MathCurious
MathCurious
Joined: Nov 27, 2020
  • Threads: 6
  • Posts: 8
December 5th, 2020 at 10:39:25 PM permalink
Hi wizards! I'm studying baccarat now and I am lost. Base on my findings in the net the possible combinations of a 1-deck baccarat of the banker winning is 6737232640, the player is 654867443 and a tie is 1372227328. But how did they come up with these numbers? I'm sure they came up with a program to count this but do you have any idea how to mathematically solve this? or do you have a program that can do this? Because I have no background in programming, if u do can you share it with me?

As always thank you to whoever responds to this query.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 106
  • Posts: 5065
December 6th, 2020 at 8:45:29 AM permalink
Quote: MathCurious

Hi wizards! I'm studying baccarat now and I am lost. Base on my findings in the net the possible combinations of a 1-deck baccarat of the banker winning is 6737232640, the player is 654867443 and a tie is 1372227328. But how did they come up with these numbers? I'm sure they came up with a program to count this but do you have any idea how to mathematically solve this? or do you have a program that can do this? Because I have no background in programming, if u do can you share it with me?


The sum of the numbers - and note the actual player number is 6,548,674,432 - is 14,658,134,400. This is 52 x 51 x 50 x 49 x 48 x 47 - in other words, this is how many possible ways the first six cards from a single deck can be dealt.

It's a little hard to do it strictly mathematically, as you have to take into account the number of cards of each value after each step.
Here is how I do it with a computer:

First, let P1 = 0, 1, 2, ..., 9
For each P1, let B1 = 0, 1, 2, ..., 9
For each B1, let P2 = 0, 1, 2, ..., 9
For each P2, let B2 = 0, 1, 2, ..., 9
For each B2, let P3 = 0, 1, 2, ..., 9
For each P3, let B3 = 0, 1, 2, ..., 9
Set an array Shoe[ ] with 10 values: Shoe[0] = 16, and Shoe[1], Shoe[2], ..., Shoe[9] each = 4
(For more than one deck, set Shoe[0] = 16 x the number of decks, and the others = 4 x the number of decks)
To determine the number of "combinations" for a particular set of six cards, do this:
Start with Combos = Shoe[P1], then reduce Shoe[P1] by 1 (for example, if the first card is an Ace, there are initially four Aces in the deck, so Combos starts at 4, and Shoe[1] is reduced to 3 as there are now only 3 Aces remaining)
Multiply Combos by Shoe[B1], then reduce Shoe[B1] by 1
Multiply Combos by Shoe[P2], then reduce Shoe[P2] by 1
Multiply Combos by Shoe[B2], then reduce Shoe[B2] by 1
Multiply Combos by Shoe[P3], then reduce Shoe[P3] by 1
Multiply Combos by Shoe[B3], then reduce Shoe[B3] by 1
Note: never mind that you may not use cards P3 or B3; since it's possible that you might use 6 cards, you have to count all possible 6-card deals.
Now, determine the result of the hand; P1 and P2 are the player's first two cards, B1 and B2 are the banker's first two cards, P3 is the player's possible third card, and B3 is the banker's possible third card.
Another note: never mind that, in reality, if the player does not take a third card but the banker does, the banker would actually get card P3 instead of B3; the results are the same.
Add up the total number of combinations (the Combos values) of the deals where the player wins, then add up the number where the banker wins, then add up the number where there is a tie.
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1398
  • Posts: 23607
December 6th, 2020 at 8:55:10 AM permalink
Quote: MathCurious

Hi wizards! I'm studying baccarat now and I am lost. Base on my findings in the net the possible combinations of a 1-deck baccarat of the banker winning is 6737232640, the player is 654867443 and a tie is 1372227328. But how did they come up with these numbers? I'm sure they came up with a program to count this but do you have any idea how to mathematically solve this? or do you have a program that can do this? Because I have no background in programming, if u do can you share it with me?



As you guessed, I wrote a program to simply loop through all the way the cards can come out. It would be very tedious to do it by hand. Baccarat is a pretty easy game to program. You should be able to do it with one introductory programming class.
It's not whether you win or lose; it's whether or not you had a good bet.
USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 6th, 2020 at 11:04:08 AM permalink
Quote: Wizard

As you guessed, I wrote a program to simply loop through all the way the cards can come out. It would be very tedious to do it by hand. Baccarat is a pretty easy game to program. You should be able to do it with one introductory programming class.



