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Dantheman
Dantheman
Joined: Apr 21, 2015
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April 29th, 2015 at 3:34:18 AM permalink
As far as I can work out the probability of hitting exactly 6 banker or player in a row in baccarat is approx 1/65?
Is this correct?
What is the correct formula to work this out?
Also what is the probability of not seeing a streak of 6 either player/banker in 65 hands? 130 hands? 200 hands?
aceofspades
aceofspades
Joined: Apr 4, 2012
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April 29th, 2015 at 6:21:35 AM permalink
Quote: Dantheman

As far as I can work out the probability of hitting exactly 6 banker or player in a row in baccarat is approx 1/65?
Is this correct?
What is the correct formula to work this out?
Also what is the probability of not seeing a streak of 6 either player/banker in 65 hands? 130 hands? 200 hands?



Is this the baccarat playing high roller "Dan the Man" from the BBC Louis Theroux documentary about Vegas?
(Fast forward to the final 10mins.)

ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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April 29th, 2015 at 6:26:43 AM permalink
For even-money bets on double-zero roulette tables, the probability of a single win is 9/19, so the probability of N wins in a row is (9/19)N. For N = 6, this is about 1 / 88.525.
For baccarat, if you ignore ties, the probability of the player winning is about 49.3176%, and the banker, about 50.6824%.

The probability of going a particular number of spins without a run of 6 wins in a row is:
68.65%, for 65 spins
45.91%, for 130 spins
29.77%, for 200 spins

For the player in baccarat, if you ignore ties (i.e. determine the probability of going through N non-tied hands without the player winning 6 times in a row):
62.74% for 65 hands
38.12% for 130 hands
22.29% for 200 hands

For the banker:
58.35% for 65 hands
32.82% for 130 hands
17.66% for 200 hands
mustangsally
mustangsally
Joined: Mar 29, 2011
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April 29th, 2015 at 6:54:25 AM permalink
Quote: ThatDonGuy

For even-money bets on double-zero roulette tables, the probability of a single win is 9/19, so the probability of N wins in a row is (9/19)N. For N = 6, this is about 1 / 88.525.

why 00 roulette? (and not 0 roulette)

and you did not point out that
6 in a row is the very next 6 spins/bets

it is at least 3 times more likely in the very next 10 spins
0.035077761
RIGHT!?

streaks RULE in gambling
cuz
all wins and losses are in streaks (length >=1)
ok
maybe after lunch i looks at your others

remember the prob of a win in a Bac shoe changes as the cards are used up
maybe a simulation would be more accurate per shoe that is

here is an accurate streak calculator i found
http://maxgriffin.net/CalcStreaks.shtml

Sally
I Heart Vi Hart
mustangsally
mustangsally
Joined: Mar 29, 2011
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April 29th, 2015 at 7:54:40 AM permalink
Quote: Dantheman

Also what is the probability of not seeing a streak of 6 either player/banker in 65 hands? 130 hands? 200 hands?

Ahhh
you used the word

EITHER

so Dons answers are not correct, in my opinion, as i read the question differently as written
either

they are close
how close you wants?

this is like asking the chance of a run of 3 Heads or Tails in 9 flips (or = either)
not just Heads, not just Tails

like here
http://wizardofvegas.com/forum/off-topic/general/14542-please-help-a-probability-math-question/

using the streak calculator i linked tooooooooooooooooooooooooo
instead of 6 for the streak (K)
use 5

instead of 65 hands
use 64 (N)

streak length -1
AND
hands -1
for EITHER run of banker or player

i knew
i knew
it was easy
==============================
edit: i think that method only works for 50/50 bets
so i will see later
easy to simulate too
================================
nows if you just want the NO Banker prob of at least a 6 run in N hands
Don be rights

all in the way the question is interpreted
trick trick
Mully
I Heart Vi Hart
98Clubs
98Clubs
Joined: Jun 3, 2010
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April 29th, 2015 at 8:58:35 AM permalink
Baccarat: 8 decks (standard game)
Banker 6 in a Row win is 1 in 107.5 (75.3% chance this will happen in any 1 shoe)
Player 6 in a Row win is 1 in 126.6 (64% chance this will happen in any 1 shoe)

Banker 6 wins BEFORE a LOSS: 1 in 34.68 (will happen twice in a standard shoe)
Player 6 wins BEFORE a LOSS: 1 in 39.7 (will happen twice in a standard shoe

Formula: chance of a win ^ 6, Banker 1/(0.4586^6), and Player 1/(0.44625^6)
chance in any 1 shoe (standard rules, 8 decks) 81/chance of a win (81 is the average # of rounds in a standard shoe)

6 BEFORE a LOSS adds the chance of a tie (0.09516) to the chance of a win.
Some people need to reimagine their thinking.
mustangsally
mustangsally
Joined: Mar 29, 2011
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April 29th, 2015 at 9:14:16 AM permalink
Quote: 98Clubs

Baccarat: 8 decks (standard game)
Banker 6 in a Row win is 1 in 107.5 (75.3% chance this will happen in any 1 shoe)

