Quote:puzzlenutAxiomOfChoice:

I should have qualified my statement to say that my calculation was for Don Johnson's rebate offer, which was 20% on a loss of $500,000 or more. My method of calculation is given in the first post of this thread. There have been many other rebate offers, including 100% rebate offers, and they must be analyzed according to their own conditions.

Ok, let's say that you play double-zero roulette. The house edge is higher than the threshold which you claim makes it impossible to beat the loss rebate. You play 1 spin, and put $100,000 bets on 5 different numbers. Then you walk away for the day (this is not the optimal strategy, but it's enough to show that your calculations are incorrect).

5 times out of 38, you win $3,100,000

33 times out of 38, you lose $500,000, and get a $100,000 rebate, so your loss is $400,000.

So every 38 spins, you are ahead $3,100,000 * 5 - $400,000 * 33 = $2,300,000. Your expected win is over $60,500 per day.

(If you complain that this is actually a $500,000 bet and not five $100,000 bets, note that you could just as easily play 1 number per spin, and quit after either 5 losses or 1 win. In this case your daily expected win would be about $2000 higher)

As long as you insist on ignoring variance, your calculations will never be correct. When playing loss rebates, variance is extremely important. The only reason that your blackjack calculations were close to correct was that the standard deviation of that game is close to 1. If you pick a game where that is not true, your calculations will no longer be close to correct.

Here are the results from the LRT for playing a number straight up on double zero roulette, $1000 wager, with a 20% loss rebate.Quote:AxiomOfChoiceOk, let's say that you play double-zero roulette.

Loss rebate %: 20.0%

Loss exit point: $37000

Win exit point: $34000

Expected win: $1756

Expected rounds: 37.9

Win percent: 49.3%

The LRT does not work well for highly skewed distributions, but it gives an idea.

Quote:AxelWolfExplain why you think this please. I think, I have to disagree, if your bet sizes are large enough. Loss rebates are far better.

A 20% loss rebate, would not be as good, only if you are making a lot of bets during a rebate session. 1 enormous bet with a 20% rebate would be a far better advantage then counting. Also the reaction from the casinos when your simply flat betting is going to be meet with open arms and encouragement. Counting will get you more heat then you can imagine.

My point that Counting is at least 3-4 times better than 20% Loss rebates (with no counting at HE) is from actual results.

I have played extensively loss rebates and kept accurate records for all sessions showing seperately the loss rebate amounts.

These were games that I both counted and got loss rebates

The % of Win amount that related to the loss rebate was around 25% of the total.

And I am talking about play in excess of 500 hours with loss rebates, so my figures are not from a small sample.

But just looking at pure math perspective.

Take the results for Optimum play as per Teliot for Don Johnson (use 1/100 of the amounts which is closer to average counters)

Average Fixed Bet $100

Loss exit point $2.600

Win Exit point $2.400

Expected Win per session $125.

Take the equivalent of above for Counting position

Spread Bet of $25 - $300 (of my head estimate as equvalent to $100 fixed)

Expected Win per hour around $75

5 hour session: 5 x $75 = $375.

That's about 3 times more for counting than for 20% loss rebates.

You might say 5 hours session is too long. My parts of the woods I would play that kind of sessions or ar least 3 hour sessions.

Even with 3 hour session the Counting would be double as good.

There was another comment from someone else about whether the loss rebate was on theoretical.

The loss rebates I got was always on actual losses.

Monte Carlo won't give a correct answer. It will just give a more accurate approximation than my result. Then again, a fair approximation in a matter of seconds seems pretty nice to me.Quote:RicardoEsteban^^ or run a Monte Carlo if you want an actual correct answer.

Quote:AxiomOfChoice

As long as you insist on ignoring variance, your calculations will never be correct. When playing loss rebates, variance is extremely important. The only reason that your blackjack calculations were close to correct was that the standard deviation of that game is close to 1. If you pick a game where that is not true, your calculations will no longer be close to correct.

Good advice! Can you calculate the optimal strategy for this game and rebate?

I calculate the variance of a single number bet in double zero roulette to be 32.237. Is this correct?

Quote:IbeatyouracesIn my book, he manipulated casino managers to give him an edge on the game. That makes him a cheater, not an AP. Same goes for Ivey.

LOOOOOL! Is a card counter also a cheater because he manipulates his bets according to the count to give him an edge??

