Contrarian theory - avoid the pass line??
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I normally play on $25 tables.
The pass line is an even money bet. Bet $25 to win $25 on a 7 or 11 (8 ways to roll those numbers) loose on a 2, 3 or 12 (4 ways to roll) the other 24 rolls set the point.
While you are getting better odds on the come out you are only getting paid even money on the established point. If the point is a 4 or 10 you are getting 1:1 when you should be getting 9:5 (place bet odds).
Yes you get the option to add odds to the pass line and get paid true odds on that portion of your bet, but you don't get to pick the number that you get odds on, the come out roll picks that number randomly for you.
Its almost like the pass line, while tempting for 1 roll, is akin to a "buy in" to get true odds on 1 random number.
Wouldn't you be better off taking the pass line bet and any odds you would have placed and spreading it out as part of a place bet strategy?
Quote: mwz524
Wouldn't you be better off taking the pass line bet and any odds you would have placed and spreading it out as part of a place bet strategy?
I guess it depends on what you mean by, "Better off."
If you mean a better expected value, then no, because a Pass Line bet has a lower house edge than any of the Place Bets. The odds bet has no house edge at all.
Lets take a $25 pass line bet and $50 odds on a point of 4 - a winner would pay $100 odds win plus $25 pass line = $125
$75 place bet on the 4 pays 9 to 5 or $135. the difference is the come out roll where you have a 22% chance to win even money but also an 11% chance of loosing your bet. (i think i figured the stats correctly 8/36 = 22.2 and 4/36 = 11.1)
Quote: mwz524When you calculate the house edge are you calculating it base don the come out roll or on the total of all rolls?
Lets take a $25 pass line bet and $50 odds on a point of 4 - a winner would pay $100 odds win plus $25 pass line = $125
$75 place bet on the 4 pays 9 to 5 or $135. the difference is the come out roll where you have a 22% chance to win even money but also an 11% chance of loosing your bet. (i think i figured the stats correctly 8/36 = 22.2 and 4/36 = 11.1)
The first thing that we should clear up is that the Pass Line bet and the Odds bets are completely separate bets. It's true that you must make a Line bet in order to make an Odds Bet, but making a Line Bet does not require that you make an odds bet.
When the house edge for the Pass Line is calculated, it is calculated based on all possibilities. Fortunately for me, I recently did this on a blog post response for the Don't Pass, so I will copy that here. The Pass bet is the same concept:
Quote:Come Out + Point Results and Outcomes:
Come Outs 2 & 3 = Automatic Win (3/36) = 0.08333333333
Come Outs 4 & 10, Point Missed, Win: (6/36) * (6/9) = 0.11111111111
Come Outs 5 & 9, Point Missed, Win: (8/36) * (6/10) = 0.13333333333
Come Outs 6 & 8, Point Missed, Win: (10/36) * (6/11) = 0.15151515151
Come Out 12, Push = 0
Come Outs 7 & 11, Loss: (8/36) = - 0.222222222
Come Outs 4 & 10, Point Made, Loss - (6/36) * (3/9) = -0.05555555555
Come Outs 5 & 9, Point Made, Loss = - (8/36) * (4/10) = -0.08888888888
Come Outs 6 & 8, Point Made, Loss = (10/36) * (5/11) = -0.12626262626
Add them together: (0.08333333333 + 0.11111111111 + 0.13333333333 + 0.15151515151) - (0.222222222 + 0.05555555555 + 0.08888888888 + 0.12626262626) = -0.01363636341
You can do the same thing with the Pass Line bet, if you wish, I'm not going to type it all out. Like I said, the general concept and how you can do it remains the same.
Anyway, if you do this for the Pass Line bets and you do it for the Place Bets, then you will come to see that the Pass Line bet has a lower house edge assuming you play any Place Bet to resolution.
The difference with Place Bets is some people look at the house edge on a, "Per roll," basis, which is also detailed on the WizardofOdds site. Pass Line bets are not looked at this way because you have no choice but to leave them up until they resolve. Most people make a Place Bet with the plan to leave it up until it resolves, though.
good for you for pondering all this, but everyone else who has done it has come to a very simple conclusion: at the moment you put your bets down, there is an inescapable house edge, and it is lower for the line bets than the place bets or buy bets. The free odds bets can't be made without first making line betsQuote: mwz524When you calculate the house edge are you calculating it base don the come out roll or on the total of all rolls?
