Cash 3 & Cash 4 are easy.  1:1000 odds out of 1000 numbers or 1:10,000

But fantasy 5 has 1:575,757 odds for 5 of 5.

I thought it was 39*38*37*36*35 which is 69,090,840.

Where does those odds come from?

2 of 5 has 1:9.6 odds.

3 of 5 has 1:103 odds

Where do they come from?

<Moved to Mathematics forum>

Please post in the appropriate forum ... thank you.

(39*38*37*36*35) / (1*2*3*4*5) = 575757

Any other odds I'm not good at.

Any given 5-number combination can be drawn in 120 different orders, hence the 1*2*3*4*5 divisor, and the true odds of 1::575,757. For instance, if 01-02-03-04-05 were to be the result of the drawing, the order of the draw could have been 03, 01, 04, 02 and 05; 02, 01, 03, 05 and 04; or any of the other 118 orders available for this set of numbers.

For more info:

http://www.johnph77.com/math/lo.html

No ads, nothing to sell, for informational purposes only.

gl

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Where does those odds come from?

2 of 5 has 1:9.6 odds.

3 of 5 has 1:103 odds

Where do they come from?

===

They come from the multiple ways of having some of the 5 winning numbers. As mentioned above, the order isnt important, but there's only 1 chance to have all 5 numbers, so there's only 1 winning combination out of the 575,757 possible combinations. There are multiple way to have  some, but not all of the winning numbers, so that gives you more than 1 chance in 575,757 of winning those prizes. How many chances is a two step calculation: (number of ways to have the right numbers) x (number of ways to have the wrong numbers). Calculating the number of combinations for right and wrong numbers is just like calculating the total possible combinations for the game: how many ways to choose r numbers from a set of n numbers. You can find an online combination calculator (which will display the combination formula when you start) here: http://www.wiley.com/college/mat/gilbert139343/java/java05_applet.html

First we'll do 4 of 5. There are 5 ways to choose 4 numbers from the 5 winning numbers. If you have 4 of the winning numbers you'll have 1 wrong number. There are 34 ways to choose 1 number from a set of 34. That gives you 5 x 34, or 170 chances out of the 575,757 possible combinations to have 4 of 5 numbers. Dividing each side by 170 simplifies to 1 in 3386.8.

For 3 of 5 it follows the same pattern.  There are 10 ways to have 3 of 5 numbers. Having 3 right means you'll have 2 wrong numbers, so we calculate how many ways there are to have 2 wrong numbers, which is 561. So, 10 ways to have 3 of the 5 times 561 ways to have 2 losing numbers means you have a total of 5610 winning combinations out of the 575,757.  5610/575,757 simplifies to 1/102.6.

It works the same for 2 of 5 and 1 of 5. I'll leave that for you to work out. Of course it will also work the same way for any other game, such as a 6 of 49. All you need to do is plug the numbers in to a combination calculator.

Most of the Pick 5's, 6's, MM, PB, etc. can be calculated using a simple nCr function on a calculator or excel which means "from n choose r" & does this nCr= n!/(n-r)!*r!

This gives you an equation that calculates if you choose r numbers from n available it has x odds...

I.E. (For Example) MegaMillions 5 out of 56 = C(56,5) = 56! / 5! (56 - 5)! = 3,819,816 & 1 out of 46 = 46, therefore 3,819,816 * 46 = 175711536

There's your outcome... Hope that helps, Pogo

Now for Pick 3's & Pick 4's it's easy just how many numbers for how many numbers... 1 out of ten (0 - 9) = 10, then for Pick 3 it would be P3 = 10^3 = 1000 or all picks (000 - 999), therefore Pick 4 would be P4 = 10^4 = 10000 or all picks (0000 - 9999)... Much easier however you can still use nCr...

nCr = 10 (C1) 1 = 10, 10 (C2) 1 = 10, 10 (C3) 1 = 10 - they are totally independent therefore using Boolean model in probability methods would be a logical AND which is multiplying mathematically, therefore it is 10 * 10 * 10 = 1000 = 10^3 = P3 - P4 = 10^4 = 10 * 10 * 10 * 10 = 10,000

Cool Deal - Haven't done that since like 1997 or 1998... Have Fun, Pogo

I did it.  I made a program in Access that calculates all possible combinations and

I made a table of that with each being a record.  There are 575757 records.

Starting from 01-02-03-04-05 and ending with 35-36-37-38-39.

It's much easier to just look at the back of the playslip!!

This isn't just for calculating the probability of a win - you can calculate the probability of any numeric occurence with this formula - you just need to know how to use it... Including but not limited to lotto, mortgage, etc. Enjoy - Pogo

That's how I know to calculate odds (math is not my strong suite). However, on the MM playslip, the odds of 5+0 are 3,904,701. I have wondered where they come up with that number, instead of 3,819,816.

The number of possible combinations in a pool of 56 numbers are 3819816 but odds are based on possible combinations in the game and when you have 46 bonus numbers the possible combinations are 46 x 3819816 = 175711536. 

Since there are 46 5of5 combinations with a bonus numbers, when the one winning 5+1 combination is removed/subtracted there are (46-1) 45 5of5 combinations left with 175711536 possible combinations and the odds of having one of them are 175711536 ÷ 45 = 3904700.8.

Thanks RJOh.