Whats the math for this?

shadroch

Well-Known Member
#1
Suppose you are playing $5 BJ,using perfect BS. With a $200 bankroll,you play until either you win 50%($100) or you lose everything.
Can someone show me what formula you would use to break down the odds of either event happening.
I'm assuming that the chances wouldn't change if you punched any BR into the equation,would it?
 

Sonny

Well-Known Member
#2
You can determine the probabilities using the binomial distribution. Here is an old post I made on another website that will give you a start. It shows you how to calculate the probability of being ahead after a certain number of hours (or hands) of play. You will see that the chances of "beating the odds" using BS alone will decrease dramatically as you play for longer periods of time (or many sessions):

So let's look at the formula for determining your probability of
being ahead. The formula is fairly straightforward. You are trying to
find out what the likelihood of being in the upper (positive) part of
the bell curve is. As I showed in my last post, your results will
vary anywhere between several standard deviations. We want to find
out the probability that we will not be on the negative side of that
fluctuation.

Our hourly EV was 1.47 and our trip EV was 51.45. Even if we have a
negative swing of –50 units we will still end up with 1.47 units for
the trip, so a little negative fluctuation is okay. We want to make
sure that we end up with more than 0 units at the end of the trip. We
start by determining what percentage of the standard deviation our
Trip Win represents:

SDs = Square Root(Number of hours) * Hourly EV / Hourly SD

So for our example above, we get:

SDs = SQRT(35) * 1.47 / 25.21 = 5.92 * 0.058 = 0.345

So our Trip Win represents 0.345 standard deviations. There are
several ways to use this number to find exactly what we want to know.

THE HARD WAY:

Remember how I mentioned that a swing of one standard deviation
occurs about 68% of the time? Well, if you are familiar with
statistics you will be able to calculate that a swing of 0.345
standard deviations will occur about 36.32% of the time. That means
that you will lose money about 36% of the time. Since the rest of
those times you will be winning, you have a 64% of being ahead after
those 35 hours.

THE EASY BUT PAIN-IN-THE-ARSE WAY:

Look up the number 0.35 (rounded up) in the Cumulative Normal
Probability Distribution chart from any statistical book (I use the
one in Blackjack Attack). You will find that it points to 0.6368.
That gives you about a 64% chance of being ahead.

THE SIMPLE WAY:

Why didn't I just mention this way first?! I don't know. I guess I'm
just a mathematical masochist. =)

Instead of just giving away the formulas and saying "plug in your
numbers here" I like to explain how the formulas work as well. If you
can understand the reasoning behind the formulas you will understand
them much better. You may even see the game in a whole new light!

Anyway, here's the formula. Plug your numbers in here:

Go into an Excel spreadsheet and type =normsdist(0.345). Better yet,
type the whole formula in there!

=normsdist(sqrt(Number Of Hours) * Hourly EV / Hourly SD)

That way you can adjust the values to see the results from any number
of hours using any EV and SD. The spreadsheet will magically spit out
the number 0.634953… (depending on how many decimal places it shows).
If you convert that cell to a percentage (Format -> Cells ->
Percentage) it will move the decimal point and add that cute little
percent sign to tell you that you have about a 64% chance of being
ahead.


You can expand on this theory to find the probability of being ahead by $X after a given amount of time and the propbability of losing $Y in that same time. The results above are for a player with an advantage, but the formula will work for a BS player as well.

-Sonny-
 
#6
shadroch said:
Suppose you are playing $5 BJ,using perfect BS. With a $200 bankroll,you play until either you win 50%($100) or you lose everything.
Can someone show me what formula you would use to break down the odds of either event happening.
I'm assuming that the chances wouldn't change if you punched any BR into the equation,would it?
Sounds to me like you are trying to play a sticky bonus, no? If so, playing $5 per hand isn't the way to do it. Putting down $100 per hand would give you the best chance of eventually being up $100.
 

shadroch

Well-Known Member
#7
Automatic Monkey said:
Sounds to me like you are trying to play a sticky bonus, no? If so, playing $5 per hand isn't the way to do it. Putting down $100 per hand would give you the best chance of eventually being up $100.
No,I don't play on-line at all.Just curious ,thats all.
 

mickpk

Active Member
#8
Sonny (and I'm being serious), fabulous explanations and I, for one, really appreciate them. Even when I already know the answer I enjoy reading well thought out answers like that. Though, I also enjoy the 'easy' way and often use the calculators referred to by QFIT.

Thanks to both. :)
 

dacium

Well-Known Member
#9
Of course you will never actually get near that number. Because of the double downs and splits you can't go on 0.5% loss per hand etc. You need to use win% win double % loose % push % and loose double %. The only time that you can use a average house edge % is over a long long amount of hands.

Because of the doubles and splits there are certain totals that you should aim for rather than just picking to win 50% of go bust.
 
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