tuttigym Joined: Feb 12, 2010
• Posts: 1039
January 16th, 2022 at 4:20:38 PM permalink
Quote: ThatDonGuy

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.

You have been very patient and accommodating. I greatly appreciate your efforts to help me. Thank you.

My simple mind looks at 36 ways to roll the dice as a starting place so that all win/loss possibilities for me mathematically are derived at that place. Therefore, since there are only two ways to roll an eleven (11), my mind interprets the probability or odds of winning at 2/36 or 5.5%. The winning of the bet is number specific and therefore as a one roll event, win or lose, dismisses any other possibilities or rolled numbers.

The Place numbers 4, 5, 6, 8, 9, & 10 again are win/loss specific although not one roll bets. Therefore, the only relevant numbers to winning or losing is that number or the 7 (after come out). All other tossed numbers during that point play are moot and do not affect the outcome of that point. My mind and the casino pay table basically states that the 4 point has 3 ways to win and 6 ways to lose (3/6) or two to one for and against with the payout at 2X for the win and 1/2 (50%) for the loss (Don't odds). That is why, for me, there is a great disconnect of what many have tried to convey to me. But it all starts with 36 ways to roll the dice.

tuttigym

tuttigym Joined: Feb 12, 2010
• Posts: 1039
January 16th, 2022 at 4:35:55 PM permalink
Quote: Mission146

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.

There you go again making up conclusions that have NOT been stated and are DISTORTIONS of my words. You need to brush up on your critical thinking and reading skills

Quote: Mission146

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.

Those are your words and conclusions, not mine.

tuttigym
Mission146 Joined: May 15, 2012
• Posts: 15308
January 16th, 2022 at 4:51:24 PM permalink
Quote: tuttigym

Quote: Mission146

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.

There you go again making up conclusions that have NOT been stated and are DISTORTIONS of my words. You need to brush up on your critical thinking and reading skills

Quote: Mission146

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.

Those are your words and conclusions, not mine.

tuttigym

I know they are my conclusions and not yours; start with the conclusions are right, so that they are more likely to be mine becomes a given.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5413
January 16th, 2022 at 8:30:47 PM permalink
Quote: tuttigym

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

No, you are not.
Expected value = the sum of (the probability of a particular result x the value of that result) over all possible results.
An easy example: betting 1 on red in double-zero roulette.
The probability of winning 1 is 18/38; the probability of losing 1 is 20/38.
EV = (18/38 x 1) + (20/38 x (-1)) = -2/38 = -1/19, or about -5.263%.

It is a little harder with the pass line in craps (BTW, I play only pass line and odds), because of the multiple conditions. However, let me borrow most of CharliePatrick's table from a previous post:
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

The probability of a particular row's result = (the number in the First Row column divided by 36) x (the number in the Second Row column divided by 165). Note that the numbers in the Win and Loss columns = the First Row value x the Second Row value, so the probability of a row's result = the number in the win (or loss, as appropriate) column / 5940, which is 36 x 165.
EV = (1 x (the sum of (each numbers in the Win column / 5940))) + (-1 x (the sum of (each number in the Loss column / 5940)) =
2928 / 5040 - 3012 / 5940 = -84 / 5940 = -7 / 495, or about 1.414%.

Note that this does not mean that every set of 5940 consecutive comeouts will have exactly 2928 wins and 3012 losses; "expected" does not mean "guaranteed." If it did, then here is a Guaranteed Surefire Cannot Miss System:
1. Track 5939 consecutive comeouts.
2. Count the number of wins; if there have been 2928, the next comeout has to be a loss, and if there have been only 2927, the next comeout has to be a win. Bet accordingly. "What if it's neither of those two?" Then you have a set of 5940 consecutive comeouts that don't have 2928 wins, but this system works on the assumption that This Cannot Happen.
3. But wait - there's more! In order for the 5940 consecutive comeouts starting with #2 to have 2928 wins, comeout #5941 has to have the same result as comeout #1. For that matter, #5942 has to have the same result as #2, #5943 has to have the same result as #3, and so on.
Ace2 Joined: Oct 2, 2017
• Posts: 1676
January 16th, 2022 at 9:30:53 PM permalink
What are the odds of twelve 7s before seven 12s ?
It’s all about making that GTA
SOOPOO Joined: Aug 8, 2010
• Posts: 8716
January 17th, 2022 at 6:06:59 AM permalink
Quote: Ace2

What are the odds of twelve 7s before seven 12s ?

