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Questions about Betting Systems for Craps |
I have a question about a series of bets in craps. 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. What are all the various odds and what-nots to go along with it? — Carter from Calgary
If the player bets $5 on the field and 5, and $6 on the 6 and 8, then he will have a net win of $2 on the 5, 6, and 8, $10 on the 2, $15 on the 12, and $5 on the other field numbers, assuming that the 12 pays 3 to 1 on the field. The player will lose $22 on a 7. On a per roll basis, the player can expect to lose 25 cents compared to $22 in bets, for a house edge of 1.136%.
This begs the question, why is this lower than the individual house edge of each bet made? It's not. The reason it seems that way is the result of comparing apples to oranges. The house edge of place bets is usually expressed as the expected loss per bet resolved. Looking at the individual bets on a per-roll basis, the house edge on the 5 is 1.11%, and on the 6 and 8 is 0.46%, according to my craps appendix 2. Comparing apples to apples, the house edge is a weighted average of the house edge on the field, 5, 6, and 8, on a per-roll basis, or (5/22)×2.778% + (5/22)×1.111% + (6/22)×0.463% + (6/22)×0.463% = 1.136%. December 8, 2008
My question is on craps. I know that the Fire Bet is a lousy bet, but I bet it anyway, when I am rolling. Well, I got lucky and hit my four points, and was on the fifth point. I had won $75, and was on my way to winning $750, if I hit the 5th point. My other bets were $5 on the pass and $20 to win $30 on the 5. Having established the 5th point, which was a 5, I realized that I had a 2/5 chance of hitting it, for a net win of $785. I also realized that I had a 3/5 chance of not hitting it, for a net win of $25. If I wanted to hedge my bets, what is the largest win I could lock in? Also, what are your thoughts on this strategy?
— Tim from Grimsby, ON
Unless life changing amounts of money are involved, I disapprove of hedging, per my seventh commandment of gambling.
I'm going to ignore the fact that if you hit the 5 you could hedge more to lock in an even larger win, and just look at this as if it ended after a 5 or 7. At this point your net will will be $785 or $50. You should start by taking down the odds bet. That will change the scenario to winning $755 or $70. Then you should lay the odds on the 5. Let b represent your lay bet against the 5. If you lose the bet, you'll have $755-$b. If you win the bet, you'll have $70 + (19/31)×$b. So, equate the two sides, and solve for b:
755-b = 70 +(19/31)×b
685 = (50/31)×b
b=424.7
That will lock in a win of $330.30. So, if rounding were not an issue, then lay $424.7 against the 5. However, rounding always is an issue, so I would lay $403 against the 5 ($390, plus $13 commission on possible win of $260). November 17, 2008
Is the "parity hedge" myth in craps true? — Craig from Los Angeles
No. I had to Google this to find out what this is. This appears to me to be an amusing urban legend about some young scientists who developed a winning craps system. The story is told at Quatloos. I would file this under other fictional stories that have become mistaken for fact, like Joshua’s missing day. As I have said hundreds of times, not only can betting systems not beat games like craps, they can’t even dent the house edge. July 11, 2008
You mentioned in one of your articles an upcoming appearance on " The Casino" (apparently, it's been cancelled). I have searched and searched to no avail in finding some kind of link to his episode. I find the idea of a story involving his advice to some young gamblers and how to most likely turn $1,000 into $5,000 quite intriguing. Please respond with some insight/leads as to how I might go about finding a copy of this episode online or purchase a video recording of it, or at the very least come across a written transcript of the episode. Thank you for your time. – Brian
Yes, there was a story taped in which some frat boys at UNLV were trying to parlay $1,000 into $5,000 to buy a high end television. They sought out my advice on how to best achieve this goal quickly. I was limited to the games at the Golden Nugget. The Nugget has 10x odds in craps, which I felt offered the opportunity to achieve the goal. It was my strategy on each come out roll to bet min(bankroll/11, (5000-bankroll)/21), subject to convenient rounding, and take the maximum odds. This way we would never go over $5,000 after a 4 or 10 win, would always have enough to take full odds, and would risk the maximum amount if we didn’t have enough to get to $5,000.
For the first bet, this formula would call for a pass line bet of $90.91, but I rounded it up to $100. Then a point was rolled, I think a 6 or 8. On the second roll the shooter sevened out. So the entire grand was lost in two rolls. It apparently didn’t make for very entertaining television and that story never made the air.
