Written by Ion Saliu on September 12, 2002 and later.
The page " Primer: Probability, Odds, Formulae, Algorithm, Software Calculator" presents the mathematics and logic of calculating the odds for various games of chance. The odds for various lotto games are presented here in more detail.
The formula of hypergeometric probability distribution is the most comprehensive calculator of odds and probabilities in lottery and lotto, including Powerball, Mega Millions, Euromillions, Keno.
The lottery games are much more diverse and complex (compared to coin tossing, or dice rolling, or even roulette spinning). The diversity widens when the Powerball-type of lotto games are taken into account. The lottery commissions set the odds as exactly k of m. Exactly is the operative keyword here. I thought it was at least. The lotto odds calculated as at least k of m lead to very different values.
The most certain element in calculating the lotto odds is total possible cases. It is known as total number of combinations C(n, k). The combination formula is widely accepted, expert or not.
This is the step that calculates the odds or probability of winning the lotto jackpot: ' 6 of 6', regardless of drawing order. The main attribute of the numeric sets known as combinations is: The numbers in the set are always sorted in ascending order; e.g. 1,2,3,4,5,6.
Calculating the odds for other lottery prizes requires a different formula. But we still need the above formula of probability to hit the jackpot. Let's calculate the odds of hitting '5 of 6' in a 6/49 lotto game. We already know the odds of getting '6 of 6'. The 6-number winning combination can be broken in C(6,5) = 6 groups of 5 lotto numbers each. Each 5-number group can be combined with 1 number from the rest of the field: 49 6 = 43. C(43,1) = 43. There is a total of 6 x 43 = 258 combinations. Therefore one winning 6-number combination offers 258 chances of '5 of 6'. Furthermore, the probability of hitting '5 of 6' is calculated by dividing total possible lotto combinations C(49,6) by total chances of '5 of 6' (258); 13,983,816 / 258 = 54,200.8 to 1 or '1 in 54,200.8'.
Let's calculate the lottery odds of hitting '4 of 6' in a 6/49 game. We already know the odds of getting '6 of 6'. The 6-number winning combination can be broken in C(6,4) = 15 groups of 4 numbers each. Each 4-number group can be combined with 2 numbers from the rest of the field: 49 6 = 43. C(43,2) = 903. There is a total of 15 x 903 = 13,545 lotto combinations. Therefore one winning 6-number combination offers 13,545 chances of '4 of 6'. Therefore, the probability of hitting '4 of 6' is calculated by dividing total possible combinations C(49,6) by total chances of '4 of 6' (13,545); 13,983,816 / 13,545 = 1,032.4 to 1 or '1 in 1,032.4'.
We can write now a general formula to calculate the odds or probability of various lotto prizes or situations. Hitting '2 of 6' is not a prize; it is just a probability situation.
Please be advised that this lottery odds formula has restrictions. There are impossible situations, such as exactly '1 of 6' in a 6/10 lottery game. The above formula leads to mathematical absurdities in restricted cases. The computer programmer must include code to avoid the mathematically absurd situations. My free software program OddsCalc.EXE does that and a lot more. OddsCalc.EXE calculates the odds of any lotto game, including Powerball, Mega Millions, Euromillions, Keno plus casino gambling and horseracing. If the lottery game draws 6 winning numbers, the program calculates the lotto odds from '0 of 6' to '6 of 6'. The lotto probabilities or winning odds are calculated two ways: exactly and at least.
You should know about this subtlety: The lottery commission pays you as at least, not as exactly! No, this not a jeu de mots. Otherwise, you would have to play for something like exactly 4 of 6. If your ticket would hit 6 of 6, you would not be paid the jackpot because you played for exactly 4 of 6! Luckily, your lottery ticket is valid for winning any prize! BRRRRRRRRAHAHAHAHA.....
We can use a more complex formula to calculate the lotto odds. The formula we used before assumes that if the game draws 6 winning numbers, the player must select 6 numbers and play 6 numbers. But the player has the liberty to select, for example, 10 numbers from the field of 49; then the player must play 6-number combinations from the 10 picks. We can use a more generalized formula to calculate the odds. What is the probability to get '4 of 6 in 10 from 49'? The first formula dealt with 3 elements (numbers): m, k, n; e.g. '6 of 6 from 49'. The hypergeometric distribution probability formula adds a 4th parameter; e.g. 'number of picks, s'. The equation becomes m of k in s from n; e.g. '4 of 6 in 18 picks from a field of 49'. The hypergeometric distribution probability reads:
The 'hypergeometric distribution probability' formula has also restrictions. As before, some cases are impossible; e.g '1 of 6 in 10 from 10'.
My freeware OddsCalc.EXE calculates the lotto odds based on the formula of hypergeometric distribution probability.
There are many more probability situations that can be calculated using formulas. Many lotto players want to calculate before generating the number of combinations of exactly '3 odd numbers and 3 even numbers'. There are 24 even numbers and 25 odd numbers in a set of 49 elements. The 24 even numbers are grouped 3 at a time. C(24, 3) = 2024 groups. The 25 odd numbers are also grouped 3 at a time. C(25, 3) = 2300 groups. The 3-number groups must be combined so that we get 6-number combinations.
Therefore, total number of '3 odd, 3 even' combinations is: 2024 x 2300 = 4,655,200.
How about '2 odd, 4 even'? C(25,2) * C(24,4) = 3,187,800.
