A question for statisticians

**basil** · 09-30-2009, 08:59 PM

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.

**bigwhiteash** · 09-30-2009, 09:27 PM

**FactoredReality** · 09-30-2009, 10:58 PM

**Shelby07** · 09-30-2009, 11:56 PM

Originally Posted by craig

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.

**Kenyth** · 10-01-2009, 01:20 PM

Originally Posted by Shelby07

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.

That would only hold true if they would let a ball win as many times as it is drawn. If a single ball can't win more than once, it is effectively not part of the exercise.

Also, odds of 1 in 200 (the simple laymans calculation) is correct from their perspective before the contest if there will be five winners out of 1000. Odds and probability are only true from a single given perspective at a specific time before other events have occurred. As soon as an altering event occurs, more information is available and the odds and probability change for the remainder of the event.

Your odds of being one of the 5 winners before the contest is 5/1000 simplified to 1/200.

Your odds of being the first drawn winner is 1/1000

After this event, you odds of being one of the remaining 4 winners is 4/999 simplified to 1/249.75

Your odds of being the second drawn are 1/999.

So on and so forth.

Your odds after this

**basil** · 10-01-2009, 01:57 PM

Originally Posted by bigwhiteash

Hey!

**bigwhiteash** · 10-01-2009, 02:18 PM

Feeling Bewindled?

**Shelby07** · 10-01-2009, 09:27 PM

Originally Posted by Kenyth

That would only hold true if they would let a ball win as many times as it is drawn. If a single ball can't win more than once, it is effectively not part of the exercise...

That is the reason that a coin toss can't be used as an example here

Originally Posted by Kenyth

...Your odds of being one of the 5 winners before the contest is 5/1000 simplified to 1/200...

As I mentioned before, I'm not sure I agree with this. It appears that "5 chances out of 1000" is not the same as "1 chance in 200" which is how the contest was originally presented.

**Kenyth** · 10-06-2009, 01:41 PM

Originally Posted by Shelby07

That is the reason that a coin toss can't be used as an example here

As I mentioned before, I'm not sure I agree with this. It appears that "5 chances out of 1000" is not the same as "1 chance in 200" which is how the contest was originally presented.

Odds presented as this are a simple proportional expression, which means that 5 chances in 1000 is equal to 1 in 200 just as it is equal to 10 in 2000. This is a simple method of trying to predict chances of an outcome with known information at a given point in time. The instant one of the chances is used, more information becomes available, and everything changes.

The confusion here is occurring because folks are trying to break it down into it's singular events and then average it back together. It doesn't work that way. In order to even be eligible for the second, third, fourth, or fifth ball you would already have lost the previous draws making those draws irrelavent to your current odds. More information is available and the odds keep changing for the worse since the numbers aren't being changed proportionally on both sides of the equation.

The fact that you had a previous chance or chances still factors in when calculating the odds for before the contest started. The odds for each individual chance don't simply average out to the original odds, because each subsequent chance assumes you lost the preceding one(s).