4 continuous OF "BANKER"

Quote: ssho88

...My simulation results(1 million shoe) shown that the average n = 39.98, n(min) = 4 and n(max) = 514.

Anyone can help to verify my results with calculations ? Is there any formula to calculate it ?

Quote: ChesterDog

I'm new to working with "transition matrices," but I just now tried them on your problem.

But instead of playing through a shoe of eight decks, my calculations shuffle after the first hand of every shoe. The Wizard gives the probability of Banker for eight decks on this page: https://wizardofodds.com/games/baccarat/basics/#toc-EightDecks.

If I use the Wizard's probability of Banker of 0.458597, which is rounded to six places, I get an average of 39.9124 for n. (I guess I did the transition matrices right because my n is close to your n.)

Quote: ThatDonGuy

I get 39.9124 as well - here's how:

This One's On Me

Let E_N be the expected number of hands needed to reach 4 in a row when you have N in a row,
p = the probability of a banker win, which I assume is 0.458597,
and q = 1 - p = the probability of anything besides a banker win

With each hand, you are one closer to 4 with probability p, and back to zero with probability q:
E_N = 1 + p E_N+1 + q E₀

E₄ = 0
E₃ = 1 + q E₀ + p E₄ = 1 + q E₀
E₂ = 1 + q E₀ + p (1 + q E₀) = (1 + p)(1 + q E₀)
E₁ = 1 + q E₀ + p (1 + p)(1 + q E₀) = (1 + p + p²)(1 + q E₀)
E₀ = 1 + q E₀ + p (1 + p + p²)(1 + q E₀) = (1 + p + p² + p³)(1 + q E₀)
E₀ (1 - q (1 + p + p² + p³)) = 1 + p + p² + p³
E₀ (1 - (1 - p)(1 + p + p² + p³)) = 1 + p + p² + p³
E₀ p⁴ = 1 + p + p² + p³
E₀ = (p⁴ - 1) / (p⁴ (p - 1))
= (1 - p⁴) / (p⁴ - p⁵)
= 39.91237153

Quote: ssho88

The results of H1, H2, H3 . . . .Hn is B,P,P . . . P,B,T,P,B,B,B,B


H1= hand 1
Hn = hand n

P= PLAYER
B=BANKER
T=TIE

What is the average value of n ?

          0         1         2         3         4
0 0.5414026 0.4585974 0.0000000 0.0000000 0.0000000
1 0.5414026 0.0000000 0.4585974 0.0000000 0.0000000
2 0.5414026 0.0000000 0.0000000 0.4585974 0.0000000
3 0.5414026 0.0000000 0.0000000 0.0000000 0.4585974
4 0.0000000 0.0000000 0.0000000 0.0000000 1.0000000

Quote: ssho88

I still can't get it, do you mind to explain it further ?

(I-S)^-1 = 
22.6086 	10.3682 	4.75485 	2.18056 	
20.428  	10.3682 	4.75485 	2.18056 	
17.8537 	8.18768 	4.75485 	2.18056 	
12.2404 	5.61339 	2.57429 	2.18056 

Quote: ThatDonGuy

I get 39.9124 as well - here's how:

This One's On Me

Let E_N be the expected number of hands needed to reach 4 in a row when you have N in a row,
p = the probability of a banker win, which I assume is 0.458597,
and q = 1 - p = the probability of anything besides a banker win

With each hand, you are one closer to 4 with probability p, and back to zero with probability q:
E_N = 1 + p E_N+1 + q E₀

E₄ = 0
E₃ = 1 + q E₀ + p E₄ = 1 + q E₀
E₂ = 1 + q E₀ + p (1 + q E₀) = (1 + p)(1 + q E₀)
E₁ = 1 + q E₀ + p (1 + p)(1 + q E₀) = (1 + p + p²)(1 + q E₀)
E₀ = 1 + q E₀ + p (1 + p + p²)(1 + q E₀) = (1 + p + p² + p³)(1 + q E₀)
E₀ (1 - q (1 + p + p² + p³)) = 1 + p + p² + p³
E₀ (1 - (1 - p)(1 + p + p² + p³)) = 1 + p + p² + p³
E₀ p⁴ = 1 + p + p² + p³
E₀ = (p⁴ - 1) / (p⁴ (p - 1))
= (1 - p⁴) / (p⁴ - p⁵)
= 39.91237153

Quote: ssho88

The results of H1, H2, H3 . . . .Hn is B,P,P . . . P,B,T,P,B,B,B,B

H1= hand 1
Hn = hand n

P= PLAYER
B=BANKER
T=TIE

What is the average value of n ?

