Is there a fixed kelly bankroll fraction where you can draw off additional funds and have a; dare I say, 0% ror or very close to the original ror. I think it would be in the 1/4th to 1/8th fixed kelly range?
Thank you for your time
Let me try to answer my own question and see if I can spark some comments.
Again, I am curious about the fixed (no resizing) kelly fraction where I can draw off additional funds and still have a very low or close to the orginal ror.
If I understand the math correctly (Chin and Ingenoso); depends on the day, these are the chances I have of never losing 50% of my fixed betting bank before doubling bank or winning to infinity:
1/4th fixed Kelly 99.2%
1/5th fixed Kelly 99.8%
1/6th fixed Kelly 100.%
Now if I draw off funds whenever I have anything over these banks I believe my chances of losing 50% of bank for 1/4th and 1/5th fixed Kelly are higher because I keep starting over every time I draw down funds. Since the 1/6th Kelly number has a 0% chance of ever losing even half bank then perhaps this is the fixed kelly number to use if one were to take out any additional funds.
Thoughts?
blackjack dark knight,
You asked:
I am curious about the fixed (no resizing) kelly fraction where I can draw off additional funds and still have a very low or close to the orginal ror.
One answer is easy: if you bet 2xKelly (or more), you can withdraw all the funds you want without changing your RoR: it'll be 100% with or without withdrawals! ;-)
On a more serious note, you later claim:
If I understand the math correctly... these are the chances I have of never losing 50% of my fixed betting bank before doubling bank or winning to infinity:
1/4th fixed Kelly 99.2%
1/5th fixed Kelly 99.8%
1/6th fixed Kelly 100.%
These numbers are not correct.
First, let's consider the case of betting WITH CONTINUOUS RESIZING: we'll get to the fixed betting later.
For 1xKelly, the chance of your BR ever falling to X% of its original value is exactly X%. Thus, you have a 90% chance of your BR someday being only 90% of its present value, and a 10% that you'll drop all the way down to 10% of your original BR.
In particular, you have a 50% chance of losing half your BR if you use 1xKelly with resizing.
Now if you bet fractional Kelly (and I'm talking about proper fractions here!) with continuous resizing, your chance of dropping to 50% will be less than 50%. For example, if you bet 1/2xKelly, you can see that that's equivalent to betting full Kelly to each of two banks, each of which is half of your original BR. You'll play with the first bank as long as it stays above 50%, but if it falls to 50%, then you'll switch to bank 2. Thus, you have a 50% chance of losing half of bank 1, then a 50% chance of losing half of bank 2, both of which must happen for you to lose 50% of your total original BR. Thus, the chance of this happening is (0.5)*(0.5) = 0.25 = 25%. Conversely, the chance of this NEVER happening is just 100% - 25% = 75%.
For 1/4xKelly with continuous resizing, think of having 4 quarter-sized banks. By similar reasoning, you can see that your chance of losing 50% overall is now (0.5)*(0.5)*(0.5)*(0.5) = (0.5)^(4) = (0.5)^(1/[1/4]) = 0.0625 = 6.25%, so the chance of this NEVER happening is 100% - 6.25% = 93.75%.
To generalize, if you bet "f"xKelly (with continuous resizing), your chance of losing 50% of your original BR will be (0.5)^(1/f), and your chance of NEVER falling to 50% will be 1 - (0.5)^(1/f).
Thus, the chance of NEVER falling to 50% of your BR for each of the cases you mentioned above will actually be the following:
With Continuous Resizing:
1/4xKelly: 100%*[1 - (0.5)^(1/[1/4])] = 93.75%
1/5xKelly: 100%*[1 - (0.5)^(1/[1/5])] = 96.875%
1/6xKelly: 100%*[1 - (0.5)^(1/[1/6])] = 98.4375%
Now with FIXED-SIZE BETTING, your RoR is higher still, so your chance of having your bank fall to 50% with 1/4xKelly is EVEN HIGHER THAN 93.75%.
This means that your statement:
Since the 1/6th Kelly number has a 0% chance of ever losing even half bank...
is incorrect. If you want to have a 0% chance of ever losing half of your BR, you have to bet 1/(infinity)xKelly, which is REALLY hard to do on a limited BR ;-)
Even more generally, if you bet "f"xKelly (again with continuous resizing), your chance of your BR falling to a fraction "X" of your original BR will be (X)^(1/f), while your chance of this NEVER happening will be 1 - (X)^(1/f).
If you'd like to know more about the effect of withdrawals on your RoR, MathProf has done some excellent work in this area. Join Green Chip and search the messages and archives.
Hope this helps!
Dog Hand
I greatly appreciate the obvious time you took in responding to my question. I am digesting the material. I was aware of what you provided and how it differed from what I have read.
I agree 100% with your statements on continuous resizing so we agree 100%, 50% of the time ;)
I think your math education and experience far surpass mine. I am being stretched just a little bit! Add on top my not so strong writing skills and fun should be had by all! :(
However, I think my statements of never losing 50% of bank with fixed betting fractions(?) are basically correct:
f(x) = 1-a^(2x-1)
a = value of bank, .5 for half bank, .8 for 80% of bank
x = inverse(?) of kelly bank, 2 for half kelly
With fixed 1/2 kelly the chance of never losing 50% of bank using the above formula = .875
To check the above statement:
The RoR of a fixed kelly bank = Losing 50% of a fixed 1/2 kelly bank = .135
So your chance of losing 50% of a fixed 1/2 kelly bank is .135 while your chance of never losing half is .875
Another example:
With fixed 1/4 kelly
chance of losing 50% of fixed 1/4 kelly = RoR of fixed 1/2 kelly = .018
chance of never losing half using the above formula = .992
final example:
With fixed 1/6 kelly my statement of never losing half probably was lost in rounding. The chance of never losing half with the above formula is: .99951171875 but not 1.00 :(
I will admit the above does not equal the chance of losing a total fixed 1/3 kelly bank of .997523186023 but is close enough in the extreme?
