i'm not sure how they distribute(divide) the prizes but i'm sure that by buying all possible combination you won't win more than you spent.]]>

So the probability of choosing all six numbers right is

(6/49)*(5/48)*(4/47)*(3/46)*(2/45)*(1/44) = 1/13,983,816.

Approximately 1 in 14 million chance.

Now there are prizes for 3 matches, 4 matches, 5 matches, 5 matches plus one bonus ball match.

To calculate the probabilities of them let us start with the case of 3 matches.

Consider the instance when you get the first 3 right and the next 3 wrong.

The probability for this to happen can be calculated like this.

(6/49) probability to get the first number right.

(5/48) to get the second right.

(4/47) to get the third right.

Now for the 4th to be wrong... there are 46 to choose from and 3 of them shouldnt come. So the probability is (46-3)/46 = 43/46

For the 5th to be wrong... there are 45 to choose from and 3 of them shouldnt come. So the probability is (45-3)/45 = 42/45

Similarly for the last to be wrong the probability is 41/44

So the probability of this particular instance is

(6/49) * (5/48) * (4/47) * (43/46) * (42/45) * (41/44) -------> (Result A)

Now consider the case when you get the first two right, third wrong, fourth right and fifth and sixth wrong. This is another instance when you get 3 matches.

So for the first to be right the probability is (6/49)

for second to be right it is (5/48)

for third to be wrong..we have 47 balls to choose from and 4 numbers remaining which shouldnt come. So we get (43/47)

for fourth to be right..we have (4/46)

for fifth to be wrong.. we have (42/45)

for sixth to be wrong.. we have (41/44)

So the probability for this kind of 3 match to happen is

(6/49) * (5/48) * (43/47) * (4/46) * (42/45) * (41/44) -------> (Result B)

Comparing results A and B we can see that they both have same probability (43*42*41*6*5*4)/(49*48*47*46*45*44)

From this we can see that all kinds of instances where we can get 3 matches have the same probability. So how many of such kinds are available? How many different combinations of 3 positions can be taken from 6? It is 6c3.. which is (6*5*4)/(3*2*1)

So the total probability is (43*42*41*6*5*4)/(49*48*47*46*45*44) * (6*5*4)/(3*2*1) which is approximately 1/57 chance.

Similarly you can find the probabilities of 4 matches and 5 matches (which are left as exercises )

Now consider the special case of 5 matches and one bonus ball match. This is almost like matching the 6 balls with only one difference. With the six ball match you have no choice but to match the available six balls. But with the bonus ball variety you have to match the bonus ball and any 5 of the other 6. So How many combinations of 5 positions can be taken from 6?

6c5 = 6

So it is six times more probable than matching all six balls.

So (6/49)*(5/48)*(4/47)*(3/46)*(2/45)*(1/44) * 6 = 1/13,983,816 * 6 = 1/2,330,636.

That is 1 in 2,330,636 chance.

Hope I am not wrong and I am not very unclear

]]>Add all the other prizes you'd win from near-perfect tickets, and you could get a pretty good profit. Surely that can't be right?

]]>To be sure that you'll win, just buy 13983816 tickets ]]>

**If you buy your lottery ticket for the Saturday draw before Friday, you are more likely to die than win.**

...OK, that wasn't a very fun fact. But it was a fact.

]]>Martyn wrote:

]]>I have to show how the odds of winning the english national lottery are worked out.6 correct balls from 49.Could someone show me how the formula is expressed.

Thanks

With Permutations you count how many different ways something can be arranged, such as abc, acb, bac ... etc. But the lottery doesn't care what order the numbers come out.

Combinations looks at how many different ways things can be chosen without concern for order - just right for lotteries

The classic formula for combinations is:

n_C_k = n! / k!(n - k)!

The "!" means "factorial". We sometimes call n_C_k "n choose k", or in your case "49 choose 6"

(Now, I give this formula without explanation because it takes a long time to explain ! You may have to look up your text book here.)

But, anyway, we should be able to work it out for the lottery:

49_C_6 = 49! / (6! x (49-6)!) = 13983816 (using my calculator)

So, the odds are nearly one in 14 Million. Ouch.

]]>Thanks]]>