Increase of Place Bets - Craps

I am a place better. What is a good progression for increasing your place bets to take advantage as a roll gets longer?

I'm a place bettor too.

Typically, I start out only placing the 6 & 8. I'd bet the minimum, let it hit a few times, then, once I'm profitable, start pressing it one unit at a time, almost every time they hit. When they get to $24 or $30, I might play the 5 & 9 rather than press, and get them going the same way. I also pause when they get to $42. There's something sexy about dropping a white and getting two greens back!


I've been toying with a different launch idea. My concept revolves around the belief / hope that a typical player will hit the other numbers once before a seven-out.

Bet more than the minimum, then DROP to the minimum after he hits, THEN start pressing if he hits it again. It brings you into profit sooner, so you can start pressing sooner. I.E. Bet $18 each on the 6 & 8 When one hits, drop them both to $6, and you're already ahead $9.

Yeah, I know often a player establishes a point and hits the 7 on the next roll, but, that's the concept.

It is not my style to regress. I will stop at a level for one hit to collect, but never decrease the bet. An example is 6 or 8 place 18 - 30 -42 - 42 - 60 - 60 -90 -90 etc..

5 or 9 place 15 -20-25 - 35 - 50 - 50 -75 etc.

Quote: docsjs

I am a place better. What is a good progression for increasing your place bets to take advantage as a roll gets longer?

What's "good"? When you progress bets, you risk losing back what you've already won, in return for a chance to win more if you win again. The same principle applies after every win. Do I want to risk what I've won in order to win more if I win again? Progressive betting results in more losing sessions, but bigger wins, compared to flat betting. Why? Because you have to win bets in a row. In flat betting, the only thing that determines your net win/loss is how many bets you win, i.e. your win-loss record; the order of the wins and losses doesn't matter. When you progress your bets, however, a WLWLWLWLWL pattern causes you to lose instead of breaking even (or winning a bit if you're getting 7 to 6, like place 6/8. The more you progress your bets, the more dependent you become on streaks of wins. Take a pure parlay, for example; if you keep doubling your bet, then you always lose the whole amount when you lose, until and unless you get up to the table maximum and cannot double your bet. Of course, nobody bets that way, but the principle is the same, just a difference of degree.

The more you progress your bets, the greater the "skew" of the results becomes, because you have fewer, but bigger wins. Try doing something like this:

flat $6 bet on 6
win two, then up one unit
up one unit every time
$18 6, then regress to $6, win two then up one

Figure out the net result for different numbers of wins, from zero to five or so, before losing the next bet. See which method does best at each number of wins. Then figure out the probability of each number of wins, because that's an important factor.

wins flat 2 then press press regress
0 -6 -6 -6 -18
1 +1 +1 -5 +15
2 +8 +2 +3 +22
etc., etc.

Probability of losing first bet .5454, probability of WL = .4545 * .5454 = .2479, etc., etc.

The more wins in a row before losing, the better an aggressive strategy will do, but the lower the probability of that happening. Overall and in the long run, the more you bet the more you are expected to lose, and the expected value is always edge * action, but the distributions of your results will be quite different.

What is "good" boils down to how much you want to win and how much you are willing to risk to win it.
Cheers,
Alan Shank

I typically start with $18 on 6 & 8, then progress to 24, 30, 42, 60, 90, 120, 150, 180, 210... I limit the increments to $30 max to recap max profits when the table it hot.

For the other numbers it's typically 15, 25, 35, 50, 75, 100... max increment $25.

I know it's agressive... but I'm playing for the opportunities where the shooter rolls for 30 minutes or more. And often times when they roll that long, there's one number they hit over and over again, so it really pays off then. There's nothing better than getting black chips on ever roll of the dice...

Quote: Phosphorous

I typically start with $18 on 6 & 8, then progress to 24, 30, 42, 60, 90, 120, 150, 180, 210... I limit the increments to $30 max to recap max profits when the table it hot.

For the other numbers it's typically 15, 25, 35, 50, 75, 100... max increment $25.

I know it's agressive... but I'm playing for the opportunities where the shooter rolls for 30 minutes or more. And often times when they roll that long, there's one number they hit over and over again, so it really pays off then. There's nothing better than getting black chips on ever roll of the dice...

