Years ago Frank used to write about the "doey/don’t" system similar to your idea. I don’t pretend to know the exact math but as I understand it, a come bet after the five count has less vig in the house favor. That is what GTC promotes and I use. It has work good for Alligator Rose and I.

Maybe someone will correct me on this if I am incorrect.

Doc

The problem with the Doey-Don’t is the house edge. If you do one come bet for $5 the house has a 7 cent edge regardless of odds placed as there is no edge on odds (7 cents is 1.4 percent). If you place a don’t come bet for $5, the house has about a 7 cent edge. If you do a come bet and a don’t come bet for $5 each (doey-don’t), the house has about a 14 cent edge (7 cents for the come, 7 cents for the don’t come).

The house extracts its edge on each and every bet. You are much better off just picking either a come or a don’t come bet for $5 and just go with that. The house has half the edge it has on a doey-don’t bet.

See, I told you one of the math experts would give us the hard facts. Thanks Stick.

Doc

Stick is correct, you will lose twice as much betting the doey-don’t for $5 each way than if you just bet $5 on the pass or don’t pass.

I realize it is counter intuitive since the 12 comes up so seldom in a random game. After all, the expected one time out of every 36 rolls seems like it hardly ever happens. But in fact you do not lose all that much with only a single pass or single don’t pass bet either. It only seems like we lose more on the pass or don’t pass because one hardly ever only makes a single wager. It is often combined with odds and other wagers. Thus it seems like we lose so much on a seven out that the pass line appears to be a high loss wager.

But the math never lies. Casinos depend on it. They build multimillion dollar buildings, pay salaries of many workers and make a tidy profit all based on the math of the game. We need to get in the habit of depending on the math the same way the casino moguls do if one expects to have a chance at being successful in this game that is stacked against us by the math. Let us look at how the math works on the pass, don’t pass and doey-don’t wagers in an attempt to show how the pass or don’t pass is really a pretty low risk wager in and of itself.

If you were to make 1980 wagers on the pass line, you could expect to win your bet 976 times and lose it 1004 times. Hence you could expect to lose only 28 pass line wagers out of 1980. As an aside, 28 divided by 1980 equals .0141 or 1.41% which is the house advantage on the pass line. So if you were to bet $1 on the pass line 1980 times you could expect to lose only $28 out of $1980 wagered.

Now if the house did not bar the twelve on the don’t pass and you made 1980 wagers on the don’t you could expect the exact opposite. You would expect to win 1004 wagers and lose 976 wagers. Since the house does not want you to have an advantage on the don’t, they bar the twelve and make it a push when the 12 comes out. You can expect to see 55 twelves in 1980 don’t pass wagers. Thus we have to take away 55 of the twelves from the 1004. You can expect to win on the don’t 949 times and lose 976 times. Out of 1980 wagers you can expect to lose 27 times on the don’t pass. As an aside, 27 divided by 1980 equals .0136 or 1.36% which is the house advantage on the don’t pass wager. So if you were to bet $1 on the don’t pass line 1980 times you could expect to lose only $27 out of $1980 wagered.

But if you were to bet the doey-don’t 1980 times you would have to bet $1 on the pass and $1 on the don’t pass. Since we know the 12 is expected to come up 55 times, you could expect to lose $55 when playing the doey-don’t 1980 times. Those of you who are astute in math may notice that you will have lost $55 out of $3960 wagered. Thus the house advantage on the doey-don’t is also a low 1.39%. But the problem is you have to make two wagers on the doey-don’t vs. only one wager on the pass or don’t pass. So even though the house advantage is about the same you have to bet twice as much on the doey-don’t and therefore can expect to lose twice as much.

You see even though it seems like the twelve hardly ever comes up and it actually does not come up all that often. After all one time out of every 36 rolls is not a frequent occurrence. The pass line or don’t pass line do not lose all that much either. A house advantage of 1.41% or 1.36% means the frequency of loss is fairly low. When Frank says to make only one wager on a random roller for the minimum he is giving you sound advice. Your frequency of loss will be as low as it is on the doey-don’t but you will only be wagering half as much. We get into trouble when we let our "intuition" take over. Our memories are selective. We remember the time we lost a pass line wager 10 times in a row or 20 times out of 25. But we tend to forget when a shooter makes 5 passes before establishing his point or makes 10 passes along with making 4 points. It is a cyclical game because that is the nature of randomness. But in the long run the math will hold up. Over the long haul of your lifetime of gaming you can expect to lose 7 cents for every $5 you ever wagered on the pass or don’t pass on random rollers. Those are the facts and they will hold up over your lifetime if you play as much as most on this board tend to play over your lifetime. Your memory or "intuition" will not alter the math of the game. So stick with the best bets and follow the advice of GTC. They have thought this through and used it in practice. It is sound advice based on mathematics and experience of some very serious individuals who collectively have developed a "best practice" set of guidelines.

"Skinny" wrote: Stick is correct, you will lose twice as much betting the doey-don’t for $5 each way than if you just bet $5 on the pass or don’t pass.
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But we tend to forget when a shooter makes 5 passes before establishing his point or makes 10 passes along with making 4 points. It is a cyclical game because that is the nature of randomness.

Thanks for the inputs. Your answers do stimulate and enhance our learning.

Which answer from Skinny brings up another confusion I always wonder about
when I read some writings by members on this site:

How can it be that a shooter ‘makes 5 passes before establishing his point
or makes 10 passes along with making 4 points’ when we seem to accept that
making a pass is repeating an established point before a seven-out?

If the shooter rolls a 7 or 11 on a come out roll (before establishing a point) the pass line bet wins and it is called a pass line winner or making a pass.

So if a shooter throws a combination of five 7’s and 11’s before establishing his point he has made 5 passes before establishing his point.

If a shooter makes 4 points he would be on 4 come out rolls for each of those points. If he throws a combination of ten 7’s and 11’s while on those 4 come out rolls he would have 10 passes in addition to making his 4 points. The point winners are also called passes. So in the latter case the shooter would have a total of 14 passes and he would still have the dice after making the 4th point. He would then be on another come out roll, his fifth.

Skinny All that talk we did earlier must have had some math to it for the bean counters changed the pay for points . It was 1 point for 20 .00 play in or out now it is 80.00 .
What do you think makes me wonder.
Good Rolling.

I guess they did not like the results at $20 and decided they had to up it to $80.

To many crap players making 7 star.

well i understand the odds and all that, but the most fun i have ever had
playing craps and actually the two years i did the best in the 35 years i have
played the game was in 2007-2008 after i read Franks book and used the dewey
donts.

Maybe the odds say that cant happen, but for me that is the truth.

Much of that may be because up to that time 99% of craps players
felt the game was 100% luck, and the dewey donts gave us a tool
to affect the results of the game.

gman