ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 17th, 2019 at 10:32:19 PM permalink
Here is an interesting question about getting 4 continuous OF "BANKER" in Baccarat games.

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 ?
ChesterDog
ChesterDog
  • Threads: 8
  • Posts: 1481
Joined: Jul 26, 2010
May 18th, 2019 at 8:25:25 AM permalink
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.)
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 18th, 2019 at 9:00:56 AM permalink
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.)



Could you please show me your calculations ? I would like to learn something from you. Tq
ThatDonGuy
ThatDonGuy
  • Threads: 117
  • Posts: 6219
Joined: Jun 22, 2011
May 18th, 2019 at 9:11:18 AM permalink
I get 39.9124 as well - here's how:

Let EN 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:
EN = 1 + p EN+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

ChesterDog
ChesterDog
  • Threads: 8
  • Posts: 1481
Joined: Jul 26, 2010
May 18th, 2019 at 9:24:20 AM permalink
Quote: ThatDonGuy

I get 39.9124 as well - here's how:


Let EN 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:
EN = 1 + p EN+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




Nice! Thanks for that explanation.
7craps
7craps
  • Threads: 18
  • Posts: 1977
Joined: Jan 23, 2010
May 18th, 2019 at 9:36:58 AM permalink
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 ?

interesting you count Ties
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
winsome johnny (not Win some johnny)
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 18th, 2019 at 10:10:49 AM permalink
I still can't get it, do you mind to explain it further ?
7craps
7craps
  • Threads: 18
  • Posts: 1977
Joined: Jan 23, 2010
May 18th, 2019 at 10:29:27 AM permalink
Quote: ssho88

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

I learned this stuff starting here
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.
winsome johnny (not Win some johnny)
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 18th, 2019 at 11:18:45 AM permalink
Quote: ThatDonGuy

I get 39.9124 as well - here's how:


Let EN 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:
EN = 1 + p EN+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





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
7craps
7craps
  • Threads: 18
  • Posts: 1977
Joined: Jan 23, 2010
May 18th, 2019 at 11:22:07 AM permalink
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 ?

say you wanted to know about this: P,P,B,B,B,T,B,P
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
winsome johnny (not Win some johnny)
ThatDonGuy
ThatDonGuy
  • Threads: 117
  • Posts: 6219
Joined: Jun 22, 2011
Thanked by
beachbumbabs
May 18th, 2019 at 11:49:41 AM permalink
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



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.

EN 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).
gordonm888
Administrator
gordonm888
  • Threads: 60
  • Posts: 5005
Joined: Feb 18, 2015
May 18th, 2019 at 12:08:02 PM permalink
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.

EN 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).



Good explanation. Interesting.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 18th, 2019 at 6:41:53 PM permalink
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.

EN 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).




NICE ! Tq
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 18th, 2019 at 9:54:57 PM permalink
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.

EN 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).




If a Tie does not break a streak in Baccarat for Banker(for example, P,B,B,T, . . . . P, B, T, B, B, T, B is considered 4 BANKER in a row), and the probability for BANKER, PLAYER and TIE is p, q and r respectively, then I think I should modify your formula :-

EN = 1 + p EN+1 + q E0 + r EN ?
Ace2
Ace2
  • Threads: 32
  • Posts: 2672
Joined: Oct 2, 2017
May 19th, 2019 at 1:59:43 AM permalink
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.
Last edited by: Ace2 on May 19, 2019
It’s all about making that GTA
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 19th, 2019 at 6:51:17 AM permalink
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.




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 ?
Last edited by: ssho88 on May 19, 2019
ThatDonGuy
ThatDonGuy
  • Threads: 117
  • Posts: 6219
Joined: Jun 22, 2011
May 19th, 2019 at 8:35:23 AM permalink
Quote: ssho88

How you got that simple formula ?


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).
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 19th, 2019 at 9:13:05 AM permalink
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).




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.
7craps
7craps
  • Threads: 18
  • Posts: 1977
Joined: Jan 23, 2010
May 19th, 2019 at 9:40:14 AM permalink
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.

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?"
winsome johnny (not Win some johnny)
OnceDear
OnceDear
  • Threads: 63
  • Posts: 7471
Joined: Jun 1, 2014
May 19th, 2019 at 9:42:24 AM permalink
Quote: 7craps


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

And there is no maximum.
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 19th, 2019 at 9:50:15 AM permalink
Quote: OnceDear

And there is no maximum.



Agreed. No maximum
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 19th, 2019 at 10:02:35 AM permalink
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?"



