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russellhq

Risk of Ruin Discussion

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I'd like to start a disscussion on the maths behind risk of ruin as from what I've read previously, it leaves me a little uncomfortable.

 

I'll start by discussing the way it's usually presented. Usually when calculating RoR, you normally start with the the answer and work backwards, so lets do that.

 

I'll use the coin flipping analogy as the game. So say we play a game where I flip a coin and if it lands on heads then you win and if it's tails you lose. Normally, before the start, you would decided on what level of risk you will accept before going broke, say it's 1 in 10,000. This works out at roughly 13 tails in a row or 0.5^13.

 

Nice and easy, we can play the game for as long as we want but if there is a run of 13 tails, then you're out.

 

But, what if, at the start of the game you have a run of 12 tails, then a head, then 2 tails. You are still out! This thought led me to the following:

 

This time, instead of ruin meaning you can no longer play, lets set ruin to be true only at then end of a set number of games. If during the course of play you pass the ruin line, you are still allowed to play to try and recover but you must stop after the set number of games.

 

Lets start out by saying we will play 27 times, what will be the chance of you being ruined?

 

To work this out, lets start with how many different permutations of the game there is (how many difference ways we can flip the coin 27 times).

 

2^27 is the answer

 

Now lets work out how combinations there are that can ruin you by the end of play.

 

If you get 0 heads during the 27 flips, then all agreed, you would be well and truly ruined. There is only 1 combination of this.

 

If you get 1 head during the 27 flips, you would still be ruined. There are 27 combinations where you can get only 1 head from 27 flips.

 

If you get 2 heads during the 27 flips, again you'll be ruined. To calculate the combinations, we use factorials: 27!/(27-2)! = Number of ways you can be ruined.

 

We keep doing this until we get to 7 heads, after 7 heads, you would always be able to recover by the end.

 

So when we add up all the combinations from 0 heads, to 7 heads:

 

0 heads = 1

1 heads = 27

2 heads = 27!/(27-2)!

3 heads = 27!/(27-3)!

4 heads = 27!/(27-4)!

5 heads = 27!/(27-5)!

6 heads = 27!/(27-6)!

7 heads = 27!/(27-7)!

 

And divide by the total number of permutations (2^27) we end up with our answer. In this case it's 1 in 100!

 

This is a lot higher than our initial assesment of 1 in 10,000!

 

This is just the basics and i've not considered Risk/Reward ratios etc (that can be added later). I just wanted to start with the basics.

 

Thoughts?

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Be interested where you go with this, a neglected subject that really should be a mandatory part of trading 101.

 

The brain is a bit sluggish at the moment but I am not sure it makes sense (to me) to divorce account size and bet size from the equation. The main parameters to RoR are expectation of outcome (50:50) in our case, bankroll and bet size. Just having difficulty getting my head around how you have abstracted things. Again you can make bet and bank constant arbitrary values but I need to think if that actually allows them to be dropped from the equation. To lazy for thinking right now :)

 

Having said that looking at 'streakiness' is a pretty valuable exercise, many trading approaches do seem to yield 'streaks' (probably due to cycling market conditions).

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Hi Blowfish.

 

Account size and bet size would be the outcome of this exercise.

 

For example, in the first method we used an arbitrary RoR of 1 in 10,000 then calculated that 13 tails in a row would be about a 1 in 10,000 chance. Therefore, we assumed we would be ruined only after 13 straight tails. So to bankroll ourselves to cover the 9,999 other events where we don't hit 13 tails, we would need a bank of 13 times our bet size. Normally though we start with a bank size, and would divide it by 13 to work out our bet size.

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  russellhq said:
Hi Blowfish.

 

Account size and bet size would be the outcome of this exercise.

 

For example, in the first method we used an arbitrary RoR of 1 in 10,000 then calculated that 13 tails in a row would be about a 1 in 10,000 chance. Therefore, we assumed we would be ruined only after 13 straight tails. So to bankroll ourselves to cover the 9,999 other events where we don't hit 13 tails, we would need a bank of 13 times our bet size. Normally though we start with a bank size, and would divide it by 13 to work out our bet size.

 

Russel,

 

Trading is negative sum when you add in all costs. As a trader you have to to do significantly better than 1 to 1 in order to overcome the costs and the obstacles that you are pointing out to make it worth one's while. It is very difficult and leads many hedge fund managers to lying, cheating, and stealing to post quarterly returns that are better than returns that would be earned by the randomly chosen hedge fund.

 

MM

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I am starting to get where you are going with your math Russell, but it only hold's true if you continue to risk the same amount after 6, 9, or even 12 losses. If you were considering Ruin to be dead broke, after 12 losses if you made another trade with the same risk you'd be risking 100% of your capital!

 

Depending on your risk model whether it be percentage based, fixed, or something else one would think you need to adjust it to be inline with your capital.

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Using analogies with coin toss for these kind of computation in trading is not much useful, and probably misleading.

 

There is a fundamental difference. While a coin can do whatever it likes, mkts don't.