Lets take a $25 pass line bet and $50 odds on a point of 4 - a winner would pay $100 odds win plus $25 pass line = $125
$75 place bet on the 4 pays 9 to 5 or $135. the difference is the come out roll where you have a 22% chance to win even money but also an 11% chance of loosing your bet. (i think i figured the stats correctly 8/36 = 22.2 and 4/36 = 11.1)
In my working days I was often accused of trying to re-invent the wheel, which seems to be what you are doing here. So I sympathize, and it can be a good exercise, but you need to realize what you are doing.
Place bets left at risk for only a single toss, then taken down immediately regardless of toss outcome, have very low House Advantages. Those H.A. are detailed by the Wizard in his Per Roll analyses.
Place bets left at risk only until resolution, then taken down immediately if they have won, have House Advantages that are greater than the H.A. for Pass and Don't Pass.
Each Place bet left at risk until it eventually loses has a House Advantage that is horrendous. Sadly, this is the usual plan. Rarely taken into account by players of Place bets, that H.A. can be calculated using the sum of the infinite series of values of possible future wins by a Place bet. Because it will have lost, at infinity the Place bet itself has a value of zero.
Quote: DeMangoSurprised no one mentioned a $75 bet on the 4 pays $150 less $3 vig
(147 * 3/9) - (75 * 6/9) = -1
-1/75 = -.0133333 or 1.33% House Edge
Better than the Pass Line, nice. At least, per bet resolved. I'm not sure which would be better per roll given x average of rolls for the pass line to resolve, but still good point.
Quote: Mission146When the house edge for the Pass Line is calculated, it is calculated based on all possibilities.
This is true for all bets - not just the Pass Line. However, the definition of "possibilities" can vary (such as "outcome per roll", "outcome per winning or losing decision", etc.) and folks quite often try to compare bets calculated using dissimilar definitions. This leads to erroneous conclusions. If this is what you were intimating then I agree. When comparing different bets, one should always strive to define them using similar terms.
Quote: Mission146Anyway, if you do this for the Pass Line bets and you do it for the Place Bets, then you will come to see that the Pass Line bet has a lower house edge assuming you play any Place Bet to resolution.
This is true but sounds a bit awkward. The Pass Line has a lower house edge than any Place bet whether figured per roll, per decision, or per come-out roll. You seem to imply that the statement is rendered true by virtue of resolving the Place bets and therefore the Pass Line bet might not have a lower house edge if the Place bets were not played to resolution.
Quote: Mission146The difference with Place Bets is some people look at the house edge on a, "Per roll," basis, which is also detailed on the WizardofOdds site. Pass Line bets are not looked at this way because you have no choice but to leave them up until they resolve. Most people make a Place Bet with the plan to leave it up until it resolves, though.
Yes, some people prefer "loss per roll" calculations with Place bets. It's therefore incumbent on them to similarly utilize "loss per roll" when comparing them to other bets.
Nothing prevents anyone from computing the edge of the Pass Line on a per roll basis.
| Bet | Loss/dec | Loss/roll | Loss/come-out |
|---|---|---|---|
| Pass Line | 1.414% | 0.419% | 1.414% |
| Don't Pass | 1.403% | 0.404% | 1.364% |
| Place 6 | 1.515% | 0.463% | 1.563% |
Another common error I see is folks comparing Don't Pass bets computed on a "per come-out" basis (1.364%) to Place bets computed on a "per decision" basis (e.g. 1.515% for Place6).
Steen
Quote: pwcrabbPlace bets left at risk for only a single toss, then taken down immediately regardless of toss outcome, have very low House Advantages. Those H.A. are detailed by the Wizard in his Per Roll analyses.
I think you're misunderstanding something here. Each bet has a house advantage which is determined by the probabilities of winning and losing and the amounts won or lost. Changing the way that you describe the edge does not change the inherent edge. For example, a Place 6 bet has an advantage that can be described as 1.52% per winning or losing decision. The fact that you can describe the same edge as 0.463% per roll doesn't change the inherent probabilities of winning or losing or the payoffs.
Suppose that I observe a particular table for a number of hours and determine that on average the dice are thrown off the table once every 37 rolls. Could I re-define the Place 6 bet edge to be 0.45% per attempted roll? Sure I could! Would it change the inherent edge defined by the house rules? No.
Quote: pwcrabbPlace bets left at risk only until resolution, then taken down immediately if they have won, have House Advantages that are greater than the H.A. for Pass and Don't Pass.
The house edge doesn't change depending on how many rolls you let bets ride!