This is 1 of those questions with lots of combination values that currently can't be done by hand unless your using my theorem. I hate to say it again guys but I know I can do this question by hand (or rather a spreadsheet but the same differences) but I think I should just wait till I publish my research. At least that's what I've been told I should do :/

But everyone should know I can do this by hand because I'm the only person who can solve the probabilities of the Royal Deck, which has far more combinations,& yet I can solve it by hand when no computer programmer can solve them!
Math is the only true form of knowledge
gordonm888
Administrator
gordonm888
Joined: Feb 18, 2015
  • Threads: 50
  • Posts: 3374
Thanks for this post from:
CrystalMath
December 6th, 2020 at 1:10:27 PM permalink
Quote: USpapergames


But everyone should know I can do this by hand because I'm the only person who can solve the probabilities of the Royal Deck game, which has far more combinations,& yet I can solve it by hand when no computer programmer can solve them!



This is incorrect. Many of us looked at the Royal Deck you invented and said that we could calculate the probabilities "by hand" as you put it.

We chose not to do it because there was no point -its not a game that exists anywhere, so who cares? - and because a few aspects of the game were not crisply defined. And, frankly, you acted as if the WOV forum mathematicians were circus animals that were supposed to spend time jumping through hoops simply because we were being challenged.

Somehow, you have confused a community-wide decision to ignore you with the idea that you are king of the world.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 24
  • Posts: 1115
December 6th, 2020 at 1:59:14 PM permalink
13^4 = 28,561 starting hands. From there you filter down the cases where player/banker take a third card

Pull out the handful of cases where the banker uses different rules than the player and thatís the house edge

This assumes infinite deck, which makes the calculations easier by several orders of magnitude yet gets you within a couple basis points of the exact answer for 8 decks
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 106
  • Posts: 5065
December 6th, 2020 at 3:30:38 PM permalink
Quote: USpapergames

This is 1 of those questions with lots of combination values that currently can't be done by hand unless your using my theorem. I hate to say it again guys but I know I can do this question by hand (or rather a spreadsheet but the same differences) but I think I should just wait till I publish my research. At least that's what I've been told I should do :/


Be careful; you're beginning to sound like David Fabian.
teliot
teliot
Joined: Oct 19, 2009
  • Threads: 41
  • Posts: 2268
December 6th, 2020 at 4:24:35 PM permalink
There is no way to answer this question without at least a rudimentary understanding of computer programming. In that sense, it has been answered. Loop through all hands.

This spreadsheet (of mine) is in the public domain and has all the code in VBA:

https://researchers.one/doc/5f848e142c77e400043aadd8

Mike has written similar code, as have many others here.
Last edited by: teliot on Dec 6, 2020
Poetry website: www.totallydisconnected.com
USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 6th, 2020 at 4:28:30 PM permalink
Quote: gordonm888

This is incorrect. Many of us looked at the Royal Deck you invented and said that we could calculate the probabilities "by hand" as you put it.

We chose not to do it because there was no point -its not a game that Lexists anywhere, so who cares? - and because a few aspects of the game were not crisply defined. And, frankly, you acted as if the WOV forum mathematicians were circus animals that were supposed to spend time jumping through hoops simply because we were being challenged.

Somehow, you have confused a community-wide decision to ignore you with the idea that you are king of the world.



1) Never said I was king of the world???

2) You didn't need any game rules to solve the hand ranking probabilities!

3) The point was to prove that I was the only person who could solve those probabilities & I put a $3,000 reward for anyone who could! I believe you were the person who asked me to bump up the reward!

4) Never assuming the people on this form would just do anything for a challenge, but if you were up for the challenge of working on the most difficult game probabilities than my challenge should have been of interest to you.

5) Nobody said they could solve the probabilities at the end of the form and all that attempted failed & quiet early because they realized how naive they were thinking the question was easy.
Math is the only true form of knowledge
USpapergames
USpapergames
Joined: Jun 23, 2020
  • Threads: 18
  • Posts: 807
December 6th, 2020 at 4:30:57 PM permalink
Quote: ThatDonGuy

Be careful; you're beginning to sound like David Fabian.



I don't know who David Fabian is but I don't see the issue with stating something that is a fact. Do I need to bring back the $3,000 challenge to prove this or should I just show you that I have the correct probabilities with the correct hand rankings?
Math is the only true form of knowledge

  • Jump to: