What's The Chances of This One? 911 NY Drawing

[MMMMMM]

ALBANY, New York (AP) -- On the first anniversary of the terrorist attacks, a date known as 9-11, the evening numbers drawn in the New York Lottery were 9-1-1.

[baggins]

wow. i bet the state had to pay out bigtime for that one. i wonder how many people played 911 yesterday.

as far as odds:

1/1000 or 999-1 against 911 coming up.

coming up on that specific date, any year? 1/365000.

coming up 911 on 11.9.02, i'm not sure what our sample size is, so i don't know how to calculate this one. if it is in infinite time, then the probability is 100%.

baggins said:

"1/1000 or 999-1 against 911 coming up.

coming up on that specific date, any year? 1/365000."

Eh? The chances of 911 coming up on any specific date, any year, are 1/1000. The only way your number would make sense is if the state only did one lotto drawing a year.

coming up 911 on 11.9.02, the chances of this are also 1/1000

The day we pick doesn't matter. The chances of drawing 911 are always 1/1000.

Matt

[cardshark]

Help me understand with a simplified example, please. (I am not claiming one way or another, I am only asking a question).

If I you roll a die, the odds of rolling, let’s say “3”, are 1/6. How about if you are about to roll the die six times, and I make a bet that you will roll “3” on the third roll; are the odds still 1/6?

By the way, baggins, I do not know how the lottery works but I don’t think the state would lose any money if many people bet the same numbers. I think that the state takes a percentage off of the entire “pot” (whatever the proper lotto terminology may be) and the winnings get distributed amongst a larger number of winners. Thus each winner ends up with a smaller piece of the pie. But I may be wrong because I don’t really know much about lottery. (Except that it is a game of chance, in theory).

[Peter]

Hey Igor,

indeed, if you roll a die six times and you bet that on the third roll the outcome will be three the probability of this is 1/6. Look at it this way: the first roll may be anything, just as the second, fourth, fifth and sixth. But the third roll must be a three. The probability for this is: 1*1*1/6*1*1*1=1/6.

So chances for the lottery to get 911 as outcome on any given day is 1/1000 (or in correct notation 1*1*1*...*1*1/1000*1*...*1=1/1000 with the 1's and dots representing the string of 1's corresponding to the dates that don't matter and the 1/1000 corresponding to the date you chose.)

Peter

[cardshark]

Yes, Peter. I also gave the math some thought after I made my post. In fact member Mike Haven started another thread <FONT color=”red”>General Gambling>>Probability>>An interesting phenomenon</FONT> and continued on the subject, so I made my reply there.

Still, skeptic as I am, I still think it smells of a fix because of some other elements involved (that can’t be put into numbers). But that’s another discussion.

The fix was on.

[BruceZ]

...on eve of aniversary. No signs of collusion or price fixing.

[Ray Zee]

if you asked this before it happened its one chance out of 999 or whatever the total numbers they use. after the fact its a moot point as it is only talked about when it happens.
just like getting aces twice in a row. its the price of getting it only once. as you only look for it after the first pair come in. but you never even think about it till it happens, which it will at some point. it only matters if before the start of the hand you exclaim i am looking to get aces in both of the next two deals. same thing with this lottery thing.

[cardshark]

Third poster, Anonymous, says it all.