Lottery Mathematics – Odds of getting other scores in choosing 6 from 49
One must calculate the total number of lottery combinations (c(49,6) = 13,983,816, as explained in the section above), and divide it by the number of those combinations which give the desired result – which equates to the number of ways one can select the winning numbers multiplied by the number of ways one can select the losing numbers.
For a score of n (e.g. if 3 of your numbers match the 6 balls drawn, then n=3), there are c(6,n) ways of selecting the n winning numbers from the 6 drawn balls. For one’s losing numbers, there are c(43,6 – n) ways to select them from the 43 losing lottery numbers. The total number of combinations giving that result is, as stated above, the first number multiplied by the second. The expression is therefore c(49,6)/(c(6,n)*c(43,6-n)). This gives the following results (remember that odds are the reciprocal of probability):
| Score | Calculation | Exact Probability | Approximate Decimal Odds |
|---|---|---|---|
| 0 | c(49,6)/(c(6,0)*c(43,6)) | 435461/998844 | 1 in 2.29 |
| 1 | c(49,6)/(c(6,1)*c(43,5)) | 68757/166474 | 1 in 2.42 |
| 2 | c(49,6)/(c(6,2)*c(43,4)) | 44075/332948 | 1 in 7.55 |
| 3 | c(49,6)/(c(6,3)*c(43,3)) | 8815/499422 | 1 in 56.66 |
| 4 | c(49,6)/(c(6,4)*c(43,2)) | 645/665896 | 1 in 1032.4 |
| 5 | c(49,6)/(c(6,5)*c(43,1)) | 43/2330636 | 1 in 54201 |
| 6 | c(49,6)/(c(6,6)*c(43,0)) | 1/13983816 | 1 in 13983816 |
References
- This article incorporates text from the Encyclopædia Britannica Eleventh Edition, a publication now in the public domain.
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