Ganchrow, how about using Kelly to wager on Phoenix Suns Over Totals?

[JDK192]

Odds of having a streak of 9 or more games during an 82 game stretch is approx 28.5%. For 9 games exactly you must start with a win 9 straight losses and end with a win, so that reduces the odds to a quarter of 28.5% (note that this adjustment is approximate since when the win streak starts on game 1 or ends on game 82 you can’t have a loss on both sides…but that’s a small effect).

[Korchnoi]

Odds of having a streak of 9 or more games during an 82 game stretch is approx 28.5%. For 9 games exactly you must start with a win 9 straight losses and end with a win, so that reduces the odds to a quarter of 28.5% (note that this adjustment is approximate since when the win streak starts on game 1 or ends on game 82 you can’t have a loss on both sides…but that’s a small effect).

makes sense. where did you get 28.5% from?

[JDK192]

Korch

My last post was kinda confusing. The odds of having a losing streak of 9 in the 82 game season isn't 28%, it's around 7%. The 28.5% comes by taking 1/2 to the power of 9 times (83-9), then 1/2 to the power of 10 times (83-10), etc... to 1/2 to the power of 82 times (83-82). The sum of that equals approx 28.5%

[VegasDave]

Actually there must be a million people that have come up with this theory. It just keeps coming up over and over.

Fortunately there is a guy like ganchrow who has the patience to explain the flaws. Horrible people like me who have been around for awhile might take one brief shot to tell you it doesn't work and if you still argue, I'd just roll my eyes and say, "Sure dude, go for it."

I'm sure there must be... it is awfully hard to come up with an original thought.

Too bad I'm majoring in sports broadcasting/psychology instead of philosophy/math, then I'd be much better suited for this topic!

I agree, Ganchrow is the man. Not that I was going to actually try it...

[Arilou]

Here's how I came up with it; it took me about a minute. I opened up a spreadsheet. Consider column A a current streak of 0 losses (with no streak of 9 losses), column B a current streak of 1, and so on. Column I is at least one string of 9 losses.

Consider row 1 to be the situation after 0 games, row 2 to be the situation after 1 game, etc.

Obviously the situation after 0 games is that you have 0 losses, so that gets the numbers 1-0-0-0-0-0-0-0-0-0. For any subsequent row (game), the nubmers are easy: The 0 column is (1-I1)/2, since half the time that you hadn't already lost your bankroll you win this game. The 1-8 columns are .5*A1, .5*B1, etc: You had lost N-1 times, and you lost again. The final column is .5*H1+I1, the chance you just lost the 9th time and the chance you'd already lost your roll. Now you can copy/paste that 2nd row down to the 83rd, and presto. You have your answer. This may not be as mathematically elegant as finding it the "right" way, and back when I was on the Math Team I would have had an issue with that, but it's almost always faster to use this method than to set up an equation.

[RickySteve]

For starters, the odds of losing 9 in a row is 511 to 1 at any given starting point. The odds of losing 9 in a row at any time in the trial is, of course, much higher.

FYP

[Ganchrow]

Here's how I came up with it; it took me about a minute. I opened up a spreadsheet. Consider column A a current streak of 0 losses (with no streak of 9 losses), column B a current streak of 1, and so on. Column I is at least one string of 9 losses.

Consider row 1 to be the situation after 0 games, row 2 to be the situation after 1 game, etc.

Obviously the situation after 0 games is that you have 0 losses, so that gets the numbers 1-0-0-0-0-0-0-0-0-0. For any subsequent row (game), the nubmers are easy: The 0 column is (1-I1)/2, since half the time that you hadn't already lost your bankroll you win this game. The 1-8 columns are .5*A1, .5*B1, etc: You had lost N-1 times, and you lost again. The final column is .5*H1+I1, the chance you just lost the 9th time and the chance you'd already lost your roll. Now you can copy/paste that 2nd row down to the 83rd, and presto. You have your answer. This may not be as mathematically elegant as finding it the "right" way, and back when I was on the Math Team I would have had an issue with that, but it's almost always faster to use this method than to set up an equation.

Yep that'd work just fine. (Except you probably made a computational error as the answer should be 7.116%.)

But very nicely done. 10 points for you as promised.

[Ganchrow]

Here's how I would have done it:
<hr>
We figure out the probability of the 9^th consecutive loss first appearing at each game from 1 through 82.

Games 1-8:
The probability of the 9^th consecutive loss first occurring at games 1-8 is obviously zero.

Game 9:
The probability of the 9^th consecutive loss first occurring at game 9 is .5^9 = 1/512.

Games 10-17:
The probability of the 9^th consecutive loss first occurring at games
10-17 is .5^10 = 1/1024. (We know that the last 10 games of the sequence would have to be 1 win followed by 9 losses -- the previous games don't matter).

Game 18:
The probability of the 9^th consecutive loss first occurring at game 18 is .5^10 * (1 - Pr(9 in a row by game 9)) = 1/1024 * (1-1/512). (We know that the last 10 games of the sequence would have to have been 1 win followed by 9 losses. The previous 8 games before that don't matter.

Game 19:
The probability of the 9^th consecutive loss first occurring at game 19 is .5^10 * (1 - Pr(9 in a row by game 10)) = 1/1024 * (1-1/1024). (We know that the last 10 games of the sequence would have to have been 1 win followed by 9 losses. The previous 9 games before that must NOT have contained a 9 consecutive game loss.

Game 20:
The probability of the 9^th consecutive loss first occurring at game 20 is .5^10 * (1 - Pr(9 in a row by game 11)) = 1/1024 * (1-1/1024). (We know that the last 10 games of the sequence would have to have been 1 win followed by 9 losses. The previous 10 games before that must NOT have contained a 9 consecutive game loss.

Game m ≥ 18:
The probability of the 9^th consecutive loss first occurring at game m (m ≥ 18) is .5^10 * (1 - Pr(9 in a row by game m - 9)) = 1/1024 * (1-1/1024). (We know that the last 10 games of the sequence would have to have been 1 win followed by 9 losses. The previous (m - 10) games before that must NOT have contained a 9 consecutive game loss.

So for 82 games we just take the summation of the first 82 terms (74 not counting the first 8 zeros).

This gives us: 7.11645651718925%, which I should note is the same answer you'd get using Arilou's rather nifty method above.

<hr>
So in general, the probability of at least one stretch of at least L consecutive losses of an event (that occurs with probability 1-q), over n trials is given (recursively) by:

Pr(L; n) = Pr(at least L Losses over n trials)
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= 0 if n < L
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= q^L if n = L
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= Pr(L; n-1) + q^L×(1-q)×(1-Pr(L; n-L)) if n > L