endermike
Joined: Dec 10, 2013
• Posts: 584
May 6th, 2014 at 9:03:27 PM permalink
So, of course I would like to distance myself from the heated rhetoric being used by Ahigh in this thread. However I do maintain there is something to his thesis.

Quote: MathExtremist

Nobody plays for X action, except perhaps those with match-play coupons. Players will typically play for a loose amount of time or until some (perhaps not predefined) win or loss event.

I agree with you. I should have been more clear when I was referring to X I was referring to X per hand/come out, not X as a total handle. On that basis (Pass/Don't), a bet which more than 2/3rds of most table makes, the odds make a large difference in the types of results experienced. In that way, if you want to have about \$25 in play on each roll, a \$5 table with 3/4/5 is the place for you (as opposed to a \$25 table). I know this applies to me. When I am at a table/machine that has less than 3/4/5 odds and a min bet less than \$5 I go max odds with multiple come bets. My utility regarding play time is complex. Generally, I do want a longer play time, however I also want a certain amount of variance in my results such that I can feel the high from gambling (hence why I would never play a doey-don't even if it would allow me to essentially play forever at a low loss rate and nearly zero variance).

Quote: MathExtremist

That's why looking at expected loss (rather than house edge) is preferable. If you flat-bet the passline for an hour, you'll average a half-bet loss. If you play passline + any amount of odds, you'll still average a half-bet loss. Saying one has a lower house edge than the other is only true because you're inflating the denominator. It's all well and good to say that you played \$5 craps with 100x odds and had a house edge of 0.021%, but that's because you divided 7.1c by a big number. Your expected loss was still 7.1c per \$5 bet.

I actually agree with you that ELPH is is an underrated measure of a games cost. However, ELPH misses any measure of variance. That is why other measures are also important. This post does a great job of giving three measures which together do a great job of comparing most any game on apples to apples terms which are approachable and effective. I highly recommend it for all, it summarizes most of my feelings on the topic a games desireabilit from players ecomonical standpoint. (Aside from wanting to add a bunch of stuff about utility theory, which they get into later)

That was my point in my post addressing RS's points (the one you quoted). I was trying to point out that the odds make a major difference on how your bankroll fluctuates, and I believe taking free odds is in all but some rare cases are a good choice. (Hence, the whole reduced juice analogy)
Quote: MathExtremist

You can use percentages in a very misleading way, as Ahigh has been amply demonstrating of late. Watch:
Place 6 EV: 1.52%
Place 8 EV: 1.52%
Make them "together":
With probability 10/16, win \$7
With probability 6/16, lose \$12
Expected loss is \$2/16 for each \$12 or 1.042%. Huh? That's a lower percentage than the individual bets.

But it gets better. Let's bet \$22 inside:
18/24, win \$7
6/24, lose \$22
Expected loss is \$6/24 for each \$22 or 1.136%. That's lower than the individual edges of any of the 4 place bets you just made.

Last one, \$32 across:
18/30, win \$7 (5,6,8,9)
6/30, win \$9 (4 or 10)
6/30, lose \$32
Expected loss is \$12/30 for each \$32 or 1.25%.

Do you think these are accurate or useful computations?

I think they are accurate. I know the combined edge is less than the average of the individual bets. (The whole appears to be better than the sum of the parts.) However, in reality the resolved bet HEs on these bets can not be combined as you listed. Many times the bets don't resolve in the standard way; they resolve via the player taking them down. This is another point which leads me to agree with another thing you have bought up "the denomenator." The denom is important because the EV is divided by it to calculate the HE. Clearly all of the investment is at risk so by that simple measure the EV calculation is spot on. However, here again is where variance is important. Each of those bets individually has higher coefficient of variation than the "hybrid" bets where they are brought down if one wins. HE is accurate but it is not the whole story.

What you are doing with the the bets which do not win but also do not lose when one of the bets does win is you are taking them down. Hence, you are avoiding the edge which occurs when those bets resolve. I of course agree you are bucking the same HE on a per roll basis. Imagine the following bet:
"\$50, 'Buy the 4, with a retreat after 10'"
In this bet:
-If a 4 is rolled: win \$48
-If a seven is rolled: lose \$50
-If a 10 is rolled: push and remove the bet (retreat)
-If a 2, 3, 5, 6, 8, 9, 11, or 12 is rolled: do nothing and roll again

Clearly this set of actions has an EV of -\$.5 upon resolution, so long as 10 is considered a resolution. Here we get to the rub, is it reasonable to consider 10 a valid resolution? Ahigh says yes, others say no. Personally, I think if someone actually plays this way, it is reasonable for them to think of it like this. I would not teach a new player this way because I think it adds layers of complexity which are unlikely to be useful initially.
Quote: MathExtremist

Would you ever tell someone that \$32 across is a "better" bet than the \$6 place bet?

