A question for statisticians

[Shelby07]

I have a question for some of you mathematicians/statisticians out there.

It struck me when I read the latest thread by Famous Smoke that they are selling 1000 tickets to their raffle and will pick 5 winners. They are saying that's a 1 in 200 chance to win. Is that really the case?

It seems to me that the best odds are actually 1 in 996.

1st draw = 1 in 1000
2nd draw = 1 in 999
3rd draw = 1 in 998
4th draw = 1 in 997
5th draw = 1 in 996

Which is right, or are they actually equivalent in some way?

[bigwhiteash]

Yeah, it can be expressed as 1 in 200 chance.

1000 / 5 = 200, so one out of every 200 people will win!
But the ACTUAL odds are as you listed.

[Eville]

I am no statistician by any means, but I think it is better than 1 in 996. You might not have any better chances than 1 in 996 for each draw, but because you have a chance to win with each draw, your overall chances to win should be much better.

[bigwhiteash]

Nope... 1 in 996 is the absolute best you can hope for.

remember your fractions?

5/1000 reduces down to 1/200, but they aren't going to separate the 1000 people into groups of 200 and draw one prize, so the odds will max out at 1/996

[Eville]

That's pretty much what I was saying. Odds for each draw max out at 1/996, but since there are 5 draws, your overall chances to win are better than 1/996.

[craig]

I have a question for some of you mathematicians/statisticians out there.

It struck me when I read the latest thread by Famous Smoke that they are selling 1000 tickets to their raffle and will pick 5 winners. They are saying that's a 1 in 200 chance to win. Is that really the case?

It seems to me that the best odds are actually 1 in 996.

1st draw = 1 in 1000
2nd draw = 1 in 999
3rd draw = 1 in 998
4th draw = 1 in 997
5th draw = 1 in 996

Which is right, or are they actually equivalent in some way?

The chances of winning a particular prize is not the same as the chance of winning any one of the five prizes. You are describing the former; Famous is describing the latter. The five prizes are identical, so Famous is correct.

[FactoredReality]

Probability and Odds are two different things.

Probability is defined as:

P(Event) = Chances For Event Happening / Total Chances

Odds are defined as:

Odds(Event) = Chances For Event Happening : Chances Against Event Happening


So P(E) = 1 / 1000 for the first drawing; P(E) = 1 / 999 for the second drawing ...

Odds(E) = (1/999) * (1/998) * (1/997) * (1/996) * (1/995) = 5/985 or 1:197


In a nutshell odds are showing you a ratio of chances to win TO chances to loose. Probability is showing you chances to win TO total possible outcomes.

[craig]

Hopefully a decent explanation of why only Famous (and me ) are right:

http://www.youtube.com/watch?v=kLmzxmRcUTo

Only the first example is the same, but the rest are related, and show one of the most common - and most dangerous - fallacies that almost everyone makes.

[mail man]

WOW, this is making my head hurt.

[Shelby07]

Hopefully a decent explanation of why only Famous (and me ) are right:

http://www.youtube.com/watch?v=kLmzxmRcUTo

Only the first example is the same, but the rest are related, and show one of the most common - and most dangerous - fallacies that almost everyone makes.

Yeah... clear as mud.

Now, tell me which of the 5 draws will give me (well, not me... I didn't buy any tickets) a 1 in 200 chance of winning.

I contend that I never will have a 1 in 200 chance of winning with only 5 draws. The odds will go up after each draw, but will only reach 1 in 996.

[Eville]

You are thinking in terms of each individual prize by its self, not all of the prizes combined into a collective, which is the case here.

[basil]

No way Craig. Regardless of interpretation, If one person buys one ticket, and the other 999 tickets are all sold, in the first round he has one chance out of 1000 to win any of the five identical prizes. After that prize is gone, the rest of the 999 individual ticket holders each have one chance in 999 to win one of the four remaining identical prizes. There is only one of five identical prizes available for each round, with only one less ticket holder in the running with each round. I haven't seen the ad, and I like Famous, but if they are describing a 1/200 chance of winning in a one-prize-per-drawing from a pool beginning with 1000, then it's misleading.

[bigwhiteash]

Yep.. like I said, after all is said and done, one out of every 200 people will have won a prize. but they won't be separating the crowd into groups of 200, will they?

[BigMacFU]

No way Craig. Regardless of interpretation, If one person buys one ticket, and the other 999 tickets are all sold, in the first round he has one chance out of 1000 to win any of the five identical prizes. After that prize is gone, the rest of the 999 individual ticket holders each have one chance in 999 to win one of the four remaining identical prizes. There is only one of five identical prizes available for each round, with only one less ticket holder in the running with each round. I haven't seen the ad, and I like Famous, but if they are describing a 1/200 chance of winning in a one-prize-per-drawing from a pool beginning with 1000, then it's misleading.

Craig is right, it's just that Famous' justification is weak. Damn straight this thing doesn't work out to 1/200 chance to win in any real sense. But, it satisfies the laws in place for drawings/lotteries I'm sure, and that's all famous is going to care about.

[craig]

No way Craig. Regardless of interpretation, If one person buys one ticket, and the other 999 tickets are all sold, in the first round he has one chance out of 1000 to win any of the five identical prizes. After that prize is gone, the rest of the 999 individual ticket holders each have one chance in 999 to win one of the four remaining identical prizes. There is only one of five identical prizes available for each round, with only one less ticket holder in the running with each round. I haven't seen the ad, and I like Famous, but if they are describing a 1/200 chance of winning in a one-prize-per-drawing from a pool beginning with 1000, then it's misleading.

Okay, let's consider a big bowl with 1000 balls in it. Each ball is labelled. You have ball number 666. We shake the bowl up. A lot (and for a lot longer than you think - which was a problem with the US Army draft in the Vietnam era, but I digress ...).

Anyhow, Hex reaches in and grabs one ball. Winnah! 1 in 1000 chance. I think we're all agreed on that?

1. Now, let's say Hex has a big hand and picks up five balls all at once when he reaches in. What are the chances that ball number 666 will be among those five balls picked? 1 in 200.

2. Now, instead of just Hex, we add four more people, Hex, Bob, Carol, Ted, and Alice. All five shove their hand in at the same time, with each person pulling out just one ball. What are the chances that ball number 666 will be one of the five picked? Still 1 in 200.

3. Oh oh, the five folks didn't exactly plunge their hands in at the same, nor did they pull their hands out at the same time. What are the chances that ball number 666 will be one of the five picked? Yup, one in 200.

Yes, the chances of Hex pulling 666 out changes, as you say. However, it doesn't matter if Hex, or Bob, or Ted, or Carol, or Alice pulls out ball number 666; you win regardless. Similarly, it doesn't matter if Bob goes first, or Hex goes first, or ... - if one of them pulls your ball, then you win.

The chances don't change - because you don't care who grabs your ball, or when, nor do any of the other 999 entrants. If you, or they, did, then that's different.

Now, the catch is, what happens if you have TWO balls in the big bowl. You can only win once - what happens if both your balls are grabbed? - well, one gets cut off. How does that affect the chances for everyone else?

I leave that as an exercise for the reader.