Steen
Quote: SteenThis is true for all bets - not just the Pass Line. However, the definition of "possibilities" can vary (such as "outcome per roll", "outcome per winning or losing decision", etc.) and folks quite often try to compare bets calculated using dissimilar definitions. This leads to erroneous conclusions. If this is what you were intimating then I agree. When comparing different bets, one should always strive to define them using similar terms.
I know that it is, the OP specifically asked if the Pass Line accounted for all possibilities is the only reason why I would even say that.
To the rest, that's exactly what I was getting at. Lots of Craps system players will define the House Edge erroneously by making a complicated series of bets while using terms like, "Expected loss," or, "House Edge," when they are referring to the house edge per roll rather than per resolved bet. I actually wish House Edge per roll was not a metric because the entire concept muddies the waters, I would just mention that you can pull your bet back (unresolved) and leave it at that. Naturally the whole thing can be pulled back if it wins, and most people are going to let the Place Bets resolve at least once.
Usually, they press it or leave the original out there when it wins, which is effectively just making a whole new Place Bet. That's another area where these snakeoil system salesmen will get you. They'll say stuff like, "Your winning potential, for only a $6 bet..." You're not making a $6 bet, though. You're making a $6 bet, then a $12 bet and so on...
Quote:This is true but sounds a bit awkward. The Pass Line has a lower house edge than any Place bet whether figured per roll, per decision, or per come-out roll. You seem to imply that the statement is rendered true by virtue of resolving the Place bets and therefore the Pass Line bet might not have a lower house edge if the Place bets were not played to resolution.
I left some room for technicality on that one because I like to be precise. If you were only going to make one craps bet your entire life, win, lose or unresolved, then the Place Bet would be a better bet than the Pass Line (because of House Edge per roll and total expected $$$ loss as a result) due to the fact that the Pass Line bet must resolve. This also assumes same or close to same dollar amount on both. So, if you were going to Place 6 for one roll and one roll only and never play Craps again as long as you live, it would be a better bet than the Pass Line.
Quote:Yes, some people prefer "loss per roll" calculations with Place bets. It's therefore incumbent on them to similarly utilize "loss per roll" when comparing them to other bets.
Nothing prevents anyone from computing the edge of the Pass Line on a per roll basis.
Bet Loss/dec Loss/roll Loss/come-out Pass Line 1.414% 0.419% 1.414% Don't Pass 1.403% 0.404% 1.364% Place 6 1.515% 0.463% 1.563%
Another common error I see is folks comparing Don't Pass bets computed on a "per come-out" basis (1.364%) to Place bets computed on a "per decision" basis (e.g. 1.515% for Place6).
Steen
Nicely done! I think Wizard has average number of rolls per Pass Line bet somewhere, but I didn't feel like going to look for it.
You should feel the same way about someone playing PL with no odds as you do about someone playing three quarters in VP for a 97% RTP vs if he put in five quarters a pull he’d have 99% RTP.
Quote: unJonI disagree with the sentiment that the odds bet in craps is an independent bet from the PL (or DP) in craps. I find it useful to use a video poker analogy. Playing the PL is like short playing VP and adding odds in full is like putting in max quarters to access the better pay table.
You should feel the same way about someone playing PL with no odds as you do about someone playing three quarters in VP for a 97% RTP vs if he put in five quarters a pull he’d have 99% RTP.
You're looking at it from a house edge and expected loss relative to your total action standpoint with the Craps, then, which is fine. It's not an invalid way of looking at it. I guess my question would be, on a table with 100x odds (and there are a few left) would you be there playing a $5 PL and $500 in odds? Most people I have seen play Craps, $500 is something more akin to their entire session buy-in, not what they plop down on one bet.
I disagree with the comparison to Video Poker because you can actually increase the expected loss on video poker by playing that way. It changes the house edge of the game whereas the house edge of the Pass Line bet itself is unchanged. At 3%, you lose $0.0225 per $0.75 whereas you lose $0.0125 per $1.25 bet. With that said, betting a single coin undeniably yields a lower expected loss per hand and expected loss per hour whilst increasing the house edge.
The Pass Line bet is just the Pass Line bet.
Anyway, I totally get where you're coming from with the effective house edge (or whatever you want to call it) of PL + Odds assuming that you are going to make a maximum Odds bet every time you have the opportunity. I again think that's a perfectly valid way to look at it (just not the way I look at it), but I don't think it's apples-for-apples with VP.
legally it is separate. You don't gotta make it and if you do make you can take your money back anytime.
the line bet is a contract bet and I wouldn't try to take your money back when there is already someone holding a stick.
originally it was a way for casinos to compete with each other, you know: free peanuts, free booze, free odds.
