house edge for betting across at craps

I've tried to calculate what the house edge would be for betting all 6 place numbers at craps. I'm not a math guy but used the following calculations and would like you good people to please point out where I might be wrong.

My first calculation is that on each roll of the dice there would be 24 ways to win (total possibility on all 6 place numbers) vs 6 ways to lose (on 7 appearing. That would provide a ratio of 24:6, which reduces to 4:1. That would, in turn, mean a player could expect, on average, 4 winning hits to every wipe-out loss of 6 units.

I've used numbers that would be put in play at a $5 minimum table. Making the minimum bet on all 6 numbers across would mean you were risking $32 on each attempt. I further calculated that with the 4&10 paying $9 per hit; and the 5,6,8 & 9 paying $7 per hit, the average profit per hit would be $7.5.

That would mean that, on average, a player could expect to win $30 (4 hits at $7.5 each) for every wipe-out loss of $32. Or, that a player could expect, on average, a loss of $2 for every $32 attempt. That's a ratio of 2:32 or 1:16. 1/16th comes out to a percentage of 6.25%, which I would take to mean as the house edge of betting all 6 numbers across.

Have I figured this out correctly? If not, please show me where I went wrong. Thanks...

Should be 5/16 * 3 /12 * 1/15 + 5/16 * 4/12 * 1/25 + 6/16 * 5/12 * 1/66.

Summation of Bet size * bet frequency * edge for placing 4, 5 and 6.

The usual way to work out House Edge for Craps bets is consider active rolls, for instance in the UK £3 on 5 would pay £4.20. So the House Edge would be (4*£4.20-6*£3)/(4+6)/£3 = £1.20/10/£3 = 4%. You can see this at https://wizardofodds.com/games/craps/appendix/2/
However you can also look at all rolls and get a lower House Edge (NB I've assumed UK odds which pay 9.5 to 5 on 4 or 10 and a typical minimum bet of £3 on each number). With multiple bets running it might be easier to look at the cost per roll given your total bet is £18.
24

Roll	Perms	Payoff	Total
2	1	£0.00	0.00
3	2	£0.00	0.00
4	3	£5.70	17.10
5	4	£4.20	16.80
6	5	£3.50	17.50
7	6	-£18.00	-108.00
8	5	£3.50	17.50
9	4	£4.20	16.80
10	3	£5.70	17.10
11	2	£0.00	0.00
12	1	£0.00	0.00
Total	36		-5.20
Average cost/roll	36	-0.802%	-£0.14
Averagecost /active roll		-1.204%	-£0.22

An alternative way to look at it is how much you have to wager to win $100 if any number (4 5 6 8 9 or 10) is thrown given the fair price would be $400. It is equivalent to making a total wager of $100 and only receiving $23.83.513c instead of a fair value of $25.
Again I've used UK place odds.

Roll	Odds	Bet Needed
4	19/10	$52.63.158c
5	7/5	$71.42.857c
6	7/6	$85.71.429c
8	7/6	$85.71.429c
9	7/5	$71.42.857c
10	19/10	$52.63.158c
Total bet		$419.54.887c

I used US place odds since I see no indication to assume otherwise. To my knowledge there are like 5 Craps tables in all of England...last year I played at 2 of them in the bottom level of the Hippodrome casino in London. Anyway it looks like the only difference is that the 4/10 pays 19 to 10 instead of 9 to 5.

My calculation is for the house edge on all money wagered. To get a per-roll edge I’d multiply that edge by 7.02 * 30/36 / 8.52.

8.52 being the average roll length until seven out. Minus 1.5 avg rolls to establish a point is 7.02 rolls that aren’t coming out after a seven out. 30/36 of those rolls have a resolved bet since only 2,3,11 and 12 have no resolution when a point had been established.

Ace2 and charliepatrick, I want to thank you both for taking the time and effort to reply to my post. Please know it is very much appreciated. I may not have been clear in my question. Or, more likely, you've both already given me the answer I was looking for, but I don't have adequate left-brain skills to recognize it. Here's the situation I was trying to address...