 

Volatility is obviously bounded, as at certain prices you dont have no more buyers or sellers. Further, mkts (especially futures) experience violent reversals. In fact GBM with reversion are pretty realistic models for these (if we let volatility change randomly).

 

You can have large dd, but can't really have "unbounded losses", unless you use stops, in which case, obviously, you do.

 

[That is another reason why all fx brokers and scammers recommend to people to use stops .]

 

 

Tom

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Hi Russel

A most interesting post.

You will flip a coin 13 times .

The odds are 1 in 10000 you will go bust.

You go bust.!!!!!!!!!!!!!!!!!!. Where did you go wrong???.

You went wrong on on the second flip

The double up.

You cant trade like that.

To stay in the Park you must minimise your losses. Therefore the moment you are wrong you are out.

START AGAIN

We wont discuss the next problem

13 start agains!!!!

Kind regards

bobc

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Hi tommaso, i can't really agree with what you are saying unless you can expand further. Markets, like any other risk based activity, must follow the laws of probability. Flipping a coin and it landing heads or tails is analogous to entering a trade and it going up or down.

 

Discussing probabilities and risk is essential to maintaining a healthy account and that is what I am attempting to do here.

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  bobcollett said:
Hi Russel

A most interesting post.

You will flip a coin 13 times .

The odds are 1 in 10000 you will go bust.

You go bust.!!!!!!!!!!!!!!!!!!. Where did you go wrong???.

You went wrong on on the second flip

The double up.

You cant trade like that.

To stay in the Park you must minimise your losses. Therefore the moment you are wrong you are out.

START AGAIN

We wont discuss the next problem

13 start agains!!!!

Kind regards

bobc

 

Hi bob, I've not mentioned wager size/risk value yet so I'm not sure where you are coming from. What I am discussing here is probabilities. Once the probability is defined, we can move onto risk amounts/stops to keep us in the game.

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  russellhq said:
Hi bob, I've not mentioned wager size/risk value yet so I'm not sure where you are coming from. What I am discussing here is probabilities. Once the probability is defined, we can move onto risk amounts/stops to keep us in the game.

 

Sorry Russel. I have jumped the Gun. Please ignore my post. Perhaps the Moderator should remove it

Kind regards

bobc

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  tommaso said:

[That is another reason why all fx brokers and scammers recommend to people to use stops .]

Tom

 

Are you saying that trading without stops is part of your strategy? You never use a stop?

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  jackb said:
Are you saying that trading without stops is part of your strategy? You never use a stop?

 

Right. Stops are designed to keep the herd around zero. And they work perfectly to that purpose.

 

Tom

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  russellhq said:
Hi tommaso, i can't really agree with what you are saying unless you can expand further. Markets, like any other risk based activity, must follow the laws of probability. Flipping a coin and it landing heads or tails is analogous to entering a trade and it going up or down.

 

Discussing probabilities and risk is essential to maintaining a healthy account and that is what I am attempting to do here.

 

 

Mkts don't have to follow anything.

 

Probability models are mere attempts made by humans to understand part of the complexity of nature. And it's certainly not a coin toss model going to help.

 

If you introduced a GBM with reversion and jump diffusion and not constant variance

maybe the model could come close to a barely acceptable start to investigate some

aspects of money management and strategy performances with proper probabilistic tools.

 

Discussing the mkts with a coin toss won't lead anywhere, and is an especially biased

and naive approach, as mkt have locally bounded volatility, which is also the reason

why the broker can take sistematically the risk to lend you money.

 

Tom

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  tommaso said:
Using analogies with coin toss for these kind of computation in trading is not much useful, and probably misleading.

 

There is a fundamental difference. While a coin can do whatever it likes, mkts don't.

 

Volatility is obviously bounded, as at certain prices you dont have no more buyers or sellers. Further, mkts (especially futures) experience violent reversals. In fact GBM with reversion are pretty realistic models for these (if we let volatility change randomly).

 

You can have large dd, but can't really have "unbounded losses", unless you use stops, in which case, obviously, you do.

 

[That is another reason why all fx brokers and scammers recommend to people to use stops .]

 

 

Tom

 

Volatility is not bound.

 

Not using stops is an incredibly dangerous approach if you are trading.

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  MightyMouse said:
Volatility is not bound.

 

.

 

Dear Mighty Mouse

Please explain to a simple farm boy from sunny South Africa how volatility is not bound.

If there are no buyers and no sellers there is no movement. if there is no movement , there is no volatility.

Kind regards

bobc

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  tommaso said:
Mkts don't have to follow anything.

 

That's exactly why we assume they can do anything.

 

Starting very simply, if I complete 100 trades in a year then each trade can either win me money or lose me money. This is analogous to flipping a coin 100 times, it'll either land heads or tails, and I'll either win or lose.

 

What you are talking about are strategies to improve our chances, I'm starting from a conservative approach where we have no edge over the probabilities.

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  bobcollett said:
Dear Mighty Mouse

Please explain to a simple farm boy from sunny South Africa how volatility is not bound.

If there are no buyers and no sellers there is no movement. if there is no movement , there is no volatility.

Kind regards

bobc

 

bobc, if there are no buyers or sellers, there is no market. Only when we have buyers and sellers, we have a market and boundless volatility.

