Craps Strategy Iron Cross

• Best Craps Strategy Iron Cross
• Craps Strategy Iron Cross Trainer
• Craps Strategy Iron Cross Template
• Craps Strategy Iron Cross Reference
• Craps Strategy Iron Cross Game

When craps players put together combination bets, there are two key things to keep in mind:

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Best Craps Strategy Iron Cross

The strategy is called the 'Iron Cross.' It involves a bet on the 5, 6, 8, and the field. I read up on this, and found that this particular bet will pay on every roll that is not a 7. I was told that this gives you the lowest house edge. The Iron Cross is a way of betting the field and place bets to win on any roll of the dice except a 7. The field already covers the 2, 3, 4, 9, 10, 11, and 12. The player will add to that place bets on the 5, 6, and 8 to cover the rest of the numbers, besides 7. At its base, the Iron Cross would mean $5 bets on the field and placing 5 and $6 bets on 6 and 8 for a total risk of $22. Bigger bettors can multiply the bets. The Iron Cross was originally designed as a one-roll, hit-and-run bet. Every number except 7 is a winner, but when the roll is 7 all bets lose. When the place bets win, the field bet loses. The Iron Cross isn't a progressive program like the Colonel's craps system, but it does take advantage of the field wager, which covers the numbers 2, 3, 4, 9, 10, 11, and 12. The Iron Cross is sometimes called the No Seven System, because the player covers all the possible numbers on the table except the dreaded 7.

• The house edge is a weighted average of the house edges on all bets in the combination.
• The overall house edge can never be lower than the best individual bet in the combination.

The second point sometimes brings raised eyebrows from veteran craps players who ask, “What about the Iron Cross?”. even if you already know how to play craps, you should keep this strategy in your mind,
The Iron Cross, sometimes called Darby’s Field, is a combination that is said to do the impossible. The component bets with the lowest house edges are place bets on 6 and 8, each at a 1.52% edge. Yet the combination is said to have a house edge of 1.14%.
How can that possibly be true? The answer is that the 1.14% figure on the Iron Cross is the house edge per roll, and the place bets in the combination are multi-roll bets. If you assume all bets are played until there is a decision, the house edge is 2.37%. If you will use this strategy wisely without making mistakes while playing craps, you have a chance to return home with some extra money in your wallet.

HOW THE IRON CROSS STRATEGY WORKS?

There are four wagers in the combination: the field, along with place bets on 5, 6 and 8.
The field is a bet that the next roll will be 2, 3, 4, 9, 10, 11 or 12. If it wins, it pays even money on 3, 4, 9, 10, 11. You get bigger payoffs on the other numbers – usually 2-1 on 2 and 3-1 on 12.
On the place bets, you’re betting the shooter will roll your number before rolling a 7. Winners are paid 7-5 on the place bet on 5 or 7-6 on either 6 or 8.
The 7-6 payoff on 6 and 8 means you should always make those bets in multiples of $6.
At its base, the Iron Cross would mean $5 bets on the field and placing 5 and $6 bets on 6 and 8 for a total risk of $22. Bigger bettors can multiply the bets.
The Iron Cross was originally designed as a one-roll, hit-and-run bet. Every number except 7 is a winner, but when the roll is 7 all bets lose.
When the place bets win, the field bet loses. However, when the field bet wins, the place bets are still in action. If you like, you can take the place bets down and walk away with the money as well as with your field winnings.

HOW THE PAYOFFS WOULD WORK FOR EACH POSSIBLE NUMBER?

• 2: You win on the field and are paid 2-1. That means you have $10 in winnings and can keep all $22 of your wagers for a $10 profit.
• 3, 4, 9, 10 or 11: You win an even-money payout on the field, giving you $5 in winnings and you keep your $22 in bets for a $5 profit.
• 5, 6 or 8: You’re paid $7 on a winning place bet. You lose $5 on the field, but keep the $17 worth of place bets, meaning you have $24 overall or a $2 profit.
• 7: You lose all $22 wagered.