A shooter has established his point. I now place a single unit wager on each of the 6 point numbers, including the point to be made. What I was trying to work out in the meatball-math I laid out in my original post is what the house edge -- expressed as a simple single percentage -- would be in that situation on each individual roll of the dice. This would be similar to craps books telling the novice player that the house edge on placing the 6 or 8 is 1.5%; or that the house edge on placing the 5 or 9 is 4%.

Again, in my naivete, I may not realize it's not possible to give a simple, single house-edge percentage on what is essentially an accumulation of 6 wagers being made at once. Whatever the case may be, I thank you both once more for the time and effort you've taken in this regard. Best of luck to you at the tables.

It will be 1/15th of what you have on the 4&10 plus 1/25th of what you have on the 5&9 plus 1/66th of what you have on the 6&8. You said the wagers will be $6 on 6&8 otherwise $5. That’s 3.90% house edge on your total wager. On a per roll basis it’s that * 30/36 = 3.25 % since 6 of 36 rolls (2,3,11,12) have no resolution.

Thanks so much, Ace2. That's what I was looking for. And thank you, too, charliepatrick. Grateful to you both.

Quote: jeremiah22

What I was trying to work out in the meatball-math I laid out in my original post is what the house edge -- expressed as a simple single percentage -- would be in that situation on each individual roll of the dice.

I agree that per roll is a better calculation
expected value = avgResolvedwager * house edge
so once we know the ev we can divide by the avg Resolved wager and get the HE per roll.

alo remember that over 100 rolls there are, on average, only 70 non-comeout rolls that you are after.
that messes things up

ok,
this can and has already been calculated B4 here, I do not have those links, so I have some actual dice rolls and rng computer dice rolls (10 million) in Wincraps
you should be able to get a real close answer as the expected he of 3.73% was very close in this simulation.
REMEMBER, NOT ALL PLACE BETS RESOLVE ON EVERY ROLL.

from wincraps classic


I get, if the wind has not clouded my brain today
from
https://wizardofodds.com/games/craps/appendix/2/
ev per roll = about -0.332976 cents... but WAIT HERE!
IS NOT THAT EVERY ROLL? COUNTING ALL COME OUT ROLLS??

different from the simulation results
-0.2361136 cents per roll
or the shown expected value
-0.2346643

REMEMBER, NOT ALL PLACE BETS RESOLVE ON EVERY ROLL.
this is not fun at all, so I say have fun with this.
you seem to have basic math skills, just remember the difference between 'per roll' and 'per bet resolved'

maybe I can find those links somewhere.
Sally

added: forgot to mention
wincraps wagers it tracks are 'resolved' wagers. meaning that when $32 is working and a 2,3,11 or 12 rolls
there is NOT any bets resolved being ADDED to the tally.
that complicates things more (for calculations) but makes some things simpler at the same time
apples and oranges

last addition:
I like this description about the 32 across method (place all numbers)

the game now is a coin flip (lol)
50% chance of winning $7 on one roll (5,6,8,9)
50% of something else on one roll split equally into 3 results
1/6 chance of winning $9 (4,10)
1/6 chance of winning $0 (2,3,11,12)
1/6 chance of LOSING $32 (7 out)

that ev = -$1/3 (per roll)
probability that any roll will be a point roll (NOT a come out roll) = 392/557
ev of $32 across = -1/3 * 392/557
about -0.234590066 cents per roll (not counting come out rolls)

Thank you for your detailed response, Sally. Very much appreciate it. I also take it by your (lol) parenthetical to the line "the game is now a coin flip" that when you say you "like this description" you are more amused by it than impressed by any validity it might have.

There’s one thing that’s not clear to me.

So the shooter first establishes his point and at that moment you make your 6 place bets. Let’s say the point is 4 and on the next roll the shooter hits his 4 so you win your place bet on that number.

Now....do you take down all $14 ($5 stake plus $9 winnings) from the 4 or just the $9 in winnings, leaving the original $5 wager ?

