tuttigym
tuttigym
Joined: Feb 12, 2010
  • Threads: 10
  • Posts: 1039
January 15th, 2022 at 1:30:08 PM permalink
Quote: DeMango

While we are at it, forgive me if this was posted before, but can we have a column for 18 yos before the seven?
link to original post


Can you or Mr. W calculate the odds of performing 495 different come out rolls each one different and distinct with each producing a different and distinct outcome as in the formula that defines the PL HA ar 1.41%?

tuttigym

p.s. That number should far exceed the 11 yo's in my estimation.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 27
  • Posts: 1676
January 15th, 2022 at 8:27:50 PM permalink
Quote: tuttigym

Quote: DeMango

While we are at it, forgive me if this was posted before, but can we have a column for 18 yos before the seven?
link to original post


Can you or Mr. W calculate the odds of performing 495 different come out rolls each one different and distinct with each producing a different and distinct outcome as in the formula that defines the PL HA ar 1.41%?

tuttigym

p.s. That number should far exceed the 11 yo's in my estimation.
link to original post

Can you rephrase? I dont understand the question
Its all about making that GTA
charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 36
  • Posts: 2599
January 16th, 2022 at 1:38:17 AM permalink
For the Pass (or Come) bet there are seventeen possible ways for the bet to be resolved (ignoring the intermediary rolls that aren't needed).
2, 3, 7, 11, 12 (n/36 n=1,2,6,2,1)
4/4 4/7 (3/36*n/9 n=3,6)
5/5 5/7 (4/36*n/10 n=4,6)
6/6 6/7 (5/36*n/11 n=5,6)
similarly for 8 thru 10
Thus there aren't 495 different outcomes but their probabilities are associated with 495 (or 1980 etc) as a common factor.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 110
  • Posts: 5413
January 16th, 2022 at 8:23:12 AM permalink
Quote: tuttigym

Quote: DeMango

While we are at it, forgive me if this was posted before, but can we have a column for 18 yos before the seven?
link to original post


Can you or Mr. W calculate the odds of performing 495 different come out rolls each one different and distinct with each producing a different and distinct outcome as in the formula that defines the PL HA ar 1.41%?

tuttigym

p.s. That number should far exceed the 11 yo's in my estimation.
link to original post


I probably could...but let me ask: why? It makes sense if you are proposing something along the lines of, "Unless you can roll every combination without repetition, the use of terms like 'expected value' and 'house advantage' are meaningless, and Martingale-style systems can, and do, work," but I prefer not to listen to such nonsense.

"Expected Value," as defined by Merriam-Webster:
"the sum of the values of a random variable with each value multiplied by its probability of occurrence"

By Investopedia:
"In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values."

But even then, "expected value" and "actual results" are quite often two different things.
An easy example: when rolling two fair 6-sided dice, the "expected value" of their sum is 7. The probability of it actually happening is 1/6.
tuttigym
tuttigym
Joined: Feb 12, 2010
  • Threads: 10
  • Posts: 1039
January 16th, 2022 at 9:10:51 AM permalink
Quote: charliepatrick

Thus there aren't 495 different outcomes but their probabilities are associated with 495 (or 1980 etc) as a common factor.
link to original post


Obviously, I am genuinely confused, so please help me understand the sentence above especially "their probabilities are associated with 495..... as a common factor." Example: The point established was with a hard 8. There are 5 ways to convert that point, correct? (4/4; 3/5; 5/3 etc) There are 6 ways to lose that point, i.e. 7-out, correct? (1/6; 6/1;2/5 etc) That 8 point alone has 11 possible different outcomes or resolutions, correct.? So the seventeen possible ways "for the Pass (or Come) bet" is incorrect. The 495 is the total of all possible resolutions one time. If that is incorrect, please resolve my confusion of your statements above. Thank you.

tuttigym

p.s. Addendum to the example above. The point is 2/6 or 8. The above resolutions also apply, i.e. 11 possible outcomes.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 110
  • Posts: 5413
January 16th, 2022 at 9:40:08 AM permalink
Quote: tuttigym

Quote: charliepatrick

Thus there aren't 495 different outcomes but their probabilities are associated with 495 (or 1980 etc) as a common factor.
link to original post


