No it's not, provided you have deep enough pockets and the other party always accepts subsequent "double or nothing" bets. Given the those conditions, double or nothing is actually a "cannot lose" gamble (also provided you have a non-zero probability of actually winning each time). This is the primary reason why there are max betting limits on blackjack and craps tables.That's also why the old, "double or nothing," bet is a loser's gamble.
Over the years I have observed that very few people have a clear concept on probability, and especially so when it relates to pool.
I want to give you a mathematical problem. You can easily solve it with a calculator, but just take a guess first before calculating.
Here is the scenario.....
The game is 9-ball. The player has made 1 ball on the break, so there are 8 balls remaining.
For each of the remaining shots, this player has a 90% chance of pocketing the ball, and a 90% chance of getting position on the next ball.
For the 9-ball, he has a 90% chance of pocketing it, and a 90% chance of avoiding a scratch.
What is the mathematical probability that he runs out?
90%?
I don't know how to solve it easily with a calculator. But if the player has a 90% chance of getting shape and the resulting shot is also 90% then I say getting out is 90% certain.
The way I read it, pocketing the ball doesn't imply automatic position, which is why the position probability was explicitly stated. I'm sticking with .9^16 (or .81^8 if you prefer). :smile:
I do agree that a good player will get out more often than .185 if there is a 90% probability of obtaining position on each ball, which implies that the balls are open, but I attribute that to the fact that a good player will make more than 90% of shots that he/she has ideal position on. If we give them a 95% chance of making each shot they have ideal position on, which I think is more realistic for a good player, their odds of getting out jump up to about 30%, which I also think is more realistic.
Aaron
The way I read it, pocketing the ball doesn't imply automatic position, which is why the position probability was explicitly stated. I'm sticking with .9^16 (or .81^8 if you prefer). :smile:
I do agree that a good player will get out more often than .185 if there is a 90% probability of obtaining position on each ball, which implies that the balls are open, but I attribute that to the fact that a good player will make more than 90% of shots that he/she has ideal position on. If we give them a 95% chance of making each shot they have ideal position on, which I think is more realistic for a good player, their odds of getting out jump up to about 30%, which I also think is more realistic.
Aaron
mohrt, work on your decimal to percentage conversions. :thumbup:
No it's not, provided you have deep enough pockets and the other party always accepts subsequent "double or nothing" bets. Given the those conditions, double or nothing is actually a "cannot lose" gamble (also provided you have a non-zero probability of actually winning each time). This is the primary reason why there are max betting limits on blackjack and craps tables.
But anyway, that's besides the point of this thread.![]()
95% to make each ball and 90% to get position: 0.285580742 or 28.6%
95% to make each ball and 95% to get position: 0.440126669 or 44.0%