Pardon my ignorance if I’m missing something obvious or customary. I never make any bets besides pass/don’t/come/don’t/odds. The Wizard doesn’t approve the others

Quote: Ace2

So the shooter first establishes his point and at that moment you make your 6 place bets. Let’s say the point is 4 and on the next roll the shooter hits his 4 so you win your place bet on that number.

Now....do you take down all $14 ($5 stake plus $9 winnings) from the 4 or just the $9 in winnings, leaving the original $5 wager ?

by default,
if one does NOT say anything to the dealer as what to do with a winning place bet, the winnings are handed off to the player and the existing bet is left alone,
set down on the table in front of the player. Seen this many many times.

The dealer may ask "same bet?" and do the same thing, but not to wait and slow the table down as the player thinks long and hard about what to do next.

on the come out roll, by default, all 'place bets to win' are OFF, meaning they can not win or lose on the come out roll.
I think the OP is after just that.

of course a small -ev is dominated by a large variance per roll
many do not get that part either

Sally

That sounds logical.

Jeremíah, please confirm you only take down winnings and you leave bets off on come out rolls.

I'm still working this out, Ace. Haven't put it into play just yet. Been writing pages and pages on how I might do this; and it's not all math-based. A lot of it has to do with what let's call "intangibles"; things that pure numbers guys would understandably snicker at and which I would never try to defend to people as statistically sophisticated as they. How one might try to catch lucky and maximize such a situation if should it occur. I'm not a blackjack guy (beyond playing some rudimentary basic strategy the few times I do sit down to play) but I remember reading where Stu Ungar said he basically "hung on, looking for one good streak." Playing around with how one might best "endure" at a dice table with both smart and minimum exposure and then capitalize on a run of table luck should one feel that it is taking place. Again, a purely subjective and absolutely defenseless call in light of the very impressive math wisdom presented here.

But to answer your question, yes, I would take down winnings as they occur; leave the place/buy bets off on the come-out rolls; then look to begin prudently pressing the hot numbers should someone take off on a run of luck. Experimental? Certainly. Laughable? Yeah, sure. But what the hell, I'm a dice player. Thanks again for your help.

Here is my answer. Please note that I'm counting pushes. So, this would answer the question at hand on a "per roll" basis.

This table shows the amount bet, what a win would pay, the number of winning and losing combinations, and the expected value per roll.

Point	Bet	Pays	Winning combinations	Losing combinations	Exp value
4	5	9	3	6	-0.083333
5	5	7	4	6	-0.055556
6	6	7	5	6	-0.027778
8	6	7	5	6	-0.027778
9	5	7	4	6	-0.055556
10	5	9	3	6	-0.083333
Total	32				-0.333333

The lower right cell shows the player can expect to lose 33 1/3 cents per roll. Compare this to the $32 bet, and you have a house edge of (1/3)/32 = 1/96 = 1.04%.

You might ask, "Wiz, how can this be, each individual bet has a higher house edge than that. It is because the house edge on place bets is traditionaly defined as the expected loss per bet resolved. If the house edge on place bets counted ties, then it would be much less.

Quote: Wizard

Here is my answer. Please note that I'm counting a pushes. So, this would answer the question at hand on a "per roll" basis.

it appears to me
OP does not have any place bets working on the come out roll.
the -$1/3 is where place bets work on the come out roll and that is a rare player indeed that wood due that.

The OP needs to understand why his method would result in a lower ev and what the probability of a point roll would be instead of all rolls.

good point showing 'per roll' and 'bet resolved'
Sally

Quote: Wizard

The lower right cell shows the player can expect to lose 33 1/3 cents per roll. Compare this to the $32 bet, and you have a house edge of (1/3)/32 = 1/96 = 1.04%.

1.04 % doesn’t seem reasonable to me.

I assume we can all agree that the house edge is 3.90 % on all money wagered, which is just the summation of all place bet edges at the 5/5/6 ratio.

So for the per-roll edge to be 1.04 %, that implies that only 1 in 3.75 (390/104) of all bets are resolved.