I am studying your loss rebate theorem. For your probability function you say house edge

u<0. I thought we were playing games with a positive house edge.The traditional formula

works with any house edge except u = 0 in which case another form is used.

If you could quote this formula with sample numbers filled in and giving a credible result

I would be most grateful.

Shouldn't win(x,b,L) = bp(x,b) - x(1 - L)(1 - p(x,b))?

It looks like you have an extra -xp(x,b) in there.

Quote:RicardoEstebanIt will be correct to an arbitrary number of significant figures dependent on sample size. Your method spits out garbage when applied to asymmetric distributions.

Here are the results of a simulation of loss rebates for single-0 roulette, $1000 bet:

Here is what the LRT says in the case of a 20% loss rebate on a $1000 straight up bet on single-0 roulette:

Loss rebate %: 20.0%

Loss exit point: $73000

Win exit point: $68000

Expected win: $3511

Expected rounds: 145.7

Win percent: 49.0%

Perhaps you should clarify what you mean by "spits out garbage." These results are fairly close, in my opinion. The loss exit points, expected win and expected rounds are very close. The win exit point, as those who play against loss rebates know, has a small effect on win-rates across a wide range of values. The LRT performs very well in this case.

If you are going to attack my work in the future, academic honesty demands that you give actual results to back up your rhetoric. I've done the research to know these things. You just attack without any facts to back up what you are saying. I will not respond to your criticisms in the future unless you accompany your statements with facts.

I suggested you sit on your hands. You did not. I see that the moderator has forced you to sit on your hands for 3 days. When you return, take time to think before you post about my work. Don't attack me personally. And if you're going to make claims that contradict or demean my work, I have no objection, but back up your claims with some good work on your end. When the original poster puzzlenut made claims about my work, I clarified his claims then posted results to prove he was wrong. The burden is on you to do the same if you make claims about my work.

These results are from the Player's point of view, hence u < 0.Quote:puzzlenutTeliot:

I am studying your loss rebate theorem. For your probability function you say house edge

u<0. I thought we were playing games with a positive house edge.The traditional formula

works with any house edge except u = 0 in which case another form is used.

If you could quote this formula with sample numbers filled in and giving a credible result

I would be most grateful.

Shouldn't win(x,b,L) = bp(x,b) - x(1 - L)(1 - p(x,b))?

It looks like you have an extra -xp(x,b) in there.

In the LRT, both b > 0 and x > 0. b is the player bankroll when he stops. x is the initial bankroll. Therefore, the win amount is (b-x) if the Player wins, and the loss amount is (-x) if he loses.

With probability p(x,b), the player wins (b-x) units. With probability (1-p(x,b)), the player loses (-x) units. But he is refunded L%, so he really only loses (1-L)(1-p(x,b))*(-x). The EV is the sum of these terms.

I wrote a C-program that quickly converges to the values of b and x that maximizes win(x,b,L). I also have a spreadsheet on my blog that can be used, but you have to try different loss-quit points to find the value that maximizes win(x,b,L).

Quote:puzzlenutGood advice! Can you calculate the optimal strategy for this game and rebate?

Not without a simulation, which I would only take the time to write and run if this deal were offered to me and I had the bankroll :)

Quote:I calculate the variance of a single number bet in double zero roulette to be 32.237. Is this correct?

I get 33.208.

I notice that your number is just 35^2/38?

Quote:AxiomOfChoice

I get 33.208.

I notice that your number is just 35^2/38?

I tried it again two different ways and got two different numbers

I tried an online calculator and got another different number.

Quote:puzzlenutI tried it again two different ways and got two different numbers

I tried an online calculator and got another different number.

There are two ways to calculate it (they give the same result):

Var(X) = E[X^2] - (E[X])^2

and

Var(X) = E[(X-u)^2] where u is the mean (generally written as the letter mu)

(E[X])^2 = (-2/38)^2 = 0.00277

E[X^2] = (1 * 35^2 + 37 * (-1)^2) / 38 = (35^2 + 37) /38 = 33.2105.

So Var(X) = 33.2105 - 0.00277 = 33.208

Using the 1st method, Var(X) = [(35+2/38)^2 + 37*(-1+2/38)^2] / 38 = 32.334 + 0.874 = 33.208

Both are the same, as expected.

How are you calculating it?

38 outcomes, one is 36, the rest are -1.