'1 odd, 5 even': C(25,1) * C(24,5) = 1,062,600.
'0 odd, 6 even': C(25,0) * C(24,6) = 134,596.
'6 odd, 0 even': C(25,6) * C(24,0) = 177,100.
'5 odd, 1 even': C(25,5) * C(24,1) = 1,275,120.
'4 odd, 2 even': C(25,4) * C(24,2) = 3,491,400.
No other combination of odd/even is possible. If we add-up the combinations of all 7 odd/even groups:
Sum-total: 13,983,816 (of course, equal to C(49,6) = total possible lotto 649 combinations).
From another perspective: There are 24 low numbers (1 to 24) and 25 high numbers (25 to 49) in a set of 49 elements. The 24 low numbers are grouped 3 at a time. C(24, 3) = 2024 groups. The 25 high numbers are also grouped 3 at a time. C(25, 3) = 2300 groups. The 3-number groups must be combined so that we get 6-number combinations. Therefore, total number of '3 low, 3 high' combinations is: 2024 x 2300 = 4,655,200.
There are 7 groups of low/high, very similar to the odd/even situation.
The calculations are more difficult (or just trickier!) if the two previous conditions act simultaneously. What is the number of combinations of exactly 3-odd, 3-high OR 3-low, 3-even? We need to take inventory of the numbers that satisfy the condition. Let's exemplify for the world-wide popular 6/49 lotto game. There are 24 low numbers (1 to 24) and 25 high numbers (25 to 49). There are 12 ODD numbers in the LOW-number group; there are 12 EVEN numbers in the LOW-number group. There are 13 ODD numbers in the HIGH-number group; there are 12 EVEN numbers in the HIGH-number group. Now, it has become really easy to calculate all possible 6-number lottery combinations for each of the four groups.
~ all-low & all-odd: C(12, 6) = 924 lotto 6/49 combinations;
~ all-low & all-even: C(12, 6) = 924 lotto 6/49 combinations;
~ all-HIGH & all-odd: C(13, 6) = 1716 lotto 6/49 combinations;
~ all-HIGH & all-even: C(12, 6) = 924 lotto 6/49 combinations.
You can see more examples for the pick-3 lottery and lotto 5/39 game. The page is dedicated to apparent lottery strategies based on sums, or odd/even, low/high lotto number grouping.
It's very important to remember that the calculations above refer to the condition EXACTLY m of k; e.g. 'exactly 5 of 6'. The lottery commissions only pay one prize of '5 of 6'. Some people expect also prizes for the '4 of 6' groups that a '5 of 6' winning combination is composed of. That's what EXACTLY stands for. If the lottery commissions were to pay for all the prizes that a winning combination is composed of, the odds would have to be calculated as AT LEAST m of k. The odds calculated by the two methods differ to a great extent. The odds of EXACTLY 1 of 6 in a lotto 6/49 game are '1 in 2.4'. The odds of AT LEAST 1 of 6 in a lotto 6/49 game are 6/49 = '1 in 8.16'.
If you thought this was difficult stuff you were right! You ain't alone in this regard. But we all can learn fairly easily, I might add. All of the above dealt with one parameter only: The probability or winning odds. Many people, especially British, have a problem with equating the odds to probability. I would say they are too puritan or picky. We semantically equate here probability to favorable odds or odds of winning. The odds of winning are expressed as '1 in SomeNumber'; the individual probability is also expressed as '1 in SomeThing', which is absolutely valid in mathematics.
In many of the world cultures, they only deal with probability or its mathematical synonym chance. They usually say the chance of winning or the probability of winning, or the probability of losing, etc. Odds came from the British passion for horseracing. The odds can be represented as hurdles to overcome.
Hold it there is more! There is one very important parameter that I introduced back in 1997: Degree of certainty. Most people confuse the probability for the degree of certainty. The layperson can look at the degree of certainty as being the probability of probability. The event of probability p has a degree of certainty DC to appear in a number of trials N. Alas! The real-life is much more complex than to be described by just the individual probability!
The individual probability would indicate that the odds of getting 'heads' (or 'tails') in 2 coin tosses is 100%. NOT! NON! The degree of certainty of getting 'heads' (favorable odds, p = '1 in 2') in 2 coin tosses is only 75%! The Ion Saliu paradox calculates with random precision that the degree of certainty of hitting one 00-roulette number in 38 spins is just about 65%. If an event has a probability p = 1/N, the degree of certainty DC tends to (1 1/e) or 0.632 to appear in N trials.
The cynics argue with me this way. Tossing heads always has a probability p = ½. Thus, getting heads after 10 trials is .5 or 50%; getting heads after 100 coin tosses is 1/2 or 50-50; getting heads after 1000 coin tosses is .5 or 50%. . . and so on ad infinitum. . . My answer is always silencing, because it is truthful to its most simplicity. While heads always has a probability p = ½, simultaneously tails always has a probability p = ½. Both elements have a right to appear and they do appear randomly equally. Has anyone witnessed 100 heads in a row? Or 100 consecutive tails? Forget about 1000 consecutives!
Read more about this fascinating stuff that is at the very core of the Universe:
The Mathematical Foundation of Fundamental Formula of Gambling (FFG).
Only Randomness is Almighty. Randomness is great!
I'm telling you....God fears mathematics, while Einstein hates gambling! And so be it for ever and never....
Doctor in Occult Science of Probabilistic Odds
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