Baccarat problem

What are the odds that any black 8 (either spades or clubs, or both) will feature in the winning Baccarat hand, if the game is dealt with a continuous shuffler? Does the number of decks play any role in this, if the cards are returned to the machine after each hand? Ties are excluded, i.e. only resolved hands are counted.
If it is not a problem, I would appreciate if anyone answering this question would also post their calculations.
Thanks!

You saw bac being dealt with a CSM? Where? I don't
know a single Asian that would play at such a table.

:EDIT: Fixed some earlier code mistakes. I feel much a bit more confident now.
:EDIT #2: Fixed one more small but meaningful mistake

So I figured I could use this an opportunity to learn to work this out by hand.

Instead I wrote a 100,000,000 hand sim because I'm lazy like that. I know that doesn't really help you much since you need to prove it, but I did it anyway so...

Player Wins: 44633911 (44.63% win rate) Banker Wins: 45772625 (45.77% win rate) Ties: 9593464 (9.59% tie rate)
One or more black eights in Winning Hands: 9719530 (9.7%)

The above assumes an infinite deck and wouldn't be 100% the same as a CSM as it doesn't take latency of redistribution into account but it's close enough. It gives an approximate black-8-in-winning-hand chance of 9.7%.

This looks fishy to me since the Wizard lists a 44.6% and 45.8% rate respectively for 8 decks so take that with a grain of salt. Also, I've been drinking a bit and it took me two edits to get to what I believe is correct.

Here's the code. It's fully commented so if anyone sees something wrong please let me know. Yes I know it's VB and yes I know the code's a bit lazy in parts but I knocked it out in 20 minutes so give me a break.

I think the line 2 + 2 = 5 might be a mistake, possibly.

Quote: buzzpaff

I think the line 2 + 2 = 5 might be a mistake, possibly.

What are you talking about? I believe that 2 = 3. Prove me wrong.

(1)

3 a - 2 a = a
and

(2)

a = b + c
Now, by inserting (2) into (1), we get:

(3)

3 a - 2 a = 3 ( b + c ) - 2 ( b + c )
After multiplication, eq. (3) yields:

(4)

3 a - 2 a = 3 b + 3 c - 2 b - 2 c
Now let us clarify the structure of eq. (4) by shifting all terms including the number 3 on the left side, and all terms including the number 2 on the right side:

(5)

3 a - 3 b - 3c = 2 a - 2 b - 2 c
Eq. (5) may be written even more clearly by the use of brackets:

(6)

3 ( a - b - c ) = 2 ( a - b - c )
Now it is very easy to see, that indeed

3 = 2

Quote: WongBo

(1)

3 a - 2 a = a
and

(2)

a = b + c
Now, by inserting (2) into (1), we get:

(3)

3 a - 2 a = 3 ( b + c ) - 2 ( b + c )
After multiplication, eq. (3) yields:

(4)

3 a - 2 a = 3 b + 3 c - 2 b - 2 c
Now let us clarify the structure of eq. (4) by shifting all terms including the number 3 on the left side, and all terms including the number 2 on the right side:

(5)

3 a - 3 b - 3c = 2 a - 2 b - 2 c
Eq. (5) may be written even more clearly by the use of brackets:

(6)

3 ( a - b - c ) = 2 ( a - b - c )
Now it is very easy to see, that indeed

3 = 2

Excellent reply.

In case anyone is lost, I was joking, and WongBo used a variant on an age-old trick to convince the mathematically challenged that any number equals any other number. The trick is at the very end, when you divide by (a - b - c), which according to (2), equals zero. Division by zero, bitches!

Quote: AcesAndEights

What are you talking about? I believe that 2 = 3. Prove me wrong.

Just my luck to have a Professor Emeritus of Mathematics on line. DAMN

Quote: AcesAndEights

The trick is at the very end, when you divide by (a - b - c), which according to (2), equals zero.

There is also the "trick" where he says:

Quote: WongBo

Now, by inserting (2) into (1), we get:

(3)

3 a - 2 a = 3 ( b + c ) - 2 ( b + c )

That's just eqn. (2) inserted into 3a-2a (i.e., not an equation at all); eqn. (1) isn't used at all.

Thank you so much for your help, I really appreciate it.
My own calculations came to 9.84%, but in all honesty, I don't have that much confidence in my probability calculations, so I had to ask for expert help.

Mini Baccarat variant delt using a continuous shuffle can be seen in some Australian casinos.

I would still appreciate if anyone has the patience to explain the method for calculating this problem.

Quote: WongBo

(1)

3 a - 2 a = a
and

(2)

a = b + c
Now, by inserting (2) into (1), we get:

(3)

3 a - 2 a = 3 ( b + c ) - 2 ( b + c )
After multiplication, eq. (3) yields:

(4)

3 a - 2 a = 3 b + 3 c - 2 b - 2 c
Now let us clarify the structure of eq. (4) by shifting all terms including the number 3 on the left side, and all terms including the number 2 on the right side:

(5)

3 a - 3 b - 3c = 2 a - 2 b - 2 c
Eq. (5) may be written even more clearly by the use of brackets:

(6)

3 ( a - b - c ) = 2 ( a - b - c )
Now it is very easy to see, that indeed

3 = 2

haha, I love it.... in my Foundations of Advanced Math, the professor showed that 2=4..

Subtract 3 from both sides... -1 = 1

Square both sides.. 1 = 1

Thus 2 = 4

But not a biconditional relationship, because you can't "undo" what you just did, without doing something mathematically illigal.. I'll never forget that.. When I tell people they rack their brain trying to prove me wrong but don't have the mathematical skills to prove it..

I am still trying to figure out Who's On First !

Quote: PopCan

:EDIT: Fixed some earlier code mistakes. I feel much a bit more confident now.
:EDIT #2: Fixed one more small but meaningful mistake

So I figured I could use this an opportunity to learn to work this out by hand.

Instead I wrote a 100,000,000 hand sim because I'm lazy like that. I know that doesn't really help you much since you need to prove it, but I did it anyway so...

Player Wins: 44633911 (44.63% win rate) Banker Wins: 45772625 (45.77% win rate) Ties: 9593464 (9.59% tie rate)
One or more black eights in Winning Hands: 9719530 (9.7%)

The above assumes an infinite deck and wouldn't be 100% the same as a CSM as it doesn't take latency of redistribution into account but it's close enough. It gives an approximate black-8-in-winning-hand chance of 9.7%.

This looks fishy to me since the Wizard lists a 44.6% and 45.8% rate respectively for 8 decks so take that with a grain of salt. Also, I've been drinking a bit and it took me two edits to get to what I believe is correct.

Here's the code. It's fully commented so if anyone sees something wrong please let me know. Yes I know it's VB and yes I know the code's a bit lazy in parts but I knocked it out in 20 minutes so give me a break.

I get 487,622,577,102,848 ( 243,813,022,855,168 Player, 243,809,554,247,680 Banker) hands out of 4,998,398,275,503,360 hands will be a winner with at least 1 black 8 or about 9.75557669129 % ( 10.7815008405171 % if you ignore ties)