[LotsOfOuts69]

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a 104-93 win the for the home team has occurred 10 times in the last 5 years.

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If this is true, then this is 10 times in 6075 games, giving a probability of 1/607.5 chance for each individual game. You could estimate the chance that this score does NOT happen for the rest of the year (approx 950 games remaining) as (1-(1/607.5))^950 = 0.209 ==> 21%

Looks like you should be GETTING 4 to 1 odds, not giving, if you are the person who says this score will not occur. Seems crazy.

Of course, this is assuming alot of things about statistical data that I cannot confirm, but for a crazy prop like this it might be a jumping off point.

Edit: It's late and this seems like a horrible way of figuring out the probability, but if you could confirm that there is 607 to 1 chance that this score occurs in an individual game then there is an 80% chance it will occur AT LEAST ONCE in 950 games.

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Another way to say it, for it to be even odds, you would need the chance for it to occur in one particular game to be (1-(1/x))^950 = 0.5 ==> where x is about 1370.

So, if its 1370 to 1 for that exact score to occur in any random game then its 50% for it to happen at least once in the rest of the season.