After crunching some more numbers, it appears that the conjecture is actually true for any wager with no house edge...an example would be betting on a number of a single die that pays 5:1 on a win. I'm quite surprised it works for this scenario since I assumed the reason you double the probability of finishing the session at a loss was related to the symmetry of the standard normal distribution.

The conjecture does not work for any bet with a house edge, but it should be a close approximation for very low-edge bets such as passline with full odds or BJ with perfect strategy (edge under half a percent).

I believe there is a way to calculate the RoR for a bet with an edge but still searching for the method. Since the expectation is for the bankroll to continuously tend down by the wager*edge%, I assume calculus will be required

Quote:Ace2As shown previously, the RoR for a given bankroll is exactly double the probability of finishing the session at a loss, provided that the wager follows the standard normal distribution...an example would be a fair coin flip that pays even money. This hasn't actually been proven so it's been named "the Ace2 conjecture".

After crunching some more numbers, it appears that the conjecture is actually true for any wager with no house edge...an example would be betting on a number of a single die that pays 5:1 on a win. I'm quite surprised it works for this scenario since I assumed the reason you double the probability of finishing the session at a loss was related to the symmetry of the standard normal distribution.

The conjecture does not work for any bet with a house edge, but it should be a close approximation for very low-edge bets such as passline with full odds or BJ with perfect strategy (edge under half a percent).

I believe there is a way to calculate the RoR for a bet with an edge but still searching for the method. Since the expectation is for the bankroll to continuously tend down by the wager*edge%, I assume calculus will be required

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Agreed. As I discussed in this post : https://wizardofvegas.com/forum/questions-and-answers/math/34502-easy-math-puzzles/276/#post847780

mean= 0, variance = X

The question here is how to prove that the risk of bankrupt is DOUBLE* the risk of loss MORE THAN certain units ?

My guess is that when mean=0, and you allow to loss more than(for example, you borrow from your friends) your bankroll before the N rounds, you have 50%* of the chance to recover(partly or fully)from your losses and not bankrupt after end of N rounds. N must be a very large number.

* This explains that why the ROR is DOUBLE compare to player with unlimited bankroll.

If you don't happen to have that issue on your nightstand, he reproduced the article in his seminal work Blackjack Attack 3, in "Chapter 8: Risk of Ruin", pages 122-125.

Hope this helps!

Dog Hand

Quote:Ace2As shown previously, the RoR for a given bankroll is exactly double the probability of finishing the session at a loss, provided that the wager follows the standard normal distribution...an example would be a fair coin flip that pays even money. This hasn't actually been proven so it's been named "the Ace2 conjecture".

After crunching some more numbers, it appears that the conjecture is actually true for any wager with no house edge...an example would be betting on a number of a single die that pays 5:1 on a win. I'm quite surprised it works for this scenario since I assumed the reason you double the probability of finishing the session at a loss was related to the symmetry of the standard normal distribution.

The conjecture does not work for any bet with a house edge, but it should be a close approximation for very low-edge bets such as passline with full odds or BJ with perfect strategy (edge under half a percent).

I believe there is a way to calculate the RoR for a bet with an edge but still searching for the method. Since the expectation is for the bankroll to continuously tend down by the wager*edge%, I assume calculus will be required

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The ROR hypothesis along with the folly of the "Ace2 Conjecture" basically infers that people who gamble to whatever their end game might be, are mindless idiots who have no self-control or discipline. There is a very small number of those addicted that would fall into such a category, but the vast majority of patrons of casinos gamble as a diversion or recreation.

The testimony of an individual that gambled away his mortage money is sad but like driving a car while stupid and causing an accident, can only fall on the shoulders of the perpetrator. Analysis is probably a waste of time and energy.

tuttigym

Quote:tuttigym

The ROR hypothesis along with the folly of the "Ace2 Conjecture" basically infers that people who gamble to whatever their end game might be, are mindless idiots who have no self-control or discipline. There is a very small number of those addicted that would fall into such a category, but the vast majority of patrons of casinos gamble as a diversion or recreation.

The testimony of an individual that gambled away his mortage money is sad but like driving a car while stupid and causing an accident, can only fall on the shoulders of the perpetrator. Analysis is probably a waste of time and energy.

tuttigym

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If you keep playing long enough, you will lose all your money.

I doubt anybody is disagreeing with that.

As I understand it, the Ace2 Conjecture attempts to address the chances of going broke when you have a number of bet units of money, and wish to play more than that number of games.

If it ever gets figured out, it should be solvable for either "I have 40 money, how many times can I probably play?" or "I want to play 127 times, how many money do I probably need?"

That sort of information can be useful to some people.

Quote:Ace2You mentioned : ...... probability of finishing the session at a loss......

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I think it should be "probability of finishing the session at a loss MORE THAN a given bankroll".

For example, mean = 0, Var = 1, N = 10000, bankroll = 52.44 units

a) Probability of finishing the session at a loss = P( -52.44 < X < 0 ) = P(Z<0) - P(Z <-0.5244) = 0.5 - 0.3 = 0.2 = 20%

b) Probability of finishing the session at a loss MORE THAN bankroll = P( X < -52.44 ) = P(Z <-0.5244) = 0.3 = 30%

ROR for a given bankroll of 52.44 units = 2 x 30% = 60%, NOT 2 x 20% = 40%.

I am not sure, what do you think ?

Quote:tuttigym

The ROR hypothesis along with the folly of the "Ace2 Conjecture" basically infers that people who gamble to whatever their end game might be, are mindless idiots who have no self-control or discipline. There is a very small number of those addicted that would fall into such a category, but the vast majority of patrons of casinos gamble as a diversion or recreation.

The testimony of an individual that gambled away his mortage money is sad but like driving a car while stupid and causing an accident, can only fall on the shoulders of the perpetrator. Analysis is probably a waste of time and energy.

tuttigym

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RoR is a common metric used in investing and gambling...it doesn't refer to addiction or losing your house. Using several inputs, it measures the risk of losing a specified amount capital/bankroll and is probably the best tool for calculating the bankroll you'll need.

But, as usual, the self-proclaimed "deer-eyed, seasoned citizen" tuttigym just doesn't get it. What's worse, he dilutes a useful thread with his senseless ramblings.

Quote:Ace2Quote:tuttigym

The ROR hypothesis along with the folly of the "Ace2 Conjecture" basically infers that people who gamble to whatever their end game might be, are mindless idiots who have no self-control or discipline. There is a very small number of those addicted that would fall into such a category, but the vast majority of patrons of casinos gamble as a diversion or recreation.

The testimony of an individual that gambled away his mortage money is sad but like driving a car while stupid and causing an accident, can only fall on the shoulders of the perpetrator. Analysis is probably a waste of time and energy.

tuttigym

link to original post

RoR is a common metric used in investing and gambling...it doesn't refer to addiction or losing your house. Using several inputs, it measures the risk of losing a specified amount capital/bankroll and is probably the best tool for calculating the bankroll you'll need.

But, as usual, the self-proclaimed "deer-eyed, seasoned citizen" tuttigym just doesn't get it. What's worse, he dilutes a useful thread with his senseless ramblings.

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The useless part of your "conjecture" as well as the RoR hypothesis and "math" is that you always seem to forget to mention that the wagers and methods remain identical throughout the exercise which shows that the player must be guilty of GWS*. (Gambling While Stupid)

Have you personally experienced a session of RoR?

tuttigym

What does that even mean ?Quote:tuttigym[

Have you personally experienced a session of RoR?

tuttigym

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