No, however I would say that the bets can "be effectively" or "appear to be" better if you take them down often. The reason for this is that those edges are based on the bet always resolving as a win or a loss. When you add extra resolution conditions which refund your investment, you change the edge on a per resolution basis if you actually follow through with such behavior.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 6th, 2014 at 9:20:48 PM permalink
Quote: MathExtremist

Would you ever tell someone that \$32 across is a "better" bet than the \$6 place bet?

Quote: endermike

No, however I would say that the bets can "be effectively" or "appear to be" better if you take them down often. The reason for this is that those edges are based on the bet always resolving as a win or a loss. When you add extra resolution conditions which refund your investment, you change the edge on a per resolution basis if you actually follow through with such behavior.

This is exactly the problem Ahigh has with his math skills.
from day 0

He almost always compares apples to oranges.
always
But says they are the same (his opinion)
per roll edges against per bet resolved edges

edges from total at risk for one roll verses edges over 6 rolls

stuff like "this bet is way better if you bet like this and not like this"

not to mention the accuracy of his simulations
too short in length and / or errors in his code

My uncle gave up on his (Ahigh) math after his website craps program constantly showed with a bankroll win goal of \$200
ending up with \$205, \$210 and \$215 while making just \$5 flat bets.

that was so funny too when I tried it out. Got a good laugh from it
He never did fix his code, just joked about it and
just removed it so no one could ever use it again

that is a great solution to most all problems

Sally
bring on the videos!
I Heart Vi Hart
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
May 6th, 2014 at 9:32:12 PM permalink
Quote: endermike

So, of course I would like to distance myself from the heated rhetoric being used by Ahigh in this thread. However I do maintain there is something to his thesis (regarding combining bets).

(Regarding \$32 across vs. \$6 place 6)
I would say that the bets can "be effectively" or "appear to be" better if you take them down often. The reason for this is that those edges are based on the bet always resolving as a win or a loss. When you add extra resolution conditions which refund your investment, you change the edge on a per resolution basis if you actually follow through with such behavior.

I think there is a fundamental distinction between "appear to be better" and "be effectively better". It is certainly clear from this thread that the combination of several bets "appear to be better" than their individual components, at least to a few people. But they are not actually "effectively better" -- the expected loss for any combination of bets is exactly the same as the expected loss for the bets individually. That's especially easy to see if you examine the expected per-roll loss as opposed to the per-resolution loss, but you can arrive at the equivalence either way as long as you do the calculations correctly (see Steen's earlier posts).

The mistakes in the per-resolution approach creep in when you start adding in handle (denominator) from bets that didn't actually resolve. I make a place 6 bet, a 6 rolls, that's a resolution for the place 6 bet. I make a place 6 and place 8 bet, a 6 rolls, that's a resolution for the place 6 bet but not the place 8. Otherwise the screwy numbers I posted earlier would be accurate, but then what's stopping you from suggesting that money you have in action elsewhere should count too? I could make a \$6 place 6 bet, lean over to the next dice table and make another \$6 place 6 bet, win with a 6 on my first bet and then take down the second. I won \$7 on \$6 in action, not \$12.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
endermike
Joined: Dec 10, 2013
• Posts: 584
May 6th, 2014 at 9:40:16 PM permalink
Quote: Buzzard

" This is a disappointing post. It misses the greater discussion, adds no value, and is below the standards I have, even for low value posts. Buzzard, I understand much of what you do is shtick, but this thread has some merit and the question of if we should consider bets which are hybrids as reasonable strategies is, at least to me, an interesting one. "

Looks like beside having moderators, we now have secret editors.

Evidently this met your high editorial standard :

" It has become pure comedy to me that you and Sally both fail here. "

Your joking aside (which in this case did amuse me), I was hoping it does not take someone in green to point out to you when you are denigrating the quality of a thread. I have often found your posts amusing (or at least inoffensive enough and short enough to be only marginally -EV). However your post in this case simply was beyond my personal threshold.

Your post added nothing to drive the discussion forward. You simply tried for humor or a swipe at Ahigh. If what I quoted was part of a larger post where you made our own opinions clear and supported them I would have never brought it up; but it wasn't. You went for a cheap joke and/or a thinly veiled swipe at Ahigh. I would hope that you have enough respect for the members of this board and the folks in this discussion to not drown out signal with noise.