(Ex

^{2}- (Ex)

^{2}/N)/N

(1296 + 37 - (-1)

^{2}/38)/38 = 35.07825485

The online calculator gets 35.078254847645

Quote:puzzlenutTeliot:

I see I did misquote your Don Johnson page No. 2 with respect to your loss stopping limit of

$500,000 and I have made a correction. I feel very comfortable with your page No. 2 but less

comfortable with your page No. 3 and your theorem.

I think that the variance or standard deviation of the game of blackjack is relevant only to

the question of whether or not either stopping limit is likely to be reached in a single

day. Please do not forget that the results of my calculation are quite close to the results

of your simulation on page 2. That should be comforting to both of us.

I have recaculated Don Johnson's strategy using a house edge of 0.29%. Using this,

p = 0.49855, q = 0.50145 and r = q/p = 1.005816869.

I'll now use your loss exit point of $2,600,000 and your win exit point of $2,400,000. For

these values n = 24. The probability of winning 24 units before losing 26 units is

w = (1 - r^26)/(1 - r^50) = 0.4838012753.

The player's expectation is

24W - 20.8(1 - w) = 0.8742971334 betting units, which is $87,429. This is better than before

but somewhat less than your calculated value of $124,999.

I'll now try to optimize the player's expectation with respect to both stopping points. I'll

call the win stopping point a for "ahead" and the lose stopping point b for "behind."

For a win stopping point of 15 the probability of winning 15 units before losing b units is

w = (1 - r^b)/(1 - r^(15 + b))

and the player's expectation is

E = 15w - 0.8b(1 - w)

which has a maximum at b = 20, so E(15,20) = 0.9406394916

I'm doing this by differentiating E with respect to b, setting the result equal to zero and

solving for b to the nearest integer.

Using b = 20, I'll optimize for a

w = (1 - r^20)/(1 - r^(20 + a)

E = aw - 16(1 - w) and the optimal value for a is 19 and E(19,20) = 0.9598383989

Optimizing for b again using a = 19

w = (1 - r^b)/(1 - r^(19 + b))

E = 19w - 0.8b(1 - w)

and the optimal value of b turns out to be 20, so our optimal stopping points are

+$1,900,000, -$2,000,000 and the expected win is $95,983.

Don has said that one of the casinos at which he has won would be happy to have him back but

would give him a 20% rebate only on losses exceeding $2,000,000. I think we can safely

advise him not to be deterred.

Not to diminish your work, impressive…….but I think the dude won 15 million because he has a huge set of balls…….that is the essence of gambling.

Quote:puzzlenutAxiomOfChoice:

38 outcomes, one is 36, the rest are -1.

(Ex^{2}- (Ex)^{2}/N)/N

(1296 + 37 - (-1)^{2}/38)/38 = 35.07825485

The online calculator gets 35.078254847645

The winning outcome is 35, not 36.

If you want to count it as 36, then you need to count the losing outcomes as 0. It's either on a "for 1" or a "to 1" basis, but you can't mix them.

http://apheat.net/2014/02/13/the-second-loss-rebate-theorem/

Here is a spread sheet that does everything. It is currently set to "Don Johnson."

http://wwwdotapheatdotnet.files.wordpress.com/2014/02/loss_rebate_theorem_full_solution.xlsx

Here is the Second Loss Rebate Theorem:

After I saw teliot equations (on the first page of this thread), I though exactly the same thing that the maximum of the function can be found by taking the 2 partial derivatives and then solving the simultaneous equations for them at zero.

My calculus being quite rusty, I used Wolfram to do the partial derivatives.

For the solving of the 2 simultaneous equations I could not make Wolfram do it and tried manually myself but failed.

I tried the special case where x=2b (ie the win and loss points are the same) and came with an equation that gives similar results.

But the equation that I came up is more complicated in any way that the Elegant one that Teliot just put on his website.

I have been looking for such equations that Eliot put for a very long time.

As far as I know these have never been published on the Internet or in any book (I do not know about beyond counting as I do not have it)

Johnson had a large edge because of the rebate. He was given a 20% rebate on any losses over

$500,000.

In order to calculate the effect of a rebate on losses you have to know how to calculate

house edge. The common definition of house edge is the ratio of the average loss to the

initial bet. If the game has a house edge of 0.0029 and management gives Don a 20% discount

on losses they probably think they are merely reducing the average loss and the house edge

by 20% and the game still is profitable for them. I disagree.

Another definition of the house edge is the probability of losing times the amount lost

minus the probability of winning times the amount won. For a unit bet this is the expected

loss, so this definition is consistent with the first one.

qL - pW = e

Since you either win or you lose, p + q = 1.

