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#1 2005-02-22 02:53:40

Martyn
Guest

calculation

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

#2 2005-02-22 09:02:27

MathsIsFun
Administrator

Online

Re: calculation

The subject is "Permutations and Combinations".

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.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#3 2006-02-07 09:31:12

paul dring
Guest

Re: calculation

;);)

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

#4 2006-02-07 09:40:42

mathsyperson
Moderator

Offline

Re: calculation

All the maths has been dealt with, so all I can do is hop in with a fun fact that might get you extra marks if you're lucky. wink

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.


Why did the vector cross the road?
It wanted to be normal.

#5 2006-02-07 17:18:55

justlookingforthemoment
Moderator
Award: Wink Sherlock

Offline

Re: calculation

So the moral of the story is... ah... don't buy your ticket before Friday? big_smile

#6 2006-02-07 18:26:49

krassi_holmz
Real Member

Offline

Re: calculation

Or other:
To be sure that you'll win, just buy 13983816 tickets wink


IPBLE:  Increasing Performance By Lowering Expectations.

#7 2006-02-08 03:34:14

mathsyperson
Moderator

Offline

Re: calculation

Hmm. That would cost you ≈14 million, but I'm sure there have been jackpots higher than that.

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?


Why did the vector cross the road?
It wanted to be normal.

#8 2006-02-08 19:43:05

Berthold
Guest

Re: calculation

In the beginning you have 49 numbers. You have to take 1 number and it should match 1 of the 6 already chosen numbers. If there is only one number to match, choosing the right number from the 49 numbers would have probability of (1/49) Since you have 6 numbers, matching one of them would have probability (6/49). Now you have 48 numbers and 5 numbers to match. So choosing a second right number would have probability (5/48) and so on.

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 tongue)

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

#9 2006-02-09 01:07:39

kempos
Full Member

Offline

Re: calculation

berthold the chances are much less than 2000000.
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.

Last edited by kempos (2006-02-09 01:10:22)

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