Lottery winning odds

The probability of winning the lottery depends on the number of possible ball combinations and we will now learn how to calculate them ourselves, and for those who don't want to calculate manually, there is an online calculator above.

Probability

— the degree (relative measure, quantitative assessment) of the possibility of a certain event occurring.

Let's start simple, we have five balls:

What is the probability of guessing one ball out of five? It equals$\frac{1}{5}$ , there are only five possible combinations for this set of numbers: either 5, or 3, or 2, or 4, or 1.

For convenience, let's denote our lotteries as "$k\text{of}n$", and when needed, we'll substitute the corresponding numbers.

Let's make our lottery rules harder — to win you need to guess "2 of 5" ($k=2,n=5$). Now the chance of guessing is $\frac{1}{10}$, since there are ten possible combinations, here they are:

It is important to note that the order in which numbers appear in each combination does not matter for winning the lottery.

In probability theory, the five balls above are actually a set of numbers from 1 to 5. A set is denoted by curly braces { }, and each individual combination is called a combination.

In combinatoricsa combination of k from n elements is a selection containing k elements chosen from a set containing n distinct elements.

In combinations, the order of elements does not matter, $\{1,2\}$ and $\{2,1\}$ are considered the same.

Now we can write all this mathematically:

We have a set of 5 balls $\{1,2,3,4,5\}$. And there are 10 combinations that can be formed from 5 by 2 balls:

The total number of combinations of n elements taken k at a time is denoted$C_n^k$(from the first letter of the French word "combinaison", meaning "combination") and is read as "the number of combinations of n elements taken k at a time". In our case$C_5^2$ — the number of combinations of 5 taken 2 at a time equals 10.

Number of combinations

The number of combinations is calculated using the formula:

$C_n^k=\frac{n!}{k!(n-k)!}$

$n!$ and $k!$ — are factorials of the corresponding numbers $n$ and $k$. The factorial of a natural number $n$ is the product of all natural numbers from 1 to $n$ inclusive. For example, the factorial of 5 equals $5!=1·2·3·4·5=120$.

Let's verify our result for the 2 of 5 lottery:

$C_5^2=\frac{5!}{2!(5-2)!}=\frac{5!}{2!·3!}=\frac{1·2·3·4·5}{1·2·1·2·3}$

Look, we can reduce the dividend and divisor by $(n-k)!$, I've highlighted with brackets to make it clearer:

$\frac{(1·2·3)·4·5}{1·2·(1·2·3)}=\frac{1·2·3}{1·2·3}·\frac{4·5}{1·2}=\frac{20}{2}=10$

Note that after we reduced the dividend and divisor, we have two numbers left in both the dividend and divisor, or more precisely $k$ numbers. In the dividend, these are the product of the two largest numbers from $n$, and in the divisor, the factorial of $k$. And if you want to calculate the probability of winning, you don't need to calculate full factorials — it's enough to multiply the $k$ largest elements from $n$ and divide by the factorial of $k$.

Probability of winning

Let's calculate the number of combinations for the "Sportloto 6 of 45" lottery:

$C_{45}^6=\frac{45!}{6!(45-6)!}=\frac{45·44·43·42·41·40}{6·5·4·3·2·1}=\frac{5,864,443,200}{720}=8,145,060$

The entire set of combinations is a full system. If you buy tickets with all combinations, you are guaranteed to win.

Now let's move on to the probability of winning. If you buy a "Sportloto 6 of 45" lottery ticket with one combination, your probability is 1 in 8,145,060. You bought 2 tickets with different combinations — your chances are 2 in 8,145,060 or 1 in 4,072,530. You bought 10 tickets but wrote the same combination on all of them — your chances are again 1 in 8,145,060. Thus, probability is the ratio of your unique combinations to the total number of combinations.

If you play a lottery where you need to correctly guess numbers in two playing fields, for example, the American lottery Powerball "5 of 69 + 1 of 26", then you need to multiply the number of combinations of "5 of 69" by "1 of 26".

$C_{69}^5=\frac{69!}{5!(69-5)!}=\frac{69·68·67·66·65}{5·4·3·2·1}=\frac{1,348,621,560}{120}=11,238,513$

$C_{26}^1=\frac{26!}{1!(26-1)!}=\frac{26}{1}=26$

$11,238,513·26=292,201,338$

"Sportloto 4 of 20"

In the Russian lottery "Sportloto 4 of 20", to win the jackpot you need to guess "4 of 20" in two fields, let's calculate the number of combinations for one field:

$C_{20}^4=\frac{20!}{4!(20-4)!}=\frac{20·19·18·17}{4·3·2·1}=\frac{116,280}{24}=4,845$

We get 4,845 combinations, the probability of guessing "4 of 20" is 1 in 4,845, but since we need to guess twice, we multiply the probabilities to get the number of combinations for two fields:

$4,845·4,845=23,474,025$

As we can see, the probability of winning "Sportloto 4 of 20" is lower than "Sportloto 6 of 45", 1 in 23 million versus 1 in 8 million.

