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How important is a knowledge of maths?

  • #41
  • Mar 25, 2007 6:56pm Mar 25, 2007 6:56pm
If I was a beginner and wanted to learn a skill to assist me
in my trading it would be programming.

Programming is essential if you want to do something
useful with all that maths. They complement each other.

Coder
  • #42
  • Mar 25, 2007 7:17pm Mar 25, 2007 7:17pm
forex or trading in general is a game of prabability, hence, you have to have a good understanding of stats, probability, and basic arithmetic.

That's all you really need in terms of maths
  • #43
  • Mar 25, 2007 10:21pm Mar 25, 2007 10:21pm
Are there any traders here that have used advanced mathematics to trade or just probe the markets? I'm thinking, Differential Equations, Fourier Analysis or Vector Algebra? It would be cool to be pointed in the right direction to how to get started formulating the market in these ways.
  • #44
  • Mar 25, 2007 11:09pm Mar 25, 2007 11:09pm
Quoting twoblink
Perhaps I'm the ignorant one, but how are you all doing MM without "math skillz"?

It makes no sense to me; charts, graphs etc.. are MATH. How is it possible that knowing that wouldn't help ????

I actually find this question almost like a fighter pilot asking "Do I need to know complex math?" and others replying "No, you don't. Reaction time is much more important than math when flying.. besides, my 9 year old son flies the simulators just fine, and he doesn't even know multiplication.."


I can only say, FX is a DIFFICULT Item, and I for one, try to give myself as much advantage as I can.. I can't imagine someone "trying to be successful" at FX and not learning math because others on a forum told them it's not really necessary..

Just ask yourself:

Do I want an advantage in FX? If so, learn math, if not, don't.
I am on your side
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  • #45
  • Mar 25, 2007 11:20pm Mar 25, 2007 11:20pm
Quoting hilmy83
forex or trading in general is a game of prabability, hence, you have to have a good understanding of stats, probability, and basic arithmetic.
A game:

four cards, 2 red, 2 black.
Shuffle and pick two,
What is the prabability of the two cards being of the same color?
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  • #46
  • Mar 25, 2007 11:32pm Mar 25, 2007 11:32pm
Quoting Samuri
A game:

four cards, 2 red, 2 black.
Shuffle and pick two,
What is the prabability of the two cards being of the same color?
umm..
here's my logic

50/50 to get either red or black
then to get the same color it would be 1/3
ex
50/50 to pick red , what's left is 1 red and two blacks
1/3 chance to pick a red

so 1/3(1/2) = 1/6??
  • #47
  • Mar 26, 2007 12:03am Mar 26, 2007 12:03am
Quoting blueruby
My 22 years as an engineer has, I feel, worked against me in trading. ---snip--- I've read the usual stack of books, not one ever said that higher math skills were a necessity, and most or all were written by psychologists, not mathmaticians.
LOL. I'm still a newbie in fx, but if there was anything that helped me become profitable after a year's work these were my 25 years as a researcher in psychology. Stuff learned in statistics and reserach methods classes, and above all - a general critical, paranoid, obsessive-compulsive scientific approach you acquire with experience. As I said earlier, I'm relying heavily on what I'm good at, although I can easily imagine people that successfully use the other side of their brains...
sans peur et sans reproche
  • #48
  • Mar 26, 2007 12:21am Mar 26, 2007 12:21am
Quoting hilmy83
umm..
here's my logic

50/50 to get either red or black
then to get the same color it would be 1/3
ex
50/50 to pick red , what's left is 1 red and two blacks
1/3 chance to pick a red

so 1/3(1/2) = 1/6??
good approach.

1/6+1/6=1/3 would be the answer, either 2 reds or 2 blacks.

Let's take another try:

Mr. and Mrs. Smith have two children, and they tell you at least one of the children is a little girl. If the birth probability of a boy or a girl is the same @ 50%, what is the probability of Smiths having the other child is also a girl?
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  • #49
  • Mar 26, 2007 6:45am Mar 26, 2007 6:45am
Quoting Samuri
good approach.

1/6+1/6=1/3 would be the answer, either 2 reds or 2 blacks.

Let's take another try:

Mr. and Mrs. Smith have two children, and they tell you at least one of the children is a little girl. If the birth probability of a boy or a girl is the same @ 50%, what is the probability of Smiths having the other child is also a girl?
well basically...

each time you have a 1/2 probability to have either gender

so since it's gonna be their third child.

1/2 for the first child to be a girl
1/2 of 1/2 for the second child to be a girl
1/2 of 1/2 of 1/2 for the third child to be a girl and so on..

so 1/2*1/2*1/2 = 12.5%
  • #50
  • Edited 7:14am Mar 26, 2007 6:50am | Edited 7:14am
Quoting twoblink
Yes, same source... why CCI, RSI, Stoch etc.. all match up..
Very often you get divergence between an indicator (such as CCI, RSI, stoch) and price. Price goes up and makes a higher peak, the indicator goes down and makes a lower peak.

