You are dealing baccarat game from a new 8 deck shoe(cut card place at 14 cards from the bottom of the shoe) and record the results of each hand, you stop once you get 4 continuous OF "BANKER". If you can't get 4 continuous BANKER when reach the cut card, then you should shuffle the cards and continue to dealt card from the new shoe, until you get 4 continuous OF "BANKER".

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 ?

OR

What the average hands you have to dealt before you get a 4 continuous OF "BANKER" ?

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: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 ?

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:ChesterDogI'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.)

Could you please show me your calculations ? I would like to learn something from you. Tq

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 E

_{0}

E

_{4}= 0

E

_{3}= 1 + q E

_{0}+ p E

_{4}= 1 + q E

_{0}

E

_{2}= 1 + q E

_{0}+ p (1 + q E

_{0}) = (1 + p)(1 + q E

_{0})

E

_{1}= 1 + q E

_{0}+ p (1 + p)(1 + q E

_{0}) = (1 + p + p

^{2})(1 + q E

_{0})

E

_{0}= 1 + q E

_{0}+ p (1 + p + p

^{2})(1 + q E

_{0}) = (1 + p + p

^{2}+ p

^{3})(1 + q E

_{0})

E

_{0}(1 - q (1 + p + p

^{2}+ p

^{3})) = 1 + p + p

^{2}+ p

^{3}

E

_{0}(1 - (1 - p)(1 + p + p

^{2}+ p

^{3})) = 1 + p + p

^{2}+ p

^{3}

E

_{0}p

^{4}= 1 + p + p

^{2}+ p

^{3}

E

_{0}= (p

^{4}- 1) / (p

^{4}(p - 1))

= (1 - p

^{4}) / (p

^{4}- p

^{5})

= 39.91237153

Quote:ThatDonGuyI get 39.9124 as well - here's how:

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 E_{0}

E_{4}= 0

E_{3}= 1 + q E_{0}+ p E_{4}= 1 + q E_{0}

E_{2}= 1 + q E_{0}+ p (1 + q E_{0}) = (1 + p)(1 + q E_{0})

E_{1}= 1 + q E_{0}+ p (1 + p)(1 + q E_{0}) = (1 + p + p^{2})(1 + q E_{0})

E_{0}= 1 + q E_{0}+ p (1 + p + p^{2})(1 + q E_{0}) = (1 + p + p^{2}+ p^{3})(1 + q E_{0})

E_{0}(1 - q (1 + p + p^{2}+ p^{3})) = 1 + p + p^{2}+ p^{3}

E_{0}(1 - (1 - p)(1 + p + p^{2}+ p^{3})) = 1 + p + p^{2}+ p^{3}

E_{0}p^{4}= 1 + p + p^{2}+ p^{3}

E_{0}= (p^{4}- 1) / (p^{4}(p - 1))

= (1 - p^{4}) / (p^{4}- p^{5})

= 39.91237153

Nice! Thanks for that explanation.

interesting you count TiesQuote:ssho88The 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 ?

Banker win = 0.458597423

this was done online with some code I made in R

https://sites.google.com/view/krapstuff/streaks

section 2r.

> runs.mean3(4,0.458597423)#Banker 4 Baccarat

the transition matrix

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

result: [1] 39.91225

in reality, a Tie does not break a streak in Baccarat for Banker or Player

I learned this stuff starting hereQuote:ssho88I still can't get it, do you mind to explain it further ?

http://www.zweigmedia.com/RealWorld/Summary6b.html

the matrix algebra tool is here: http://www.zweigmedia.com/RealWorld/fancymatrixalg.html

A is only for the distribution probabilities after N trials (not needed for any calculations right now)

A=[0.5414026, 0.4585974, 0.0000000, 0.0000000, 0.0000000

0.5414026, 0.0000000, 0.4585974, 0.0000000, 0.0000000

0.5414026, 0.0000000, 0.0000000, 0.4585974, 0.0000000

0.5414026, 0.0000000, 0.0000000, 0.0000000, 0.4585974

0.0000000, 0.0000000, 0.0000000, 0.0000000, 1.0000000]

S is from A

S=[0.5414026, 0.4585974, 0.0000000, 0.0000000

0.5414026, 0.0000000, 0.4585974, 0.0000000

0.5414026, 0.0000000, 0.0000000, 0.4585974

0.5414026, 0.0000000, 0.0000000, 0.0000000]

(I-S)^-1

result:

(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

sum the top row (all columns)

22.6086+10.3682+4.75485+2.18056

mean (average): 39.91221

I found learning how to do Markov chains (and the data they can give you) slow and frustrating.

Once I got, I got it.

some learn this stuff real fast, others, never learn it.

Quote:ThatDonGuyI get 39.9124 as well - here's how:

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 E_{0}

E_{4}= 0

E_{3}= 1 + q E_{0}+ p E_{4}= 1 + q E_{0}

E_{2}= 1 + q E_{0}+ p (1 + q E_{0}) = (1 + p)(1 + q E_{0})

E_{1}= 1 + q E_{0}+ p (1 + p)(1 + q E_{0}) = (1 + p + p^{2})(1 + q E_{0})

E_{0}= 1 + q E_{0}+ p (1 + p + p^{2})(1 + q E_{0}) = (1 + p + p^{2}+ p^{3})(1 + q E_{0})

E_{0}(1 - q (1 + p + p^{2}+ p^{3})) = 1 + p + p^{2}+ p^{3}

E_{0}(1 - (1 - p)(1 + p + p^{2}+ p^{3})) = 1 + p + p^{2}+ p^{3}

E_{0}p^{4}= 1 + p + p^{2}+ p^{3}

E_{0}= (p^{4}- 1) / (p^{4}(p - 1))

= (1 - p^{4}) / (p^{4}- p^{5})

= 39.91237153

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

say you wanted to know about this: P,P,B,B,B,T,B,PQuote:ssho88The 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 ?

still 4 Bankers in a row with the Tie in there.

the average would be less (of course)

and it requires a slightly different transition matrix

https://sites.google.com/view/krapstuff/streaks

section 2r. still works

> 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