Quote: Mission146You're looking at it from a house edge and expected loss relative to your total action standpoint with the Craps, then, which is fine. It's not an invalid way of looking at it. I guess my question would be, on a table with 100x odds (and there are a few left) would you be there playing a $5 PL and $500 in odds? Most people I have seen play Craps, $500 is something more akin to their entire session buy-in, not what they plop down on one bet.
I disagree with the comparison to Video Poker because you can actually increase the expected loss on video poker by playing that way. It changes the house edge of the game whereas the house edge of the Pass Line bet itself is unchanged. At 3%, you lose $0.0225 per $0.75 whereas you lose $0.0125 per $1.25 bet. With that said, betting a single coin undeniably yields a lower expected loss per hand and expected loss per hour whilst increasing the house edge.
The Pass Line bet is just the Pass Line bet.
Anyway, I totally get where you're coming from with the effective house edge (or whatever you want to call it) of PL + Odds assuming that you are going to make a maximum Odds bet every time you have the opportunity. I again think that's a perfectly valid way to look at it (just not the way I look at it), but I don't think it's apples-for-apples with VP.
I hear that. And defer to you on VP as I don’t play it so see that it’s not a perfect analogy.
I have no problem with someone taking less than full odds so long as they aren’t putting money at risk elsewhere during the roll. I have an issue with mainly to types of betting:
1) Someone at the $25 craps table betting PL with no odds while there are spots at the $15 table next to it.
2) Someone not taking full odds but making place bets. I get that the person wants to hedge so he can win on more numbers but it gets to me.
Quote: Mission146I left some room for technicality on that one because I like to be precise. If you were only going to make one craps bet your entire life, win, lose or unresolved, then the Place Bet would be a better bet than the Pass Line (because of House Edge per roll and total expected $$$ loss as a result) due to the fact that the Pass Line bet must resolve. This also assumes same or close to same dollar amount on both. So, if you were going to Place 6 for one roll and one roll only and never play Craps again as long as you live, it would be a better bet than the Pass Line.
I agree but it's an awkward and misleading comparison. You're comparing one Place bet versus one Pass Line bet but restricting the Place bet to just one roll while waiting for the Pass Line to resolve which takes on average 3.375 rolls. I would maintain that comparing the loss per roll of a Place bet with the loss per decision of a Pass Line bet is unfair.
Ordinarily, it doesn't matter whether you pull the Place bet in one roll or not because on average it would lose more per roll (and more per decision) than the Pass Line bet. However, because you're making just one bet for your entire life then the expected value of the bet is irrelevant. In this case, the only thing that matters is the probability of winning or pushing in one roll. In other words, the probability of walking away with money (regardless of the amount) after one roll.
To make this a fair comparison we can either allow both bets to resolve (average 3.27 rolls for Place and 3.375 rolls for Pass Line) or we can restrict both bets to just one roll.
In the first case, the Pass Line has a higher probability of returning money:
Place 6
-- probability of winning = 5/11 = 45.45%
Pass Line (if bet on come-out roll)
-- probability of winning = 244/495 = 49.29%
In the second case, the Place bet has a higher probability of returning money because if the Pass Line doesn't win in one roll then it must be abandoned as a loss (in other words you can't walk away with it).
Place6
-- probability of winning = 5/36
-- probability of pushing = 25/36
-- probability of walking with money after one roll = 30/36 = 83.33%
Pass Line (if bet on come-out roll)
-- probability of winning = 8/36
-- probability of pushing = 0
-- probability of walking with money after one roll = 8/36 = 22.22%
So you have almost a 4x better chance of walking with money in one roll on the Place6 bet. But this is soley due to non-resolving pushes being included in the equation. If you were to ask what's the chance of just winning either bet in one roll, then the Pass Line would be better (8/36 Pass Line versus 5/36 Place). In my opinion, including pushes in craps confuses the issue and can lead to erroneous conclusions.
If pushes matter then what about the player who observes the action but never bets? His chance of walking with money is 100%. A craps genius!
Personally, I've never met anyone who wanted to make just one lifetime bet for just one roll, so the issue is somewhat moot.
Steen
Expected Return for a $6 wager on Place 6 after one projected roll:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) = ( $5.9722 )
Expected Loss = ( $0.0278 )
House Advantage = ( 0.4630 % )
Expect Return for a $6 wager on Place 6 after two projected rolls:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) [ ( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) ] = ( $5.9491 )
Expected Loss = ( $0.0509 )
House Advantage = ( 0.8488 % )
The number of projected rolls is chosen at leisure. For most players of Place bets, that number is infinity
Just for grins, extending the equation above, after how many projected rolls does the H.A. closely approximate the standard presentation, which implicitly assumes exposure until resolution followed by wager removal?
