Can you profit from trading if you know the next percentage change will be smaller or larger than the last one?
I think being strong at maths is underrated in trading. You don't need more than GCSE level to trade profitably.. BUT it helps. It will speed up your understanding of trading and strategy building. And helps with coding up strategies and testing things if you go that route
Thank Fugazi. @ Jimmibit. Well, I might beg to differ on it being “just theoretical.” I’ve actually shown quite a few practical examples. Just because we might not make money straight away doesn’t mean it’s not practical, right?
So obviously this works—and that’s without a shadow of a doubt. If you’re not getting those results, then respectfully, you’re probably not applying it correctly.
But here’s the real question: why can’t we just slap this onto odds, or any price we’re trying to speculate on? Isn’t that the obvious next step?
If you’ve noticed, we talked about two types of distributions—normal and uniform—like in the example with dice. This method works well with those two. But not with the price of a match, a stock, or any other market price. Or maybe it does?
So maybe the question we should ask is: what do those two distributions (normal and uniform) have in common that others don’t?
And when you look closely, I think the answer is pretty straightforward—normal and uniform distributions are symmetrical and it is very easy to see mean value straight away, while price distributions tend to be log-normal and very asymmetrical… but that isn’t to say they don’t have mean and median values? So where do we go from here? Just a thought.
Cheers
So obviously this works—and that’s without a shadow of a doubt. If you’re not getting those results, then respectfully, you’re probably not applying it correctly.
But here’s the real question: why can’t we just slap this onto odds, or any price we’re trying to speculate on? Isn’t that the obvious next step?
If you’ve noticed, we talked about two types of distributions—normal and uniform—like in the example with dice. This method works well with those two. But not with the price of a match, a stock, or any other market price. Or maybe it does?
So maybe the question we should ask is: what do those two distributions (normal and uniform) have in common that others don’t?
And when you look closely, I think the answer is pretty straightforward—normal and uniform distributions are symmetrical and it is very easy to see mean value straight away, while price distributions tend to be log-normal and very asymmetrical… but that isn’t to say they don’t have mean and median values? So where do we go from here? Just a thought.
Cheers
At this stage, you're in the classic Dunning Kruger. You're about to find out this isn't easy soon enough.Flatino wrote: ↑Thu Apr 24, 2025 6:29 pmThank Fugazi. @ Jimmibit. Well, I might beg to differ on it being “just theoretical.” I’ve actually shown quite a few practical examples. Just because we might not make money straight away doesn’t mean it’s not practical, right?
So obviously this works—and that’s without a shadow of a doubt. If you’re not getting those results, then respectfully, you’re probably not applying it correctly.
But here’s the real question: why can’t we just slap this onto odds, or any price we’re trying to speculate on? Isn’t that the obvious next step?
If you’ve noticed, we talked about two types of distributions—normal and uniform—like in the example with dice. This method works well with those two. But not with the price of a match, a stock, or any other market price. Or maybe it does?
So maybe the question we should ask is: what do those two distributions (normal and uniform) have in common that others don’t?
And when you look closely, I think the answer is pretty straightforward—normal and uniform distributions are symmetrical and it is very easy to see mean value straight away, while price distributions tend to be log-normal and very asymmetrical… but that isn’t to say they don’t have mean and median values? So where do we go from here? Just a thought.
Cheers
HOWEVER, youre analytical and like problem solving and have the maths ability to follow through with strategies. You will go far so long as you stay open minded to some of the really good traders on here.

Basically, it's not really about being a mind-reader or anything crazy like that. It’s just about using the clue you get in a smart way.
Think about any situation with a range of possibilities, people's heights, ages, temperatures, whatever. Someone picks two different ones from that range and shows you just one.
The number (or whatever value) they show you kind of sets a marker. Everything else that could have been picked is either higher or lower than that marker.
If the number they show you is really low, or really high It just makes sense that most of the other possible numbers are in the other direction.
If someone picks two people. If they point out someone who's only 1.52m, the other person they picked is probably taller, right? Because most heights are taller than that.
If the number is somewhere in the middle that's when it gets trickier. There are roughly the same number of possibilities higher and lower, so it’s more of a toss-up, harder to guess right.
So yeah, maybe figuring out the exact chances is where it feels a bit complicated. But the basic idea is just common sense, if you see a really high or really low number, you've got a pretty good hint about where the other one might be.
It's nothing fancy, really. You're just using the piece of information you do have to make a better guess about the one you don't have, knowing the overall range. It doesn't mean you'll always be right, but it definitely helps your odds compared to just guessing randomly.
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1. Does a price have a known "probability distribution" with a mean & a variance?Flatino wrote: ↑Thu Apr 24, 2025 6:29 pm
And when you look closely, I think the answer is pretty straightforward—normal and uniform distributions are symmetrical and it is very easy to see mean value straight away, while price distributions tend to be log-normal and very asymmetrical… but that isn’t to say they don’t have mean and median values? So where do we go from here? Just a thought.
Cheers
2. Is the price of a selection independent of other selections?
3. Does the price of a selection represent value to another market participant & is that market participant involved in the market with enough funds to take that value?
4. Related to 3) the timing of when value traders enter / leave a market
I'd argue that it is not possible (in general) to "model" the price direction for an individual selections movement "in real time" with any form of accuracy You might as well flip a coin.
(the exception being when you get in folks value range(s) & if they have the money the price will stick / reverse till they max out the liability they are willing to risk)
I suppose my point is an individual price movement is just that.. i.e. an "observation" that may be in the 1% or 99% percentile of whatever model you are using and / or you have no way of really knowing if the model you are using is the right one at that time due to market dynamics (or just one large player entering / leaving market).