Over 99%? Just a guess….
tuttigym Joined: Feb 12, 2010
• Posts: 1039
January 17th, 2022 at 6:34:18 AM permalink
Quote: Mission146

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.

You posted the above. I DID NOT "MAINTAIN" or infer what you wrote. You did NOT state what YOUR conclusion was. The post was dishonest and mis-leading and sought to discredit covertly.

Quote: Mission146

I know they are my conclusions and not yours; start with the conclusions are right, so that they are more likely to be mine becomes a given.

What does that mean???

tuttigym
Dieter Joined: Jul 23, 2014
• Posts: 3209
Thanks for this post from: January 17th, 2022 at 7:02:51 AM permalink
Quote: tuttigym

Quote: Mission146

I know they are my conclusions and not yours; start with the conclusions are right, so that they are more likely to be mine becomes a given.

What does that mean???

tuttigym

I'm going to conclude it means Mission is reasonably confident in his 15th grade math skills.
May the cards fall in your favor.
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5413
January 17th, 2022 at 7:13:43 AM permalink
Quote: Ace2

What are the odds of twelve 7s before seven 12s ?

P(12 7s, 0 12s) = (6/7)^12 = 6^12 / 7^12 = 6^12 / 7^18 x 7^6
P(12 7s, 1 12, ending with a 7) = C(12,1) (6/7)^12 (1/7)^1 = 12 6^12 / 7^13 = 6^12 / 7^18 x 12 x 7^5
P(12 7s, 2 12s, ending with a 7) = C(13,2) (6/7)^12 (1/7)^2 = 78 6^12 / 7^14 = 6^12 / 7^18 x 78 x 7^4
P(12 7s, 3 12s, ending with a 7) = C(14,3) (6/7)^12 (1/7)^3 = 364 6^12 / 7^15 = 6^12 / 7^18 x 364 x 7^3
P(12 7s, 4 12s, ending with a 7) = C(15,4) (6/7)^12 (1/7)^4 = 1365 6^12 / 7^16 = 6^12 / 7^18 x 1365 x 7^2
P(12 7s, 5 12s, ending with a 7) = C(16,5) (6/7)^12 (1/7)^5 = 4368 6^12 / 7^17 = 6^12 / 7^18 x 4368 x 7
P(12 7s, 6 12s, ending with a 7) = C(17,6) (6/7)^12 (1/7)^6 = 12,376 6^12 / 7^18 = 6^12 / 7^18 x 12,376

P(12 7s before 7 12s) = 6^12 / 7^18 x (117,649 + 201,684 + 187,278 + 124,852 + 66,885 + 30,576 + 12,376)
= 741,300 x 6^12 / 7^18

tuttigym Joined: Feb 12, 2010
• Posts: 1039
January 17th, 2022 at 7:35:25 AM permalink
Mr. TDG: Your hard work is evident and appreciated. Your math IQ is many levels above mine. If I were in the same room with you having this conversation with the charts, I could ask questions, point by point, that would help me to fully understand what you are trying to convey.

I am sorry. I just cannot grasp the totality of the concept. It is not your fault; it is my deficiency.

Having said that, let me ask:
1. The roulette example does not show a positive edge to the player (+50%) from the start, i.e., 18 chances to win vs 38 chances to lose. My example in craps is 30 chances to win vs only 6 chances to lose or a positive possible outcome favoring the player at 5 to 1. While there are no guarantees as we both have stated in our examples, My mind clearly believes my example favors the player greatly over your example of winning a bet. So, the straight up question is if given these two choices for a one wager win, which would you choose?

2. The table (thank you CharleyPatrick), for me, is even more confusing. I do not know where the "second roll" numbers come from and what purpose it is they serve. The Come Out natural winners and losers are one roll bet outcomes while the point bets can be finalized with no definitive number of rolls. Hense my confusion there.
How does one come up with 5940 rolls exactly?

3. The "Win"/"Lose" columns really baffle me because I do not know how they are determined based on the number of rolls exhibited.

As I said, a conversation at a table with a beverage of choice and a relaxed, informal atmosphere would, more than likely, allow me to more fully understand the concepts.

Thank you for your patience and endurance.

tuttigym