The Standard Presentation is ( 5 / 11 ) ( $7 + 6 ) = ( $5.9091 )
Expected Loss = ( $0.0909 )
House Advantage = ( 1.5152 % )
Quote: pwcrabbSteen suggests that "The house edge doesn't change depending on how many rolls you let bets ride!" However, Steen himself offers a calculation of per-roll loss for a Place bet on 6. Exposure for multiple rolls would seem to have consequences for the house edge, and of course it does so. His suggestion is patently silly. The theoretical probabilities of course never change, but duration of exposure can change and does so at player discretion, with attendant implications for house edge.
Expected Return for a $6 wager on Place 6 after one projected roll:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) = ( $5.9722 )
Expected Loss = ( $0.0278 )
House Advantage = ( 0.4630 % )
Expect Return for a $6 wager on Place 6 after two projected rolls:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) [ ( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) ] = ( $5.9491 )
Expected Loss = ( $0.0509 )
House Advantage = ( 0.8488 % )
The number of projected rolls is chosen at leisure. For most players of Place bets, that number is infinity
Just for grins, extending the equation above, after how many projected rolls does the H.A. closely approximate the standard presentation, which implicitly assumes exposure until resolution followed by wager removal?
The Standard Presentation is ( 5 / 11 ) ( $7 + 6 ) = ( $5.9091 )
Expected Loss = ( $0.0909 )
House Advantage = ( 1.5152 % )
Unfortunately, I have to catch a plane right now so I can't give you a full answer until tomorrow. In the meantime, I'll let you re-think your nonsense here and post a correction.
I'll give you a couple hints: your calculations and your definitions are off.
:)
Steen
Quote: pwcrabbSteen suggests that "The house edge doesn't change depending on how many rolls you let bets ride!" However, Steen himself offers a calculation of per-roll loss for a Place bet on 6. Exposure for multiple rolls would seem to have consequences for the house edge, and of course it does so. His suggestion is patently silly. The theoretical probabilities of course never change, but duration of exposure can change and does so at player discretion, with attendant implications for house edge.
Expected Return for a $6 wager on Place 6 after one projected roll:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) = ( $5.9722 )
Expected Loss = ( $0.0278 )
House Advantage = ( 0.4630 % )
Expect Return for a $6 wager on Place 6 after two projected rolls:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) [ ( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) ] = ( $5.9491 )
Expected Loss = ( $0.0509 )
House Advantage = ( 0.8488 % )
The number of projected rolls is chosen at leisure. For most players of Place bets, that number is infinity
Just for grins, extending the equation above, after how many projected rolls does the H.A. closely approximate the standard presentation, which implicitly assumes exposure until resolution followed by wager removal?
The Standard Presentation is ( 5 / 11 ) ( $7 + 6 ) = ( $5.9091 )
Expected Loss = ( $0.0909 )
House Advantage = ( 1.5152 % )
I don’t know anyone that makes a place bet to infinity rolls. That would take a long time. Most players place bet until a seven out, which is very different than infinity.
Hear! Hear!Quote: unJonMost players place bet until a seven out, which is very different than infinity.
Sometimes math types consider such things as those history making rolls but who here has ever been present for one or expects to be.
Many tables are Point then SevenOut, sometimes it goes far longer. I think 8 rolls is the average but variance is high. No one needs to calculate for an infinitely long roll.
Quote: pwcrabbSteen suggests that "The house edge doesn't change depending on how many rolls you let bets ride!" However, Steen himself offers a calculation of per-roll loss for a Place bet on 6. Exposure for multiple rolls would seem to have consequences for the house edge, and of course it does so. His suggestion is patently silly. The theoretical probabilities of course never change, but duration of exposure can change and does so at player discretion, with attendant implications for house edge.
Expected Return for a $6 wager on Place 6 after one projected roll:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) = ( $5.9722 )
Expected Loss = ( $0.0278 )
House Advantage = ( 0.4630 % )
Expect Return for a $6 wager on Place 6 after two projected rolls:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) [ ( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) ] = ( $5.9491 )
Expected Loss = ( $0.0509 )
House Advantage = ( 0.8488 % )
What's patently silly is your confusing percentage lost versus actual dollar amount lost.
Yes, your actual expected dollar loss will increase with each event (that should be obvious to anyone) but your percentage expected loss relative to your wagers remains the same! It does not increase as you seem to think.