We simplify this by replacing p with 1 - q and solve for q

q = (e + W)/(L + W)

To take into the account the loss rebate We replace L by 0.8L and associate a new house

edge, f, with it.

0.8qL - pW = f

We replace p With 1 - q

f = 0.8Lq - W(1 - q)

Substitute (e + W)/(L + W) for q

f = (e(4L + 5W) - LW)/5(L + W)

substitute 0.0029 for e, substitute 1 for L and substitute the average win in blackjack for

W. I'm not sure what that is, but if it is 1.1 the house edge will now be -0.10213. In other

words Don Johnson will have an edge over the house of 10.2% because of the rebate. I am

confident that the algebra is correct because I used a computer algebra system.

From this calculation there is no obvious stopping point except that he should not stop with

a loss unless it is large enough to qualify for the rebate. He had to bet the limit in order

to make that possible.

Let's reason it out using nothing but intuition. Since the house edge is so small and when

you win you usually win the amount of your bet, blackjack is very close to being a coin

tossing game. In a coin tossing game half of the days you will win and half of the days you

will lose and on average you will win and lose the same amount, but if on the days you lose

you get a 20% rebate, your average gain will be 10% per day of the average amount won or lost.

As regards your post, without considering the variance, you will never get an accurate result. I think you should read my work carefully. You can try the spread sheet given in the Second Loss Rebate Theorem and compare it to your work. In particular, I derive the edge for Don Johnson.

Here is LRT3:

http://apheat.net/2014/02/16/the-third-loss-rebate-theorem/

Quote:RicardoEstebanLOL, your Gaussian garbage hasn't solved anything.

This seems overly harsh. I'd be hesitant to trust this formula for a game with exceedingly rare large payouts (eg, slots or video poker) or for situation where the number of hands to play was very small, but, I'd expect it to be pretty accurate if we are talking about hundreds of hands of blackjack or spins of roulette (playing even-money bets)

Quantifying an upper bound on the error would be useful, though. Any takers?

As for the criticism above, Teliot has already said that these formulas only apply where the game approximates a normal distribution.

For games with skewed distributions these will not apply as accurately.

Many Casino games with high variance, the distributions are skewed , for example poker variants and video poker.

ie player loses 1 unit or 2-3 units in games with raise decisions and can win upto 100 units or more.

But for casinos games with low variance (ie BJ, baccarat) any skewness would be small and the formulas work fine.

The other practical issue that the formulas do not solve is the playtime decision when the playtime figure becomes big.

In the Don Johnson figures analysis the expected playtime is 479 bets, ie around 5 hours (say 100 bets per hour)

If you decrease the HE in the formulas say to -0,1% the expected playtime becomes 4040 bets, ie around 40 hours.

Obviously, practically this is not optimum.

I assume that the rebate is given per day (24 hours period), so even if you could play 40 hours, the rules of the rebate mean that you will be better to stop at some time before the 24 hour limit.

If you also put a practical limit of say 8 hours play per day (so as not to get exhausted and start making mistakes) then:

the Maximum Playtime becomes an extra Variable.

So the next Step to develop these formulas is by also having as additional variable the maximum Playtime.

The formulas already calculate Average Expected Playtime, so it might be possible for these formulas to be developed by this extra variable.

And additional the maximum playtime variable can be used as dynamic variable in the sence that you can have different Win/Loss figures as play time progresses.

For example say that we put as maximum playtime 800 bets (8 hours) and that the existing calculation for Don Johson does not change (or changes very little)

Say after he plays 4 hours his position is at zero.

Since his expected playtime at this point is for another 5 hours but he can play maximum another 4 hours, then his optimum win/loss exit will be different than these figures, ie they will reduce from 26 units lost and 24 units won.

As he approaches the maximum time limit of 8 hours these will be reduced even further in such a way that he would rarely play the full 8 hours.

Practically such calculations can be made for 1 hour (or 0.5 hour intervals).

And the other issue that such formulas can solve is when the EV is positive for the player.

Telio's formulas do not work for positive EV, for the simple reason that in such case the answer is infinity.

But formulas that also put maximum playtime as variable might also work for postive Evs.

Someone playing a positive Ev game + rebates has two sources of win: the positive Ev and the Rebate.

As his day win/loss moves further away from zero then the value of the rebate reduces.

