Adding Players Edges...

Couldn't find the answer to this one, so figured I'd ask it here...

If you make a Pass Line bet and the Players Edge is -1.41, and let's say you make a Place bet of 8 after a point is made (not an 8, of course) - which is -1.52, do you just add these 2 numbers to come up with the house edge? (Ex. -1.41+-1.52=-2.93)

Thanks :)

Quote: chipsndice

Couldn't find the answer to this one, so figured I'd ask it here...

If you make a Pass Line bet and the Players Edge is -1.41, and let's say you make a Place bet of 8 after a point is made (not an 8, of course) - which is -1.52, do you just add these 2 numbers to come up with the house edge? (Ex. -1.41+-1.52=-2.93)

Thanks :)

You need a weighted average based on the amount of each bet. So for example, if your pass line bet was $5 and your place 8 was $6, then you have $11 total in action. So to get your total house edge, you multiply the edge on the pass line by 5/11 and the edge on the place 8 by 6/11, and add those two numbers together, like so:

-0.0141*(5/11)+-0.0152*(6/11) = −0.0147

So your combined house edge is -1.47%. Another way to think about is to leave the percentages behind, and just calculate your total expected loss. This is easier as you don't have to worry about the weighted average, you just take each expected loss separately and add them together. The end result is that you just leave off the two denominators of 11 in the equation above, leaving you with:

-0.0141*5+-0.0152*6 = -0.1617

So your total expected loss on these two bets is is $0.1617, or about 16 cents. This the same result if you take your combined house edge from the first equation (-0.0147) and multiple it by your total action ($11).

Quote: chipsndice

Couldn't find the answer to this one, so figured I'd ask it here...

If you make a Pass Line bet and the Players Edge is -1.41, and let's say you make a Place bet of 8 after a point is made (not an 8, of course) - which is -1.52, do you just add these 2 numbers to come up with the house edge? (Ex. -1.41+-1.52=-2.93)

Thanks :)

No, if you just want to know the overall House Edge on the two bets (Mean) you would add them and divide by two assuming you are risking the same $$$ amount on each.

The overall House Edge on your action would be ~1.465%, in this case.

If you are betting different amounts (such as $5 PL and $6 Place Eight) then you would either have to use an expected loss formula.

Expected Loss Formula

5 * .0141 = .0705
6 * .0152 = .0912

(.0912 + .0705)/11 = 0.0147 or 1.47%

Expressed in full:

((5*.0141)+(6*.0152))/11 = .0147 or 1.47%

I spread it out at first just so it would be clear what I was doing.

Quote: Mission146

Expressed in full:

((5*.0141)+(6*.0152))/11 = .0147 or 1.47%

I spread it out at first just so it would be clear what I was doing.

Strictly speaking, this is also wrong.

You will always make a Pass Line bet, but you will only Place Eight on 19/36 occasions.

36/36 - 12/36 (Come Out Winners/Losers) - 5/36 (Eight is the Point) = 19/36

Thus:

((5*.0141)+(19/36 * 6 * .0152))/((5+(19/36*6)) = 0.01452653061 or 1.452653061%

PROOF

For every $5 on PL, you will bet (19/36 * 6) = 3.16666666667 on Place Eight

Going back to the original formula:

((5*.0141)+(3.16666666667*.0152))/8.1666666667 = 0.01452653061 or 1.452653061%

Quote: chipsndice

Couldn't find the answer to this one, so figured I'd ask it here...

If you make a Pass Line bet and the Players Edge is -1.41, and let's say you make a Place bet of 8 after a point is made (not an 8, of course) - which is -1.52, do you just add these 2 numbers to come up with the house edge? (Ex. -1.41+-1.52=-2.93)

because the house edge is an average so to speak
I agree with Mission with his first opinion
you would be much closer after you divide by 2.

another one of my opinions,
AcesAndEights
answered some one else's question, not yours.

This will happen a lot when one wants to know about combined house edges because it forces one to calculate all the possible outcomes to get an average bet
and that could be per roll or per bet resolved.