The reason I ignored Ahigh's poor tone was that he was an active participant in the argument. His quote was part of a larger post where he tried to make a point clear and not simply taking a pot shot where it could be done. I agree his tone was not cordial but, I suggest he has some mitigation as he had received some significatnly hostile blowback in this thread and so the tone shifted as it inevitably does. I'm not going to parse who said what first. My issue was that you have no substantiated side in the argument. You simply choose to make a post which sowed disharmony for what I can only assume was your own amusement without any +EV opinions.
endermike
Joined: Dec 10, 2013
• Posts: 584
May 6th, 2014 at 10:08:14 PM permalink
Quote: Steen

Allow me to tailor an excerpt from WinCraps' help file:

Craps bet are not synergistic. The erroneous conclusion of Ahigh's argument comes from not expressing the EV as a function of the action. When one of his Buy 4 or Buy 10 bets wins, the amount won is solely a function of the amount that was wagered on the winning bet and not on the other bet. For instance, if a 10 hits then the \$49 win is a function of the \$25 that was wagered on Buy 10 and not the \$50 that was wagered on both. Yet the argument treats the full amount wagered on both bets as if it had a bearing on each winning outcome. This is wrong.

Intuitively, since both Buy bets have EV's of -1.33%, we know that a combined wager should also have an EV of 1.33%. To arrive at this (per decision) figure we simply need to divide our average loss by our average action.

The average loss is figured as follows:

6/12 chance to win \$49 = \$24.50
6/12 chance to lose \$50 = -\$25.00
----------------------------------------
Average loss per decision = -\$0.50

The average action is figured as follows:

3/12 chance of rolling a 4 and producing \$25 worth of action
3/12 chance of rolling a 10 and producing \$25 worth of action
6/12 chance of rolling a 7 and producing \$50 worth of action
-----------------------------------------------------------------------
Average action per decision = \$37.50

Hence, the EV per dollar of action = -\$0.50/37.50 = -1.33%.

By the way, MathExtremeist is correct that the generic Buy 4 or 10 bet EV is -1.67%. This is figured on the correct vig of 5%. As we know, the casinos often round down fractional vigs, so rather than collect the correct \$1.25 vig for a \$25 Buy 4 bet, some casinos bend a little and only collect \$1 vig thus reducing the house advantage (in this case) to -1.33%

Steen

I must say this is a very strong argument. It very convicing as a unification of the two viewpoints. However, I would say that it may not be fully consistent with how HE's are calculated in general.
Quote: The WoO Craps page, emphasis added

Defining the House Edge

Craps has a lot of different kinds of bets. Some always resolve in one roll and others may take many rolls. The standard definition of the house edge is the ratio of the expected player loss to the initial wager. Almost every legitimate gambling writer counts pushes in this calculation. However, in craps it often takes many rolls to resolve a bet, with the player being allowed to take down the bet at any time until it wins or loses.

craps page

The standard definition of the house edge is the ratio of the expected player loss to the initial wager. Maybe we could say that in this case since the bet is such a semi-degenerate (in the math sense) case where only half of the bet will ever get action in a win even though all the bet can lose we should redefine HE as the E(loss)/E(total action)=-.5/37.5=-1.33%. But as we all know, that is clearly the Element of Risk.

So I think we have reached the beginning of some agreement. In the Element of Risk sense the "hybrid buy" has the same EOR as the combination of its' component bets. However, in the House Edge sense the "hybrid buy" has a HE of 1% which is less than the average of its component parts. (Many would say this is because it is doing some psuedo-hocus-pocus with how it gets paid on its' initial bet). I would take the view we have come upon an excellent example of where using HE is probably the inferior measure compared to EOR.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 6th, 2014 at 11:21:12 PM permalink
Quote: ThatDonGuy

No - that's what I was asking. I didn't notice it at the end of Catlin's article. I wasn't expecting a method to count rolls to be headed "House Edge."

hehe
Don Catlin has a few articles like that

This paper was also in my notes
http://www2.math.uu.se/~sea/kurser/stokprocmn1/slumpvandring_eng.pdf
Theorem 5 on page 10

Sally
I Heart Vi Hart
Ahigh
Joined: May 19, 2010
• Posts: 5195
May 7th, 2014 at 12:10:11 AM permalink
I apologize for any hostile tone. I think those that were targets of mine took it in stride and I appreciate that.
aahigh.com
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 7th, 2014 at 12:12:24 AM permalink
My first answers for the OP questions
Quote: mustangsally

First off,
Bankroll \$1000 and \$10 flat bets (bankroll units = 100. So it does not now matter what your \$ unit is. bankroll of \$100 and flat bet of \$1 is equal to a Bankroll of \$2500 and \$25 flat bets and on and on)

final summary (100 units into 110 units)
\$1000 into \$1100
90.410687% from Bold Play
88.406245% with 345X odds
74.2149768% with 0 odds