To illustrate the "continuum of progressiveness", so to speak, I wrote a computer program to compare some different betting strategies for placing the 6. We will call them "the four guys". Everyone starts at $18. Guy #1 then regresses twice before going back up, then regresses again. Guy #2 increases his bets a unit at a time, but only every other bet. Guy #3 progresses a unit at a time every bet. Guy #4 doubles first and then adds $18 each bet. So, their bet series are:
Guy #1 18,12,6,12,18,24,18,12,6,12,18
Guy #2 18,18,24,24,30,30,36,36,42,42,48
Guy #3 18,24,30,36,42,48,54,60,66,72,78
Guy #4 18,36,54,72,90,108,126,144,162,180,198

My program examines the net outcome for each of these "Guys" after different numbers of hits, always assuming that the next bet is lost. So, they all lose $18 if they lose the first bet. If they win the first, then lose, only Guy #1 has a profit, since he locked up his initial win. Here is a chart of all four Guys' net results for zero through 10 hits, along with the probabilities of each number of hits:

num hits	guy 1	guy 2	guy 3	guy 4	prob
0	-18	-18	-18	-18	0.545455
1	9	3	-3	-15	0.247934
2	29	18	19	9	0.112697
3	30	46	48	54	0.051226
4	38	68	84	120	0.023285
5	53	103	127	207	0.010584
6	87	132	177	315	0.004811
7	114	174	234	444	0.002187
8	134	210	298	594	0.000994
9	135	259	369	765	0.000452
10	143	302	447	957	0.000205

"Mr. Conservative" is still ahead after two hits, but then falls behind fast. It only takes Guy #4 three hits to take the overall lead, but he doesn't net a profit until two hits, so he's losing money almost 80% of the time.

The program also figured the weighted net outcome for each guy for zero through ten hits:

Guy 0 weighted net is -0.473997
Guy 1 weighted net is -0.674662
Guy 2 weighted net is -0.835687
Guy 3 weighted net is -1.349305

This simply reflects the fact that the expectation is always edge * action; it also favors the lower bettors because it does not consider streaks longer than ten, which can occur.

Within the next few days, I will also run a few thousand two-hour sessions for each "Guy" and examine the differences in variance among them.
Cheers,
Alan Shank

I like to try and minimize against the early 7 out, but get to a decent bet size as fast as possible.

Here is one scheme I have used. Assume the point is 6.
I will bet $80 inside ($25 on 5&9, $30 on 8).
If one of the numbers hits, I put $45 down and say press all the insides.
I now have $50--point--$60--$50 (5--6--8--9)
If one of them hits again, I put $10 on the table and say press all the insides.
I now have $75--point--$90--$75.
My total outlay so far is $80+$45+$10 = $135, and 2 inside numbers have hit.
I will get paid $105 if any number rolls again, so 2 more hits and I am plus $75.
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This is another variation on a similar scheme.
Bet $18 on 6 & 8. (You have $36 invested.)
If one of them hits for $21, add $3, and press them both to $30. (You have $39 invested.)
If one of them hits, take the $35. (You now have $4 invested.)
If one hits for $35, press each to $42. (you get $11 back, you are plus $7.)
If one hits for $49, press each to $54. (you get $25 back, you are plus $32.)
4 numbers have rolled so far. You will get paid $63 for each new hit.

I will add the 5 & 9 as well, doing the same press with the 5/9 as a pair (press in $10 units)
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Both of these are pretty aggressive press strategies. They require 4 hits to get back to even, but if a roll gets 5 or 6 hits, you are getting paid a decent amount. On a cold table, both of these pressing routines will deplete your bank in a hurry.

Goat's probabilities chart above shows a 2.3% chance of 4 numbers hitting. I won't argue with the number, but it sure 'seems' to me that at least 4 numbers are rolled more often than that, otherwise I would never last more than an hour on a table...... ;-)

Quote: RaleighCraps

I like to try and minimize against the early 7 out, but get to a decent bet size as fast as possible.