1) If P,T . . . B,B,B,B is 4 in a row, average rounds dealt is 39.91

2) If P,T . . . B,T,B,B,T,T,B is STILL 4 in a row, average rounds dealt is 31.74( my simulations results and may not correct !)

But Ace2 formula shown that the results for scenario 2) is 28.703, which not match my results, I suspect he not include TIE in his calculations. If he include TIE in his calculations, T. The total average rounds = 28.703 + T, so :-

T/(28.703 + T) = 0.095156, so T = 3.0184, average rounds = 28.703 + T = 31.72, match my results ?

0.095156 = PROBABILITY OF TIE

There is advantage play if you have the info of average rounds
7craps
7craps
  • Threads: 18
  • Posts: 1977
Joined: Jan 23, 2010
May 19th, 2019 at 10:06:20 AM permalink
Quote: ssho88

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

Many Asian Baccarat players have thought this over the years.

It IS laughable.

Your statement IS 100% false.
good luck
winsome johnny (not Win some johnny)
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 19th, 2019 at 10:16:34 AM permalink
Quote: 7craps

Many Asian Baccarat players have thought this over the years.

It IS laughable.

Your statement IS 100% false.
good luck



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

I am a maths guy, do you think that I am chasing the streak or bet according the PATTERN of the previous results ? You are 100% wrong !
ThatDonGuy
ThatDonGuy
  • Threads: 117
  • Posts: 6219
Joined: Jun 22, 2011
May 19th, 2019 at 10:35:30 AM permalink
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.


He never said how he got the formula - just that it existed.

And yes, you can rewrite the formula in your way if ties do not break the streak. You can also then subtract r EN from both sides to get:
EN (1 - r) = 1 + p EN+1 + q E0
then divide both sides by (1 - r) (or by (p + q), since p + q + r = 1 in your formula) to get:
EN = (1 + p EN+1 + q E0) / (1 - r)
unJon
unJon
  • Threads: 14
  • Posts: 4574
Joined: Jul 1, 2018
May 19th, 2019 at 11:18:25 AM permalink
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.

.

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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Ace2
Ace2
  • Threads: 32
  • Posts: 2672
Joined: Oct 2, 2017
May 19th, 2019 at 1:34:52 PM permalink
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.
It’s all about making that GTA
OnceDear
OnceDear
  • Threads: 63
  • Posts: 7471
Joined: Jun 1, 2014
May 19th, 2019 at 1:52:21 PM permalink
Quote: ssho88

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

Remarkably, no-one here cares. Go use your knowledge: Exploit your advantage: Get tremendously rich and then report back.
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
Ace2
Ace2
  • Threads: 32
  • Posts: 2672
Joined: Oct 2, 2017
May 19th, 2019 at 2:10:34 PM permalink
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.

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.

I realize my logic might be incorrect on the last part since if we are starting with a sample that contains 4 consecutive bankers, that’s not random and maybe I can’t just apply the 1 / .904544 average adjustment factor to add in ties. I haven’t fully thought it through yet but regardless I think it would still be a very accurate estimate
It’s all about making that GTA
unJon
unJon
  • Threads: 14
  • Posts: 4574
Joined: Jul 1, 2018
May 19th, 2019 at 2:14:11 PM permalink
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.



Yes exactly.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
lilredrooster
lilredrooster
  • Threads: 232
  • Posts: 6505
Joined: May 8, 2015
May 19th, 2019 at 2:15:01 PM permalink
Quote: ssho88

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




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
Please don't feed the trolls
Ace2
Ace2
  • Threads: 32
  • Posts: 2672
Joined: Oct 2, 2017
May 19th, 2019 at 2:39:14 PM permalink
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 ?

See my latest post. Adding ties into the count, my calculation now jives with your simulation. 28.703 / 0.904544 = 31.732.
It’s all about making that GTA
ThatDonGuy
ThatDonGuy
  • Threads: 117
  • Posts: 6219
Joined: Jun 22, 2011
May 19th, 2019 at 3:33:58 PM permalink
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.


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

Quote: ThatDonGuy


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

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 + ... + EN, where E is the expected number of trials to get one result of that type" came from.
Ace2
Ace2
  • Threads: 32
  • Posts: 2672
Joined: Oct 2, 2017
May 19th, 2019 at 4:57:10 PM permalink
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 + ... + EN, 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.