If the indicator is just a moving average, how could this be?

It cannot, it's mathematically impossible.
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  • #51
  • Edited 7:09am Mar 26, 2007 6:55am | Edited 7:09am
Quoting Samuri
good approach.

1/6+1/6=1/3 would be the answer, either 2 reds or 2 blacks.

Let's take another try:

Mr. and Mrs. Smith have two children, and they tell you at least one of the children is a little girl. If the birth probability of a boy or a girl is the same @ 50%, what is the probability of Smiths having the other child is also a girl?
Hilmy would be right IF you asked, "what are the odds of drawing two reds".

The first draw is 50/50 probability. The second draw is 1/3.

1/2 * 1/3 = 1/6. That would be the answer.

However, the question was, "what are the odds of getting two the same color", so the first draw is unimportant. It will either be red or black, then what you're concerned with is, what are the odds the second draw will be that same color? There is only one left of that color in the three cards, so the odds are 1 in 3.

The answer to the Smith question is 50/50 or 1/2. The gender of the next kid has nothing to do with gender of the previous kids. If you were drawing kids from a pool where there were 1/2 and 1/2 girls and boys, then you need to do the math since the odds change as you take from the pool (as with the cards).
  • #52
  • Mar 26, 2007 7:20am Mar 26, 2007 7:20am
RSI divergence. Proponents of the indicator would point at the rightmost example on the chart - when there was a strong divergence between price movement and RSI (price ran up while strength did not increase proportionately) then there was a huge move downward afterward.
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  • #53
  • Mar 26, 2007 7:27am Mar 26, 2007 7:27am
Maybe I did not put it clear. The question was what is the odds of Smiths having two girls? (preliminarily they tell you that one of the two is a girl)

This is kind of conditional probability, it mostly will beat your first thought.
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  • #54
  • Mar 26, 2007 7:39am Mar 26, 2007 7:39am
Quoting Samuri
Maybe I did not put it clear. The question was what is the odds of Smiths having two girls? (preliminarily they tell you that one of the two is a girl)

This is kind of conditional probability, it mostly will beat your first thought.
If it is like a fair coin being flipped I think there is always a 50% chance regardless of the prior results. So you have a 100 change on the first half of the double and 50% on the second.
  • #55
  • Mar 26, 2007 7:48am Mar 26, 2007 7:48am
From nvwine's post:

Yes each individual flip is 50/50.... but when considering a group of winners or losers, law of averages applies. Two Smith daughters is 50% x 50% = 25%

Otherwise, you have a 50% chance of flipping heads infinitely.
  • #56
  • Mar 26, 2007 7:48am Mar 26, 2007 7:48am
Quoting Samuri
Maybe I did not put it clear. The question was what is the odds of Smiths having two girls? (preliminarily they tell you that one of the two is a girl)

This is kind of conditional probability, it mostly will beat your first thought.
50/50 on the first, 50/50 on the second, just like a coin toss.

So the chances of two girls would be 1/2 * 1/2 = 1/4.

If you look at the possible outcomes, this makes it easy to see:

B B
B G
G B
G G

1 in 4 possible outcomes that there will be two girls.
  • #57
  • Mar 26, 2007 7:51am Mar 26, 2007 7:51am
Quoting blueruby
50/50 on the first, 50/50 on the second, just like a coin toss.

So the chances of two girls would be 1/2 * 1/2 = 1/4.

If you look at the possible outcomes, this makes it easy to see:

B B
B G
G B
G G

1 in 4 possible outcomes that there will be two girls.
Don't we already know the first is a girl?
  • #58
  • Mar 26, 2007 7:52am Mar 26, 2007 7:52am
Quoting Samuri
Maybe I did not put it clear. The question was what is the odds of Smiths having two girls? (preliminarily they tell you that one of the two is a girl)

This is kind of conditional probability, it mostly will beat your first thought.
Well, I still didn't get the question right. nvwine is the winner.

If they have a girl alreay then there's 100% chance on the first. The second is 50/50, so 1 * 1/2 = 1/2.

50% chance of having 2 girls.

The trading lesson here is not the statistical problem, it's the reading of the problem.

I answered what I thought I saw, not what the question said. There's a good lesson for me.
  • #59
  • Mar 26, 2007 8:01am Mar 26, 2007 8:01am
Quoting hilmy83
each time you have a 1/2 probability to have either gender
Not to mess up your neatly calculated math, but there is a slightly higher probability that any random child will be born a boy.

http://www.indexmundi.com/world/sex_ratio.html

Just trying to make life complicated.

EDIT:// Ohh and this might help http://mathforum.org/library/drmath/view/67124.html
  • #60
  • Mar 26, 2007 8:02am Mar 26, 2007 8:02am
The point is the two girls were born already.

Under the condition that they tell you one of the two children is a girl, the odds of they having two girls is then 1/3.
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