Here is one scheme I have used. Assume the point is 6.
I will bet $80 inside ($25 on 5&9, $30 on 8).
If one of the numbers hits, I put $45 down and say press all the insides.
I now have $50--point--$60--$50 (5--6--8--9)
If one of them hits again, I put $10 on the table and say press all the insides.
I now have $75--point--$90--$75.
My total outlay so far is $80+$45+$10 = $135, and 2 inside numbers have hit.
I will get paid $105 if any number rolls again, so 2 more hits and I am plus $75.
-----------------------------------------------------------------------------------

This is another variation on a similar scheme.
Bet $18 on 6 & 8. (You have $36 invested.)
If one of them hits for $21, add $3, and press them both to $30. (You have $39 invested.)
If one hits for $35, press each to $42. (you get $11 back, you have $28 invested.)
If one hits for $49, press each to $54. (you get $25 back, you have $3 invested.)
3 numbers have rolled so far. You will get paid $63 for each new hit, so you will be plus $60 if a 4th 6 or 8 is rolled.

I will add the 5 & 9 as well, doing the same press with the 5/9 as a pair (press in $10 units)
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Both of these are pretty aggressive press strategies. They require 4 hits to get back to even, but if a roll gets 5 or 6 hits, you are getting paid a decent amount. On a cold table, both of these pressing routines will deplete your bank in a hurry.

Goat's probabilities chart above shows a 2.3% chance of 4 numbers hitting. I won't argue with the number, but it sure 'seems' to me that at least 4 numbers are rolled more often than that, otherwise I would never last more than an hour on a table...... ;-)

Alan's table shows the probability of just one number, the 6 or 8, repeating exactly N times -- i.e. N repeats followed by a 7. So on the row for 4 hits, the probability is 5/11 * 5/11 * 5/11 * 5/11 * 6/11, or more compactly (5/11)^4 * 6/11.

The inside numbers have 18 combinations. For this situation, you only care about those 18 and the 6 combinations for the 7. The probability of exactly 4 hits is (18/24)^4 * 6/24 = 0.079 or 7.9%. There is a 31.6% probability of at least 4 hits -- this should be more in line with your perception.

Quote: goatcabin

Within the next few days, I will also run a few thousand two-hour sessions for each "Guy" and examine the differences in variance among them.

I hope you were simply wording that badly. You made it sound like you're going to have each guy run independent simulations.

Have all 4 players betting on the same 'shooter', at the same time.

And, if I may suggest an additional strategy:
18, 12, 12, 18, 18, 24, 24, 30, 30.... Press 6 every other roll thereafter.

On six and eight the distribution possibilities are:

Lose initial bet: 6/16 = 37.50%
Hit once = 23.4375%
Hit twice = 14.6484%
Hit three = 9.1553%
Hit four = 5.7220%
Hit five = 3.5763%
Hit six = 2.2352%
Hit seven = 1.397%
Hit eight = 0.8731%
Hit nine = 0.5457%
Hit ten = 0.3411%
Hit more than ten = 0.5684%

So you need to look at your betting progression on six and eight only that way. You can start big and regress or you can bet small and progress. A $12 flat strategy ends up with a HA of 2.778% and an expected value of -$.66667. A $12 press $12 strategy ($12 -> $18 -> $24 has an EV of -$1.22174. In the end, the EV is the betting value x the HA, pure and simple.

Quote: seattledice

The inside numbers have 18 combinations. For this situation, you only care about those 18 and the 6 combinations for the 7. The probability of exactly 4 hits is (18/24)^4 * 6/24 = 0.079 or 7.9%. There is a 31.6% probability of at least 4 hits -- this should be more in line with your perception.

Thanks Seattle, for correcting my misreading of the posted information.
Can you show me the math to get to the 31.6% probability for 4 inside numbers? I'm sure it has been shown before, but this time I am going to print it out, so I have it for instant referral.
And yes, 31% of the time to get 4 numbers is more of a number I can agree with.

0 inside numbers = 6/24 = 25%
1 or more inside number = 18/24 = 75%. Exactly one number = 75% - 56.25% = 18.75%
2 or more inside number = (18/24)^2 = 56.25%. Exactly two numbers = 14.06%
3 or more inside numbers = (18/24)^3 = 42.19%. Exactly three numbers = 10.55%
4 or more inside numbers = (18/24)^4 = 31.64%.

There are 18 ways to make an inside number, 6 ways to big red. Therefore the odds of hitting an inside number before big red is 18/24.

