Lotto

Pigman II

Simple one here. Which gives you a higher chance of winning.

1) Buying 2 'unique' (ie no 2 tickets have same 6 numbers) tickets for 1 lotto draw
2) Buying 1 ticket for 2 seperate lotto draws.

I reckon they're both exactly the same but someone told me there is a difference in the odds there somewhere. So is there? If so then where?

Thanks

whyamihere?

I would think option 1 has slightly better chances of you winning. i emphasize slightly because it would be neglible.

Reduce the problem to a lotto of 6 numbers (instead of 42) and each ticket has 3 numbers (instead of 6) because its easier and the principle is the same.

Option 1.
Possible tickets: 6C3=20
Choose two different tickets.
Chance of first ticket winning=1/(6C3)=1/20
Chance of second ticket winning=1/(6C3-1)=1/19
(minus one, since tickets are different, possible outcomes of lotto is reduced by one since it cannot be the same permutation as the first ticket)
total prob=(1/20)x(1/19)

Option 2.
Prob is 1/(6C3) for both draws.
Total prob=(1/20)^2

In original problem the difference is 1.9062923x10^(-7)

Pigman II

Thanks for the reply whyamihere. However the 1/20 x 1/19 doesn't seem to make sense tho. ie if you had 1 ticket then your odds of winning are 1/20 but if you have two tickets your odds are now 1/380?

Also In Option2 Isn't (1/20)^2 the prob of you winning BOTH draws? I'm just looking for the prob of 'winning' ie winning either draw (or both draws together).

I wonder would the prob of that be " 1/20 + 1/20 - (1/20)^2 " ?

red_fox

That should be: 1-{(1-(1/20))x(1-(1/19))}

The product being the overall chance of losing, ie chance of losing with the first ticket AND second ticket. 1-that gives overal chance of winning.