SOOPOO Joined: Aug 8, 2010
• Posts: 8716
January 17th, 2022 at 8:53:32 AM permalink
Quote: ThatDonGuy

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

I am very proud of my ‘more than 99%’ guess! So it is ever so slightly more than 99%!
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5412
January 17th, 2022 at 9:41:38 AM permalink
Quote: tuttigym

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?

You are confusing expected value with probability of winning.
Let me give you another example: suppose the casino had this game - you bet \$1000, and you roll a 6-sided die; if you roll a 1-5, you win \$1, but if you roll a 6, you lose your \$1000. You have a 5/6 chance of winning, but obviously you don't play it as the risk far outweighs the reward.
This is what Expected Value is - a comparison of risk versus reward.
If you are interested in a bet with a high chance of winning, then go to the roulette wheel and bet 1 on each number from 1 to 35. You have a 35/38 chance of winning on a double-zero wheel; you gain 35 on the winning bet and lose 34 on the 34 losing bets combined, so you win 1 on the total bet. True, if any of the other three numbers come up, you lose 35, but that doesn't seem to be important to you; only the fact that you will win 92% of the time does.

As to which I would choose: you will have to refresh my memory as to which bet you are referring to that has 30 chances to win. Given the choice between the Pass Line in craps (EV = -1.414%) and pretty much any bet in double-zero roulette (EV = -5.263%), I will choose craps every time.

If you are interested in determining the probability of starting with some bankroll B and being able to get X ahead without busting the entire bankroll first, that is a completely different discussion called the Gambler's Ruin Problem.

Quote: tuttigym

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?

The "second number" is the probability, divided by 165, that this bet wins or loses.

Quote: tuttigym

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

Remember when I said that expected value is the sum of every product of (the probability of an event happening and the result of that event)?
Let's use rolling a point of 4 as an example.
The probability of rolling a 4 on a comeout is 1/12.
Once you have a point of 4, the only numbers that matter are 4 and 7. There are 3 ways to roll a 4 and 6 ways to roll a 7, so the probability of rolling a 4 before a 7, which wins the bet, is 3 / (3 + 6) = 1/3, and the probability of rolling a 7 before a 4, which loses the bet, is 6 / (3 + 6) = 2/3.
The portion of the EV that comes from winning with a point of 4 = the probability of rolling, and then making, a point of 4, multiplied by the value of the win. The probability = 1/12 x 1/3 = 1/36, and the value is 1, so the product is 1/36.
Similarly, the portion of the EV that comes from losing with a point of 4 = the probability of rolling, and then missing, a point of 4, multiplied by the value of the loss. The probability = 1/12 x 2/3 = 1/18, and the value is -1, so the product is -1/18.
Now, do this for every comeout roll, keeping in mind that some probabilities - for example, winning with a comeout roll of 2 - are zero.
This is what Charlie's chart does, but it multiplies the numbers by constant values so that they all appear as integers.
The "first number" is the probability of rolling that number in the comeout, multiplied by 36.
The "second number" is the probability that, given that the particular point number for that row has been established, you will win or lose the bet, multiplied by 165. For example, for a point of 4, the "second number" for a win is 1/3 x 165 = 65, and the "second number" for a loss is 2/3 x 165 = 110. 165 was chosen because it is the smallest number that, when you multiply all of the "make or miss the point" probabilities by it, turns them all into integers.
The "win" number equals the second number if the row represents a win, and zero if it represents a loss.
The "loss" number equals the second number if the row represents a loss, and zero if it represents a win.
The reason those two columns exist is to show what the sums of the winning and losing results are.

I will try expressing it in a different way:

Each result is the product of three values: the probability of rolling that number on the comeout, the probability of winning or losing, as appropriate for that result, once the point has been established, and the value of that result (1 for a win, or -1 for a loss; if you are calculating Don't Pass, it would be -1 for a win and 1 for a loss, but remember that a comeout of 12 has a result of 0).
Winning with a comeout of 2: 1/36 x 0 x 1 = 0
Losing with a comeout of 2: 1/36 x 1 x (-1) = -1/36
Winning with a comeout of 3: 1/18 x 0 x 1 = 0
Losing with a comeout of 3: 1/18 x 1 x (-1) = -1/18
Winning with a comeout of 4: 1/12 x 1/3 x 1 = 1/36
Losing with a comeout of 4: 1/12 x 2/3 x (-1) = -1/18
Winning with a comeout of 5: 1/9 x 2/5 x 1 = 2/45
Losing with a comeout of 5: 1/9 x 3/5 x (-1) = -1/15
Winning with a comeout of 6: 5/36 x 5/11 x 1 = 25/396
Losing with a comeout of 6: 5/36 x 6/11 x (-1) = -5/66
Winning with a comeout of 7: 1/6 x 1 x 1 = 1/6
Losing with a comeout of 7: 1/6 x 0 x (-1) = 0
Winning with a comeout of 8: 5/36 x 5/11 x 1 = 25/396
Losing with a comeout of 8: 5/36 x 6/11 x (-1) = -5/66
Winning with a comeout of 9: 1/9 x 2/5 x 1 = 2/45
Losing with a comeout of 9: 1/9 x 3/5 x (-1) = -1/15
Winning with a comeout of 10: 1/12 x 1/3 x 1 = 1/36
Losing with a comeout of 10: 1/12 x 2/3 x (-1) = -1/18
Winning with a comeout of 11: 1/18 x 1 x 1 = 1/18
Losing with a comeout of 11: 1/18 x 0 x (-1) = 0
Winning with a comeout of 12: 1/36 x 0 x 1 = 0
Losing with a comeout of 12: 1/36 x 1 x (-1) = -1/36
The total EV is the sum of these 22 numbers, which is -7/495.

