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Duritz
11th January 2005, 08:32 AM
Hey systems punters - can one of you answer a question for me? It's been bugging me for quite some time. How do you work out the probability of a given run of outs of a system based on it's strike rate, and therefore what is a reasonable bank to have to support any potential run of outs for a given method?

For example, say you've got a method with a 20% strike rate. What is the % stake of your bank you should use (level stakes) whereby any reasonable run of outs can be endured? And what runs of outs can one expect with a 20% strike rate? Or 40%? Or 10%?

What's the calculation punters?!

Your help appreciated.
Duritz

La Mer
11th January 2005, 08:58 AM
For example, say you've got a method with a 20% strike rate. What is the % stake of your bank you should use (level stakes) whereby any reasonable run of outs can be endured? And what runs of outs can one expect with a 20% strike rate? Or 40%? Or 10%? Duritz

The answer is not as simple as stating with a 20% strike-rate there is an expectation of X being the longest run of outs. It really depends on the number 'events' that the 20% strike-rate is being tested on. For example, there is a 56% expectation of a longest run of outs in any 100 'events' with a 20% strike-rate, but over 1000 'events' there is a 52% expectation of a longest run of outs of 25. Over 10,000 there is a 55% expectation of a run of outs of 35.

Of course there would be greater expectations of runs of outs of lesser amounts.

darkydog2002
11th January 2005, 11:58 AM
Have a look at the thread PROPUNTERS.
I believe Neil has something there on losing runs.
Cheers.
darky.

BJ
11th January 2005, 12:32 PM
If you are losing 50%, your chance of losing 10 in a row is:
1/(.5 * .5 * .5 * .5 * .5 * .5 * .5 * .5 * .5 * .5) = 1 in 1024.

If you are winning 20% and losing 80%, your chances of losing 10 in a row is:
1/(.8 * .8 * .8 * .8 * .8 * .8 * .8 * .8 * .8 * .8) = 1 in 9.3

You cannot workout a longest losing run, because there is no such thing. Whilst there is a chance of losing the next race, the figure is infinate.

KennyVictor
11th January 2005, 12:49 PM
And how meaningful is this "Longest run of outs" anyway? If you calculate your maximum run of losses is likely to be 20 and you give yourself a safety net by betting a twenty-fifth of your bank - what happens when you have a run of 15 losses a win or two and another 15 losses? Your bank is still going to be caput.
When I computer model my systems against past data I record the high point of my bank and the worst following low point and work out the equivalent number of bets I lost from my peak to my trough. I have no idea how you would work this out from a percentage strike rate though.

KV

system
11th January 2005, 04:23 PM
mate with 20% strike rate i would suggest betting flat stake 1% of bank.

woof43
11th January 2005, 06:28 PM
To find your longest losing run of losses to a 99% confidence level do as follows;

=sum(1-(Losing % ^(to the power of) N (N=number of Losses) this will provide a % when you get to 99% this should equate to your longest losing run

example win%= .092 Losing % .908

=sum(1-(.908^5losses)=.382
=sum(1-(.908^17losses)=.806

Cheers

Duritz
12th January 2005, 11:56 AM
Thanks all, I really appreciate that.

Duritz.

Duritz
12th January 2005, 12:00 PM
So using that formula, I can expect a run of outs of about 20? If I therefore bet 5% of the bank at some point I will do the bank, even if it is winning! perhaps I should take the advice and do the 1% thing.

system
12th January 2005, 12:11 PM
tp start off with 1% is good at flat stake then down track you can have a look at staking plan,s or increasing bet % too say 2%

BJ
12th January 2005, 10:02 PM
If you are betting a percentage of the bank, it will never run dry....... Only if you bet 100% of the bank can you wipe it out otherwise your bets just get smaller and smaller until it is not worth betting, but it will never run dry.....

Bhagwan
12th January 2005, 10:46 PM
Longest run of expected Run of outs based on 10,000
This question has come up a number of times before.
So, here it is again.


10%-65
15%-42
20%-31
25%-24
30%-20
35%-16
40%-14
45%-12
50%-10
60%-9
70%-5
80%-5
90%-3

system
13th January 2005, 01:37 AM
guys i dont like this run of outs stuff cause 1 lousy winner breaks up the outs so 24 out 1 win 24 out i think this is bad news.yes ive had a couple of drinks but still 1 win breaks the run of outs but doesnt put your bank back in the black,or save you from going bust.

DR RON
13th January 2005, 05:32 PM
Instead of worrying about runs of outs, would anyone be able to produce figures for expected runs of winners???

Would it be just as likely that in 10000 races the longest run of outs would be followed by the longest run of winners???