Your average bet , as I see how you worded your question, could never be $11.

Now Mission also gave a few opinions and stated one with a proof.
Maybe to add more weight to his opinions. I do not know.
Did his opinion (the proof one) answer your question?

well, in my opinion,
because you did not state if your place 8 was or was not working on any come out roll
I think his answer is that your place 8 is always working on the come out roll.
Not a common occurrence at all.
This is not the default rules of Craps, place bets are off on the come out roll,
so one would need to know exactly how you would play this out.

I think my opinion might be closer than the other opinions so voiced so far
because I would calculate every possible outcome and that would be per roll and per bet resolved.
Hmmm, the pass line and the place 8 resolve at different rates too.


my question to you is why do you want to know what the combined house edge is
instead of the total expected loss (expected value) per decision or per roll?
and another would be what do you think would be your average bet?

here is another opinion too
WinCraps (both versions) shows this for not working the place 8 on the come out roll for the combined edge


You, OP, are in for many different opinions here

you can also look here for some good information
https://wizardofodds.com/games/craps/appendix/2/

fun stuff huh?
I agree
Sally

you can add the expected values if they are both in $ units..

Quote: chrisr

you can add the expected values if they are both in $ units..

true to a point, in my opinion,
if the EVs are calculated the same way.
In Craps may bets resolve on one roll
many others one or more as many know.

so adding the EVs from a Field bet to an EV for a Place 5 bet that is calculated per roll will be different from
adding the EVs from a Field bet to an EV for a Place 5 bet that is calculated per bet resolved.

If
expected value = house edge * (avg bet * # of bets)
Then
house edge = expected value / (avg bet * # of bets)

(avg bet * # of bets) or action
can be very different on a per roll and a per bet resolved basis on what is also counted as action.
Some count total action as all bets at risk and not any bet that resolves.
It really makes no difference except for mixing up apples and oranges, very easy to do.
This is where one has to be very careful, not only doing the math but
also from the conclusions gained from the results, but most are not careful by this point.

Sally

You can't really add, or average, the edges for two reasons.
First, the bets can have different termination points.
Second, since you wait until a point is made before placing the 8, the edge on the pass bet changes to making the point before the 7 (which is -9.09 if the point is 6 or 8, -20 if 5 or 9, and -33.33 if 4 or 10).

You can calculate the "potential house edge" on the combined bets, by looking at the possible results:
Natural
Crap out
Make point, then 8, before 7
Make 8, then point, before 7
Make point, then 7 before 8
Make 8, then 7 before point
7 before either point or 8
Note that, except for the first two, each one has to be calculated separately for each of the five points besides 8 (although, of course, once you calculate 4 and 5, the numbers are the same for 10 and 9).

Here's how I calculate it:

Natural: 8/36 x +1 = +2/9
Crap out: 4/36 x -1 = -1/9
(I treat the place bet as a push in both cases)

Point 4 or 10: (1/6 - there are 3+5+6 = 14 rolls)
4, then 8, before 7: 3/14 x 5/11 x +13/6
8, then 4, before 7: 5/14 x 3/9 x +13/6
4, then 7, before 8: 3/14 x 6/11 x +1/6
8, then 7, before 4: 5/14 x 3/9 x +1/6
7, before 4 or 8: 6/14 x -2

Point 5 or 9: (2/9 - there are 4+5+6 = 15 rolls)
5, then 8, before 7: 4/15 x 5/11 x +13/6
8, then 5, before 7: 5/15 x 4/10 x +13/6
5, then 7, before 8: 4/15 x 6/11 x +1/6
8, then 7, before 5: 5/15 x 6/10 x +1/6
7, before 5 or 8: 6/15 x -2

Point 6: (5/36 - there are 5+5+6 = 16 rolls)
6, then 8, before 7: 5/16 x 5/11 x +13/6
8, then 6, before 7: 5/16 x 5/11 x +13/6
6, then 7, before 8: 5/16 x 6/11 x +1/6
8, then 6, before 5: 5/16 x 6/11 x +1/6
7, before 6 or 8: 6/16 x -2

Point 8: (5/36) no place bet (I treat this as a push on the place bet)
8 before 7: 5/11 x +1
7 before 8: 6/11 x -1

The totals are:
Natural: 0.2222222222
Craps: -0.1111111111
4/10: -0.0548340548
5/9: -0.0107849327
6: -0.0147306397
8: -0.0170499639
The sum is -0.0170499639 per come-out (based on pass line and place bets of 1, and assuming the casino has chips in denomination of 1/6).
If you make the pass line and place bets both $6, you lose 0.102299784 per come-out.

