Friday, November 07, 2008

My Failed Mathematics Logic or Is It?

Did my odds of winning the jackpot get bigger after the drawing?

I was arguing with my friend last night that my odds of holding a winning ticket went up after the jack pot drawing.

Here is my logic

Before the jackpot drawing my odds of holding a winning ticket were 1 in 10 million.

The day after the drawing I figure since the winning set of numbers was picked and only 500,000 people played my odds of holding the winning ticket improved to 1 in 500,000 instead of the 1 in 10,000,000 before the drawing.

Sounds reasonable right?

22 napkins, 18 charts later I was still unconvinced that my odds did not improve after the drawing. I know that my overall chances are still the same regardless but since a winner was determined by a much narrower data set shouldn't my odds of holding the winning ticket mathematically go up?

This is reason # 2541523 I miss the chips @ 11:00 crew....Rocket, G-Off am I crazy?


Rocketstar said...

I miss that too man...

"by a much narrower data set "
-- I think the way to look at it is that you are playing against the entire possible set of 6 or 7? numbers. You're playing against the house and the number of players that are playing alongf with you against the house doesn't affect your chnaces of winning since they can use the exact same mubers as you are.

If there was a rule that each set of winning numbers could only be held by one person, then I think that math would work.

I keep buying one when i get gas hpoing and praying to almighty god.

~Sheila~ said...


Not my strong point. I'm lucky I got out of College Alebra alive.

All I know is that if you played twice in one day...yes. you increase your odds.

That's all the help you will get from me.

Geoffrey said...

I agree with Rocket... I think you're mixing up two separate events. Let's represent our sample space as set A = {all possible numbers that could be drawn}. Let's also suppose that set B = {all numbers that have been purchased} and set C = {all numbers that Brian has purchased}. C is a subset of B, and B is a subset of A.

Because we know the number of possible outcomes in the sample space, we can apply the classical probability formula for calculating likelihood that a given event will occur: P[event] = (number of possible outcomes in which the event occurs) / (total number of outcomes in the sample space).

Now let's look at two events... The first event, represented as Z, is *someone* will win. The second event, represented as Y, is Brian wins. The probability of event Z is P[Z] = n(B) / n(A) = 500,000 / 10,000,000 = .05. The probability of event Y is P[Y] = n(C) / n(A) = 1 / 10,000,000 (not so good).

The way more ticket sales would increase your odds would be if the size of the sample space were equal to the number of tickets sold, i.e. n(A) = n(B), like in a raffle...

I think we need a chips @ 11 reunion...

Brian in Mpls said...

I am up for that:) Let me know when. I think me and Chow Do are going today