But at least that's realistic, let's look at the rules of the Russian lottery "Russian Loto":

"Russian Loto"

Barrels numbered from 1 to 90 are loaded into a bag. The host draws barrels one by one and announces their numbers. In round 1, tickets where 5 numbers in any of the six horizontal rows matched the barrel numbers drawn from the bag first win. In round 2, tickets where all 15 numbers in any field matched the barrel numbers drawn from the bag first win. If on the fifteenth turn, all fifteen numbers of one of the two playing fields of your ticket (upper or lower) match the barrel numbers drawn from the bag — you win the Jackpot.

It turns out that on the 15th turn we need to "guess" "15 of 90". The word guess is in quotes because we don't choose the numbers in this lottery, unlike others — in "Russian Loto" the numbers are already chosen. Let's estimate the probability of guessing "15 of 90":

$C_{90}^{15}=\frac{90!}{15!(90-15)!}=45,795,673,964,460,820$

Since one ticket has two playing fields (upper and lower), the probability of winning the Jackpot when purchasing one ticket is:

$\frac{45,795,673,964,460,820}{2}=22,897,836,982,230,410$

Wikipedia helped me find this word — quadrillion. The probability of winning the jackpot in Russian Loto is one in twenty-two quadrillion. Remember the wheat and chessboard problem? This number is of the same order, maybe about 200 times smaller. It's an astronomical number, unreal.

When you play a regular lottery, for example "6 of 45", you fill in a ticket and your combination participates in the draw. In Russian Loto, you don't fill in a ticket — you buy a ticket with an already chosen combination of numbers. It would be fair if you could choose one combination out of 45 quadrillion, but you can't, since nobody will ever be able to print that many tickets for a single draw.

But let's continue evaluating probabilities. Next lottery is "Sportloto 5 of 36". The rules tell us the following:

"Sportloto 5 of 36"

Choose five or more numbers in the range from 1 to 36 in field 1 and one or more numbers in the range from 1 to 4 in field 2. By guessing 5 numbers in field 1 and 1 number in field 2, you win the jackpot. By guessing only 5 numbers in field 1, you win the "prize" category.

To win the jackpot, you need to guess "5 of 36 and 1 of 4", let's see:

$C_{36}^5=\frac{36!}{5!(36-5)!}=\frac{36·35·34·33·32}{5·4·3·2·1}=\frac{45,239,040}{120}=376,992$

$C_4^1=\frac{4!}{1!(4-1)!}=\frac{4}{1}=4$

$376,992·4=1,507,968$

One in one and a half million, there's a chance. Let's look at the probability of winning a prize by guessing "5 of 36":

$C_{36}^5=\frac{36!}{5!(36-5)!}=\frac{36·35·34·33·32}{5·4·3·2·1}=\frac{45,239,040}{120}=376,992$

The chances are even higher.

"6 of 36"

Next lottery is "6 of 36", here you can't choose your own combination — you'll have to buy what's offered. Let's see:

$C_{36}^6=\frac{36!}{6!(36-6)!}=\frac{36·35·34·33·32·31}{6·5·4·3·2·1}=\frac{1,402,410,240}{720}=1,947,792$

"Sportloto 7 of 49"

$C_{49}^7=\frac{49!}{7!(49-7)!}=\frac{49·48·47·46·45·44·43}{7·6·5·4·3·2·1}=\frac{432,938,943,360}{5,040}=85,900,584$

"Rapido" lottery

Now let's move on to exotic lotteries. The "Rapido" lottery. The rules say that to win the jackpot:

You need to guess 8 unique numbers from 1 to 20 in the first part of the playing field and one number from 1 to 4 in the second part.

We get "8 of 20 and 1 of 4"

$C_{20}^8=\frac{20!}{8!(20-8)!}=\frac{20·19·18·17·16·15·14·13}{8·7·6·5·4·3·2·1}=\frac{5,079,110,400}{40,320}=125,970$

$C_4^1=\frac{4!}{1!(4-1)!}=\frac{4}{1}=4$

$125,970·4=503,880$

The probability of winning "Rapido" is 1 in 503,880.

"Zodiac" lottery

In the "Zodiac" lottery you need to guess 4 numbers: the first — from 1 to 31 inclusive, the second — from 1 to 12 inclusive, the third — from 0 to 99 inclusive, and the fourth — from 1 to 12 inclusive. We get probabilities of 1 in 31, 1 in 12, 1 in 100 (since 0 to 99 inclusive), and again 1 in 12. We multiply these probabilities:

$31·12·100·12=446,400$

The probability of winning the jackpot in the "Zodiac" lottery is 1 in 446,400.

"Duel" lottery

A draw combination consists of four numbers: two numbers (in the range from 1 to 26) for the first field and two numbers (in the range from 1 to 26) for the second field.

To win the jackpot we need to guess "2 of 26 and 2 of 26":

$C_{26}^2=\frac{26!}{2!(26-2)!}=\frac{26·25}{2·1}=\frac{650}{2}=325$

$C_{26}^2=\frac{26!}{2!(26-2)!}=\frac{26·25}{2·1}=\frac{650}{2}=325$

$325·325=105,625$

The probability of winning the jackpot in the "Duel" lottery is 1 in 105,625.