Note that 495 does not appear in any of the 22 numbers themselves; it just happens that when you add them up and reduce the fraction to lowest terms, the denominator is 495.

tuttigym Joined: Feb 12, 2010
• Posts: 1039
Thanks for this post from: January 17th, 2022 at 2:55:40 PM permalink
Mr.TDG: Yes, you are correct. I was confusing EV with probability of winning. My personal ultimate approach mathematically (4th grade style) is probability of winning. I try to create betting patterns that give me an edge over the house without regard to the EV. I basically disregard EV, even though I know the actual loss potential of the event, to concentrate or focus on the table, the shooters, and the play. With that being stated, my play goes to both sides (right side and "dark side) and will include hedging. It does not work all the time, but with a moderate buy-in, I can play for a period of time that is comfortable and rewarding. So, thank you for ending my confusion.

Rhetorical question: You stated that you are basically a right side bettor (PL+odds). Your math acumen is off the charts in my estimation. Mr W. is a "Dark" side player. Which of you gets the edge? Have you ever played along side one another?

Your explanation of point play, for me, is probability which allows me the luxury of using my 4th grade arithmetic without regard and the "confusion" of EV. I believe many players shy away from the game and the real simplicity it can offer if they can stay "out of the weeds" of HA/HE which in my view are misleading with the extremely low percentages that are offered within the confines of this forum.

Charlie's chart using integers is just too much information, and, for me as well as the vast majority of potential players, would have them running and screaming into the night.

Again, thank you for your hard work, patience, due diligence, and humility. It is much appreciated.

tuttigym
tuttigym Joined: Feb 12, 2010
• Posts: 1039
January 17th, 2022 at 3:01:47 PM permalink
Quote: Dieter

I'm going to conclude it means Mission is reasonably confident in his 15th grade math skills.

Mr. Dieter: This forum is extremely lucky to have you, a certified "MSL" * member in its ranks.

Thank you.

tuttigym

*Mission 146 as a Second Language
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5412
January 17th, 2022 at 3:09:07 PM permalink
Quote: tuttigym

Rhetorical question: You stated that you are basically a right side bettor (PL+odds). Your math acumen is off the charts in my estimation. Mr W. is a "Dark" side player. Which of you gets the edge? Have you ever played along side one another?

He has the edge, as Don't Pass is a slightly better bet than Pass. I don't like playing DP because I am turned off by the fact that odds bets with DP pay less than even money.

As for whether we have played alongside each other: no. I usually don't come across the other forum regulars while I'm in Vegas, especially as I don't gamble that much - for example, the last time I was here, the only gambling I did was in VP, and even then, it was less than an hour.
Ace2 Joined: Oct 2, 2017
• Posts: 1675
January 17th, 2022 at 3:20:16 PM permalink
I have a similar feeling about laying odds. 5:6 is close to even money and 2:3 isn’t so bad, but I don’t like laying 2:1, especially when I have a large flat bet
It’s all about making that GTA
unJon Joined: Jul 1, 2018
• Posts: 3469
January 17th, 2022 at 4:03:13 PM permalink
Funny. I do too and recognize it as irrational but still feel it. I always think now that I’ve survived the come out, my DP is +EV, why should I dilute it with odds . . .

. . . of course I then go make a don’t come.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Mission146 Joined: May 15, 2012
• Posts: 15308
January 18th, 2022 at 6:00:25 AM permalink
Quote: tuttigym

Quote: Dieter

I'm going to conclude it means Mission is reasonably confident in his 15th grade math skills.

Mr. Dieter: This forum is extremely lucky to have you, a certified "MSL" * member in its ranks.

Thank you.

tuttigym

*Mission 146 as a Second Language