Probability - or, when is it better to stay in bed?

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tomorton

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Let's suppose we have a strategy with a 60% win rate. The probability of a win on Day 1 must therefore be 60% and obviously the probability of a loss must be 40%.

Now let's suppose D1 is a winner. The probability of D2 being a second consecutive winner must surely have decreased to (.60 x .60) x 100 = 36%. The probability of D2 being a loser is not increased by D1 being a winner, it remains 40%, so the odds for the trader are now slightly negative but worth trading.

Suppose D2 is a winner also, so now the probability of D3 being a third consecutive winner must surely be decreased even further to (.60 x .60 x .60) x 100 = 21.6%. The odds of a loss are still 40%, but I am now almost twice as likely ((40 / (40 + 21.6) x 100 = 64.9%) to lose on D3 as to win.

Is the logical conclusion therefore with this system to stay out of the market on any day following two consecutive winners?
 
By that rationale, you'd definitely need a system with negative expectancy to get up each morning with a smile on your face.

It, can only get better....
 
tomorton said:
Now let's suppose D1 is a winner. The probability of D2 being a second consecutive winner must surely have decreased to (.60 x .60) x 100 = 36%.
Click to expand...

If you throw a coin and it comes down heads, does that mean that there is only a 25% chance of the next throw coming down heads?
 
tomorton said:
Suppose D2 is a winner also, so now the probability of D3 being a third consecutive winner must surely be decreased even further to (.60 x .60 x .60) x 100 = 21.6%.
Click to expand...
The probability of having three consecutive winners as you describe is indeed 21.6%. This can be calculated as you have shown before any trades were taken.

After having two wins on the board however, the probility of having three consecutive wins for the day rises to 60%, as that is the probability of any trade being a winner, and each trade is an independent event.

I assume you're just mucking about with you logic but I'm in a bitey mood :cheesy:
 
Opportunity doesn't come along on command. You should stay in bed when there is no opportunity and get up extra early when there is. I have gotten up extremely late for almost the entirety of my life. Now I stumble out of bed early with my eyes closed to take my final snooze in front of the screen. The opportunity at market open can be quite marvelous. Although it's not available every morning, the thought of it might gets me up and running every day.

A 60% probability seems a bit low. I'd want a 100% of the good stuff when they are there.
 
Why would the probability on D1 be related to the probability on D2? IMO if you have a 60% win rate then every day you should wake up thinking you have a 60% chance of a winner...
 
mb325 said:
Why would the probability on D1 be related to the probability on D2? IMO if you have a 60% win rate then every day you should wake up thinking you have a 60% chance of a winner...
Click to expand...

Actually, you should wake up thinking you have a 100% chance of winning, but nobody is perfect so you'll end up winning less.
 
tomorton said:
Is the logical conclusion therefore with this system to stay out of the market on any day following two consecutive winners?
Click to expand...

No, please sign up for a basic maths course at your local adult education centre.
 
I can accept that each individual trade is indepenent of the one before, as has been pointed out. If there is a 60% probability of something happening, that does not change from the viewpoint of that individual event. But we are linking events here, as we aggregate the proceeds from trades. If it's 60% that one thing will happen, then it cannot be 60% that two unrelated things will both occur, the probability of two out of two is less than 60% (it is 36%).

I have a practical reason for asking, as I currently conduct 1 trade per day using a system that has about a 60% win rate. If I get two consecutive wins, surely the probability of a third in an unbroken series is not 60%?
 
Hey tomorton, yes if you had 2 days of wins its 'likely' there may be a bad day but it can't be determined by maths, in reality you could flip a coin 100 times and get heads 100 times, its just unlikely... if you skipped days you are just as likely to miss a win as a lose, and if its a 60% win rate then you are working against yourself... the 60% is on the day, and the day before has no bearing on the next day.

Also the fact that the system had 5 wins in a week just 3 weeks ago kinda proves that consecutive wins doesn't necessarily mean a loss, i'll pop in your thread tomorrow anyway, good luck this week :)
 
tomorton said:
i have a practical reason for asking, as i currently conduct 1 trade per day using a system that has about a 60% win rate. If i get two consecutive wins, surely the probability of a third in an unbroken series is not 60%?
Click to expand...