Expected value of $44 inside, infinite rolls, no pressure, -$2.00, 4.55% HA

Quote: boymimbo

0 inside numbers = 6/24 = 25%
1 or more inside number = 18/24 = 75%. Exactly one number = 75% - 56.25% = 18.75%
2 or more inside number = (18/24)^2 = 56.25%. Exactly two numbers = 14.06%
3 or more inside numbers = (18/24)^3 = 42.19%. Exactly three numbers = 10.55%
4 or more inside numbers = (18/24)^4 = 31.64%.

There are 18 ways to make an inside number, 6 ways to big red. Therefore the odds of hitting an inside number before big red is 18/24.

Expected value of $44 inside, infinite rolls, no pressure, -$2.00, 4.55% HA

Thanks boy. It really is simple when you put it down, I'm not sure why I can't get this set enough in my mind to be able to calc it myself. Hoping osmosis will step up..........

If the HA on the 6/8 is 1.52%
and the HA on the 5/9 is 4.00%

How can the HA on the 5,6,8,9 be 4.55%? I would expect it to fall roughly midway between the 1.52 and the 4.00, meaning around 2.75%.

Quote: DJTeddyBear

Quote: goatcabin

Within the next few days, I will also run a few thousand two-hour sessions for each "Guy" and examine the differences in variance among them.

I hope you were simply wording that badly. You made it sound like you're going to have each guy run independent simulations.

Have all 4 players betting on the same 'shooter', at the same time.

And, if I may suggest an additional strategy:
18, 12, 12, 18, 18, 24, 24, 30, 30.... Press 6 every other roll thereafter.

Yes, of course, they will all experience the same decisions. I have enough scenarios here, I think. This is not really a matter of specific scenarios, but rather a way to get a picture of what progression and regression do to the shape of the graph of possible outcomes.

BTW, I extended the program to 20 consecutive hits, which essentially removes any bias toward the more conservative strategies. Note the probabilities get below 10 to the minus 6th power.

num hits	guy 1	guy 2	guy 3	guy 4	prob
0	-18	-18	-18	-18	0.545455
1	9	3	-3	-15	0.247934
2	29	18	19	9	0.112697
3	30	46	48	54	0.051226
4	38	68	84	120	0.023285
5	53	103	127	207	0.010584
6	87	132	177	315	0.004811
7	114	174	234	444	0.002187
8	134	210	298	594	0.000994
9	135	259	369	765	0.000452
10	143	302	447	957	0.000205
11	158	358	532	1170	0.000093
12	180	408	624	1404	0.000042
13	221	471	723	1659	0.000019
14	255	528	829	1935	0.000009
15	282	598	942	2232	0.000004
16	302	662	1062	2550	0.000002
17	303	739	1189	2889	0.000001
18	311	810	1323	3249	0.000000
19	326	894	1464	3630	0.000000
20	348	972	1612	4032	0.000000

Guy 0 weighted net is -0.413625
Guy 1 weighted net is -0.543476
Guy 2 weighted net is -0.639012
Guy 3 weighted net is -0.916979

Cheers,
Alan Shank

Quote: RaleighCraps

Quote: boymimbo

0 inside numbers = 6/24 = 25%
1 or more inside number = 18/24 = 75%. Exactly one number = 75% - 56.25% = 18.75%
2 or more inside number = (18/24)^2 = 56.25%. Exactly two numbers = 14.06%
3 or more inside numbers = (18/24)^3 = 42.19%. Exactly three numbers = 10.55%
4 or more inside numbers = (18/24)^4 = 31.64%.

There are 18 ways to make an inside number, 6 ways to big red. Therefore the odds of hitting an inside number before big red is 18/24.

Expected value of $44 inside, infinite rolls, no pressure, -$2.00, 4.55% HA

Thanks boy. It really is simple when you put it down, I'm not sure why I can't get this set enough in my mind to be able to calc it myself. Hoping osmosis will step up..........

If the HA on the 6/8 is 1.52%
and the HA on the 5/9 is 4.00%

How can the HA on the 5,6,8,9 be 4.55%? I would expect it to fall roughly midway between the 1.52 and the 4.00, meaning around 2.75%.

It's because the HA is based on "per decision" where as this "HA" is based on decision to leave the bet up until it dies.

-On a per decision basis for the 6/8, the HA is (7/6*10/16 - 6/16*2) = 1.042% with a EV of ($.25) on $24.
-On a per decision basis for the inside bet, the HA is (18/24*14 - 44*6/24)/44 = 1.136% with an EV of ($.50) on $44.
-On a $64 across bet, on a per decision basis, the HA is (18 x 6/30 + 14 x 18/30 - 64 x 6/30)/64 = 1.25% with an EV of ($.80).

