[MightyMouse]

Originally Posted by tommaso »

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.

[bobcollett]

Originally Posted by MightyMouse »

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

[russellhq]

Originally Posted by tommaso »

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.

[russellhq]

Originally Posted by bobcollett »

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.

[Ingot54]

Originally Posted by russellhq »

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 ...

[jgiegerich1]

Originally Posted by russellhq »

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

[russellhq]

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...

[jyork13]

Originally Posted by russellhq »

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?