quick probability question

The recent article by Marilyn v.S. said that if you bought 2 tickets in a lottery that had a 1-in-80million chance of winning…you would have a 2-in-80million chance of winning.

That doesn’t seem right, although it’s probably a good approximation. What’s the more precise answer? Better yet, what’s the equation used to calculate the answer?

An easier example would be flipping a coin where you “win” if you get tails. You have a 1-in-2 chance of winning. But if you flip twice, you do not have a 2-in-2 chance (100%) because we all know that it’s totally feasible to get heads twice in a row.

That’s not a correct analogy.

The coin example describes two successive occurances, both with two possible outcomes.

The two lottery tickets describe two possible outcomes of one single occurance with 80 million possible outcomes.

See the difference? They can’t be compared directly like that. It would appear that Mar… that woman is quite correct in saying that the possesion of two lottery tickets in a lottery with 80 million outcomes (all with equal possibility) constitutes a chance of 2 in 80 million to win that lottery.

There is only one drawing in lottery. The analogy with a coin flip would be if you called heads or tails for a single flip. The probability of head or tails coming up in one flip is 2/2 or 100%.
Marilyn happens to be correct.

Well, just to be a pain in the ass…

Marilyn is right assuming the two lottery tickets have different number combinations. Since many folx buy tix by letting the computer generate random numbers for them, there is a 1/80 million chance that the tickets will have the same number. So it’s ever so very slightly less than the 2 in 80 million.

(Analogy: If you bought 80,000,000 computer generated numbers, you clearly wouldn’t have a 1-in-1 chance, as the computer almost certainly missed some combinations.)


You are a pain in the ass and you are correct. I sort of assumed that if you bought two tickets, you would buy different combinations. If the two tickets are chosen at random, then the probability of a win would be
2 out of 8 million minus 1 out of (8 million)[sup]2[/sup]

Assuming two different tickest the probability of a win is:
Assuming two independent random tickest the probability of a win is:

D’OH! Yes, that makes sense now. Thanks.

Just to get totally anal about this, your calculation assumes that the computer generated numbers are actually random, rather than the pseudo-random numbers generated by most algorithms. Your chances of getting the same numbers “at random” twice depends on how good the random number generator is.

You could go on with this logic ad infinitum. To put a stick up the stick up the stick up our collective asses, you would have to check the manufacturing flaws in the chip, etc. I think the “2 in 80 million” number is good enough for 99999999999999999999999999999999999999.9999999999999999999999999999999999999999% of us, assuming of course that…

ah I’d better shut up now.

D’oh…and I even previewed.

I meant 99.99999999999999999999999999999999999999999% :slight_smile:

As I recall from my distant studies of statistics you can’t add the two together. I seem to remember that you multiply the odds of the non event. Thus in two coin tosses the odds of not getting heads in two tosses would be .5*.5=.25 so your chances of getting a head in 2 tosses are 3 in 4.

Applying this to the lottery tickets the chances of two successive losses are 79,999,999.0000000125 in 80 million so the chances of winning are 1.9999999875 in 80 million.

In the coin toss example I should have said your chances of getting AT LEAST one head in two tosses= 3 in 4

I obviously need to review more carefully before I hit send, the chances of the 2 successive losses is 79,999,998.0000000125.

In fact I understand that there are some folks who have tried to take advantage of that wrinkle by trying to get enough people to buy enough non-duplicated tickets that they can beat the odds. Jeez… I think it would be easier to just work for a living.

Ned wrote:

You multiply the probabilites of non-events when the events are not mutually exclusive. In the case of two (different) tickets to the same lottery, they are mutually exclusive events (both tickets couldn’t win), so you can simply add the probabilites. All the way up - if you bought 80 million tickets in this lottery, you would have a 100% chance of winning.

This isn’t the how to blow up a cat thread is it?

Thought not. Sorry. 'Bye.

I said it was distant :slight_smile:

I suppose the multiplication theory holds if you let the machine pick though you could recalculate as soon as you got the ticket.

So, anyone else wanna rephrase, re-parameter, re-circumstance, or re-bullshit my perfectly correct, spot-on, and well worded answer??

Bastards! :wink:

I’d just change “That’s not a correct analogy.” to “That’s not an accurate analogy.” :slight_smile:

Countdown to thread closure…

For two independent events: to get the probatility that at least one event will occur, add the probability of each occurring, then subtract the probability that they both will occur. For the coin toss it’s 0.5 + 0.5 - 0.25 = 0.75 for a head. For the lottery it’s 1/80m + 1/80m - 1/(80m*80m). If it’s the lottery I’m thinking about, the odds are exactly 1:80089128.00, but if you buy two random tickets 1:40044564.25.