Expanded bet

We've already learned how to calculate the probability of winning the lottery , and for convenience, I've made an online probability calculator . Today, we'll learn how to calculate the probability when using expanded bets.

Расчёт вероятности при использовании развернутых ставок производится по следующей формуле:

\(P(A) = \frac{ C_{M}^{m}C_{N - M}^{n - m} }{C_{N}^{n}}\)

To make it easier to understand this formula, imagine that we are playing the «M out of N» lottery, marking n numbers on the ticket and hoping to correctly guess m numbers.

Another important note! If bonus balls are drawn from the main drum during the lottery draw, the probability is calculated using a different formula.

Let's calculate the expanded bet

Let's consider the «6 out of 45» lottery. We mark 10 numbers on the ticket and aim to guess 5.

\(P(A) = \frac{C_{6}^{5}C_{45-6}^{10-5}}{C_{45}^{10}} = \frac{C_{6}^{5}C_{39}^{5}}{C_{45}^{10}}\)

We've already learned how to find the number of combinations :

\(C_{6}^{5} = \frac{6!}{5!(6-5)!} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} = \frac{720}{120} = 6\)

\(C_{39}^{5} = \frac{39!}{5!(39-5)!} = \frac{35 \cdot 36 \cdot 37 \cdot 38 \cdot 39}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} = \frac{69,090,840}{120} = 575,757\)

\(C_{45}^{10} = \frac{45!}{10!(45-10)!} = \frac{36 \cdot 37 \cdot 38 \cdot 39 \cdot 40 \cdot 41 \cdot 42 \cdot 43 \cdot 44 \cdot 45}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10} = \frac{11,576,551,623,436,800}{3,628,800} = 3,190,187,286\)

\(P(A) = \frac{C_{6}^{5}C_{45-6}^{10-5}}{C_{45}^{10}} = \frac{C_{6}^{5}C_{39}^{5}}{C_{45}^{10}} = \frac{6 \cdot 575,757}{3,190,187,286} \approx \frac{1}{923} \approx 0,108 \%\)

The probability of guessing 5 numbers in the «6 out of 45» lottery, marking 10 numbers on the ticket, is approximately 1 in 923.

The probability of winning secondary prizes

This formula also allows calculating the probability of winning secondary prizes with a standard bet. Let's calculate the probabilities of guessing one, two, and three numbers in one field in «4 out of 20».

Marking 4 numbers in one field, the chance to correctly guess one number is approximately 1 in 2:

\(P(A) = \frac{C_{4}^{1}C_{20-4}^{4-1}}{C_{20}^{4}} = \frac{C_{1}^{4}C_{16}^{3}}{C_{20}^{4}} = \frac{4 \cdot 560}{4,845} \approx \frac{1}{2} \approx 46,233 \% \)

To correctly guess two numbers is almost 1 in 7:

\(P(A) = \frac{C_{4}^{2}C_{20-4}^{4-2}}{C_{20}^{4}} = \frac{C_{4}^{2}C_{16}^{2}}{C_{20}^{4}} = \frac{6 \cdot 120}{4,845} \approx \frac{1}{7} \approx 14,861 \% \)

Three numbers is 1 in 76:

\(P(A) = \frac{C_{4}^{3}C_{20-4}^{4-3}}{C_{20}^{4}} = \frac{C_{4}^{3}C_{16}^{1}}{C_{20}^{4}} = \frac{4 \cdot 16}{4,845} \approx \frac{1}{76} \approx 1,321 \% \)

And to correctly guess 4 numbers is 1 in 4,845:

\(P(A) = \frac{C_{4}^{4}C_{20-4}^{4-4}}{C_{20}^{4}} = \frac{C_{4}^{4}C_{16}^{0}}{C_{20}^{4}} = \frac{1 \cdot 1}{4,845} \approx \frac{1}{4,845} \approx 0,021 \% \)

Expanded bets for the "4 out of 20" lottery

Keep in mind that the probability is calculated for only one field out of two. If you want to get the probability for two fields, you need to multiply the probability of one field by the probability of the other. This way, you will find the overall probability.

Marking 5 numbers in one field, trying to guess 4 numbers, the probability is 1 in 969:

\(P(A) = \frac{C_{4}^{4}C_{20-4}^{5-4}}{C_{20}^{5}} = \frac{C_{4}^{4}C_{16}^{1}}{C_{20}^{5}} = \frac{1 \cdot 16}{15,504} \approx \frac{1}{969} \approx 0,103 \% \)

Marking 6 numbers in one field, trying to guess 4 numbers, the probability is 1 in 323:

\(P(A) = \frac{C_{4}^{4}C_{20-4}^{6-4}}{C_{20}^{6}} = \frac{C_{4}^{4}C_{16}^{2}}{C_{20}^{6}} = \frac{1 \cdot 120}{38,760} \approx \frac{1}{323} \approx 0,310 \% \)

But perhaps it's easier to show all this in a table:

Expanded bets for the «6 out of 45» lottery

Expanded bets for the «7 out of 49» lottery