The reason why the HA is lower is because there are alot more decisions. But the EV does indeed go up with the more money bet.




The HA is actually lower on a per decision basis but the EV is higher.

Thank you. I think it is actually starting to sink in.
With 4 bets active, the HA is lower than it would be with just one single bet, but the lower HA is offset by the act that there is now more money at risk (4 bets), so the house is expected to collect more of my money overall.

Quote: goatcabin

Quote: DJTeddyBear

Quote: goatcabin

Within the next few days, I will also run a few thousand two-hour sessions for each "Guy" and examine the differences in variance among them.

I hope you were simply wording that badly. You made it sound like you're going to have each guy run independent simulations.

Have all 4 players betting on the same 'shooter', at the same time.

And, if I may suggest an additional strategy:
18, 12, 12, 18, 18, 24, 24, 30, 30.... Press 6 every other roll thereafter.

Yes, of course, they will all experience the same decisions. I have enough scenarios here, I think. This is not really a matter of specific scenarios, but rather a way to get a picture of what progression and regression do to the shape of the graph of possible outcomes.

So, I finished writing the program and ran 10,000 sessions of 60 bets each. The overall W-L percentage was .453978, so the sample was slightly unfavorable. Each player experienced exactly the same bet results, since they were standing right next to each other, making the same bet but each in his own amount. It takes about 200 rolls to resolve 60 place bets on the six, so these sessions would take somewhere in the neighborhood of two hours.

parameter	Guy 1	Guy 2	Guy 3	Guy 4
mean net result	-$14.95	-$19.23	-$22.24	-$31.35
median net result	-$14	-$26	-$37	-$81
standard deviation	$130	$166	$200	$323
winning sessions	4599	4342	4223	3965
breakeven sessions	6	47	19	12
losing sessions	5395	5611	5758	6023
lost more than $300	154	384	659	2051
lost more than $500	3	5	17	257
won more than $300	63	357	618	1424
won more than $500	0	22	128	614
biggest win	$416	$641	$894	$2112
biggest loss	-$509	-$579	-$644	-$786
skew	-.06	.25	.50	1.14

The idea here is to isolate the effect of progressing/regressing bets by looking at just one bet at a time.
Cheers,
Alan Shank

Quote: goatcabin

To illustrate the "continuum of progressiveness", so to speak, I wrote a computer program to compare some different betting strategies for placing the 6. We will call them "the four guys". Everyone starts at $18. Guy #1 then regresses twice before going back up, then regresses again. Guy #2 increases his bets a unit at a time, but only every other bet. Guy #3 progresses a unit at a time every bet. Guy #4 doubles first and then adds $18 each bet. So, their bet series are:
Guy #1 18,12,6,12,18,24,18,12,6,12,18
Guy #2 18,18,24,24,30,30,36,36,42,42,48
Guy #3 18,24,30,36,42,48,54,60,66,72,78
Guy #4 18,36,54,72,90,108,126,144,162,180,198
Cheers,
Alan Shank

I have quoted the original post, to bring the betting patterns closer to the new results.
Awesome analysis Alan. There is a lot of insight that can be gleaned from studying these results.

The stat I find the most interesting is the biggest loss. Guy 1 lost $509, while Guy 4 lost only slightly more at $786.
However, on the biggest win, Guy 1 was $416, compared to Guy 4 at $2112.
So, Guy 4 got 5x the money on the biggest win, and only lost 1.5x as much on the worst loss?
Of course, Guy 4 also had a much greater number of losses over $500

Quote: RaleighCraps

Quote: goatcabin

To illustrate the "continuum of progressiveness", so to speak, I wrote a computer program to compare some different betting strategies for placing the 6. We will call them "the four guys". Everyone starts at $18. Guy #1 then regresses twice before going back up, then regresses again. Guy #2 increases his bets a unit at a time, but only every other bet. Guy #3 progresses a unit at a time every bet. Guy #4 doubles first and then adds $18 each bet. So, their bet series are:
Guy #1 18,12,6,12,18,24,18,12,6,12,18
Guy #2 18,18,24,24,30,30,36,36,42,42,48
Guy #3 18,24,30,36,42,48,54,60,66,72,78
Guy #4 18,36,54,72,90,108,126,144,162,180,198
Cheers,
Alan Shank

I have quoted the original post, to bring the betting patterns closer to the new results.
Awesome analysis Alan. There is a lot of insight that can be gleaned from studying these results.

