# Odds question (math...sigh...)

Lottery/raffle odds.

if the odds of one particular ticket hitting are 1/830 (550 prizes, 460,000 tickets), what is the formula for more tickets? What if you have 20 tickets? It’s 20 out of 830, but what is the formula for turning that into 1 out of ______?

thanks.

20/830=1/x

x = 830/20

Not exactly. If that were the case, then buying 830 tickets would yield a 100% chance of winning a prize. While that would be true if there were one prize and 830 different choices for tickets, but it’s bviously not true when you have 550 prizes and 660,000 tickets.

I used to know how to calculate the specific odds of an event like this occurring but it’s been years and it’s late now so I can’t figure it out.

830/20 is close enough if you just want a reasonable estimate. But it is not the exact answer.

Someone paid \$40 dollars to have basically the same question answered.
I was going to follow the example given and calculate the final answer for this situation, but I don’t know what the hell this means:

It means that you have a 99.999941% chance of losing each draw. Multiply the probability by itself 4255 times (one for each draw), and the chance of losing 4255 draws is 77.8566%.

By the way, I stand by my answer before. 20 out of 830 is 1 out of 830/20. I may not know probability too well, but I do know fractions.

460,000 - 550 = 459,450/460,000 chance, or a 0.9988043478260869 chance, that we won’t get a winner if we get one lottery ticket.
We raise this fraction to the 20th power, representing taking 20 tickets.
0.9988043478260869 ^ 20 = 0.976356
So we have roughly a 97.6% chance of not having a winner in 20 tickets. Therefore, we have roughly a 2.3643% chance of having at least one winner.
By the way, 20/830 is pretty damn close. It gives 2.4096%, just 0.0453% off.

Ahh, thanks.

But your formula wouldn’t have been as accurate had the OP asked what the chances would be if 400 tickets were purchased. Using your formula we would come up with a 48% chance of winning but using the correct formula we’d only be at about 38%. It’s certainly handy for estimating in a lot of circumstances but it’s not the correct formula, which is what the OP was asking for.

The expected number of wins for one ticket is 11/9200 (about 1/837). If a contestant can win multiple times, the expected number of wins for 20 tickets is 220/9200 - about 0.023913. However, their chances of winning anything at all are slightly smaller: 1 - (1 - 11/9200)[sup]20[/sup], about 0.023643, just like WoodenTaco said.

99.9941% (You slipped the decimal when converting to percent.)

The problem, though, is that Stoid asked the wrong question.

Her questions were:

1. if the odds of one particular ticket hitting are 1/830 (550 prizes, 460,000 tickets), what is the formula for more tickets?
2. What if you have 20 tickets?
3. [Assuming] It’s 20 out of 830… what is the formula for turning that into 1 out of ______?

You answered question three, but the OP’s assumption that the answer for 20 tickets is 20/830 is incorrect.

Question 2 has already been answered; question 1 is just an extension of that. Taking Punoqllads’s formulation, the odds for any number of tickets is: [1 - (1 - 11/9200)[sup]n[/sup]], where n is your number of tickets, and 11/9200 is the odds of a single ticket winning.

You realize you guys are making my head hurt, right?

Sadly, this is exactly why so many people do play the lottery and other forms of gambling. The math involved is pretty basic. (Stripped down, you just divide and then multiply.) But people have a tendency to run off screaming when confronted with it.

As the saying goes, lotteries are a tax on stupid people. No offense meant, Stoid. It’s sad, that’s all.

If the odds are very very low of one ticket winning, then buying 20 tickets will increase your chances by 20-fold. If the odds of winning with one ticket aren’t that low, then you need to to the more advanced math.

Others have said it already, but maybe this will clearer for how to do the math.

1. Take the probability of losing with one ticket. For example, 98% exactly.
2. Multiply this by itself as many times as you buy tickets. For example, if you buy 12 tickets, multiply 98% by itself 12 times, also known as raising to the 12th power. (98%) ^ 12 = 78.5%. This is the odds of losing with 12 tickets.
3. Take that percentage, subtract it from 100%. 1 - 78.5% = 21.5%. This is your odds of winning with 12 tickets.

Let’s apply this to your OP.

1. The probabilty of losing is 829/830, or 99.88%.
2. If you buy 20 tickets, the probably of losing on all of them is (829/830)^20 = 97.6%
3. 100% - 97.6% = 2.4% chance of winning, as others have said.

Because 1/830 is pretty bad odds, in this case, it’s almost the same as 20/830. I’ll do one more example, just so you can see the difference.

Example: Roll an ordinary 6-sided die. If it’s a six, you win a prize. You roll it 15 times, what are the odds of winning?

1. Odds of losing on one roll = 5/6.
2. 15 rolls = (5/6)^15 = 6.5%
3. 100% - 6.5% = 93.5% of winning at least once in those 15 rolls. This is obviously much different than multiplying 1/6 by the 15 rolls, which gets you 15/6, or 250%, a figure that doesn’t even make sense.

:smack: I caught that right after the edit window. I guess the Dope is the wrong place to think it would have slipped by.

Yeah, I know. I should have been more explicit in admitting my ignorance in my last post. Like I said, probability isn’t my strong suit. Actually, I know enough probability to have answered this question, but the “20 out of 830 equals 1 out of ____” stood out at me before I thought through the rest of the OP. I apologize for that.