For a simple example, imagine you have to choose two numbers and you can pick numbers from 1 to 5. Your odds of choosing the two “correct” numbers (the winning numbers) would be defined as 5!2!×3!{\displaystyle {\frac {5!}{2!\times 3!}}}. This would then be solved as 5×4×3×2×12×1×3×2×1{\displaystyle {\frac {5\times 4\times 3\times 2\times 1}{2\times 1\times 3\times 2\times 1}}}, which is 120÷12{\displaystyle 120\div 12}, or 10. So, your odds of winning this game are 1 in 10. Factorial calculations can get unwieldy, especially with large numbers. Most calculators have a factorial function to ease your calculations. Alternately, you can type the factorial into Google (as “55!” for example) and it will solve it for you.
Other games may have you choose 5 or 6 numbers, or more, from a larger or smaller pool of numbers. To calculate the odds of winning, you simply need to know the number of winning numbers and the total number of possible numbers. [3] X Research source
To calculate your odds of choosing the final Powerball correctly, you would complete the same equation using the values for the Powerball (1 number out of 26 possible numbers). Since you’re only picking 1 number here, you don’t necessarily have to complete the entire equation. The answer will be 26 because there are 26 different ways 1 number can be chosen from a set of 26 unique numbers.
So, your odds of choosing the first five numbers and the Powerball correctly and winning the jackpot are 1 in 292,201,338.
To win the second prize, you would have to guess the Powerball incorrectly. If you calculated your odds of winning the jackpot, you know that your odds of guessing the Powerball correctly are 1 in 26. Therefore, your odds of guessing the Powerball incorrectly are 25 in 26. Use the same equation with these values to determine your odds of winning the second prize: 111,238,513×2526{\displaystyle {\frac {1}{11,238,513}}\times {\frac {25}{26}}}. When you complete this calculation, you’ll see that your odds of winning the second prize are 1 in 11,688,053. 52.
For example, you might use the Powerball values to determine your odds of correctly guessing 3 of the 5 chosen numbers from the set of 69 unique numbers. Your equation would look like this: 5!3!×(5−3)!×(69−5)!((69−5)−(5−3))!×(5−3)!{\displaystyle {\frac {5!}{3!\times (5-3)!}}\times {\frac {(69-5)!}{((69-5)-(5-3))!\times (5-3)!}}} The result of this equation tells you the number of ways that 3 numbers can be chosen correctly out of 5 numbers. Your odds will be that number out of the total number of ways 5 numbers can be chosen correctly.
In the previous example, your odds of guessing 3 of the 5 chosen numbers in Powerball would be 20,160 in 11,238,513.
For example, if you wanted to calculate your odds for getting only 3 of the 5 numbers correct and getting the Powerball incorrect, your equation would be 20,16011,238,513×2526{\displaystyle {\frac {20,160}{11,238,513}}\times {\frac {25}{26}}}, or 1 in 579. 76. On the other hand, your odds for getting 3 of the 5 numbers correct and getting the Powerball correct would be 20,16011,238,513×126{\displaystyle {\frac {20,160}{11,238,513}}\times {\frac {1}{26}}}, or 1 in 14,494. 11.
If you’re calculating odds for Powerball or a similar game, don’t forget to multiply your result by the Powerball value.
To return to the Powerball example, the expected return of a single $2 ticket would be around $1. 79 at the high end and as little as $1. 35 at the low end. Keep in mind that “expected return” is a term of art used in statistics. Your actual payout will almost always be much less than the expected return you calculate.
Calculating the odds can help you determine which lottery games have the best expected benefit. For example, at one time, the New York Lottery had a $1 Take Five ticket with an expected value that equaled its cost. If you played this game, you could expect to break even over time. [13] X Research source
For example, if your overall chances of winning are 1 in 250,000,000, your chances of losing on one play are 249,999,999÷250,000,000{\displaystyle 249,999,999\div 250,000,000}, which is equal to a number very close to 1 (0. 99999. . . ). If you play twice, that number is squared ((249,999,999÷250,000,000)2{\displaystyle (249,999,999\div 250,000,000)^{2}}), representing a movement slightly away from 1 (and therefore a better chance of winning).
For example, if you had a 1 in 250,000,000 chance of winning on one play, it would take roughly 180 million plays to reach 50-50 odds of winning. At this rate, if you bought ten tickets a day for 49,300 years, you would have a 50 percent chance of winning. Additionally, if you finally reached 50-50 odds, you still wouldn’t be guaranteed a win if you bought two tickets on that day. Your overall odds of winning would still remain roughly 50% for each of those tickets.