Given a maximum playtime per day there is a win/loss exit point that maximizes day wins.

ie if he continues to play upto the maximum playtime the extra Ev that he will get (from the positive Ev) is less than the value of rebate (if he is losing at that time) or opportunity cost of he is winning.

you are an idiot. if said casino isn't intelligent enough to do the math, well then they get crushed. your stance on this is ludicrous.Quote:IbeatyouracesIn my book, he manipulated casino managers to give him an edge on the game. That makes him a cheater, not an AP. Same goes for Ivey.

Quote:bdc42you are an idiot. if said casino isn't intelligent enough to do the math, well then they get crushed. your stance on this is ludicrous.

Easy big fella. This was the first ibya post that I didn't agree with but the guy is obviously not an idiot.

I am used to being followed around by people who have nothing more to contribute than saying mean-spirited things about my work. Go for it. But it appears IOTTCO that your agenda is to do exactly this and little more with your new account here on WoV.Quote:RicardoEstebanLOL, your Gaussian garbage hasn't solved anything.

Thank you for your excellent suggestions for further pursuit. I'll look at the playtime boundary condition question (and the +EV question) the next time I get inspired. Good stuff.Quote:AceTwo... So the next Step to develop these formulas is by also having as additional variable the maximum Playtime ...

Both Teliot's and the classical gambler's ruin formulas give the probability of reaching one goal or another in an unlimited number of plays and thus do not necessarily give useful results where there is a practical limit on the number of plays due to the facts that the player must rest and that the loss rebate offer expires every midnight.

The number of hands that optimizes the value of the game can be calculated in the following way:

Without the loss rebate the value of the game is -e*n, where e is the house edge and n is the number of hands played. That could be offset by a loss rebate of L which kicks in at a loss threshold of t chips.

In Don's case the loss threshold was 5 chips. If the threshold is an odd number it can be reached exactly only if n is an odd number and if it is an even number it can be reached exactly only if n also is an even number, so I shall assume n is odd and then the possible qualifying losses are 5, 7, 9, etc. If we make the condition that n and t are both even or both odd, we can use the expression (n - t)/2 to calculate the number of wins needed to realize a net loss of t. If I wanted to include the possibility that one number was odd and the other even I would use FLOOR((n - t)/2).

The number of wins needed to generate qualifying losses will be 0 to (n - t)/2 and the amounts of these losses will range from t to n, n corresponding to 0 wins. We multiply each of these losses by their respective probabilities and sum them over this range and multiply by L to get the rebate.

R = L*∑(BINOMIAL_DENSITY(k, n, p)·(n - 2·k), k, 0, (n - t)/2)

The value of the game, then, with the rebate is

V = R - e*n

Here are a few values calculated using L = 0.2, t = 5, e = 0.00263636 and p = 0.4986818

This is Teliot's house edge of 0.0029 adjusted for a variance of 1.1 for the game of blackjack.

n V

101 0.5028

201 0.6120

251 0.6309

301 0.6361

351 0.6310

401 0.6177

501 0.5727

601 0.5089

It appears that the optimal number of daily hands for this game is around 300.

The whole show is worth a view, but the game rules are discussed at minute 36.

Secrets of a Vegas Whale

Quote:bdc42you are an idiot. if said casino isn't intelligent enough to do the math, well then they get crushed. your stance on this is ludicrous.

Once a sleepy-eyed administrator wakes up, you may be not be sharing your opinion with us for a few days.

Many casino's do loss rebates for Hi rollers, in the same way casino's do 2-1 bj promos.So based on your thinking if you play a 2-1 promo you are cheating. Or a 5x cash back on VP.If the management sets the rules how can it be cheating?Quote:IbeatyouracesIn my book, he manipulated casino managers to give him an edge on the game. That makes him a cheater, not an AP. Same goes for Ivey.

Quote:BuzzardOnce a sleepy-eyed administrator wakes up, you may be not be sharing your opinion with us for a few days.

Despite no announcement, his suspension came and went a week ago.

Quote:Mikey75Did any of you actually watch the video about Don Johnson? He states in it that one of the rules he got the casinos to change for him was for them to stand on soft 17. Just about everywhere I have played stands on soft 17!! It's actually more advantageous to hit a soft 17. Did anyone else catch that?

Huh?

S17 is good for the player to the tune of about 0.2%

Quote:IbeatyouracesAnd the fact that some high roller had to negotiate to get a S17 game? Something doesn't jive with his story.