\$1000 into \$2000 (100 units into 200 units)
49.2929% from Bold Play
46.885% with 345X odds
5.5804887% with 0 odds
Sally Oh

The 0 odds has always bothered me, now that Steen posts here, about being a *fair* comparison.
many like to make unfair comparisons to prove their points.

so for this we need to see the average unit bet the 345x odds player makes
I know that number = 34/9 or 3 7/9
so again the odds taker is betting way way more than the 0 odds player.

if we let the 0 odds player flat bet 4 units (rounded up)
the results are

final summary (100 units into 110 units)
\$1000 into \$1100
90.410687% from Bold Play
88.406245% with 345X odds (34/9 unit avg bet)
87.3714650% with 0 odds (4 unit avg bet)
74.2149768% with 0 odds (1 unit avg bet)

\$1000 into \$2000 (100 units into 200 units)
49.2929% from Bold Play
46.885% with 345X odds (34/9 unit avg bet)
33.0236003% with 0 odds (4 unit avg bet)
5.5804887% with 0 odds (1 unit avg bet)

Sally
I Heart Vi Hart
Ahigh
Joined: May 19, 2010
• Posts: 5195
May 7th, 2014 at 12:19:42 AM permalink
Good job
aahigh.com
Steen
Joined: Apr 7, 2014
• Posts: 126
May 7th, 2014 at 2:10:36 PM permalink
Quote: endermike

I must say this is a very strong argument. It very convicing as a unification of the two viewpoints. However, I would say that it may not be fully consistent with how HE's are calculated in general.

Quote: The WoO Craps page, emphasis added

Defining the House Edge

Craps has a lot of different kinds of bets. Some always resolve in one roll and others may take many rolls. The standard definition of the house edge is the ratio of the expected player loss to the initial wager. Almost every legitimate gambling writer counts pushes in this calculation. However, in craps it often takes many rolls to resolve a bet, with the player being allowed to take down the bet at any time until it wins or loses.

craps page

The standard definition of the house edge is the ratio of the expected player loss to the initial wager. Maybe we could say that in this case since the bet is such a semi-degenerate (in the math sense) case where only half of the bet will ever get action in a win even though all the bet can lose we should redefine HE as the E(loss)/E(total action)=-.5/37.5=-1.33%. But as we all know, that is clearly the Element of Risk.

"Action" is entirely consistent with the standard definition of "initial wager". In fact, when evalutating the HA of a single bet, action is exactly the same as the intial wager. However, it's important to remember that the initial wager is limited to the amount risked on the bet in question and not on other bets. This requirement is often overlooked when players attempt to evaluate the HA of multiple simultaneous bets.

I like to use the term "action" because it because it more clearly defines the "initial wager" to be the amount risked on the resolved bet or bets, and not the amount on other unresolved bets.

Does it really make sense to you that simultaneous Buy 4 / Buy 10 bets could possibly have a lower HA than either bet alone? By limiting the "initial wager" you use in your calculations to the amount wagered on the resolved bet in question (the action) you can find the right answer.

Consider this: Mr. Whale likes to play \$1000 Buy 4 and is accustomed to the \$50 vig for a HA of 1.67%. He reads Ahigh's post and decides to give it a go by simultaneously betting Buy 10. However, since he prefers Buy 4, he elects to wager a pitance on Buy 10. He figures (vig on win only):

--------------------
Total initial wager \$1002

3/12 chance to win \$1950
3/12 chance to win \$3 (assume house takes min vig of \$1)
6/12 chance to lose \$1002
------------------------------------------------------
Average loss per decision \$12.75

EV = -\$12.75/\$1002 = -1.272%

Wow! Mr. Whale is elated. By adding a trival \$2 bet to Buy 10, he thinks he's lowered the HA from 1.67% to 1.272% even while paying a tremendously high vig% on Buy 10. Are you still buying it?

Now, regarding the counting of pushes ... I personally don't count them in craps but I'm speaking in terms of single wagers and not multiple wagers that might cancel each other. Most authors I've seen address this issue talk about counting or not counting the Bar 12 on the Don'ts during the come-out roll. But if a 12 is a push and must be counted for the Don'ts then what about bets like say the Place 6?

The Place 6 (when working) pushes for any non-6 or 7. Do you ever see these same authors count these pushes for the Place 6?

In truth, I don't think it really matters as long as you're consistent and the reader understands what the numbers are representing. For a person who counts pushes, the same average outcome is spread out over all decisions including pushes rather than just those decisions which show a net gain or loss.

Steen