Obviously, I am genuinely confused, so please help me understand the sentence above especially "their probabilities are associated with 495..... as a common factor." Example: The point established was with a hard 8. There are 5 ways to convert that point, correct? (4/4; 3/5; 5/3 etc) There are 6 ways to lose that point, i.e. 7-out, correct? (1/6; 6/1;2/5 etc) That 8 point alone has 11 possible different outcomes or resolutions, correct.? So the seventeen possible ways "for the Pass (or Come) bet" is incorrect. The 495 is the total of all possible resolutions one time. If that is incorrect, please resolve my confusion of your statements above. Thank you.

tuttigym

p.s. Addendum to the example above. The point is 2/6 or 8. The above resolutions also apply, i.e. 11 possible outcomes.
link to original post


The overall probability of a pass line bet winning is 244 / 495. This does not mean that there are 495 distinct combinations of comeout rolls and results.
That is kind of like saying that the probability of tossing exactly two heads and two tails with four coin tosses is 3/8, so there are 8 possible outcomes.
Here's where 495 comes from:
The probability of winning with a comeout of 4 is 1/12 x 1/3 = 1/36
The probability of winning with a comeout of 5 is 1/9 x 2/5 = 2/45
The probability of winning with a comeout of 6 is 5/36 x 5/11 = 25/396
The probability of winning with a comeout of 7 is 1/6
The probability of winning with a comeout of 8 is 5/36 x 5/11 = 25/396
The probability of winning with a comeout of 9 is 1/9 x 2/5 = 2/45
The probability of winning with a comeout of 10 is 1/12 x 1/3 = 1/36
The probability of winning with a comeout of 11 is 1/18
The least common multiple of the denominators is 1980
The sum of these four numbers = 976 / 1980 = 244 / 495
You can't use 495 as a common denominator as, for example, the probability of winning with a comeout of 11 is 27.5 / 495, but you can't have 27.5 comeouts.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 36
  • Posts: 2599
January 16th, 2022 at 10:42:29 AM permalink
Firstly the assumption most people will make is that it doesn't matter which combination of dice make any given total, so there are, for instance, 3 ways to make a 4, 4 ways to make a 5, and 5 ways to make a 6.

This means at the resolution stage
(a) 4 or 10: there are 3 ways to make the point and 6 ways to make the seven-out, so the denominator is 9. As the numerators are 3 or 6, it can be simplified to 1/3 and 2/3 = 55 and 110 / 165
(b) 5 or 9: similarly 4 to make 5, 6 to make 7, denominator is 10; simplified to 2/5 and 3/5 = 66 and 99 / 165.
(c) 6 or 8: similarly 5 to make 6, 6 to make 7, denominator is 11; simplified to 5/11 and 6/11 = 75 and 90 / 165.

The chance of natural win with 7 is 6/36, 11 is 2/36.
The chance of craps lose with 2 is 1/36, 3 is 2/36, 12 is 1/36.
The chance of point being 4 is 3/36.
The chance of point being 5 is 4/36.
The chance of point being 6 is 5/36.

First RollOutcomeFirst RollSecond RollWinLose
Natural 7Win6165990
Natural 11Win2165330
Craps 2Lose1165165
Craps 3Lose2165330
Craps 12Lose1165165
Point 4Win355165
Point 4Lose3110330
Point 5Win466264
Point 5Lose499396
Point 6Win575375
Point 6Lose590450
Point 8Win575375
Point 8Lose590450
Point 9Win466264
Point 9Lose499396
Point 10Win355165
Point 10Lose3110330
Totals594029283012
Pass line (Win)1.414%2928
Don't Pass (Wins)2847
Don't Pass (Ties)1.364%165
You can see that to keep all the numbers whole, and if you care about which point you need, the denominator is 5940 (which is 36*165).

If you worried about specific ways to roll numbers, then you would use more rows in the table, but it would still add up to 36*165.
tuttigym
tuttigym
Joined: Feb 12, 2010
  • Threads: 10
  • Posts: 1039
January 16th, 2022 at 3:25:30 PM permalink
Quote: ThatDonGuy

I probably could...but let me ask: why? It makes sense if you are proposing something along the lines of, "Unless you can roll every combination without repetition, the use of terms like 'expected value' and 'house advantage' are meaningless, and Martingale-style systems can, and do, work," but I prefer not to listen to such nonsense.


First, thank you for the reply and the definitions plus sources. They are very helpful for me, and I do have questions which will follow. Next, "why"?: I am fully aware that any system that is repeated as a basis for one's play will ultimately fail at some point and create losses which can be extreme without discipline and touting such is not only nonsense but sometimes dangerous for the gullible and uninformed.