I have heard that the AC rules are pretty bad. In Vegas S17 is the norm for high-limit games, but if the casino he wanted to play at didn't normally offer S17 (even at high limits), then it makes sense that he would ask for it.

Also, there might be a psychological aspect to it. By spending time negotiating about rules, it might distract them from the loss rebate, which was the real money-maker. With the deal that he got, he did not need good rules. He could have made money in a 6:5 game.

blackjack and baccarat and that the house may safely offer this rebate for slots or roulette

without fear of being taken advantage of...."

I am not agreed with this statement. Loss rebate is more useful for Roulette or any game with reasonable edge(<10%) and high variance !

Quote:BuzzardOnce a sleepy-eyed administrator wakes up, you may be not be sharing your opinion with us for a few days.

bdc42 already did his time for this quote, Buzz...if you're going to critique the moderating, you'll have to keep up better than this.

Quote:teliotHere are the results from the LRT for playing a number straight up on double zero roulette, $1000 wager, with a 20% loss rebate.

Loss rebate %: 20.0%

Loss exit point: $37000

Win exit point: $34000

Expected win: $1756

Expected rounds: 37.9

Win percent: 49.3%

The LRT does not work well for highly skewed distributions, but it gives an idea.

As long as your formula can take consideration of full payoff table of the game, then you are not worry about the highly skewed distributions. I think ONLY variance without full payoff is not good enough. Simulation will give very good results.

In any case, any one who is a legitimate AP is going to be some technique to minimize or play with an advantage before the addition of the rebate. In Johnson's case it is my understanding he was using confederates to eat cards. You can't just ignore something like that to make your formula fit.

Smart whales shoving out money and playing short sessions, has been going on forever. Baccarat revenue from the whole strip is often negative because of it.

The difference with Don Johnson is that he decided to tell the casinos exactly what he was doing. Not a smart move.

Quote:GBV

The difference with Don Johnson is that he decided to tell the casinos exactly what he was doing. Not a smart move.

Must have missed that. What was he doing? In the Bloomberg story there is no mention of "counting" or any other sort of advantage. As I also mentioned, calling the 20percent a "rebate" is incorrect. That is a discount on a paid loss. Huge difference, maybe not in the "math" we have seen but in reality for sure.

Quote:NokTangMust have missed that. What was he doing? In the Bloomberg story there is no mention of "counting" or any other sort of advantage. As I also mentioned, calling the 20percent a "rebate" is incorrect. That is a discount on a paid loss. Huge difference, maybe not in the "math" we have seen but in reality for sure.

I don't recall where I heard it (maybe on the GWAE radio show? or maybe or bj21.com), but I heard the same thing as GBV, that he was using confederates to eat cards in low counts and he was mostly only playing positive counts. So he was at an advantage even before the loss rebate came into play.

Obviously I was not there and this could have been a story someone started. But it is plausible.

Quote:beachbumbabsbdc42 already did his time for this quote, Buzz...if you're going to critique the moderating, you'll have to keep up better than this.

Deleted.

Quote:AcesAndEightsI don't recall where I heard it (maybe on the GWAE radio show? or maybe or bj21.com), but I heard the same thing as GBV, that he was using confederates to eat cards in low counts and he was mostly only playing positive counts. So he was at an advantage even before the loss rebate came into play.

Obviously I was not there and this could have been a story someone started. But it is plausible.

Seems odd, with those stakes, they would continue to allow him to play at various casinos if counting with helpers was involved. I think the Wynn story recently is most interesting. Not the disco incident but the fact he was playing at the Wynn "recently" in Las Vegas at high stakes.

Quote:beachbumbabsbdc42 already did his time for this quote, Buzz...if you're going to critique the moderating, you'll have to keep up better than this.

you are quoting me a lot. i am wondering what they said?

Quote:bdc42you are quoting me a lot. i am wondering what they said?

Here I am quoting you again.lol... On the 17th of Feb, you called someone something, I can't recall what it was, but it was a personal insult at another member, which is not allowed. So you were banned for 3 days (you may not have known that you were banned if you didn't try and post during that time). Anyway, because I didn't make a big issue out of banning you, the guys missed that you got popped for it, and were coming across your insult days later, and trying to push me into banning you. Which I had already done. Got it? lol..

btw: if you haven't, you might want to read the forum rules here, and if you haven't noticed it, you're in good company on the ban list here. The point where you made the insult is hyperlinked next to your name, and the other comments would follow from that.