6/49 game, six numbers are drawn from a range of 49 and if the six numbers on a ticket match the numbers drawn, the ticket holder is a [[Progressive jackpot|jackpot]] winner—this is true no matter in which order the numbers appear. The probability of this happening is 1 in 13,983,816.
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| This small [[Probability|chance]] of winning can be demonstrated as follows:
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| Starting with a bag of 49 differently-numbered lottery balls, there are 49 different but equally likely ways of choosing the number of the first ball selected from the bag, and so there is a 1 in 49 chance of [[prediction|predicting]] the number correctly. When the draw comes to the second number, there are now only 48 balls left in the bag (because the balls already drawn are not returned to the bag) so there is now a 1 in 48 chance of predicting this number.
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| 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 <math>{49!\over (49-6)!}</math>. This works out to a very large number, 10,068,347,520, which is much bigger than the 14 million stated above.
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| The last step is to understand that 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[[factorial|!]] 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)!), or more generally as
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| :<math>{n\choose k}={n!\over k!(n-k)!}</math>. | |
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| This function is called the [[combination]] function; in [[Microsoft Excel]], this function is implemented as COMBIN(''n'', ''k''). For example, COMBIN(49, 6) (the calculation shown above), would return 13,983,816. For the rest of this article, we will use the notation <math>{n\choose k}</math>. "Combination" means the group of numbers selected, irrespective of the order in which they are drawn.
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| An alternative method of calculating the odds is to never make the erroneous assumption that balls must be selected in a certain order. 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
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| :<math>{n\choose k}={49\choose 6}={49\over 6} * {48\over 5} * {47\over 4} * {46\over 3} * {45\over 2} * {44\over 1} </math>
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| :
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| 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.
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| ==Odds of getting other possibilities in choosing 6 from 49==
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| One must divide the number of combinations producing the given result by the total number of possible combinations (for example, <math>{49\choose 6} = 13,983,816</math>, as explained in the section above). The numerator equates to the number of ways one can select the winning numbers multiplied by the number of ways one can select the losing numbers.
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| For a score of ''n'' (for example, if 3 of your numbers match the 6 balls drawn, then ''n'' = 3), there are <math>{6\choose n}</math> ways 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 <math>{43\choose 6-n}</math> ways. The total number of combinations giving that result is, as stated above, the first number multiplied by the second. The expression is therefore <math>{6\choose n}{43\choose 6-n}\over {49\choose 6}</math>.
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| This can be written in a general form for all lotteries as:
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| <math>{K\choose B}{N-K\choose K-B}\over {N\choose K}</math>, where <math>N</math> is the number of balls in lottery, <math>K</math> is the number of balls in a single ticket, and <math>B</math> is the number of matching balls for a winning ticket.
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| The generalisation of this formula is called the [[hypergeometric distribution]] (the HYPGEOMDIST() function in most popular spreadsheets).
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| This gives the following results:
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| {| class="wikitable" cellspacing=3
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| !Score
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| !Calculation
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| !Exact Probability
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| !Approximate Decimal Probability
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| !Approximate 1/Probability
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| |-style="background: #e3e3e3;"
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| | 0
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| | <math>{6\choose 0}{43\choose 6}\over {49\choose 6}</math>
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| | 435,461/998,844
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| | 0.436
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| | 2.2938
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| |-
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| | 1
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| | <math>{6\choose 1}{43\choose 5}\over {49\choose 6}</math>
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| | 68,757/166,474
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| | 0.413
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| | 2.4212
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| |-style="background: #e3e3e3;"
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| | 2
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| | <math>{6\choose 2}{43\choose 4}\over {49\choose 6}</math>
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| | 44,075/332,948
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| | 0.132
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| | 7.5541
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| |-
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| | 3
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| | <math>{6\choose 3}{43\choose 3}\over {49\choose 6}</math>
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| | 8,815/499,422
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| | 0.0177
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| | 56.66
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| |-style="background: #e3e3e3;"
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| | 4
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| | <math>{6\choose 4}{43\choose 2}\over {49\choose 6}</math>
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| | 645/665,896
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| | 0.000969
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| | 1,032.4
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| |-
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| | 5
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| | <math>{6\choose 5}{43\choose 1}\over {49\choose 6}</math>
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| | 43/2,330,636
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| | 0.0000184
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| | 54,200.8
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| |-style="background: #e3e3e3;"
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| | 6
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| | <math>{6\choose 6}{43\choose 0}\over {49\choose 6}</math>
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| | 1/13,983,816
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| | 0.0000000715
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| | 13,983,816
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| |-
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| |}
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| ==Pick8-32 Odds and Calculations==
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| The Pick8-32 Lottery game by Trillion Coins<ref>http://TrillionCoins.com</ref> implements a lotto game ticket where 8 numbers from 01 to 32 are selected in any order and can be repeated. The odds of a play ticket matching all 8 numbers is simple to calculate and is illustrated by the following math.
