mustangsally
Joined: Mar 29, 2011
• Posts: 2463
January 5th, 2018 at 9:39:47 AM permalink
in PEPEZAGA'S BLOG
"What are the odds of 3 straight ties showing up in a regular big table baccarat?"

my understanding is an 8 deck shoe is used at big table Bacc.
I guess 6 deck could also be used but I do not know really. My experience is at mini-Bacc tables.
*****
I found trying to calculate the probability of a streak of a certain length
from a Bacc shoe (because to many the shuffled 'shoe' is a closed universe)
works well with both the Banker and Player bets (card removal without replacement)
but falls short with the Tie bet, so I used 10 million shoe simulation(s) to get
my results.
*****
at least 2 in a row
so these qualify
BTTP (this is 1 run of 2 in a row)
BTTTB
PTTTTB
48.6% chance that any 8 deck shoe (13 card ribbon) has at least 1 run>=2
Hey, that is almost half!
"see a Tie, bet the Tie!" is heard at many Bacc tables
run>=2 distribution
`      group        middle      freq  freq/100----------------------------------------------0.5 <= x < 0.50     0.00   5139077    51.39%0.50 <= x < 1.50     1.00   3493834    34.94%1.50 <= x < 2.50     2.00   1104761    11.05%2.50 <= x < 3.50     3.00    227116     2.27%3.50 <= x < 4.50     4.00     32885     0.33%4.50 <= x < 5.50     5.00      2324     0.02%5.50 <= x < 6.50     6.00         3     0.00%`

at least 3 in a row
so these qualify
BTTP nope! (fake out)
BTTTB yes
PTTTTB yes
5.9% chance that any 8 deck shoe (13 card ribbon) has at least 1 run>=3
that about 1 in 17 shoes (I love shoes)
run>=3 distribution
`      group        middle      freq  freq/100----------------------------------------------0.5 <= x < 0.50     0.00   9406751    94.07%0.50 <= x < 1.50     1.00    575032     5.75%1.50 <= x < 2.50     2.00     17445     0.17%2.50 <= x < 3.50     3.00       772     0.01%`

at least 4 in a row
so these qualify
BTTP no
BTTTB no
PTTTTB yes
PTTTTTP yes (may never see this one)
about a 0.55% chance that any 8 deck shoe (13 card ribbon) has at least 1 run>=4
(another reason to not like the number 4)
that about 1 in 180 shoes (too many shoes)
run>=4 distribution
`  group        middle      freq  freq/100----------------------------------------------0.5 <= x < 0.50     0.00   9943112    99.43%0.50 <= x < 1.50     1.00     56502     0.57%1.50 <= x < 2.50     2.00       386     0.00%`

hope this helps out
PEPEZAGA this was 4U too!

Sally
I Heart Vi Hart
BackgammonKid69
Joined: Jan 5, 2018
• Posts: 4
January 5th, 2018 at 12:15:12 PM permalink
Big table or small table, each still uses 416 cards, so there is no difference in odds. In 45 years of playing casino Baccarat, I have seen 5 ties once, and also a shoe with 23 ties in it. The tie bettors (of which I am not one) cleaned up and probably rode that memory all the way to the Poor House. If you must bet the tie, put a red chip on it for the dealers after a nice session. And while I'm here, let me also disparage the 'Dragon' bet. The Dragon bet will hit 30 to 1 ONLY when there is no money on it. True story.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
January 5th, 2018 at 3:20:58 PM permalink
Quote: BackgammonKid69

Big table or small table, each still uses 416 cards, so there is no difference in odds. In 45 years of playing casino Baccarat, I have seen 5 ties once, and also a shoe with 23 ties in it.

