A question for statisticians

[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.

[basil]

I disagree. And if I'm seeing it right, Famous is convinced too. Just looked on the site at the raffle ad(s) and if they ever said anything about a 1/200 chance, it's gone now.

There's just no mathematical, ethical, and evidently legal, way to justify a 1 in 200 statement in that contest. Semantically? Maybe, but Famous' attempt has been pulled down.

[ashauler]

The only thing I've seen in this whole thread that makes sense to me is the guy in craigs video talking about head and tail. Neither of which I can argue with. ;)

[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.

[ndv21]

what happens if both your balls are grabbed? - well, one gets cut off.

I understand your point completely, but I like my balls right where they are at!

[Shelby07]

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.

I'd agree if you had 5 bowls with 1-200 in bowl 1, 201-400 in bowl 2, 401-600 in bowl 3, etc. Then and only then would you cut the odds down to 1 in 200. Each draw would give a different set of people a chance of 1 in 200, but the'd only have 1 chance. It seems that, for these purposes anyway, 1 chance in 200 is very different than 5 chances in 1000.

Here's another way of thinking about it. One chance in 200 means that after 200 draws I would be guaranteed to win, no matter what number I had. But in your scenario with 1000 balls in the bucket there would still be 800 losers after 200 draws.

[craig]

I'd agree if you had 5 bowls with 1-200 in bowl 1, 201-400 in bowl 2, 401-600 in bowl 3, etc. Then and only then would you cut the odds down to 1 in 200. Each draw would give a different set of people a chance of 1 in 200, but the'd only have 1 chance. It seems that, for these purposes anyway, 1 chance in 200 is very different than 5 chances in 1000.

Here's another way of thinking about it. One chance in 200 means that after 200 draws I would be guaranteed to win, no matter what number I had. But in your scenario with 1000 balls in the bucket there would still be 800 losers after 200 draws.

Okay, I'll buy the "if I buy 200 tickets then I have to win" point. That's not how Famous advertised it IIRC, but if they worded it so that you and others drew that conclusion, then they were wrong.

Note, however, that the statement does not hold true for the chances published for any lottery or game of chance, or in probability and statistics. It is usually defined in terms of what happens on average; in the long run. You flip a fair coin; chances are 1 in 2 that you'll get a head in one flip. The odds don't guarantee that you'll get a head - and only one head - if you flip a coin twice (or flip two coins simultaneously).

[basil]

At the risk of seeming obtuse, that dog still ain't gonna hunt. There's no pre-designation saying "we're going to see if #666 one of five balls randomly chosen". I mean, what are you going do with the other four? Can't hardly tell folks mine was arbitrarily picked to test the five drawn (and by default, theirs were not). All five of those balls chosen are winners and they were pulled out of a bowl of 1000-996.

And if I have multiple entries, two of my numbers are pulled, three others win, and Famous throws my extra back in the bowl. Now everyone left has a 1 in 996 chance, not the 200 that was advertised, and Famous has made a false claim. If my extra prize is thrown out, then in retrospect the chances were what, 1 in 250? Again, a deviation from the original claim. They can't hardly pull 4.98 (to maintain a 1 in 200 chance) balls out and say since there's only one prize left, 3.98 of you guys are SOL.

The only way to change the odds from 1 in 1000, 1 in 999, etc., is to buy more than one ticket.

[Shelby07]

Okay, I'll buy the "if I buy 200 tickets then I have to win" point. That's not how Famous advertised it IIRC, but if they worded it so that you and others drew that conclusion, then they were wrong.

Note, however, that the statement does not hold true for the chances published for any lottery or game of chance, or in probability and statistics. It is usually defined in terms of what happens on average; in the long run. You flip a fair coin; chances are 1 in 2 that you'll get a head in one flip. The odds don't guarantee that you'll get a head - and only one head - if you flip a coin twice (or flip two coins simultaneously).

The analogy doesn't hold. If you flip a coin twice you have a chance of getting 2 heads because after you get the first one the heads are back in play. In the situation where you draw 2 balls out of a bucket, the first ball is not in play after it is drawn. So 2 draws will always yield one of each ball.

In the case of 2 balls in the bucket you have a 2 in 2 chance of winning. In the case of the coin being flipped twice you always have a 1 in 2 chance of getting heads no matter how many times you flip the coin. Now if you're saying that they will draw 5 times (or any number of times for that matter) from a bucket containing 1000 balls and they will throw the winning ball back in the bucket after each draw, then you never have better than 1 in 1000 chance of winning.