After the first two wins have happened why should their outcome affect the third trade? Its the same as I mentioned before, if you toss a coin and it come up heads twice, the chances of it coming up again afterwards is still 50%. What has happened before has no bearing on the outcome.

I'm also interested when you think that you could start trading again so that the odds are back in your favour, after 3 days, 4 days? How would waiting a couple of days before you traded again improve your odds?
 
tomorton said:
I can accept that each individual trade is indepenent of the one before, as has been pointed out. If there is a 60% probability of something happening, that does not change from the viewpoint of that individual event. But we are linking events here, as we aggregate the proceeds from trades. If it's 60% that one thing will happen, then it cannot be 60% that two unrelated things will both occur, the probability of two out of two is less than 60% (it is 36%).

I have a practical reason for asking, as I currently conduct 1 trade per day using a system that has about a 60% win rate. If I get two consecutive wins, surely the probability of a third in an unbroken series is not 60%?
Click to expand...
If you knew that you were going to get 3 wins over the next 5 working days, which is of course a 60% win rate, and you got those 3 wins on the first 3 days, then you would be well advised to go fishing for the remainder of the week, or whatever takes your fancy.
But you don't know that. You can't predict in what pattern the wins and the losses will ocurr, as they are not linked to previous events (trades). If they were linked to previous trades you could potentially change your system to weed them out, thus increasing the probability of your system. As they are independent, the chance of a third win remains at 60%.
 
Of course the trade on D3 has a 60% chance of being a winner. If you asked me today how likely it is that the trade on D3, Wednesday, will be a winner, I would have to say it's 60%. But that's not what we're trading. I aim to run this trade on every day, so effectively trading on the probability that every consecutive trade will be a winner. So who would say on the evening of D7, a week on Tuesday, supposing we have had 7 straight wins, D1-7 inclusive, 'Oh, I absolutely must get a trade on tomorrow, as it's 60% likely it will be a winner'? It just can't be equally likely that we will get an 8th consecutive win as 1 single win on a randomly chosen day.
 
tomorton said:
Let's suppose we have a strategy with a 60% win rate. The probability of a win on Day 1 must therefore be 60% and obviously the probability of a loss must be 40%.

Now let's suppose D1 is a winner. The probability of D2 being a second consecutive winner must surely have decreased to (.60 x .60) x 100 = 36%. The probability of D2 being a loser is not increased by D1 being a winner, it remains 40%, so the odds for the trader are now slightly negative but worth trading.

Suppose D2 is a winner also, so now the probability of D3 being a third consecutive winner must surely be decreased even further to (.60 x .60 x .60) x 100 = 21.6%. The odds of a loss are still 40%, but I am now almost twice as likely ((40 / (40 + 21.6) x 100 = 64.9%) to lose on D3 as to win.

Is the logical conclusion therefore with this system to stay out of the market on any day following two consecutive winners?
Click to expand...

This misguided thought process is well known and is called the gamblers fallacy. :LOL:
 
tomorton said:
It just can't be equally likely that we will get an 8th consecutive win as 1 single win on a randomly chosen day.
Click to expand...
But it is, and Jason101 has been able to recall the name of the this incorrect way of thinking and it would certainly be in your interest to research this issue further to correct it. The last thing you want to be doing is to start skipping trades.

It would be pretty easy to devise a test. Flip a coin, every time you get say three heads in a row, then record what the fourth coin flip turns out to be. Do this enough times and you will find that the fourth flip is equally likely to be heads as tails. I can assure you that you will not get significantly more tails than heads due to the previous three flips being heads - providing your sample size is sufficiently large.
 
tomorton said:
It just can't be equally likely that we will get an 8th consecutive win as 1 single win on a randomly chosen day.
Click to expand...

Actually, I think I phrased this badly. I've heard of Gambler's Fallacy before but I'm not convinced it applies, the reason being I won't be betting on the result of the 8th consecutive event, I'm betting on 8 identical consecutive results of 8 different but consecutive events. Any one trade in the series has a 60% probability of being a win. But surely the entire series of 8 does not have a 60% probability of producing 8 wins?
 
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