Just trying to keep positive.

woof43
13th January 2005, 06:53 PM
Dr. Ron you have maybe touched on one of the most important aspect of wagering that is called Deviation.

Once you get to your second bet in any wagering situation your finances fall into one of three categories:
1. YOU HAVE A PROFIT.
2. YOU HAVE A LOSS.
3. YOU ARE EVEN.
The key question is: if I am in a losing or winning streak, does the probability of a reversal of this trend increase or decrease, as my number of bets increases?
Take a look below at this insightfull chart on deviation and coin flipping.
First, when you flip a coin 10 times, you'll get one of 1024
possible combinations of heads and tails. Assuming the coin is
unbiased, each combination is as likely as any other. Your
0110001011 is no more or less apt to occur than 1010101010 or
1111111111. Although yours doesn't seem peculiar in any way, it's
still just one out of 1024 equally probable combinations and
therefore has a chance of 1/1024 -- just like any of the others.

Second, when you flip a coin 10 times, the "expected" number of
heads and tails is five each. But "expected" is a statistical
term, and by no means tells what will actually happen. In fact,
based purely on the ways any given number of heads and tails can
be arranged, you can find the following probabilities:
===========================
Chance of various numbers
of heads in 10 flips
no of ways probability
heads
0 1 0.098%
1 10 0.977%
2 45 4.394%
3 120 11.719%
4 210 20.508%
5 252 24.609%
6 210 20.508%
7 120 11.719%
8 45 4.395%
9 10 0.977%
10 1 0.098%
===========================

The list shows that, in 10 flips, five heads can occur most often
-- 252 out of 1024 ways. This is why five is the "expected"
value. But four or six heads aren't all that unlikely, with 210
ways each. And so on, along the line. There's nothing mystical
behind these figures. Just a matter of counting arrangements.
Zero heads can occur one way: 0000000000. One can occur 10 ways:
1000000000, 0100000000, 0010000000, 0001000000, 0000100000,
0000010000, 0000001000, 0000000100, 0000000010, and 0000000001.
You can try it yourself with other numbers of heads, but the
enumerations get messy so you might prefer to take me at my word.

Of course, what the vast majority of people will fail to see is the whole truth about the fact, that the most probable outcome of ten flips of a coin is 5 heads and 5 tails.
An important aspect of DEVIATION IS THAT WHILE YOU ARE EXPERIENCING A POSITIVE DEVIATION, you can and will have LOSING STREAKS. Of course the same principle applies to what will happen during a NEGATIVE DEVIATION. THE MOST IMPORTANT ISSUE CONFRONTING THE US IS HOW DO YOU COMPENSATE FOR THESE DEVIATIONS, AND IN THE PROCESS MAKE CONSISTENT, PROFITS.
Cheers

Duritz
13th January 2005, 07:26 PM
Rosenkrantz and Guildenstern would've wiped out their bank, that's for sure.........

Anyway, the point of my original thingo was trying to work out what was a reasonable percentage of a bank to bet on a system with a strike rate of 20%. System suggested starting out at 1% of bank, sounds good. Even with a run of outs of 31, as bhagwan suggested was the maximum, the bank wouldn't be gone, and even if all of the nightmares that keep system up at night drinking ;P came true - ie 31 losers, 1 winner, 31 losers, 1 winner, 31 losers (and if I hadn't ABANDONED it by then!) then the bank would STILL not be gone. Inglorious, sure, but enduring!

Cheers.
Duritz.

system
14th January 2005, 03:17 AM
well said duritz and that coin thing well i might have to read that few times to get the whole meaning(even without a drink).

chiller
14th January 2005, 07:55 AM
Hello Duritz,
Maybe you could work out the figure by using the Kelly Method...
Percentage of winner minus the percentage of losers/ divided by average price. So 20 - 80( at say 4.6) = 20-17.39 ( say 3 or 3.5)
For some reason this doesn't look right, so I am sure that I will hear about it.

KennyVictor
14th January 2005, 02:42 PM
THE MOST IMPORTANT ISSUE CONFRONTING THE US IS HOW DO YOU COMPENSATE FOR THESE DEVIATIONS, AND IN THE PROCESS MAKE CONSISTENT, PROFITS.
Cheers

I would imagine that at any point in time there is a statistically safe sized bet (being a proportion of your bank) which would depend only on your strike rate and your average return which would give you a (say) 99.99% chance of never wiping out your bank. This assumes that on level stakes you can come out in front because otherwise I suggest that whatever your betting strategy the bank is going to dissappear eventually.