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If you're into automation you can model the next tick for a random walk scenario, but beyond that is difficult/ neigh on impossible. But it's an approach that's totally independent of market types so if you eventually crack it, you can use it on every market, on every selection, 247365.sionascaig wrote: ↑Sun Apr 27, 2025 8:58 amI'd argue that it is not possible (in general) to "model" the price direction for an individual selections movement "in real time" with any form of accuracy You might as well flip a coin.
Eg. Queue depletion aka order flow imbalance , comparing new orders - cancellations - traded, on each side to see which empties first. A bit like PIQ but using rates of change. There's a load of info on it as it's not surprising people have tried maths to unlock markets rather than just guesswork.
I suppose in the manual trading world directional scalping would be the equivalent and whether you know it or not this is what you're doing in your head if you're any good at it.....
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Thanks Shaun...ShaunWhite wrote: ↑Sun Apr 27, 2025 1:18 pmIf you're into automation you can model the next tick for a random walk scenario, but beyond that is difficult/ neigh on impossible. But it's an approach that's totally independent of market types so if you eventually crack it, you can use it on every market, on every selection, 247365.sionascaig wrote: ↑Sun Apr 27, 2025 8:58 amI'd argue that it is not possible (in general) to "model" the price direction for an individual selections movement "in real time" with any form of accuracy You might as well flip a coin.
Eg. Queue depletion aka order flow imbalance , comparing new orders - cancellations - traded, on each side to see which empties first. A bit like PIQ but using rates of change. There's a load of info on it as it's not surprising people have tried maths to unlock markets rather than just guesswork.
I suppose in the manual trading world directional scalping would be the equivalent and whether you know it or not this is what you're doing in your head if you're any good at it.....
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Screenshot_20250427_121445_YouTube.jpg
In that context you can develop a "model" & I probably didn't express myself very well.
For something like Tennis it is absolutely critical if you want to know what the price should be for a particular score & hence what the movement is likely to be in either direction.
For say a horses market though the indicators you mention above would be much more interesting to follow / measure rather than attempt to develop a random walk model.
In one of my earlier posts, I mentioned that we can apply this type of "prediction" to goals or, more precisely, to goal difference. The individual distribution of goals for each club typically follows a Poisson distribution, which is highly asymmetrical. However, by shifting our perspective slightly, we can uncover a degree of symmetry.
When we combine two Poisson distributions, one for goals scored and one for goals conceded we arrive at a new distribution that is much more symmetrical. This resulting distribution is known as the Skellam distribution. The Skellam distribution is the probability distribution of the difference between two independent Poisson-distributed variables.
The Skellam distribution models the goal difference (GD), which is calculated as:
Goal Difference (GD) = Goals Scored−Goals Conceded
To ensure consistency in analysis, we always calculate the goal difference as:
Goal Difference=Goals by Team A−Goals by Team B
This applies regardless of home or away status—Team A (the first team listed) is always treated as the reference team. (Alternatively, you can reverse the roles and use Team B as the reference; the hit rate remains unchanged. How beautiful is that?)
While not perfectly symmetrical like a normal distribution, the Skellam distribution is sufficiently symmetric to offer valuable predictive insights in football analytics.
In this context, the midpoint or median (also referred to as the k value)—of the Skellam distribution is 0, representing a balance between goals scored and goals conceded.
Let’s now explore what can happen when we apply our strategy using 0 as the reference point.
We will compare each match’s goal difference result to 0:
• If the goal difference is greater than 0 (i.e., the home team won), we predict that the next result will be equal to or less (i.e., same or smaller goal difference).
• If the goal difference is less than 0 (i.e., the home team lost, negative number), we predict that the next result will be equal to or greater than (i.e., same or less negative goal difference).
• If the goal difference is exactly 0 (i.e., a draw), we treat it as a neutral outcome and skip it (or take a guess and have 50:50 chance of being correct)
Please find attached an example based on Liverpool (for no particular reason). It could be any club or any league you prefer. And once you see it, you can’t unsee it—lol In this example we would be correct 90% of the time. But in the long run the average should hover around 70% like previously proven
When we combine two Poisson distributions, one for goals scored and one for goals conceded we arrive at a new distribution that is much more symmetrical. This resulting distribution is known as the Skellam distribution. The Skellam distribution is the probability distribution of the difference between two independent Poisson-distributed variables.
The Skellam distribution models the goal difference (GD), which is calculated as:
Goal Difference (GD) = Goals Scored−Goals Conceded
To ensure consistency in analysis, we always calculate the goal difference as:
Goal Difference=Goals by Team A−Goals by Team B
This applies regardless of home or away status—Team A (the first team listed) is always treated as the reference team. (Alternatively, you can reverse the roles and use Team B as the reference; the hit rate remains unchanged. How beautiful is that?)
While not perfectly symmetrical like a normal distribution, the Skellam distribution is sufficiently symmetric to offer valuable predictive insights in football analytics.
In this context, the midpoint or median (also referred to as the k value)—of the Skellam distribution is 0, representing a balance between goals scored and goals conceded.
Let’s now explore what can happen when we apply our strategy using 0 as the reference point.
We will compare each match’s goal difference result to 0:
• If the goal difference is greater than 0 (i.e., the home team won), we predict that the next result will be equal to or less (i.e., same or smaller goal difference).
• If the goal difference is less than 0 (i.e., the home team lost, negative number), we predict that the next result will be equal to or greater than (i.e., same or less negative goal difference).
• If the goal difference is exactly 0 (i.e., a draw), we treat it as a neutral outcome and skip it (or take a guess and have 50:50 chance of being correct)
Please find attached an example based on Liverpool (for no particular reason). It could be any club or any league you prefer. And once you see it, you can’t unsee it—lol In this example we would be correct 90% of the time. But in the long run the average should hover around 70% like previously proven
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