When I first saw “consecutive win” problems a few years ago, it was immediately obvious there was a pattern. For instance, in the case of p = 1/6 the answers are 6, 42 and 258 for 1, 2, and 3 consecutive wins. The first value is clearly 6, then it’s easy to see the pattern: the second value is the first value plus 6^2, then the third value is the second value plus 6^3 and so on. This works for any p value and any number of consecutive wins. Once the wins get over about 3, it gets much easier to use the geometric series formula than to sum up the power series.

I realize you know what a geometric series is. I was answering the OP and thought he/she was asking about that
Last edited by: Ace2 on May 19, 2019
It’s all about making that GTA
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 19th, 2019 at 5:06:30 PM permalink
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




Did I promise anything when I ask for help to verify my simulation results ? I may share it ONLY to those(ThatDonGuy, Chesterdog, 7craps) that helped me upon request but definitely not in the public forum. Fair ?

The funny thing here is main contributors here did not ask anything from me but you are trying . . . .
Last edited by: ssho88 on May 19, 2019
lilredrooster
lilredrooster
  • Threads: 232
  • Posts: 6505
Joined: May 8, 2015
May 20th, 2019 at 1:00:14 AM permalink
Quote: ssho88

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



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
Please don't feed the trolls
ssho88
ssho88
  • Threads: 53
  • Posts: 657
Joined: Oct 16, 2011
May 20th, 2019 at 1:11:49 AM permalink
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




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 !
lilredrooster
lilredrooster
  • Threads: 232
  • Posts: 6505
Joined: May 8, 2015
May 20th, 2019 at 1:15:44 AM permalink
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 !



okay, my bad - I missed that - sorry
Please don't feed the trolls
ThatDonGuy
ThatDonGuy
  • Threads: 117
  • Posts: 6219
Joined: Jun 22, 2011
May 20th, 2019 at 4:20:35 PM permalink
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 + ... + EN, 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.


Since I have nothing better to do (and I don't), here's a proof of the Ace-onacci Series:


Back to the "state" method - let EK be the expected number of trials needed to get to N consecutive results of a particular type given that you are currently at K consecutive, where p = the probability of the event happening and q = 1 - p.
EK = 1 + q E0 + p EK+1

EN = 0
EN-1 = 1 + q E0 + p EN = 1 + q E0
Assume EN-K = (1 + q E0) (1 + p + p2 + ... + pK-1);
EN-(K+1) = 1 + q E0 + p EN-K
= 1 + q E0 + p (1 + q E0) (1 + p + p2 + ... + pK-1)
= 1 + q E0 + (1 + q E0) (p + p2 + ... + pK-1 + pK)
= (1 + q E0) (1 + p + p2 + ... + pK-1 + p(K+1)-1)
Since it is true for K = 1 and it is true for K + 1 if it is true for K, it is true for all positive integers K by induction

Let K = N:
E0 = EN-N = (1 + q E0) (1 + p + p2 + ... + pN-1)
= (1 + q E0) (1 - pN) / (1 - p)
= (1 - pN) / (1 - p) + q E0) (1 - pN) / q (since q = 1 - p)
= (1 - pN) / (1 - p) + E0) (1 - pN)
E0 (1 - (1 - pN)) = (1 - pN) / (1 - p)
E0 = (1 - pN) / (pN (1 - p))
= (1 - pN) / (pN - pN+1)

Now, a quick proof that the expected number of trials for an event with probability p is 1 / p:
Let E be the expected number
E = p + 2 q p + 3 q2 p + 4 q3 p + ...
= p (1 + 2 q + 3 q p + 4 q2 p + ...)
= p (1 + q + q2 + q3 + ...)2
= p / (1 - q)2 (since -1 < q < 1)
= p / p2 = 1 / p

Finally:
E0 = (1 - pN) / (pN - pN+1)
= 1 / pN x (1 - pN) / (1 - p)
= (1 + p + p2 + p3 + ... +pN-1) / pN
= 1 / pN + p / pN + p2 / pN + ... + pN-1 / pN
= 1 / pN + 1 / pN-1 + 1 / pN-2 + ... + 1 / p
= (1 / p)N + (1 / p)N-1 + (1 / p)N-2 + ... + (1 / p)
= EN + EN-1 + EN-2 + ... + E
where E = 1/p is the expected number of trials to get a result once

Ace2
Ace2
  • Threads: 32
  • Posts: 2672
Joined: Oct 2, 2017
May 20th, 2019 at 4:36:40 PM permalink
Nice proof. I never did one, I only recognized the pattern.
It’s all about making that GTA
  • Jump to: