The Iron Cross Craps
The Iron Cross Craps System The Iron Cross craps system has been around for a long time. It’s more powerful now because many casinos pay double on a field bet for a pair of aces rolled (a two) and triple for boxcars (a twelve), compared to the even money they used to pay 50 years ago. The Meaning Of Iron Cross Craps. Posted on November 29, 2019 November 29, 2019 by FRCH. Typically the Program Basics.Like virtually all gamblers, casino craps members are usually hunting for a completely new system. Your The form of iron Mix is not a developing technique such as Colonel’s snake eyes product, however it lets you do make use of. 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.
In this video we discuss the Iron Cross strategy and various ways of playing with small money and when/where to increase your bet. This is a great strategy b. The Plus Ultra is a legendary shield which doubles Moze’s HP and gives her a 20% chance to absorb a bullet and increase Iron Bear’s cooldown rate for a few seconds. This makes it an ideal choice for Iron Bear builds, not only because of the improved action skill cooldown, but also because Iron Bear’s armor scales with Moze’s HP.
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%.
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%.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.
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
Determ(B) / Determ(A) = apx. 9.65599148388557
If you’re confused by the Craps table, you’re not alone. It’s actually really easy to grasp, but the overall layout and the chaos at the Craps table makes newbies feel overwhelmed. In previous posts, we’ve told you to stick to one or two bets. But if you’re getting bored with that, you might want to employ a betting strategy. And the Iron Cross is one of the best strategies in the book.
Iron Cross is great if you’re a seasoned player looking for bigger gains, but it can also help steer new players in the right direction. Once you figure out how this strategy works (and don’t worry, we’ll walk you through it), feel free to practice at Palace of Chance.
For the Iron Cross Strategy to work, you’ll need to be comfortable placing more than one bet at the same time. If you’re familiar with betting on multiple areas of the Roulette table, Iron Cross isn’t all that different. You’re basically spreading your bet, but the payout potential can be impressive. That said, you must also be comfortable losing a hefty chunk of change. If you’re not comfortable with the possible swings in both directions, we recommend you stay away. But if you like to win big when you do and minimize your losses, Iron Cross might just be the best bet you can make.
All about the Iron Cross System
Many casinos today pay double on a field bet when you roll a pair of Aces and pay out triple when you roll a 12. That’s better than the even money they used to hand out a half-century ago, which makes the Iron Cross Craps System even more lucrative. More on that in a second. With the increased payouts, the house edge drops on average to 2.78%, but it gets even lower when you involve Iron Cross .
The most important thing to know about Iron Cross is that this betting system doesn’t just involve a bet in the field area of the Craps table. It also involved placing a bet on 5, 6, and 8 simultaneously. With this combination of bets, you’re covering almost every possible roll with the exception of 7. That’s because field wins on 2, 3, 4, 9, 10, 11, and 12.
With these numbers, it seems like that you should be able to win every single time. But that’s not the case. Remember the field bet doesn’t lose sometimes because it loses on 5, 6, 7, and 8. That means while you’ll win with a place bet on 5, 6, and 8, you’ll lose your field bet. We’ll show you how to maximize these bets for more winning opportunity.
How to use the Iron Cross System to your benefit
With the Iron Cross system, you’ll place four bets. Here’s a step by step way to make this betting system work for you:
- Find a Craps table that has a low minimum. Remember, online Craps tables let you wager as little as $1 per bet. And you can even play for free, so there’s technically nothing to lose.
- For this example, let’s pretend you’re working with a little more than the minimum. Start by placing $5 on the number 5.
- Then, place a $6 bet on the number 6.
- Now, place a $6 on the number 8.
You’ve now got $22 committed. If the shooter sevens out, you’ll lose that $22. But here’s the best part. If she rolls any other number besides a 7, you’re actually sitting on a profit.
Remember, if the place bet numbers 5, 6, or 8 end up hitting, you’ll lose that $5 bet on the field but you’ll win $7 on the place bet you made. At this point, you’ll actually be up $2. And if a field number ends up hitting, you’ll win $5 at the very minimum, and more if you hit 2 or 12.
The Iron Cross works in your favor when the dice are hot, though you will end up with a loss if 7 keeps hitting every few rolls. And overall, there’s a house edge of just 2.48% involved when you employ this strategy, which is way better than most other betting systems.
Stay away from Big 6 or 8
To be clear, you want to stay away from a Big 6 or Big 8 bet. The biggest mistake you could make is putting your money down on Big 6 or 8 thinking that you’re covered with your place bets. Never bet on Big 6 or 8. Not when you use this strategy. Not ever, really.
While the word “big” might make you assume that your bankroll is going to see a big boost, that’s not exactly what’s going to happen. The house advantage for Big or 8 hovers around 9%. The house advantage drops significantly when you bet on the actual numbers.
Craps Iron Cross Strategy
Try Iron Cross now
Ready to put Iron Cross to the ultimate test? Sign up for a free Palace of Chance account. All we need is a few pieces of information to get started. As soon as you’ve got a free account, you can play Craps for as long as you like, without having to risk real money. We’ll set you up with a play-money account, which gets you a bunch of pretend chips. Go ahead and use your free stack to see how well Iron Cross works for you. If you don’t have much luck with it, walk away. No loss, no pain. But if it does work, make that first deposit and you’ll be able to play for real money (and win real cash while you’re at it).
Playing The Iron Cross In Craps
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