Consecutive Streak Probabilities

I am looking to get some help with some numbers. I play baccarat and enjoy playing a few different martingale variations to keep things interesting. I am currently set up where I gain 1 unit on streaks of 1-4 and on streaks of 5+ (5+ = 5 or more) I lose. Nothing too exciting but it keeps me from getting too bored while playing. I profit on all shoes with 0 or 1 streak of 5+, basically break even on 2 streaks of 5+ (win/lose a few depending how streaky or choppy the shoe is) and I lose on shoes that have 3 or more streaks of 5+.

What I am looking for specifically is how often a streak of 5+ is immediately followed by a streak of 5+. So, a player 5+ streak followed by a banker 5+ streak or vise vera. I am not sure if this can simply be calculated? My guess is that the most accurate numbers will have to come from a simulation (paging @mustangsally). I also wonder the best way to represent the data; happens 1 in x.xx 5+ streaks or 1 in x.xx shoes. Finally, will this also include multiple 5+ streaks (ie. 3,4,5 consecutive streaks) or would that be a separate statistic?

If anyone can simulate this and breakdown the distribution in an easy to read format I would be very grateful!

Since baccarat is pretty close to a coin flip, I’ll approximate this using a coin flip. The expected waiting time for any pre-specified string of 10 flips is 2^(10+1) - 2 = 2,046 flips. Since baccarat isn’t exactly 50/50, this will be a bit less frequent. I guesstimate 1 in 2,400 baccarat hands will meet the 5/5 b/p~ requirement, ignoring ties.

Quote: Ace2

Since baccarat is pretty close to a coin flip, I’ll approximate this using a coin flip. The expected waiting time for any pre-specified string of 10 flips is 2^(10+1) - 2 = 2,046 flips. Since baccarat isn’t exactly 50/50, this will be a bit less frequent. I guesstimate 1 in 2,400 baccarat hands will meet the 5/5 b/p~ requirement, ignoring ties.

Thank you for the approximation! If I round up to 80 hands per shoe, given your numbers, I could expect to see this 1 in 30 shoes. I was expecting it to be more frequent but confirmation bias can work for or against you without numbers to back it up.

Is there an issue with this method where the "10 flips" would start at the end of one shoe and carry on to the next? Maybe this was figured into your rounding up?

Thanks again, looking forward to how this compares to an accurate baccarat simulation!

The estimate I gave was for expected waiting time for a string that you pre-select: like 5 banker then 5 player. But 5 player 5 banker would not be counted as success.

Going back to coin flip example. 2046 flips is the expecation for a selected string. But if the string can start with heads or tails then the expectation will be 2046 - 2^9 = 1534 flips. Call it 1800 hands of baccarat since it’s not exactly 50/50. This assumes the string can happen anytime, which means it could span over 2 shoes.

I should be able to calculate an exact number later today, but I’m in Vegas right now shooting dice. Should just be a matter a setting up a few linear equations and solving for x.

Quote: falecf4

Thank you for the approximation! If I round up to 80 hands per shoe, given your numbers, I could expect to see this 1 in 30 shoes. I was expecting it to be more frequent ...

I used a Markov chain for this.
for a fair coin flip (because it was mentioned)
I get 0.125217791 (about 1 in 8) for 82 flips

for Baccarat
it can be Banker 1st, then Player PLUS Player 1st then Banker
(This is for 8 deck, regular Baccarat - average 82 hands per shoe)
banker 1st: 0.060511807
player 1st: 0.060648413
total: 0.12116022 (about 1 in 8.25)

close to one simulation I did,
but that code in Excel that I used, would have to be changed to answer your question about multiple times during a shoe.
I have other projects that need to be completed first.
It is interesting, from the coding part, so it might be done sooner than 'later or never'.

Does that 1 in 8 feel more in line with what you have seen during your play?

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Quote: Ace2

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I am going to change my answers as I saw I left out a few states in my transition matrix that needed to be considered. (exactly what happens when in a hurry)

the fair coin: 0.128247034 (about 1 in 7.8)
82 flips
this is for a run of at least 5 followed immediately by the opposite run of at least 5.

For Baccarat (taking in account Banker, Player and Tie, 8 deck)
for 82 rounds
0.115287937 (about 1 in 8.67)
B5.P5 = 0.05783694
P5.B5 = 0.057450996

final answer before attempting a simulation

Quote: 7craps

I am going to change my answers as I saw I left out a few states in my transition matrix that needed to be considered. (exactly what happens when in a hurry)

the fair coin: 0.128247034 (about 1 in 7.8)
82 flips
this is for a run of at least 5 followed immediately by the opposite run of at least 5.

For Baccarat (taking in account Banker, Player and Tie, 8 deck)
for 82 rounds
0.115287937 (about 1 in 8.67)
B5.P5 = 0.05783694
P5.B5 = 0.057450996

final answer before attempting a simulation

This looks more in line with what I would expect. I was hoping it would be spaced out more but I suspect variance here will determine whether I'm having a good day or a bad one.

If you do get into programming this can you do a 5+ vs a 6+? It'd be nice to see the disparity.