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  russellhq said:
bobc, if there are no buyers or sellers, there is no market. Only when we have buyers and sellers, we have a market and boundless volatility.
Then it may follow that the more players there are, the lower the volatility, as the likelihood of any one of 10,000 players accepting a price becomes a higher probability. The more players ... the more liquidity.

 

The only spanner in those works then, is market sentiment and herd mentality.

We saw this over the past 10 days - herds and sentiment!

 

Has anything really changed?

Why was the DOW good value on TUESDAY, but stank on THURSDAY?

 

No reason at all - the market was spooked by fear, that's all - value didn't change.

 

Back to the topic ...

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  russellhq said:
I'd like to start a disscussion on the maths behind risk of ruin as from what I've read previously, it leaves me a little uncomfortable.

 

I'll start by discussing the way it's usually presented. Usually when calculating RoR, you normally start with the the answer and work backwards, so lets do that.

 

I'll use the coin flipping analogy as the game. So say we play a game where I flip a coin and if it lands on heads then you win and if it's tails you lose. Normally, before the start, you would decided on what level of risk you will accept before going broke, say it's 1 in 10,000. This works out at roughly 13 tails in a row or 0.5^13.

 

Nice and easy, we can play the game for as long as we want but if there is a run of 13 tails, then you're out.

 

But, what if, at the start of the game you have a run of 12 tails, then a head, then 2 tails. You are still out! This thought led me to the following:

 

This time, instead of ruin meaning you can no longer play, lets set ruin to be true only at then end of a set number of games. If during the course of play you pass the ruin line, you are still allowed to play to try and recover but you must stop after the set number of games.

 

Lets start out by saying we will play 27 times, what will be the chance of you being ruined?

 

To work this out, lets start with how many different permutations of the game there is (how many difference ways we can flip the coin 27 times).

 

2^27 is the answer

 

Now lets work out how combinations there are that can ruin you by the end of play.

 

If you get 0 heads during the 27 flips, then all agreed, you would be well and truly ruined. There is only 1 combination of this.

 

If you get 1 head during the 27 flips, you would still be ruined. There are 27 combinations where you can get only 1 head from 27 flips.

 

If you get 2 heads during the 27 flips, again you'll be ruined. To calculate the combinations, we use factorials: 27!/(27-2)! = Number of ways you can be ruined.

 

We keep doing this until we get to 7 heads, after 7 heads, you would always be able to recover by the end.

 

So when we add up all the combinations from 0 heads, to 7 heads:

 

0 heads = 1

1 heads = 27

2 heads = 27!/(27-2)!

3 heads = 27!/(27-3)!

4 heads = 27!/(27-4)!

5 heads = 27!/(27-5)!

6 heads = 27!/(27-6)!

7 heads = 27!/(27-7)!

 

And divide by the total number of permutations (2^27) we end up with our answer. In this case it's 1 in 100!

 

This is a lot higher than our initial assesment of 1 in 10,000!

 

This is just the basics and i've not considered Risk/Reward ratios etc (that can be added later). I just wanted to start with the basics.

 

Thoughts?

 

I have been reading a good book on this subject. The Universal Principles of Successful Trading, by Brent Penfold. He says the most important leg of the 3 legged stool is Risk Management and Risk of Ruin. He even has a computer program associated with this book that helps to calculate risk of ruin situations. I would like any comments on this book if anyone has read or is reading this book.

Jim

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I've been doing a bit more research as this puzzle has been a real head scratcher, but I think I've made more progress.

 

Let's look at the following problem:

 

A trader has a win rate of 55% and on average his losing traders lose $1000 and his wining traders win $1200.

 

Over the next year he expects to execute 100 trades. What would the size of his bank need to be such that he would only be ruined once in 10,000 years.

 

My calculations say he would need $23,000 to cover the risk...

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  russellhq said:
I've been doing a bit more research as this puzzle has been a real head scratcher, but I think I've made more progress.

 

Let's look at the following problem:

 

A trader has a win rate of 55% and on average his losing traders lose $1000 and his wining traders win $1200.

 

Over the next year he expects to execute 100 trades. What would the size of his bank need to be such that he would only be ruined once in 10,000 years.

 

My calculations say he would need $23,000 to cover the risk...

 

Can you post your calculations on that?

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  MightyMouse said:
Volatility is not bound.

 

Not using stops is an incredibly dangerous approach if you are trading.

 

 

Stops make the loss unbounded. It's bounded only if taken a stop you quit trading forever.

All scammers and brokers push hard on stop because this way your loss is unbounded.

 

If one does some serious reasearch (and i mean "serious", not te coin toss) and simulations, it is easily seen that strategies which use stop at trade level are sistematically unprofitable.

 

If you see seemingly good backtested results, it's just curve fitting.

 

I have been trading live for years showing and publishing live results of the order of 10% monthly, and all my traders have good returns. I never used stops.

 

Clearly, it requires capital. But anyway small traders are wiped out in any case. Just matter of time.

 

Tom

 

see for instance:

Forums - Robotrading: CT + Trending Strategy on folios of futures

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