The stat I find the most interesting is the biggest loss. Guy 1 lost $509, while Guy 4 lost only slightly more at $786.

Well, that's 54% more; I would hardly call that "slightly more".

Quote: RaleighCraps

However, on the biggest win, Guy 1 was $416, compared to Guy 4 at $2112.
So, Guy 4 got 5x the money on the biggest win, and only lost 1.5x as much on the worst loss?
Of course, Guy 4 also had a much greater number of losses over $500

Look at the standard deviation figures, Raleigh. That is the key. The reason the graph of Guy 4 is right skewed, not symmetrical, is that he progresses only on wins, returning to $18 after each loss, so he doesn't lose very many big bets. For the opposite reason, Guy 1's is slightly left-skewed due to the regressions. Note also that Guy 4 had a net loss in over 60% of his sessions.

BTW, I created formulae for each "Guy's" betting pattern so they could extend indefinitely.
Cheers,
A

Quote:

Well, that's 54% more; I would hardly call that "slightly more".

Quote: RaleighCraps

However, on the biggest win, Guy 1 was $416, compared to Guy 4 at $2112.
So, Guy 4 got 5x the money on the biggest win, and only lost 1.5x as much on the worst loss?
Of course, Guy 4 also had a much greater number of losses over $500

Look at the standard deviation figures, Raleigh. That is the key. The reason the graph of Guy 4 is right skewed, not symmetrical, is that he progresses only on wins, returning to $18 after each loss, so he doesn't lose very many big bets. For the opposite reason, Guy 1's is slightly left-skewed due to the regressions. Note also that Guy 4 had a net loss in over 60% of his sessions.

BTW, I created formulae for each "Guy's" betting pattern so they could extend indefinitely.
Cheers,
A

Yes, the loss is 54% greater, but it still was only $280 more. So $280 more at risk, and a $1700 greater profit on the biggest win. And of course, way more losing sessions overall.
But I think most of all this shows that to be happy with your craps sessions, you really need to decide what kind of results you are looking for.

++ Are you looking for a big killing (3x or 4x your buyin)?
Aggressively presss, play full odds, etc. Very strong likely hood you are going to end the session losing money.

++Are you a grinder, looking to play for a few hours, and end up close to even?
Look at using a variation of a regression system, and keep your odds bets low to reduce the variance.

I find this thread fascinating reading. It really is insightful into how different betting patterns will affect the outcome of a 2-3 hour session.

Quote: RaleighCraps

Yes, the loss is 54% greater, but it still was only $280 more. So $280 more at risk, and a $1700 greater profit on the biggest win. And of course, way more losing sessions overall.
But I think most of all this shows that to be happy with your craps sessions, you really need to decide what kind of results you are looking for.

Yes, that is exactly what, in my view, people should do, and running simulations like this is exactly the way to do it. It's an extension of the same idea as knowing the probabilities of the dice combinations, or the HA, variance, skew, etcc. of individual bets. If I play this way, I have a n% chance to bust my $1000 bankroll, I have a n% chance to come out ahead, I have a n% chance to win $500, etc. etc. You can see the "shape of possible futures", just like you can with the "perfect 36" pyramid or the "perfect 1980". When you go out and actually play, the dice select for you one of those possible outcomes.

Cheers,
Alan Shank

Alan;
I wonder if you would consider cross posting your material on some of the dice boards?

Quote: DeMango

Alan;
I wonder if you would consider cross posting your material on some of the dice boards?

What boards? I also post on crapsforum.com
Cheers,
Alan Shank

Just posted these on another thread:
http://axispower.proboards12.com

http://diceinstitute.proboards55.com

The latter especially deals in practical theory, the former in dice seminars.

Quote: DeMango

Just posted these on another thread:
http://axispower.proboards12.com

http://diceinstitute.proboards55.com

The latter especially deals in practical theory, the former in dice seminars.

I respectfully ask you not to post any of my stuff on those sites.
Thanks,
Alan Shank

No I would not. I wondered if you would.