Two questions I can anticipate being asked would be (1) why did I have them bet the pass as opposed to the don’t pass, and (2) why didn’t I bet $91 on the line and $910 on the odds, adding the extra dollar out of my own pocket. To answer the first question, I think that for purposes of going for a quick big win the pass line is better. While the overall house edge is less on the don’t pass, I felt it would have taken more rolls to achieve the $5,000 goal, thus exposing more money to the house edge. To answer the second question, there is not much difference between 9x odds and 10x odds and I thought it would look better on television to be betting only black chips, at least to start.
July 31, 2006
The
American Mensa Guide to Casino Gambling has the
following "anything but seven" combination of craps bets
that shows a net win on any number except 7. Here's how much
MENSA advises to bet in the "Anything but 7" system:
- 5- place $5
- 6- place $6
- 8- place $6
- field- $5
- total= $22
They claim the house edge is 1.136%. How is that
possible if every individual bet made has a higher house
edge?
Good question. To confirm their math I made the
following table, based on a field bet paying 3 to 1 on a
12. The lower right cell does shows an expected loss of
25 cents over $22 bet. So the house edge is indeed .25/22
= 1.136%.
|
Mensa Anything but Seven Combo
|
|
Number
|
Probability
|
Field
|
Place 5
|
Place 6
|
Place 8
|
Win
|
Return
|
|
2
|
0.027778
|
10
|
0
|
0
|
0
|
10
|
0.277778
|
|
3
|
0.055556
|
5
|
0
|
0
|
0
|
5
|
0.277778
|
|
4
|
0.083333
|
5
|
0
|
0
|
0
|
5
|
0.416667
|
|
5
|
0.111111
|
-5
|
7
|
0
|
0
|
2
|
0.222222
|
|
6
|
0.138889
|
-5
|
0
|
7
|
0
|
2
|
0.277778
|
|
7
|
0.166667
|
-5
|
-5
|
-6
|
-6
|
-22
|
-3.666667
|
|
8
|
0.138889
|
-5
|
0
|
0
|
7
|
2
|
0.277778
|
|
9
|
0.111111
|
5
|
0
|
0
|
0
|
5
|
0.555556
|
|
10
|
0.083333
|
5
|
0
|
0
|
0
|
5
|
0.416667
|
|
11
|
0.055556
|
5
|
0
|
0
|
0
|
5
|
0.277778
|
|
12
|
0.027778
|
15
|
0
|
0
|
0
|
15
|
0.416667
|
|
Total
|
1
|
|
|
|
|
|
-0.25
|
The reason the overall house edge appears to be less
than the house edge of each individual bet is because the
house edge on place bets is generally measured as
expected player loss per bet resolved.
However in this case the player is only keeping the place
bets up for one roll. This significantly reduces the
house edge on the place bets from 4.00% to 1.11% on the 5
and from 1.52% to 0.46% on the 6 and 8. For you purists
who think I am inconsistent in measuring the house edge
on place bets as per bet resolved (or ignoring ties) then
I invite you to visit my craps
appendix 2 where all craps bets are measured per roll
(including ties). July 11,
2004
I am a novice, just starting to play. My question
concerns the "Five Count Doey/Don't" System. The way I
understand the system:
- Wait until the shooter establishes a
point.
- Play both come/don't come (same amount). Until you
have a maximum of four numbers
- After the shooter has rolled five times without
rolling a 7, take odds on all your numbers on the front
side.
The rationale: Limit your exposure until you find a
"qualified" (five rolls without a 7) shooter. Only betting
the odds so there is no "house edge"! Can you compare this
system with just playing pass/come and taking the odds? -
Don from Little Rock, Arkansas
As I stated in the other craps strategy question
you are only mixing another house edge bet into the game
by betting on both the pass and don't pass, or come and
don't come. It is also not going to help to wait until a
shooter hits five points. The probability of making a
point is the same for me and you as it is for somebody
who just threw 100 points in a row. In other words, the
past does not matter. As I stated to the person who asked
the other question (whom I think may also be you) don't
make opposite bets, just stick to either the do or don't
side and always back up your bets with the odds.
Jan. 2, 2002
I'd like your thoughts on this craps strategy. I think
it's a Patrick system for playing don't pass. Bet one unit
on both pass and don't pass. Then lays odds on the don't
side. You can stop here or then make a don't come bet. After
the dc travels, take the odds off your don't pass bet (if
you don't like to lay odds). So now you have a unit on the
don't come that pretty much got there with less risk. I know
you can never get the advantage over the house, but this
seems like a great way to play the don't side. You eliminate
the sevens on the come out roll. And only get hurt by the
12; or the 11 on your don't come bet. P.s. Your site is the
greatest.
Thanks for the compliment on my site. The best
thing I can say about this system is that it composed of
low house edge bets. Yes, a 12 will lose the pass bet and
push the don't pass on the come out roll, this is where
the house edge is. By making the pass bet you are
increasing the overall house edge. If you're afraid
losing you shouldn't be playing at all. Never hedge your
bets. So my advice is to stick to just the don't pass and
laying odds. Yes, you'll lose some on the come out roll.
However if you don't lose on the come out roll the don't
pass bet will usually win. Jan.
2, 2002
I love to play craps and would like your opinion on a
conventional method of play. Pass line and two come bets
with full double odds or with one come bet? Does having
three different bets working superior to two? - Richard from
Binghamton, USA
As long as you are backing up your pass and come
bets with full odds it doesn't make any difference how
many come bets you make. However it does help to keep the
odds on your come bets turned on in the come out roll.
March 11, 2001
In Craps, could one gain an advantage over the house
by making both a Pass and Don't Pass bet (one unit) and then
playing the Don't Pass odds. Although the occasional 12
would steal 1 unit here and there, it seems that the 7 would
have an advantage over the point. At triple odds one could
take 3x on the 4&10, 2x on the 5&9 and 1x on the
6&8. ñ Jon Moriarty from Danville, New Hampshire
No combination of bets can give the player an
advantage. In your case the player would win more often
on the come out roll but would lose often on the
subsequent rolls. The probability of making a point
ranges from 1/3 to 5/11. Dec.
31, 2000
What is the better system, or which gives me the
better chance to win on craps? On the come out roll, I bet
$10 on the don't and $10 on the do, and then when a point
comes out I lay full odds against the number. Or is it
better to just play the don't pass, and then lay the odds. I
think getting passed the come out roll will increase my
chances of winning. ñ Ray from Plainfield, USA
The better system is to bet on the don't pass
only and take full odds. Yes, betting on both does
increase you chances of winning on any one bet. However
you are suffering a higher combined house edge by betting
on both the pass and don't pass and it will cost you in
the long run. Oct. 5,
2000
Q: I like your site very much. It is very informative.
Thanks for putting out your thoughts.
I noticed a betting strategy for craps suggested at
Crappers Delight called "classic regression". In it he
suggests, placing a 6 and 8, after a point is established.
Then taking it down after one of them is hit. He said there
are 10 combined ways to make the 6 and 8, but only 6
combined ways to make the 7. It sounds logical, but I've
seen where you are able to show, that what appears logical
on the surface is not so bright once it is analyzed.
What are your thoughts on this strategy and what would
the true odds be, if you did take the bets down after one
hit? - Michael Andrews
This is similar to a question I got last
week. Yes, it is true that there are 10 ways to roll
a 6 or 8 and 6 ways to roll a 7. However one must not
look at the probabilities alone but weight them against
the payoffs. The place bet on the 6 and 8 pays 7 to 6
odds when fair odds would pay 6 to 5. By making 6 unit
place bets on the 6 and 8 and taking the other down if
one wins the probability of winning 7 units is 62.5% and
the probability of losing 12 units is 37.5%. If the
player must cover both the 6 and 8 then the place bet is
the way to go. This rate of return isn't bad but could be
better. For the player who puts a priority on minimizing
the overall house edge, the best strategy is to make
combinations of pass, don't pass, come, and don't come
bets, and always take the maximum allowable odds.
June 18, 2000
Q: Have you ever heard of the craps pro? I can
honestly say that I have been playing this craps method for
a year now and over the long run I have been winning 60% of
my bets with the casino paying me odds on the place bets.
I'm one of the crazies out here that believe that this
works. I have tried many systems and spent thousands of
dollars believing there must be a way to beat the casino
over the long haul. Nothing worked untill I purchased this
method. I'm a true believer and my wallet shows it, or maybe
hundreds of hours at the table is not enough time to prove
it to be true I'll let you know how I do over the next year.
If you haven't heard of this method than research it and
prove me wrong. -Jim of Belluvue, US
A: Although I have tested a lot of systems I
don't need to test any of them to know they are all
worthless. No system can ever pass the test of time, it
is a mathematical fact. It is not unusual to win for a
while with a system but if you keep playing the odds will
eventually catch up to you and you will fall behind. If
you want me to test your system you'll have to pay $100
per hour for my time. In the unlikely event it wins over
a billion bets you won't have to pay me and I'll
publically admit I was wrong.
Feb. 12, 2000
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