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| :<math>{32\over 8} * {32\over 7} * {32\over 6} * {32\over 5} * {32\over 4} * {32\over 3} * {32\over 2}* {32\over 1}={109951162776\over 40320}</math>
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| :
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| Which means there is a 1 in 27,269,633 chance of matching all 8 numbers. The Trillion Coins lottery is a pay out on every drawing lotto game so players have much better overall odds of winning and the odds are dependent on the total number of tickets that are played.[need math with # players]. With this game you can win simply by matching more numbers than anyone else.
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| ==Powerballs And Bonus Balls==
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| Many lotteries have a [[powerball]] (or "bonus ball"). If the powerball is drawn from a pool of numbers ''different'' from the main lottery, then simply multiply the odds by the number of powerballs. For example, in the 6 from 49 lottery, if there were 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). Another example of such a game is [[Mega Millions]], albeit with different jackpot odds.
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| 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 should be calculated using the method shown in the "[[Lottery Math#Odds of getting other scores in choosing 6 from 49|other scores]]" section above (in other words, treat the powerballs like a mini-lottery in their own right), and then multiplied by the odds of achieving the required main-lottery score.
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| If the powerball is drawn from the ''same'' pool of numbers as the main lottery, then, for a given target score, one must calculate the number of winning combinations, including the powerball. For games based on the [[Lotto 6/49|Canadian lottery]] (such as the [[National Lottery (United Kingdom)|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"), and there is an extra prize for matching 5 balls and the bonus ball. As described in the "[[Lottery Math#Odds of getting other scores in choosing 6 from 49|other scores]]" section above, the number of ways one can obtain a score of 5 from a single ticket is <math>{6\choose 5}{43\choose 1}</math> or 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). So, there are 258/43 = 6 ways of achieving it. Therefore, the odds of getting a score of 5 and the powerball are <math>{6}\over {49\choose 6}</math> = 1 in 2,330,636.
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| 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 <math>{258 \cdot {{42}\over {43}}}\over {49\choose 6}</math> = 3/166,474 (approximately 55,491.33) for obtaining a score of 5 without matching the powerball.
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| Using the same principle, to calculate the odds of getting a score of 2 and the powerball, calculate the number of ways to get a score of 2 as <math>{6\choose 2}{43\choose 4}</math> = 1,851,150 then multiply this 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, the probability of obtaining the score of 2 and the bonus ball is <math>{172,200}\over {49\choose 6}</math> = 1025/83237. This gives approximate decimal odds of 81.2.
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| The general formula for <math>B</math> matching balls in a <math>N</math> choose <math>K</math> lottery with one bonus ball from the <math>N</math> pool of balls is:
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| <math>{K-B\over N-K}{K\choose B}{N-K\choose K-B}\over {N\choose K}</math>
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| The general formula for <math>B</math> matching balls in a <math>N</math> choose <math>K</math> lottery with zero bonus ball from the <math>N</math> pool of balls is:
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| <math>{N-K-K+B\over N-K}{K\choose B}{N-K\choose K-B}\over {N\choose K}</math>
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| The general formula for <math>B</math> matching balls in a <math>N</math> choose <math>K</math> lottery with one bonus ball from a separate pool of <math>P</math> balls is:
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| <math>{1\over P}{K\choose B}{N-K\choose K-B}\over {N\choose K}</math>
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| The general formula for <math>B</math> matching balls in a <math>N</math> choose <math>K</math> lottery with no bonus ball from a separate pool of <math>P</math> balls is:
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| <math>{P-1\over P}{K\choose B}{N-K\choose K-B}\over {N\choose K}</math>
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| ==Minimum number of tickets for a match==
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| It is a hard, in most cases open, mathematical 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 that can guarantee a ticket with at least 2 matches is 100.<ref>{{Cite journal|author=[[Zoltán Füredi|Z. Füredi]], [[Gabor Szekely|G. J. Székely]], and Z. Zubor|title=On the lottery problem|journal=Journal of Combinatorial Designs|volume=4|issue=1|year=1996|pages=5–10|publisher=Wiley}} [http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1520-6610(1996)4:1%3C5::AID-JCD2%3E3.0.CO;2-J/abstract]</ref>
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| ==References==
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| {{Reflist}}
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| ==External links==
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| *[http://mathdl.maa.org/convergence/1/?pa=content&sa=viewDocument&nodeId=217&bodyId=146 Euler's Analysis of the Genoese Lottery] at [http://mathdl.maa.org/convergence/1/ Convergence]
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| *[http://probability.infarom.ro/lottery.html Lottery Mathematics]
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| [[Category:Combinatorics]]
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| [[Category:Lotteries]]
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| [[Category:Games (probability)]]
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