The basis for the PL HA (1.41%), as you know is what I call the rule of 495. While I understand how the math is derived, I am confident that it cannot be performed at the tables because, I believe, the odds are so long as to make this constant referral nonsense, meaningless, and totally misleading. Further, in my experience, I have never witnessed or heard of any player actually stepping up to the table, buying in, and just playing the PL as the exclusive wager for one's basis of play. If the actual odds of performing the rule of 495 were known, perhaps players might be better informed as to the reality that the HA for the game is considerably higher as it is really played. In short, the term "house advantage" is quite real but highly diluted with the droning of such unreal low percentages 1.41% just does not happen at the table.


Quote: ThatDonGuy

"Expected Value," as defined by Merriam-Webster:
"the sum of the values of a random variable with each value multiplied by its probability of occurrence"


To be honest, this definition I do not comprehend. My personal math IQ is pretty insufficient.

Quote: ThatDonGuy

By Investopedia:
"In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values."


OK better, I think maybe. So, if I have a five to one advantage of winning a bet, regardless of size or regardless of win amount, can that be construed as a +EV? So the dice example is a -EV because it occurs only 18+% of the time. Am I understanding that correctly?

tuttigym
Mission146
Mission146
Joined: May 15, 2012
  • Threads: 133
  • Posts: 15308
January 16th, 2022 at 3:49:29 PM permalink
Quote: tuttigym

Quote: ThatDonGuy

I probably could...but let me ask: why? It makes sense if you are proposing something along the lines of, "Unless you can roll every combination without repetition, the use of terms like 'expected value' and 'house advantage' are meaningless, and Martingale-style systems can, and do, work," but I prefer not to listen to such nonsense.


First, thank you for the reply and the definitions plus sources. They are very helpful for me, and I do have questions which will follow. Next, "why"?: I am fully aware that any system that is repeated as a basis for one's play will ultimately fail at some point and create losses which can be extreme without discipline and touting such is not only nonsense but sometimes dangerous for the gullible and uninformed.

The basis for the PL HA (1.41%), as you know is what I call the rule of 495. While I understand how the math is derived, I am confident that it cannot be performed at the tables because, I believe, the odds are so long as to make this constant referral nonsense, meaningless, and totally misleading. Further, in my experience, I have never witnessed or heard of any player actually stepping up to the table, buying in, and just playing the PL as the exclusive wager for one's basis of play. If the actual odds of performing the rule of 495 were known, perhaps players might be better informed as to the reality that the HA for the game is considerably higher as it is really played. In short, the term "house advantage" is quite real but highly diluted with the droning of such unreal low percentages 1.41% just does not happen at the table.



(Bold added, by me, for emphasis)

Okay, so what you are maintaining is that, because players do not exclusively play the Pass Line...that changes the House Edge of the Pass Line.

Simply put, if the House Edge of the, 'Any 7,' bet is 16.67%, then playing Pass Line AND Any 7 (as one example) increases the House Edge against you on your total action, but no amount put on Any 7 bets changes the House Edge of the Pass Line bet taken alone.

(Middle part removed by me as irrelevant)

Quote: ThatDonGuy

By Investopedia:
"In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values."


Quote: Tuttigym

OK better, I think maybe. So, if I have a five to one advantage of winning a bet, regardless of size or regardless of win amount, can that be construed as a +EV? So the dice example is a -EV because it occurs only 18+% of the time. Am I understanding that correctly?

tuttigym
link to original post



This is just more of the usual from you. Probability of winning being greater than 50% does not yield a positive expectation. It's the same question as betting $18 on black at Roulette and then individually betting 12 red numbers at $1 apiece. You have $30 in total action, the probability that you will profit on the single spin is 30/38 (double zero assumed), but that does not make the bet +EV.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1423
  • Posts: 24357
January 16th, 2022 at 3:53:06 PM permalink
Quote: ThatDonGuy

Quote: Wizard

The probability of n 12's before a 7 is (6/7)^n*(6/7).


Er, I think you mean (1/7)^n * 6/7.

Also note that this is the probability of rolling exactly n 12s before a 7.

The probability of "rolling n 12s before a 7" is (1/7)^n, which is also 1 / 7^n.
link to original post



Yes, you're right, to both.
It's not whether you win or lose; it's whether or not you had a good bet.

  • Jump to: