Recent content by JJR

Wolf, the math you're doing is for winning 3 in a row. Losing 3 in a row would be (.51)^3 = .132651 1/.132651 is approximately a 1 in 7.5 chance of losing 3 in a row. 6.5 x 5 = 32.5 vs. $35 loss is a net loss of $2.5 Sorry, but you would need an advantage over the house for the...

What are the chances of getting to break even or above break even? My guess is 50/50. I understand getting exactly equal numbers of heads and tails is uncommon, but what I really meant was at or above break even. I'm sure for a binomial distribution of a fair game of 50/50 half of the numbers...

Carry it out a few trillion more trials. Eventually it has to break even or the game is biased. It has to, or the true odds for this biased game is 49.999%.

The "given" is simply that the game is 50/50. That's the known. That the game will have equal amounts of wins and losses. The game is not a 49.999% game or a 50.0001% game it is known that the game is 50/50. That's a given. It's inherent that it will eventually have equal amounts of wins...

Well, it will obviously be uncommon to have exactly equal amounts of wins and losses in a million trials. You will most likely be above or below that line during most of the trials. But, "eventually" you will, of course, at some point have equal amounts of wins and losses. That's inherent...

Well, I do apologize. I did have a misunderstanding of the Gamblers Fallacy. I thought it only applied to the next trial in a series. That after losing 4 in a row you're chances of winning the next trial are still just 50/50. I didn't know they were saying a 50/50 game of chance won't have...

I don't understand why there's an argument that a 50/50 game will break even eventually. That's what it's suppose to do, unless it's a biased game. The point that you could be down 100 thousand, well you could be up 100 thousand the entire time also, it's a 50/50 game. But, at some time it...

Well, actually the D'alembert assumes that a 50/50 game such as a coin toss, will result in equal amounts of wins and losses, heads vs tails. 500 million heads vs 500 million tails. You're suggesting that a 50/50 game will result in a loss of 100 thousand at some point in a billion trials. I...

There's just one problem with what you stated. The D'alembert with a straight 50/50 game of chance would actually win. By your logic, the D'Alembert should just break even on a 50/50 game of chance, since every increased bet is just subject to the same 50% chance of winning. The end result...

LOL Good advice. Well, if anyone out there gets curious and puts the system through a simulation let me know the results. My email address is [email protected].

I don't know, I'm still getting good results with this system. Has anyone else tried playing it yet?

Well, I think the D'alembert Variation I've developed would by itself keep the game about even. As I said before, the D'alembert would win if you could get it back to the original 1 unit bet. The HA makes it at some point impossible to return to the original 1 unit bet, but the D'Alembert is a...

Well by "scalps" I mean the instances where you subtract 2 units. The D'alembert, of course, would work if you could get the betting all the way back down to 1 unit. The problem with the D'alembert is that the HA causes the betting to get out of control and you at some point will not be able...

The biggest difference with this system is that it keeps the betting down, by subtracting the 2 units after 2 wins in a streak, you're taking about 10 "scalps" per 100 hands. So, the betting can't really get out of control on you. The only concern I have is how much are the "scalps" going to...

Has anyone tried this system yet? What do you think if you have?