> runs.mean3(4,0.458597423,0.095155968)#Banker 4 Baccarat Tie
          0          1          2          3         4
0 0.5414026 0.45859742 0.00000000 0.00000000 0.0000000
1 0.4462466 0.09515597 0.45859742 0.00000000 0.0000000
2 0.4462466 0.00000000 0.09515597 0.45859742 0.0000000
3 0.4462466 0.00000000 0.00000000 0.09515597 0.4585974
4 0.0000000 0.00000000 0.00000000 0.00000000 1.0000000
[1] 31.72111

Quote: ssho88

Please explain further this : With each hand, you are one closer to 4 with probability p, and back to zero with probability q:
EN = 1 + p EN+1 + q E0

Could you please explain with a sketch or graphically ? I eagerly to understand/learn this. Tq

Quote: ThatDonGuy

There are five "states", each representing how many consecutive Banker results you are currently at.
You start at state 0.
If the first result is Banker (which happens with probability p; in this case, p = 0.458597), then you are at state 1 (since you have "one Banker result in a row"), and if it is Player or Tie (which happens with probability 1 - p, which I call "q"), you stay at state 0.
When you are at state 1, if the next result is Banker, you go to state 2, and if it is Player or Tie, you go back to state 0 (since you now have "zero Banker results in a row").
The same goes for states 2 and 3. When you are at state 4, you are done.

E_N is the expected number of hands needed to get to state 4 from state N.
If you are at state N, the expected number to get to state 4 from there is 1 (the next hand dealt) + p times the expected number from state (N + 1) (because you will be at state N + 1 with probability p) + q times the expected number from state 0 (because you will be back at state 0 with probability q).

Quote: ThatDonGuy

There are five "states", each representing how many consecutive Banker results you are currently at.
You start at state 0.
If the first result is Banker (which happens with probability p; in this case, p = 0.458597), then you are at state 1 (since you have "one Banker result in a row"), and if it is Player or Tie (which happens with probability 1 - p, which I call "q"), you stay at state 0.
When you are at state 1, if the next result is Banker, you go to state 2, and if it is Player or Tie, you go back to state 0 (since you now have "zero Banker results in a row").
The same goes for states 2 and 3. When you are at state 4, you are done.

E_N is the expected number of hands needed to get to state 4 from state N.
If you are at state N, the expected number to get to state 4 from there is 1 (the next hand dealt) + p times the expected number from state (N + 1) (because you will be at state N + 1 with probability p) + q times the expected number from state 0 (because you will be back at state 0 with probability q).

Quote: ThatDonGuy

There are five "states", each representing how many consecutive Banker results you are currently at.
You start at state 0.
If the first result is Banker (which happens with probability p; in this case, p = 0.458597), then you are at state 1 (since you have "one Banker result in a row"), and if it is Player or Tie (which happens with probability 1 - p, which I call "q"), you stay at state 0.
When you are at state 1, if the next result is Banker, you go to state 2, and if it is Player or Tie, you go back to state 0 (since you now have "zero Banker results in a row").
The same goes for states 2 and 3. When you are at state 4, you are done.

E_N is the expected number of hands needed to get to state 4 from state N.
If you are at state N, the expected number to get to state 4 from there is 1 (the next hand dealt) + p times the expected number from state (N + 1) (because you will be at state N + 1 with probability p) + q times the expected number from state 0 (because you will be back at state 0 with probability q).

Quote: Ace2

Ignoring ties, banker has a 0.506824 chance of winning the next hand. Take the reciprocal for x = 1.97301 average hands to get one banker win.

Though using a series of states/equations is fine, there’s a simple formula for this type of problem which, for 4 consecutive binary events is:

(x^(4+1) - x) / (x -1) = 28.703 expected hands for 4 consecutive banker wins.

If ties are counted then x = 1 / .458597 = 2.18056 average hands to get one banker win. Plug x into the formula and you get 39.912 expected hands for 4 consecutive banker wins.

This calculation assumes that these streaks can cross shoes. If that’s not the case, there would be a material adjustment.**It also assumes that the hands are independent, which they are not. However I believe that is a minuscule, effectively nil effect for an 8 deck shoe, and I don’t think it could be feasibly calculated (only simulated like most blackjack stuff is). For nearly all practical purposes, 6+ decks is the same as infinite deck.

**Counting ties, I calculate that you have a 0.877297 chance of getting the 4 wins in 80 hands. Divide that into 39.912 for an average of 45.494 hands to get the 4 wins if they can’t span over shoes, assuming an average shoe lasts 80 hands.

Quote: ssho88

How you got that simple formula ?

Quote: ThatDonGuy

Start with the last value from my solution:
(1 - p⁴) / (p⁴ - p⁵)
= (1 - p⁴) / (p⁴ (1 - p))
= (p (1 - p⁴)) / (p⁵ (1 - p))
= (1 - p⁴) / p⁵ * p / (1 - p)
= (1/p⁵ - 1/p) / (1/p - 1)

Let x = 1/p; this becomes (x⁵ - x) / (x - 1).

Quote: ssho88

If a Tie does not break a streak, my simulation results is 31.74, which is slightly higher compare to your 28.703.

Quote: 7craps

this all lead to
"why do you want to know the average?"

Quote: OnceDear

And there is no maximum.

