Every now and again, someone asks how the odds for the various games are calculated. Although Camelot publish what the odds are for each prize level, many of you maths buffs out there would like to see just how they've arrived at those figures.

Using the examples below you'll be able to work out the odds for any lottery game, so grab hold of a calculator and we'll take a trip down fiddly sum lane.

A Few Introductory Examples - Steve's Pretend Lottery

For our calculations we'll be dealing with various combinations of numbers. Before we start dealing with the different combinations involved in 49 numbers, let's start off with something simpler so we can visualise what's going on.

Let's see how we'd calculate the odds for Steve's Pretend Lottery, where your ticket consists of 2 numbers picked from a pool of 5 and both must match the 2 drawn numbers, in any order, to win the jackpot.

We'll start by looking at the different pairs we can get from the total pool of numbers (1, 2, 3, 4, 5):

Note that half of the above pairs contain the same numbers but in a different order (1,5 and 5,1 for example).When the winning numbers are drawn, the first number can be any one of 5, whilst the second one can only be taken from the remaining 4. However the mathematics behind this isn't quite so simple as 1/5 x 1/4. Let's see what's going on:

m = matching balls = 2
t = total range    = 5

Using factorials:
                t! = 1 x ... (t-1) x t = 1x2x3x4x5 = 120

To get the total number of ordered pairs, which we know to be 20, we can use the following equation:

                       t!   = 5x4x3x2x1 = 5x4 = 20
                     (t-m)!     3x2x1

This might seem rather elaborate but this will help illustrate the more complicated functions that we'll be using. The above equation doesn't take into account the fact that for Steve's Pretend Lottery - like the National Lottery games, the order that the balls were drawn doesn't matter. The first drawn number could have appeared in either the first or second position, whilst the second ball had to be in the second position. Therefore, if we take the total combinations and divide that by the number of different combinations our winning numbers can form (shown as m!), we can find the number of winning combinations.

            C(t,m) =     t!   
                     m!x(t-m)!
                   = 5x4x3x2x1
                     2x1x(5-2)!
                   = 5x4x3x2x1
                     2x1x3x2x1
                   = 120 / 12
                   = 10

The above function allows us to pick out m numbers from a range of t. So for Steve's Pretend Lottery the chances of a ticket's two numbers matching the jackpot are 1 in 10.

National Lottery Odds

Using the previous function we can now calculate how many 6-number combinations are available in the National Lottery. Let's assign some letters to the figures we'll be using:

d = drawn numbers    = 6
t = total numbers    = 49

            C(t,d)   =     t!   
                       d!x(t-d)!
                     =   49! 
                       6!x43!
                     = 49x48x47x46x45x44
                          6x5x4x3x2x1
                     = 13,983,816

With 13,983,816 6-number combinations, the jackpot odds are 1 in 13,983,816

As all the numbers on the ticket are needed to get the jackpot, this will be the easiest result to calculate. However to calculate the lower prize combinations we have to reconsider the C function. We'll have to expand the function to recognise the winning numbers we pick in addition to the remaining numbers on our ticket, as the probability is calculated not just on the numbers that match but also on the numbers that don't.

Sounds confusing, so let's look at a new function that makes use of what we've learned so far to calculate the odds of getting a 3-Ball win:

         d = drawn numbers
         t = total range of numbers
         m = matching numbers
    C(d,m) = ways of matching m numbers from d drawn numbers
C(t-d,d-m) = combinations missed ticket numbers from the undrawn numbers
    C(t,d) = total number of combinations

Pm = C(d,m) x C(t-d,d-m)    C(6,3) =      6!        C(43,3) =      43!    
           C(t,d)                    3! x (6-3)!              3! x (43-3)!
   = C(6,3) x C(49-6,6-3)          =  720                   = 43x42x41
           C(49,6)                   6 x 6                     3x2x1
   = C(6,3) x C(43,3)              = 20                     = 12341
         13983816

Pm = 20x12341
     13983816
   = 1/57 (rounded up)

The 4-Ball prize odds can be calculated by substituting the relevant figures, you can even check the Pm function against the 6-Ball prize (0! = 1). However don't try the equation for the 5-Ball prize because there's something else we have to take into account - the Bonus Ball.

The 5-Ball prize is not just paid out for getting 5 matches, in fact it's for 5 matches and for not matching the Bonus Ball. If we work through the equation for a 5-Ball match, we'll eventually end up with the following result:

   Pm =   258   
        13983816

We'll leave it at that stage to avoid any decimals just yet. Once the 6 main balls have been drawn there are 43 left in the pool. The chance of drawing the bonus ball at this stage is 1 in 43, and so we combine the 5-Ball probability with this figure to get the results we're after:

P(5 Ball+bonus) =    258x1       P(5 Ball-bonus) =   258x42   
                  13983816x43                      13983816x43
                = 1/2330636                      = 1/55492 (rounded up)

Finally, if you want to know the overall odds of winning, just add all the winning probabilities together:

 1 +   1  +   1   +    1    +     1    = 1/54 (rounded up)
57   1033   55492   2330636   13983816

Thunderball Odds

The odds for this game are easy to handle once we've seen how to deal with the Bonus Ball in the National Lottery game. As the Thunderballs are drawn from a different pool, the odds for prizes needing the Thunderball are always multiplied by 1/14, whilst those without use 13/14. Let's start by finding out the number of 5-ball combinations we can get in a range of 34 numbers.

d = drawn numbers    = 5
t = total numbers    = 34

            C(t,d)   =     t!   
                       d!x(t-d)!
                     =   34! 
                       5!x29!
                     = 34x33x32x31x30
                         5x4x3x2x1
                     = 278,256

We can use this figure in all our P() functions, but it also illustrates the way that the presence, or lack, of the Thunderball affects the odds. Initially from this result we can work out the odds for the jackpot and 5-ball win.

P(Jackpot) =     1        P(5 Ball no TB) =    13    
             278256x14                      278256x14
           = 1/3895584                    = 1/299661 (rounded up)

Using the previous examples you shouldn't have any trouble finishing off the remaining odds for the Thunderball game.

Lottery Extra Odds

Not much point having a separate section for the Extra odds, as your chance of winning is simply the same as for the jackpot on the National Lottery - 1/13983816.

Official Odds from National Lottery Literature