To restate and see if I am correct?
chances of never losing 50% of fixed fractional kelly banks:
fixed 1/2 kelly 87.5%
fixed 1/3 kelly 96.9%
fixed 1/4 kelly 99.2%
fixed 1/5 kelly 99.8%
fixed 1/6 kelly 99.951171875%
Isn't math fun, sigh :(
"So your chance of losing 50% of a fixed 1/2 kelly bank is .135 while your chance of never losing half is .875.
If you intended for the two numbers, above, to add to 1.000, you missed by a little bit! :-)
"Another example: With fixed 1/4 kelly chance of losing 50% of fixed 1/4 kelly = RoR of fixed 1/2 kelly = .018 chance of never losing half using the above formula = .992.
See above!!
Don
I have to work hard to reach math challenged! :)
It's very good the major bj authors (you included) cleaned up the math; simplified it, for us unwashed masses.
However, I am curious about various issues and I do have a little bit of money involved.
Yes, I did try to sneak those results by. I think I will go with, not statistically signifigant or close enough?
Kinda like a stand 16 vs 8 deal ;)
So is the formula not quite accurate?
My application?
My final conclusions?
See all the above! :)
Let's try again! I am in so deep I can't even open my mouth to say how deep I am! :)
Can you guys give me these?:
fixed kelly ror = 13.53%
fixed 1/2 kelly ror = 1.83%
Now, stepping into it:
Approximate formula for determining chances of losing a certain percentage of a fixed (not resizing) Kelly bank:
f(x) = 1-a ^ (2x-1)
a = chances of losing bank .5 for half, .8 for losing down to 80%.
x = inverse of kelly fraction, 2 for fixed 1/2 kelly, 4 for fixed 1/4 kelly.
with formula, chance of never losing 50% of fixed 1/2 kelly = .875
chance of losing fixed kelly bank .1353
.875 + .1353 = 1.0103
So, the formula gives 87.5% chance when the more correct? answer should be approx. 86.47%? It's pretty close.
with formula, chance of never losing 50% of fixed 1/4 kelly = .9921875
chance of losing fixed kelly 1/2 bank .0183
.9921875 + .0183 = 1.0104875
with formula, chance of never losing 20% of fixed 1/5 kelly = .865782272
chance of losing fixed kelly bank .1353
.865782272 + .1353 = 1.001082272
Aren't these close enough? Validate the approximate formula? If not, why not?
Is there another easier, and I emphasize very easy formula?
f(x) = 1-a ^ (2x-1)
Approximate formula for never losing to a, as a varies from .5 (50%) to .8 (80%) of bank and x is equal to inverse fixed kelly fraction 1 (kelly) thru 6 (1/6 kelly).
Who told you this formula is accurate? Did you read it somewhere? Tell us where. Did you derive it with mathematical logic? Show the derivation.
f(x) = 1-a ^ (2x-1)
Approximate formula for never losing to a, as a varies from .5 (50%) to .8 (80%) of bank and x is equal to inverse fixed kelly fraction 1 (kelly) thru 6 (1/6 kelly).
So with a = .8 and x = 1, you have a 20% chance of losing 20% of your bankroll by betting full Kelly with no resizing. The actual figure from standard formulas is 67.032%. I am going by your interpretation of a and x given in previous examples. I admit, your language is fuzzy so it's hard to be sure, but if you change the interpretation, then those previous examples will be way off.
You don't just throw some letters together and think you've got something, if two or three examples come close.
ETF
Number 2 reason is to much time on my hands
Disclaimer - I have never been to a Star Trek convention and I have been kissed recently ;)
I did mention who wrote the formula in an earlier post.
Formula from
bjmath.com
risk formulas for proportional betting
section 4.1 to halve and halve not
William Chin, Marc Ingenoso
I believe section 1 and 2 consider continous resizing
I believe section 3 and 4 consider fixed betting
The possible exception is 3.5 that concludes poker players can be closer to continuous resizing in betting
I almost certainly did a bad job interpreting/explaining their work. I probably should have asked what the board thought of their paper. However, I thought I basically understood it and the posts ran on. LOL
Thank you for the source.
I believe section 1 and 2 consider continous resizing
I believe section 3 and 4 consider fixed betting
The problem with that is the formula from section 4:
1) f(x) = 1 - a^(2x - 1)
comes directly from the formula in section 2:
2) P(a) = a^(2/k - 1)
As they explain in the article, x = 1/k, and f(x) is the probability of never reaching a, while P(a) is the probability if ever reaching a. In other words, f(x) = 1 - P(a). With those two facts, you can very quickly derive 1) from 2) (or vice versa).
They both refer to proportional betting with resizing, just as Dog Hand tried to tell you.
In general, you should assume anything to do with Kelly betting involves resizing, unless the opposite is clearly specified.
In case anyone wants to see the article, it is here:
http://www.bjmath.com/bjmath/proport/riskpaper1.pdf
ETF
I follow your reasoning. I had thought of it myself. However, it does not seem to match with the charts in the following paper which led to my confusion.
http://www.bjmath.com/bjmath/proport/riskpaper1.pdf
chart 4.1 and 4.2
these charts consider continuous resizing?
A specific example.
I believe you would give the 1/2 kelly (continuous resizing) risk of losing 50% of bank as .5 * .5 = .25? The chance of it never happening is .75?
In chart 4.2 at infinity the chance of never losing half is .875 so the chance of losing half is .125
In the text section 2.2 they state the probability of hitting a fraction of ones bank at 1/2 kelly is a^3.
What am I missing?
