Need help with streak probabilities

Hi guys, just a question that has me confused when it comes to calcualting the probability of losing streaks. I'm using Dealer adv: 52.51%, Player: 47.49.

I've done up a spreadsheet using what the Wiz has suggested usingthe Dealer figure of 52.51%; however it seems to me that I should be using the Player ratio of 47.49% instead, as this is what you would be using when trying to estimate winnings over the long term. I.e. for every $100 wagered I would expect around $95 in return. (Actually as a side note, I would like to know how to more accurately calculate estimated winnings/losses if anyone has the time to educate me!)

However when using the same 47.49% in the Probability calculation, it dramatically increases the number of hands required to hit the losing streak, which seems incorrect given the losing bias towards the player.

ie. Number of Hands for a 10 losing streak, using: =POWER((52.51/100) * n) where n = number of losing hands:
Wizards calculation (I believe) using 52.51%: p= 0.001593750, or 627 hands
My new fandangled calcualtion using 47.49: p= 0.000583474, or 1714 hands

So I'm just wanting confirmation in a way that I actually understand as to why 52.51% is used instead. Logic says this is correct but I don't understand why. Thanks!

Hi ozzidazza

Welcome to the forum.

I don't play that game myself, but let's see....

If the probability of result of player, in any one hand, is 47.49% then on average, you will have a result of player once in (1/0.4749)=2.1057 hands.
similarly for banker. . .
If the probability of result of 'banker', in any one hand, is 52.51% then on average, you will have a result of banker once in (1/0.5251)=1.9044 hands.

'banker' is more likely than 'player' so you won't have to wait as long for a one card streak of 'banker' as you would for 'player'. That's intuitive.

Now lets look at a streak of 'banker' two cards long. . .
If the probability of result of 'banker', in any one hand, is 52.51% then the probability of two concecutive hands of 'banker' is 0.5251x0.5251 = 0.27573 and you can expect to see a result of 'banker ','banker' in 1/0.27573 = 3.6267 trials*

for a streak of 10 hands of banker, the probability is 0.5251x0.5251x0.5251x0.5251x0.5251x0.5251x0.5251x0.5251x0.5251x0.5251 or 0.5251^10 = 0.0015937 or about 1 in 627 trials*

If you are betting on 'player' then a losing streak would be a streak of 'banker', so use the 'banker' probability calculations.
If you are betting on 'banker' then a losing streak would be a streak of 'player', so use the 'player' probability calculations.

As to your expected losses, just take the amount of action you place. ie the sum of what you wager, and multiply it by the house edge of 0.0106 You cannot estimate your winnings as a positive number unless you are wagering a negative number. I.e You have to be the casino, not the player. No amount of messing about with streaks will change that.

* a trial begins when you say it begins. E.g 'I'm going to start looking for a streak of 10 NOW'. You can do that 41 times in a session where you plan to observe 50 games. Think about it.

Excellent, I do follow your logic! Just how I needed it explained, thank you very much.