The answer to the second of the two questions above is simply to select suffiecient winners that you would make a profit on even stakes and the bank will eventually grow.
The answer to the first half may be able to be supplied by a statastician of the calibre of woof43. Given a 20% strike rate and a small positive POT can you tell us what proportion of our bank can be bet to give us a 99.99% chance of never wiping out?
Assuming this figure can be calculated my strategy would be to keep that same size bet during the losing streak because statistics has told us we'll bounce back and increase it in proportion to the growth of the bank as we make a profit (again because we are statistically safe). The other alternative would be to decrease the bet as we encounter a losing streak which I suppose would be safer but would cause you to take longer to bounce back.
Is there such a safe bet? or have I been talking complete bollocks?

KV

woof43
14th January 2005, 04:54 PM
Hi,
What is the best method of betting or determining your bankroll size? Is it the Kelly or half Kelly? Or something else?
THEORETICALLY: the kelly
AS PRESENTED IN LEADING HANDICAPPING and WAGERING BOOKS: the half kelly.
REALITY: SOMETHING ELSE!!!
The starting point to consistent profits is to identify 20/20 Factors that is 20%POT and a 20% S/R,
Using only 1 or 2 factors on their own or combined, once these 20/20 are identified the kelly criterion provides the best known method to increase your bankroll, but it also carries substantial risk for eroding your bankroll.
So you need to find a balance between Risk and Reward.

The KELLY CRITERION provides for MAXIMUM BANKROLL GROWTH over any extended Profit Determination Periods, it does not provide for MINIMUM RISK. Let me explain:
1. PURE KELLY: REQUIRES YOU TO DETERMINE YOUR POSITIVE EDGE, AND BET THIS PERCENTAGE OF YOUR BANKROLL ON EACH WAGER. The primary problem with this strategy is the uneven payoffs.
2. HYPER-KELLY: REQUIRES YOU TO DETERMINE YOUR POSITIVE EDGE ON EACH BET, DIVIDE THE POSITIVE EDGE BY YOUR ANTICIPATED ODDS, AND WAGER THIS PERCENTAGE OF YOUR BANKROLL ON EACH WAGER. This strategy requires you to make your computations before each race, based upon your estimate of each entrant's probability of winning and the tote board odds. The primary problem with this and the following strategy is the accuracy of your Probability estimates, and their reliance on Tote Board Odds.
3. HALF-KELLY: THE SAME AS HYPER-KELLY, BUT YOU WAGER NO MORE THAN 1/2 THE OPTIMUM KELLY PERCENTAGE.
4. INCREMENTAL FLAT BETTING:IT IS BASED UPON YOUR HISTORIC WIN PERCENTAGE AND AVERAGE ODDS. COMPUTE YOUR POSITIVE EDGE PRIOR TO EACH VISIT TO THE TRACK, AND DIVIDE THE POSITIVE EDGE BY YOUR AVERAGE ODDS. BET NO MORE THAN 1/2 OF THIS PERCENTAGE OF YOUR BANKROLL ON EACH WAGER FOR THE DAY. REPEAT THIS PROCESS EVERY TIME YOU GO TO THE TRACK. This is a method that is more adaptable to Greyhound Betting.
Optimum Method
To find that balance you need to know, when during your betting series the extent of your positive expectation or the negative expectation (win, loss series and the bet number), within this betting series, (look at the probabilities of flipping a coin as posted before) with this knowledge you will be able to adjust the size of the Kelly Opt. and apply this to your next bet. Always remembering that you need to have a higher avg bet size on winning bets then on losing bets.

Duritz
15th January 2005, 12:10 AM
Hear that pop? I think my brain just imploded...

Bhagwan
15th January 2005, 02:26 AM
I knew that.

woof43
15th January 2005, 12:04 PM
Hi,
I am posting a simple spreadsheet that makes it possible to test a number of parameters, make particular note that the cells are all unprotected also it would be benificial to look at the formulas used in the constructing of the wager size.
If you enter your own Strike rates an avg. payoffs in the identified columns then run tests as per the instructions on the sheet. You will be able to observe the differences that occur when you change certain parameters, you will also note that it uses a Bankroll and Session Bankroll an it allows a weekly drawdown based on 100 bets per week.
Follow the instructions carefully on the sheet, take note of cells D6&D8 an record the variances that occur in the amount of winners per 100 series even though the S/R has not changed.

darkydog2002
15th January 2005, 04:02 PM
The late Arn Rogers of the original TURF MONTHLY used to say that betting 1 % of your bank on any given day was good business and he was no betting chump.
Cheers.
Darcy.

system
15th January 2005, 07:04 PM
geez wolf i dont know about the rest of the boys but my head is hurtin,i think im bleeding from me ears.