The most frequent roll is the lose 7, with six of 36 two-dice combinations giving that total. The house retains an edge, but how an edge is a matter of perspective.
The Iron Cross was originally designed for players who wanted to make one last bet on their way out of the casino. They were betting the next roll would be anything but 7. In the days when even $1 would buy ham and eggs in Las Vegas, any win at craps would give the bettor enough to dine out.
They might want to know the house edge for just the next roll. The house edge on the field bet with the above listed payouts is 2.78%, while the house edges per roll on the place bets are 1.111%on 5 and 0.463%on 6 and 8.
Because you bet in multiples of $6 on the and 8 and in multiples of $5 on 5 and on the field, the 6 and 8 are given 1.2 times the weight of the other bets in calculating an overall house edge.
That one-roll edge on the Iron Cross is 1.14%. Note that it isn’t really lower than the best component bet in the combination, as has been claimed. The 0.463% one-roll edge on 6 and 8 and 1.11%on 5 are lower.
Some players see that house edge and think the Iron Cross is a viable bet for periods of longer than one bet. They need to know the overall edge of each craps bet is played to a decision. Under those conditions, the edges on each component are 2.78%on the field, 4%on the place bet on 5 and 1.52% on place bets on 6 and 8.
The overall house edge of 2.37% on the Iron Cross, played to decision on all pieces, higher than the 1.52% on 6 and 8.
Either way you look at it, the Iron Cross is no miracle bet. It’s a better combination than most that involve one-roll bets, but it has a higher overall edge than its lowest-edge component. That’s just as it must be – it can’t defy the math.

The Iron Cross is a way of betting the field and place bets to win on any roll of the dice except a 7. The field already covers the 2, 3, 4, 9, 10, 11, and 12. The player will add to that place bets on the 5, 6, and 8 to cover the rest of the numbers, besides 7. The following table shows what the math looks like with a $5 field bet, $5 place bet on the 5, and $6 place bets on the 6 and 8.

Iron Cross

Dice Total	Win	Combinations	Probability	Return
2	10	1	0.027778	0.277778
3	5	2	0.055556	0.277778
4	5	3	0.083333	0.416667
5	2	4	0.111111	0.222222
6	2	5	0.138889	0.277778
7	-22	6	0.166667	-3.666667
8	2	5	0.138889	0.277778
9	5	4	0.111111	0.555556
10	5	3	0.083333	0.416667
11	5	2	0.055556	0.277778
12	15	1	0.027778	0.416667
36	1.000000	-0.250000

The lower right cell of the table shows an expected loss of $0.25. The total amount bet is $22. This makes the overall house edge $0.25/$22 = 1/88 = 1.14%.

At this point you may be wondering how this house edge can be lower than the house edge of each individual bet. The answer is because the house edge of 1.52% placing the 6 and 8 and 4.00% placing the 5 is based on per bet resolved. If define the house edge on place bets on a per roll basis, then the house edge placing the 6 or 8 is 0.46% and on the 5 is 1.11%.

We can get at the 1.14% house edge by taking a weighted average of all bets made, as follows:

($5*2.78% + $5*1.11% + $12*0.46%)/22 = $0.25/$22 = 1.14%.

Craps Strategy Iron Cross Trainer

Be wary of casinos that pay only 2 to 1 for 12 on the field bet. Insist on getting the full 3 to 1. The short pay doubles the house edge on that bet from 2.78% to 5.56%.

As to my opinion, compared to most games, 1.14% is a pretty good bet. However, you could do much better in craps. For example, with 3-4-5x odds, making the pass and come bets, with full odds, you can get the house edge down to 0.37%. Doing the opposite, betting the don't pass and don't come, plus laying full odds, results in a house edge of 0.27%.

Craps Strategy Iron Cross Template

While this could be solved with a long and tedious Markov chain, I prefer an integral solution. I explain how to use this method in my pages on the Fire Bet and Bonus Craps.

Craps Strategy Iron Cross Reference

Imagine that instead of significant events being determined by the roll of the die, one at a time, consider them as an instant in time. Assume the time between events has a memory-less property, with an average time between events of one unit of time. In other words, the time between events follows an exponential distribution with a mean of 1. This will not matter for purposes of adjudicating the bet, because events still happen one at a time.

Per the Poisson distribution, the probability that any given side of the die has been rolled zero times in x units of time is exp(-x/6)*(x/6)⁰/0! = exp(-x/6). Poisson also say the probability of any given side being rolled exactly once is exp(-x/6)*(x/6)¹/1! = exp(-x/6) * (x/6). Thus probability any side has been rolled two or more times in x units of time is 1 - exp(-x/6)*(1 + (x/6)). The probability that all six sides have been rolled at least twice is (1 - exp(-x/6)*(1 + (x/6)))⁶. The probability that at least one side has not been rolled at least twice is equal to:

We need to integrate that over all time to find how much time will go by, on average, where the desired goal has not been achieved.

Fortunately, we can use an integral calculator at this point. For the one linked to, put 1- (1 - exp(-x/6)*(1 + x/6))^6 dx = apx. 24.1338692 in the text box following 'Calculate the integral of' and under custom, set the bound of integration from 0 to ∞.

The answer is 390968681 / 16200000 = apx. 24.13386919753086

This question is asked and discussed in my forum at Wizard of Vegas.