I left out 5 in a row
shoe probability as it is small,
but we can get a close calculation anyways as to the average number of hands and shoes it would take to see one such run=5.
(I think actual wait time is longer due to drawing cards without replacement
but this is a start point.)
*****
p(Tie)=0.095156
1 / (p^5) = 128,179.658970 attempts = A
each attempt is 1.105154257 hands long
(1+p+p^2+p^3+p^4) = B

A*B=141,658.30 hands
@80 per/shoe = 1,771 shoes
my Mom has about that many shoes!

of course
once Tie has hit 4 times in a row
the probability the next hand is a Tie is still 0.095156
*****
When I played with Mom mini-Bacc tables, once a Tie hit, the electricity level went out of this world.
It was fun to watch.
I remember when it hit 2 times in a row and everyone bet for it and won on the 2nd Tie!
6 face cards I do remember that!

ok...
sim data
10 million shoes
# of Ties per shoe data. 8 deck, 13 card ribbon
Tie per shoecountcount/10,000,0001 in
030950.00030953,231.0
1208860.0020886478.8
21048410.010484195.4
32975510.029755133.6
46110510.061105116.4
59591810.095918110.4
613041890.13041897.7
715066650.15066656.6
814846900.14846906.7
912483940.12483948.0
109399660.093996610.6
116608810.066088115.1
124036000.040360024.8
132391040.023910441.8
141195500.011955083.6
15576400.0057640173.5
16236180.0023618423.4
1793040.00093041,074.8
1827020.00027023,701.0
1919370.00019375,162.6
207700.000077012,987.0
2110.000000110,000,000.0
22+3840.000038426,041.7
`grouped dataitems:                  10000000   minimum value:                 0.00first quartile:                6.00median:                        8.00third quartile:                9.00maximum value:                22.00mean value:                    7.77midrange:                     11.00range:                        22.00interquartile range:           3.00mean abs deviation:            2.12sample variance (n):           7.05sample variance (n-1):         7.05sample std dev (n):            2.66sample std dev (n-1):          2.66`

Sally
I Heart Vi Hart
CyrusV
Joined: Aug 8, 2015
• Posts: 102
January 7th, 2018 at 8:17:19 AM permalink
In 15 years of playing this game, I witnessed 6 ties in a row once, it was a 6 deck game and the casino never paid out a cent as nobody was playing it.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
January 8th, 2018 at 8:21:39 AM permalink
Quote: CyrusV

In 15 years of playing this game, I witnessed 6 ties in a row once, it was a 6 deck game and the casino never paid out a cent as nobody was playing it.

I guess I have played with the type of players that bet on the Tie.
6 in a row in 6 deck is something!
*****
6 in a row in a 6 deck shoe
we can get a close calculation to the average number of hands and shoes it would take to see one such run=6.
p(Tie)=0.095069
1 / (p^6) = 1,354,460.8 attempts = A
each attempt is 1.105055813 hands long
(1+p+p^2+p^3+p^4+p^5) = B

A*B=1,496,754.8 hands
@60 per/shoe = 24,946 shoes
maybe happens every other day somewhere?
Sally

*****
and might as well place some 6 deck data here too.
(I think it can be found in other corners of the net)

again, I used 10 million shoe simulation(s) to get
my results.
*****
at least 2 in a row
38.5% chance that any 6 deck shoe (13 card ribbon) has at least 1 run>=2
run>=2 distribution
`            group        middle      freq  freq/100----------------------------------------------0.5 <= x < 0.50     0.00   6145127    61.45%0.50 <= x < 1.50     1.00   3060883    30.61%1.50 <= x < 2.50     2.00    687594     6.88%2.50 <= x < 3.50     3.00     96477     0.96%3.50 <= x < 4.50     4.00      9259     0.09%4.50 <= x < 5.50     5.00       660     0.01%`

at least 3 in a row

about a 4.5% chance that any 6 deck shoe (13 card ribbon) has at least 1 run>=3
that about 1 in 22 shoes
run>=3 distribution
`            group        middle      freq  freq/100----------------------------------------------0.5 <= x < 0.50     0.00   9553577    95.54%0.50 <= x < 1.50     1.00    436291     4.36%1.50 <= x < 2.50     2.00     10132     0.10%`

at least 4 in a row
about a 0.45% chance that any 6 deck shoe (13 card ribbon) has at least 1 run>=4
run>=4 distribution
`  group        middle      freq  freq/100----------------------------------------------0.5 <= x < 0.50     0.00   9955424    99.55%0.50 <= x < 1.50     1.00     44576     0.45%`
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ThatDonGuy
Joined: Jun 22, 2011