HUH?
by these sim data from me Dad
http://wizardofvegas.com/forum/gambling/tables/5928-baccarat-streaks-question/

the chance of NO 6 run or longer in an 8 deck shoe (13 or 14 card ribbon) is about 56.44%
so 43.56% chance there is at least 1 such run
(a tie does not break the streak)

you say different
"75.3% chance this will happen in any 1 shoe"

"some ONE IS way off!"
says Bill Clinton

streak probabilities are challenging to calculate (computers calculate)
easy to do something different (like the number of runs per N trials)

Sally

added:
OK
i say now 98Clubs is way off (way way off) and my Dad is about 2% off (could not find his sim data either)

here what i gots

Don, finish your homework, please. (trash goes out tonight)
thank you

my sim data
8 deck,13 card ribbon
      group        middle     freq  freq/100
--------------------------------------------
-0.5 <= x < 0.50 0.00 544738 54.47% <<< no run of 6 Banker
0.50 <= x < 1.50 1.00 349382 34.94%
1.50 <= x < 2.50 2.00 93018 9.30%
2.50 <= x < 3.50 3.00 11887 1.19%
3.50 <= x < 4.50 4.00 937 0.09%
4.50 <= x < 5.50 5.00 38 0.00%


this is close to me calculated value in Excel
i C Sally says

automate R code next project
I Heart Vi Hart
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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April 29th, 2015 at 10:12:01 AM permalink
Quote: 98Clubs

Formula: chance of a win ^ 6, Banker 1/(0.4586^6), and Player 1/(0.44625^6)
chance in any 1 shoe (standard rules, 8 decks) 81/chance of a win (81 is the average # of rounds in a standard shoe)


Not quite. You are assuming that each group of 6 hands (hands 1-6, hands 2-7, hands 3-8, ..., hands 76-81) is independent of the others. If hands 1-5 are player wins and hand 6 is a bank win, then none of the groups 1-6, 2-7, 3-8, 4-9, 5-10, and 6-11 have player wins.
mustangsally
mustangsally
Joined: Mar 29, 2011
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April 29th, 2015 at 10:44:21 AM permalink
Quote: ThatDonGuy

Not quite. You are assuming <snip>

looks to me You are assuming
we have to each assume, i assume, what exactly the OP wants...
exactly...
exact

either
a Banker / Player run per N hands played

and the run = exactly 6 and no more, no no more than 6
BBB BBB P = counts as an OP success
but this does not count as an OP success
BBB BBB BB P

what really is OP after?
i knows streaks are in the OPs blood
he said in another thread

streaks rule!
Sally
I Heart Vi Hart
mustangsally
mustangsally
Joined: Mar 29, 2011
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April 29th, 2015 at 11:21:04 AM permalink
Quote: mustangsally

using the streak calculator i linked tooooooooooooooooooooooooo
instead of 6 for the streak (K)
use 5

instead of 65 hands
use 64 (N)

streak length -1
AND
hands -1
for EITHER run of banker or player

i knew
i knew
it was easy
==============================
edit: i think that method only works for 50/50 bets
so i will see later
easy to simulate too
================================

yep
i was 1st wrong
i knew it after reading

ok
the probability of NO Banker or Player run of 6 or longer in 65 Banker/Player hands
is abouts
[1] 0.3516415

that means the prob of 1 or the other (either) is abouts
[1] 0.6483585
> ##############################################
> # Biased/unbiased recursion of heads OR tails
> ##############################################
>
> N = 65 # number of flips
> m = 6 # length of run (must be > 1 and <= N/2)
> p = 0.506825 # P(heads)
>
> prob = rep(0,N)
> h = rep(0,N)
> t = rep(0,N)
>
> h[m] = p^m
> t[m] = (1-p)^m
> prob[m] = h[m] + t[m]
>
> for (n in (m+1):(2*m-1)) {
+ h[n] = (1-p)*p^m
+ t[n] = p*(1-p)^m
+ prob[n] = prob[n-1] + h[n] + t[n]
+ }
>
> for (n in (2*m):N) {
+ h[n] = ((1-p) - t[n-m] - prob[n-m-1]*(1-p))*p^m
+ t[n] = (p - h[n-m] - prob[n-m-1]*p)*(1-p)^m
+ prob[n] = prob[n-1] + h[n] + t[n]
+ }
>
> # m heads OR m tails
> prob[N]
[1] 0.6483585
> 1-prob[N]
[1] 0.3516415

that was using BruceZ code (R code)
from that same link (hey, He is no longer over at 2+2. maybe he is about to release his own website!)

but I had to check this in Excel using transition matrix (sure)
yep
right on!

oh yes
130

200

simulation time too?

Sally is good
"be good"

added for fun B4 dinner
for no run of 6 either B or P
in Baccarat (Tie does not break the streak) and counting all hands - Ties included
me matrix shows:
65 hands played
B or P run


130 hands played
B or P run


200 hands played
B or P run
I Heart Vi Hart

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