For example, if the expected loss on a Place6 bet is given as 0.463% per roll then it remains the same for each roll. All you need do is multiply this constant by the size of your bet and the number of rolls to get the expected actual dollars lost:
--- One roll = 0.463% * $6 * 1 = $0.0278
-- Two rolls = 0.463% * $6 * 2 = $0.0556 (not $0.0509 as you showed)
---- 36 rolls = 0.463% * $6 * 36 = $1
Similarly, if the expected loss on a Place6 bet is given as 1.515% per winning or losing decision (let's just call these decisions) then it remains the same for each decision. All you need do is multiply this figure by the size of your bet and the number of decisions to get the expected actual dollars lost:
--- One decision = 1.515% * $6 * 1 = $0.0909
-- Two decisions = 1.515% * $6 * 2 = $0.1818
---- 11 decisions = 1.515% * $6 * 11 = $1
Notice that the expected loss in 36 rolls is equal to the expected loss in 11 decisions. This is because on average there are 11 decisions per 36 rolls.
Regardless of how you choose to describe your expected loss, the inherent advantage held by the house does not change (and again, I'm expressing advantage as a percentage of the player's wager which is how it's typically expressed in craps.) Run some simulations if you don't believe me. It would be folly to think that a bettor could change the inherent house advantage through a scheme of betting patterns.
"Mr Boxman, how much does a $6 Place 6 pay?"
"It pays $7 every time a 6 rolls sonny." (very old boxman)
"But I'm only playing for 2 rolls now, then 3 rolls later, and maybe one roll after dinner."
"Thank you for telling me that son. Our payoffs are dependent on the number of rolls that you stay up. We have different payoffs for each bettor depending on his/her betting pattern so we appreciate finding out in advance. Hey boss, can we get a fill over here? This guy is knocking our advantage all to hell!"
Quote: pwcrabbThe number of projected rolls is chosen at leisure. For most players of Place bets, that number is infinity
They play for infinity? Wow, I guess I better play more often then. It'll take me at least a few years to hit infinity. :-)
Quote: pwcrabbJust for grins, extending the equation above, after how many projected rolls does the H.A. closely approximate the standard presentation, which implicitly assumes exposure until resolution followed by wager removal?
The Standard Presentation is ( 5 / 11 ) ( $7 + 6 ) = ( $5.9091 )
Expected Loss = ( $0.0909 )
House Advantage = ( 1.5152 % )
You asked about rolls but then gave your answer in dollars and percentage. Do you have a dollars-to-rolls conversion factor that helps answer the question? :-)
Steen
If it survives for eight at-risk rolls then Hallelujah, because the probability of that occurrence is ( 30 / 36 ) raised to the exponent ( 8 ). You may use your own calculator.
In theory that Place bet could survive without limit. The probability is always greater than zero that Big Red will not occur at that table until some far distant roll. You may choose your own large exponent and see for yourself. The probability diminishes to zero only when the exponent is infinity.
Craps bets that are not contract bets may be removed at any time. Rather than in years, their projected investment lifetimes are measured in rolls. The applicable discount factor for any particular projected future roll will be based upon the probability of not having experienced Big Red on projected prior rolls.
For an imminent next-roll future, the per-roll percentage cost of a Place bet is stated above by several forum participants. Evaluated anew after each survived roll, the next imminent roll presents an identical prospect. If the future is always only one time period away then there is no need to compound the discount factor. If the future is always only one time period away then costs and revenues remain constant and may be simply summed.
For a future period that is multiple time periods away, however, the discount factor must be compounded. Both the costs and the revenues of that future period must be discounted back to present value by multiplying them by the discount factor raised to the applicable exponent. Imminent, near future, and far future costs and revenues may then be appropriately compared.
Quote: pwcrabbThe analysis technique known as Net Present Value (NPV) is used by money managers to evaluate investment projects.
Seriously?
I state that the house edge doesn't change depending on how many rolls you let bets ride!
You reply that my suggestion is patently silly and now offer up "Net Present Value" as your proof? Nonsense.
House Advantage has been well defined and used for hundreds of years. It's not invalidated nor supplanted by Net Present Value! HA is a distillate of probabilities and outcomes which provides a fundamental metric for evaluating bets. Does it tell you everything? Of course not. There are many other metrics you can use to analyze bets but none invalidate HA.
HA represents a probable future value of a present wager, not the probable present value of a future wager.
Knock yourself out if you feel NPV adds something meaningful to your craps analysis but I've not seen anyone else use it so I can't imagine you'll have much company. I for one would be interested to know what nuggets of wisdom you've found with it.
Steen
Quote: DeMangoSteen, do you know what you can’t fix?
I'm all ears.
LOL. not even close on the evQuote: pwcrabbExpect Return for a $6 wager on Place 6 after two projected rolls:
( 5 / 36 ) ( $7 ) + ( 30 / 36 ) [ ( 5 / 36 ) ( $7 ) + ( 30 / 36 ) ( $6 ) ] = ( $5.9491 )
Expected Loss = ( $0.0509 )
House Advantage = ( 0.8488 % )
not even close on the HE
Place bet 6 per roll ev = 5/36*7 + 6/36*-6 + 25/36*0 = -1/36
HE = ev/bet resolved(action or handle) = -1/36 / $6 = -1/216
1 roll is fineQuote: pwcrabbJust for grins, extending the equation above, after how many projected rolls does the H.A. closely approximate the standard presentation, which implicitly assumes exposure until resolution followed by wager removal?
ev 1 roll = -1/36
HE = -1/36 /$6 = -1/216
ev 2 rolls = -2/36
HE = -2/36 /$12 = -1/216
ev 3 rolls = -3/36
HE = -3/36 /$18 = -1/216
all 9 possible outcomes for 2 rolls
| event | prob | handle | edge | ev |
|---|---|---|---|---|
| w,w | 0.019290123 | 12 | - 1/216 | -0.001071674 |
| w,l | 0.023148148 | 12 | - 1/216 | -0.001286008 |
| w,t | 0.096450617 | 12 | - 1/216 | -0.005358368 |
| l,w | 0.023148148 | 12 | - 1/216 | -0.001286008 |
| l,l | 0.027777778 | 12 | - 1/216 | -0.00154321 |
| l,t | 0.115740741 | 12 | - 1/216 | -0.006430041 |
| t,w | 0.096450617 | 12 | - 1/216 | -0.005358368 |
| t,l | 0.115740741 | 12 | - 1/216 | -0.006430041 |
| t,t | 0.482253086 | 12 | - 1/216 | -0.026791838 |
| total >> | 1 | . | total >> | -0.055555556 |
| . | . | . | edge >> | -0.00462963 |
all 27 possible outcomes for 3 rolls
| event | prob | handle | edge | ev |
|---|---|---|---|---|
| w,w,w | 0.002679184 | 18 | - 1/216 | -0.000223265 |
| w,w,l | 0.003215021 | 18 | - 1/216 | -0.000267918 |
| w,w,t | 0.013395919 | 18 | - 1/216 | -0.001116327 |
| w,l,w | 0.003215021 | 18 | - 1/216 | -0.000267918 |
| w,l,l | 0.003858025 | 18 | - 1/216 | -0.000321502 |
| w,l,t | 0.016075103 | 18 | - 1/216 | -0.001339592 |
| w,t,w | 0.013395919 | 18 | - 1/216 | -0.001116327 |
| w,t,l | 0.016075103 | 18 | - 1/216 | -0.001339592 |
| w,t,t | 0.066979595 | 18 | - 1/216 | -0.005581633 |
| l,w,w | 0.003215021 | 18 | - 1/216 | -0.000267918 |
| l,w,l | 0.003858025 | 18 | - 1/216 | -0.000321502 |
| l,w,t | 0.016075103 | 18 | - 1/216 | -0.001339592 |
| l,l,w | 0.003858025 | 18 | - 1/216 | -0.000321502 |
| l,l,l | 0.00462963 | 18 | - 1/216 | -0.000385802 |
| l,l,t | 0.019290123 | 18 | - 1/216 | -0.00160751 |
| l,t,w | 0.016075103 | 18 | - 1/216 | -0.001339592 |
| l,t,l | 0.019290123 | 18 | - 1/216 | -0.00160751 |
| l,t,t | 0.080375514 | 18 | - 1/216 | -0.00669796 |
| t,w,w | 0.013395919 | 18 | - 1/216 | -0.001116327 |
| t,w,l | 0.016075103 | 18 | - 1/216 | -0.001339592 |
| t,w,t | 0.066979595 | 18 | - 1/216 | -0.005581633 |
| t,l,w | 0.016075103 | 18 | - 1/216 | -0.001339592 |
| t,l,l | 0.019290123 | 18 | - 1/216 | -0.00160751 |
| t,l,t | 0.080375514 | 18 | - 1/216 | -0.00669796 |
| t,t,w | 0.066979595 | 18 | - 1/216 | -0.005581633 |
| t,t,l | 0.080375514 | 18 | - 1/216 | -0.00669796 |
| t,t,t | 0.334897977 | 18 | - 1/216 | -0.027908165 |
| total >> | 1 | . | total >> | -0.083333333 |
| . | . | . | edge >> | -0.00462963 |
I see same HE at each roll because the probabilities never change and the payoff never changes (for a unit wager). Others that use their math method may show something way different.
thought you would know that one, starts with st****!Quote: SteenI'm all ears.
One potentially useful nugget for Steen is the observation is that the next Big Red, with its probability each roll of ( 1 / 6 ), and with its entire future evaluated only in a constantly changing NOW, can be projected as likely to appear after 3.801784 future rolls. Likelihood is defined as probability greater than ( 0.5 ). As each roll evolves from future to past, it becomes irrelevant to the analysis and is discarded.
A corollary to this observation is the implication that half of all future Reds will require from one to 3.801784 tosses to appear and the other half will require more tosses, with no upper limit. If we use the approximation of four tosses then we are including roughly 51 % of all future Reds. This forecast in no way invalidates the fact that overall the average non-Red interval is five tosses. One may easily evaluate a sizeable sample of Reds to test whether the sample divides nearly equally into "short" and "long" types.
Can this division of future Reds into two equally likely types be useful? It can be useful if we consult clustering theory, which describes the clusters in which events of defined probability are likely to be distributed across future time. In a simple problem like this binary model, for a selected cluster size ( N ) of favorable outcomes, e.g. nine consecutive "short" Sevens requiring four or fewer rolls to appear, the proportion over infinity of all future Sevens that is represented by clusters of size ( N ) is equal to:
{ ( N ) / [ ( 2 ) ^ ( N + 1 ) ] }
In practical terms, such clustering of "short" and "long" Sevens is variously experienced as Craps tables that are "cold," "hot," or "choppy." One can state affirmatively that such clusters will in fact appear and in what proportions. However, no one can rationally forecast the crucial matter of WHEN they will appear.
Quote: pwcrabbDeMango gets full credit for his subtle lowbrow Ad Hominem argument.
In practical terms, such clustering of "short" and "long" Sevens is variously experienced as Craps tables that are "cold," "hot," or "choppy." One can state affirmatively that such clusters will in fact appear and in what proportions. However, no one can rationally forecast the crucial matter of WHEN they will appear.
Thank you. Just trying to keep my name out of red.
Streaks happen, you cannot predict the beginning or the end. I’m sure others will reinforce the faulty arguments.
No arguments about your faulty math?
I do realize that no one knows when a humungous roll is starting anymore than they know when it will be Point,SevenOut for a while.
I've never noticed much difference in our winnings.
Quote: pwcrabbOne potentially useful nugget for Steen is the observation is that the next Big Red, with its probability each roll of ( 1 / 6 ), and with its entire future evaluated only in a constantly changing NOW, can be projected as likely to appear after 3.801784 future rolls.
I agree there's a 50% chance of seeing a 7 within 3.801784 rolls but that seems to be a simple matter of probability and not one of Net Present Value.
As I'm sure you know, the easiest way to figure this is to take 1 minus the probability of not rolling a seven in X rolls.
1-(5/6)^X
For example, the prob of one or more 7's in 3 rolls would be:
1-(5/6)^3 = 1 - 125/216 = 42.13%
Here's a table of the first 10 rolls:
| Roll | pNo7% | p7% |
|---|---|---|
| 1 | 83.33 | 16.67 |
| 2 | 69.44 | 30.56 |
| 3 | 57.87 | 42.13 |
| 4 | 48.23 | 51.78 |
| 5 | 40.19 | 59.81 |
| 6 | 33.49 | 66.51 |
| 7 | 27.91 | 72.09 |
| 8 | 23.26 | 76.74 |
| 9 | 19.38 | 80.62 |
| 10 | 16.15 | 83.85 |
To get the 50% figure you set (5/6)^X = 0.5 and solve for X
ln = natural log
ln(5/6)^X = ln(.5)
X = ln(.5) / ln(5/6) = 3.801784
As for the unpredictable clustering, I wouldn't call that a nugget of wisdom since it's is of absolutely no use.
Steen
The clustering idea emerges from chaos theory. In three dimensional space, it describes the distribution of galaxies. On a two dimensional dropcloth, it describes the distribution of random paint spattering. In one dimensional time, it describes the distribution of Big Red.
Discovering the utility of knowledge of the distribution of Big Red is not intuitively obvious. In light of the pushback that has emerged thus far on this thread, I will not elaborate.