A Monte Carlo shows that about 0.0145 is lost for each $1 bet.

Quote: mustangsally

another one of my opinions,
AcesAndEights
answered some one else's question, not yours.

Strictly speaking, you're correct, as I didn't take into account the fact that the bets start and end at different points during the game.

But you can't deny that if you make a $5 pass line bet, and a $6 place-8 bet, your total expected loss is in fact $0.1617.

This is why I like to focus on total expected loss when dealing with craps, since the house edge can get confusing when making multiple bets, especially when you start talking about per-roll edge or per-resolution edge. Just figure out what bets you'll be making and at what frequency, and you can calculate your expected loss per unit time, given an estimate of rolls per unit time.

And of course you should completely ignore the standard deviation and variance. Because everyone loses their expected loss exactly!

(That last line was just a joke, really sally don't yell at me!)

(post deleted - it was a correction to my earlier post before I realized there is no time limit to edit my posts on this forum)

Wow, thanks everyone!
I'm really getting into the statistics side of things, and I'm finding it rather fun - though leaves my head spinning most of the times.

Sorry for any confusions.
I was hoping it would be easy enough to apply a Pass/Place formula to any bets, though I did see a couple of cool explanations on how to do that.
It's going to take me a while to go through everyone's answers in detail and with pen/paper (or most likely - Excel LOL) until I get things right.

To elaborate on what I meant, I just want to do a typical Pass Line bet (no odds), and Place the 6 and 8.
And if 4,5,9, or 10 become the point, still place the 6 and 8. (Still no odds)
The 6/8 would be placed after the point is established and will play out until it's resolved after 7'ing out.
The 6/8 won't be working during C.O. as is typical.
That's one beast I'm going to be trying to figure out and put into an Excel doc eventually.
With your answers, it looks like I can accomplish this.
I do want the house edge(s), but expected loss sounds very interesting as well.
Thanks to everyone, I think I'll be able to get both!

Then the next monster is trying to figure out how pressing affects this type of typical play.
All this, while just starting to learn hardcore statistics that you math geniuses seem to already have down pat! (This forum is phenomenal!)
Seriously - thanks!

Quote: chipsndice

I just want to do a typical Pass Line bet (no odds), and Place the 6 and 8.
And if 4,5,9, or 10 become the point, still place the 6 and 8. (Still no odds)

The 6/8 would be placed after the point is established and will play out until it's resolved after 7'ing out.
The 6/8 won't be working during C.O. as is typical.

That's one beast I'm going to be trying to figure out and put into an Excel doc eventually.

I would start by using Excel unless you do not normally use it.
This example of your playing system is easier to understand what you are trying to accomplish in my opinion.

Play until 7 out and start again.
So this can be calculated on a per shooter basis too.

But you still might need to be more specific
"The 6/8 would be placed after the point is established and will play out until it's resolved after 7'ing out"
so does this mean after a place bet win you keep the same bet and take the winnings?
so you could win 20 times on your two place bets before a 7 out and you would still have $6 on each of them.

Sally

Thanks again Sally,

Excel is actually one thing I am an expert it ;)

I just want to assume 1 shooter for now - since it may be easier (?).

"so you could win 20 times on your two place bets before a 7 out and you would still have $6 on each of them"
You're right... That's why I don't want to worry about pressing just yet.
I just want to get the basics down of same bet until 7'ing out. So stays $6 till the end.
Once I have that, then I want to try and tackle pressing/regressing, etc. (That's a war for another day)