I believe that 30 of 36 bets are resolved, which would put the per-roll edge at 3.25%. The only time that bets aren’t resolved is when a 2,3,11,or 12 is rolled (6 out of 36 combinations). Come out rolls are insignificant since bets either haven’t been placed (new shooter) or place bets are off (same shooter just hit point). So come out rolls are not ties, they simply don’t exist since nothing is ever wagered unless a point has been established (original poster confirmed). Every other number - 4 through 10, 30 of 36 combinations, is a resolved bet for this scenario.

Or I am missing something fundamental here..

Quote: Ace2

1.04 % doesn’t seem reasonable to me.

or my or others simulation results
when one has as total action bets resolved, not $$$ risked. IF $$$ risked (bets on the table) mattered why not include all money in the rail too?

Quote: Ace2

I assume we can all agree that the house edge is 3.90 % on all money wagered, which is just the summation of all place bet edges at the 5/5/6 ratio.

I think many others have show that to be 3.73% but what is more important really,
HE or EV?

Quote: Ace2

So for the per-roll edge to be 1.04 %, that implies that only 1 in 3.75 (390/104) of all bets are resolved.

the Wizard used $32 at risk, for what reason, still remains unknown. for 10 million rolls, that would mean that $32 million was resolved bets. do knot C that happening.

Quote: Ace2

Or I am missing something fundamental here..

I think not. IF the math for this combinations of bets was so easy to do, everyone would do it. and of course they all do.

now what are the results and what do they mean and are there any simulations that match any calculated results.

it is also a rare place bettor that keeps the wagers at the same unit all the time. Press to win more $$$.
Press to lose more $$$.
Sally

Quote: jeremiah22

then look to begin prudently pressing the hot numbers should someone take off on a run of luck.

I would suggest getting your plan together and simulate it. that would give you results from many trying your system to see what kind of range of wins and losses to expect.
MOST ALL place bettors I have seen think they win most all the time and only lose because of
1) poor money management,
2) bet management (not pressing their bets when the shooter does get hot, or taking down their bets B4 the shooter gets cold - locking up a profit)
3) betting on poor shooters (should be on the don't side for those)

well, have fun no matter what you get done here
Sally

had to add this
flat-betting across place bets
this was calculated (Markov chain)

the more bets you make, the GREATER the chance U have of showing a net loss over X # of equal wagers.
not good I say

Quote: mustangsally

I think many others have show that to be 3.73% but what is more important really,

I now agree with that number for this scenario. The 3.90 % would be correct if place bets were taken down individually as they won. But as I learned that is not common practice nor what the original poster does.

So the place bets start out at a $5/5/6 ratio but the subsequent wins and additional bets will come at a 3/4/5 ratio, slightly skewing the edge down since a higher % (5/12ths not 6/16th) of new wagers will go to lower edge bets 6&8 and a lower % (3/12ths not 5/16th) of new wagers will go higher edge bets 4&10.

It’s already been shown that on average a shooter’s turn lasts 8.53 rolls, and exactly 6 of those rolls, on average, are made with an established point. The residual 2.53 rolls are come out rolls which have no action in this scenario. Incidentally you can get to that number by knowing that on avg it takes 1.5 rolls to establish a point. We also know that on average a shooter hits 0.68 points, so he has (1 + .68) * 1.5 = 2.53 comeout rolls. Therefore we know that our bettor averages 6 active rolls per turn, one of which is the final seven out and one of which (6/36 * 6) is a 2,3,11 or 12 which is a tie. The remaining 4 rolls are wins and, for example, 3/24 * 4 * $5 = $2.50 of new wager gets added on to the intial $5 placed on the 4.

Crunch the other numbers and the combined bet ratio (initial plus additions) is $7.50 / 8.33 / 11.00 = $ 26.83 per side. Apply the individual place bet house edges (7.50 * 1/15 + 8.33 * 1/25 + 11.00 * 1/66) / 26.83 = 3.73% combined.

So the per roll number would be 3.73% * 5/6 = 3.11%.