Quote: 7craps

Does not matter if the actual 'average' of the distribution of 31.72 or 31.74 or X

B,B,T,B,T,B is still 4 in a row Banker

and the probability that you see 'at least one' 4 Bankers in a row (with or without Ties)
is NOT 100%
not even 90%,
80% or 70% (pathetic)

in 31 hands: 0.629434727 (62.94%)
in 32 hands: 0.642174489 (64.22%)

this all lead to
"why do you want to know the average?"

Quote: ssho88

There is advantage play if you have the info of average rounds

Quote: 7craps

Many Asian Baccarat players have thought this over the years.

It IS laughable.

Your statement IS 100% false.
good luck

Quote: ssho88

I know we can get that simple formula by substitute x=1/p. My question is how can he get that formula WITHOUT your EN formula ?

BTW, If a Tie does not break a streak in Baccarat for Banker, can I re-write your formula as below :-

EN = 1 + p EN+1 + q E0 + r EN ?

where the probability for BANKER, PLAYER and TIE is p, q and r respectively.

Quote: Ace2

Ignoring ties, banker has a 0.506824 chance of winning the next hand. Take the reciprocal for x = 1.97301 average hands to get one banker win.

Though using a series of states/equations is fine, there’s a simple formula for this type of problem which, for 4 consecutive binary events is:

(x^(4+1) - x) / (x -1) = 28.703 expected hands for 4 consecutive banker wins.

If ties are counted then x = 1 / .458597 = 2.18056 average hands to get one banker win. Plug x into the formula and you get 39.912 expected hands for 4 consecutive banker wins.

This calculation assumes that these streaks can cross shoes. If that’s not the case, there would be a material adjustment.**It also assumes that the hands are independent, which they are not. However I believe that is a minuscule, effectively nil effect for an 8 deck shoe, and I don’t think it could be feasibly calculated (only simulated like most blackjack stuff is). For nearly all practical purposes, 6+ decks is the same as infinite deck.

**Counting ties, I calculate that you have a 0.877297 chance of getting the 4 wins in 80 hands. Divide that into 39.912 for an average of 45.494 hands to get the 4 wins if they can’t span over shoes, assuming an average shoe lasts 80 hands.

Quote: ssho88

It's fine, but you don't know what I am doing and I can't disclose it here.

Quote: unJon

.

I believe your formula is wrong when ties do not break the streak because it is ignoring ties, but really the ties still need to be “counted” to get to the average number of hands you see before 4 bankers.

Quote: Ace2

Depends how you count. Remember that ties are ignored, so if we say that an average of 29 hands are needed ignoring ties, that means ties aren’t counted.

I think you’re asking to ignore ties (meaning they can’t break a streak) but do include them in the hand count. So in this case we take the 28.703 expected hands and divide by .904544 to get 31.732 expected hands when ties are “ignored but still counted”.

.904544 of all hands are not ties.

Quote: ssho88

what I am doing and I can't disclose it here.

Quote: ssho88

How you got that simple formula ?

If a Tie does not break a streak, my simulation results is 31.74, which is slightly higher compare to your 28.703.
I guess you NOT add in the rounds for TIE ?

Quote: Ace2

Google “geometric series” and you’ll see this and other formulas. I’ve used it many times since this is such a common problem (consecutives). If, for instance, someone asks the expected number of rolls to get 3 consecutive 7’s with 2 die (1 in 6 chance each roll) the answer is 6 + 6^2 + 6^3 = 258 which can also be calculated as (6^(3+1) - 6) / (6 - 1).

There are many formulas for series that can make a calculation much easier and in some cases make a calculation possible.

Quote: ThatDonGuy

E0 p4 = 1 + p + p2 + p3
E0 = (p4 - 1) / (p4 (p - 1))

Quote: ThatDonGuy

Gee, I'll have to remember that one...

I think what ssho88 (and myself, for that matter) wants to know is, where "the expected number of trials to get N consecutive results of a particular type is E + E² + E³ + ... + E^N, where E is the expected number of trials to get one result of that type" came from.

Quote: lilredrooster

you ask for assistance in very difficult and complicated math equations and then you say you can't disclose what you're doing here

those that answered you did a lot of work for you

what did you do for them?

zilch

Quote: ssho88

The funny thing here is main contributors here did not ask anything from me but you are trying . . . .

Quote: lilredrooster

I didn't ask you for anything and you don't have to worry - I never will
the other guys didn't ask because that is not in their nature


just thought after they did all that work for you, you might have considered saying thank you

Quote: ssho88

Please read carefully my early replies( post #3 and # 9), I have said "Tq" to ChesterDog and ThatDonGuy, and yet I shared my advantage play info with 7craps !

Quote: Ace2

Quote: ThatDonGuy

I think what ssho88 (and myself, for that matter) wants to know is, where "the expected number of trials to get N consecutive results of a particular type is E + E² + E³ + ... + E^N, where E is the expected number of trials to get one result of that type" came from.

I guess that series is my discovery, though I’d be shocked if no one ever saw this pattern before. For now I’ll call it the Ace-Onacci series.