## Thursday, July 14, 2011

### Euromillion odds

While the jackpot was so high the probability of winning were discussed, however in keeping with we don't have time and maths is boring nothing was really done to explain why the odds were want they are. Let me remedy that.

The pick is 5 numbers from 50 and 2 numbers from 11. For the first number picked there are 50 balls of which 5 match my numbers the odds of the first pick being one of mine is therefore 5 in 50 or 5/50.

For the second pick there are only 49 balls left; if the first draw matched one of my numbers (which is what I'm trying to calculate) then I only have 4 numbers left to match. That's 4 in 49 or 4/49.

This continues onward 3/48, 2/47 and finally 1/46.

Now I want all these things to happen so I have to multiply these figures together, the easiest way to show that is to mash them together like this:

(5*4*3*2*1)/(50*49*48*47*46)

The lucky stars work in the same way I have 2 numbers out of eleven 2/11 and given one being picked I'm left with 1 from 10 or 1/10. Again as I want these to both happen I need to multiply them together and then as I want 5 numbers AND 2 lucky stars multiply the lot:

(5*4*3*2*1)/(50*49*48*47*46) * (2*1)/(11*10)

The answer is 1/116,531,800 which is the odds given by the Euromillions page.

But what about something a little more difficult? How about calculating the odds for 5 numbers and only 1 lucky star?

The five numbers I've  already been done and remains (5*4*3*2*1)/(50*49*48*47*46) it's the lucky stars that have changed.

Again the chances of either of the 2 numbers matching from the 11 is equal to 2/11 however in the second case we don't want our number to be drawn. Of the 10 numbers left 9 aren't ours so the odds are 9/10

Multiply everything up and we get 1/12,947,978 which... doesn't match the odds posted.

We've missed out an option. The first pick matched and the second didn't, but what if the first pick didn't and the second pick did?

Again 2 numbers from 11 but this time we want the odds that it doesn't match that makes 9/11; for the second pick we do want it to match which makes 2/10.

So if we imagine a two world scenario in World One we get all five numbers and the first lucky star; in World Two we get all five numbers and the second lucky star.

As we want either of these to happen we add the odds

(5*4*3*2*1)/(50*49*48*47*46) * (2*9)/(11*10)
+
(5*4*3*2*1)/(50*49*48*47*46) * (9*2)/(11*10)

and the result is 1/25,895,956 which still doesn't match up with the odds posted. Where's the flaw?

The answer is that the first five numbers picked are independent of the lucky stars so rather than have two worlds arranged as above the split only occurs after the five numbers are matched. In other words

Five numbers match AND the first lucky star OR the second lucky star:

(5*4*3*2*1)/(50*49*48*47*46)
*
(2*9)/(11*10) + (9*2)/(11*10)
=
1/6,473,989 (with some rounding)

Noting what is and isn't dependent is important otherwise it can result in some strange odds. Consider a wheel with ten numbers on it, pick a number, spin the wheel. The odds of winning are 1/10.

Now add another identical wheel, doing the same thing the odds of me winning on the first wheel are 1/10 and the odds for the second wheel are also 1/10. For me to win on the first wheel AND the second wheel I multiply them together and get 1/100. So what are my odds on winning on the first wheel OR the second wheel?

Add them together and it appears to be 2/10. Seems plausible, but make it ten wheels and my odds of winning become 1/1 I can't lose! Except of course I can.

They're independent from each other, a number appearing on one wheel doesn't preclude it from appearing on another - they could all point to number 5 or alternate between 5 and 6 or display 2,4,3,2,6,7,3,8,2,10. No winner for the picker of numbers 1, 5, or 9 there.

As an exercise consider the odds for the outcome of every state the wheel can be in for a lose/win situation and calculate the odds of winning if at least one wheel picks your number.