I have a two-part question.

For part 1, given:

• x + y + z = 1
• x^2 + y^2 + z^2 = 4
• x^3 + y^3 + z^3 = 9

What is x^4 + y^4 + z^4 ?

For the second part, what is the answer to the general case when:

• x + y + z = a
• x^2 + y^2 + z^2 = b
• x^3 + y^3 + z^3 = c

Question 1: 97/6 = apx. 16.166666

Question 2: a⁴/6 + (4/3)ac - a²b + b²/2

This question is raised and discussed in my forum at Wizard of Vegas.

You start with a fair 6-sided die and roll it six times, recording the results of each roll. You then write these numbers on the six faces of another, unlabeled fair die. For example, if your six rolls were 3, 5, 3, 6, 1 and 2, then your second die wouldn’t have a 4 on it; instead, it would have two 3s.

Next, you roll this second die six times. You take those six numbers and write them on the faces of yet another fair die, and you continue this process of generating a new die from the previous one.

Eventually, you’ll have a die with the same number on all six faces. What is the average number of transitions from one die to another (or total rolls divided by 6) to reach this state?

Let's label the initial die with letters instead of numbers, to avoid confusion. Let's label each possible die state with letters. For example, AAABBC would mean three of one letter, two of another, and one of a third. The initial state would obviously be ABCDEF.

Let E(ABCDEF) be the expected number of rolls from state ABCDEF.

E(ABCDEF) = 1 + [180 × E(AAAAAB) + 450 × E(AAAABB) + 300 × E(AAABBB) + 1800 × E(AAAABC) + 7200 × E(AAABBC) + 1800 × E(AABBCC) + 7200 × E(AAABCD) + 16200 × E(AABBCD) + 10800 × E(AABCDE) + 720 × E(ABCDEF)]/46656

Building on the number of combinations of going from one state to another, the following transition matrix shows how many ways there are for going from each initial state (left column) to each new state. This took a few hours to construct properly, by the way.

Transition Matrix A

State Before	AAAAAA	AAAAAB	AAAABB	AAABBB	AAAABC	AAABBC	AABBCC	AAABCD	AABBCD	AABCDE	ABCDEF
AAAAAB	15,626	18,780	9,750	2,500	-	-	-	-	-	-	-
AAAABB	4,160	13,056	19,200	10,240	-	-	-	-	-	-	-
AAABBB	1,458	8,748	21,870	14,580	-	-	-	-	-	-	-
AAAABC	4,098	12,348	8,190	2,580	7,920	10,080	1,440	-	-	-	-
AAABBC	794	5,172	8,670	5,020	6,480	17,280	3,240	-	-	-	-
AABBCC	192	2,304	5,760	3,840	5,760	23,040	5,760	-	-	-	-
AAABCD	732	4,464	4,140	1,680	7,920	14,400	2,520	4,320	6,480	-	-
AABBCD	130	1,596	3,150	1,940	5,280	16,800	3,600	4,800	9,360	-	-
AABCDE	68	888	1,380	760	3,960	11,520	2,520	7,200	14,040	4,320	-
ABCDEF	6	180	450	300	1,800	7,200	1,800	7,200	16,200	10,800	720

I won't go into a long lecture on matrix algebra, except to say let's say matrix B is as follows:

Matrix B

State Before	AAAAAB	AAAABB	AAABBB	AAAABC	AAABBC	AABBCC	AAABCD	AABBCD	AABCDE	ABCDEF
AAAAAB	-27876	9750	2500	0	0	0	0	0	0	-46656
AAAABB	13056	-27456	10240	0	0	0	0	0	0	-46656
AAABBB	8748	21870	-32076	0	0	0	0	0	0	-46656
AAAABC	12348	8190	2580	-38736	10080	1440	0	0	0	-46656
AAABBC	5172	8670	5020	6480	-29376	3240	0	0	0	-46656
AABBCC	2304	5760	3840	5760	23040	-40896	0	0	0	-46656
AAABCD	4464	4140	1680	7920	14400	2520	-42336	6480	0	-46656
AABBCD	1596	3150	1940	5280	16800	3600	4800	-37296	0	-46656
AABCDE	888	1380	760	3960	11520	2520	7200	14040	-42336	-46656
ABCDEF	180	450	300	1800	7200	1800	7200	16200	10800	-46656

The answer is the determinant of matrix B to that of matrix A:

Determ(A) = 1,461,067,501,120,670,000,000,000,000,000,000,000,000,000,000

Determ(B) = 14,108,055,348,203,100,000,000,000,000,000,000,000,000,000,000

Craps Strategy Iron Cross Game

Determ(B) / Determ(A) = apx. 9.65599148388557