Lets say a soccer team has x% chance of winning.
How much chance is expected for a draw?
Assume everything else average.
How much chance is expected for a draw?
Assume everything else average.
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This would depend on many many variables. Some teams and leagues have a strong draw bias (French for instance). You also cannot calculate the chance of a draw, with knowing only the chance of one team winning.
Best way of calculating the draw generally is to calculate the expected goals for both teams feed these into a diagonally inflated poisson which which can then be used to calculate the odds of each score line, you then need to add up the probabilities of all draw lines.
Excel can be used to calculate a normal poisson, but you will need to manually adjust the figures as poisson will undervalues the chance of the draw.
Take the Arsenal -v- Manchester United game today as an example
Arsenal are expected to score 1.75 goals
Man Utd are expected to score 1.45 goals
So to get the chances of Arsenal scoring 1 goal we need to do =poisson(1,1.75,0) and then the same for Man Utd =poisson(1,1.45,0)
We can then find the chances of a 1-1 draw by finding the product of those two, which is around 0.11, which equates to around 9.7 in European Odds. Obviously to find the chances of a draw, rather than just the 1-1, you need to find the probabilities of 2-2, 3-3 etc and add these up and make some manual adjustments
Just a starting point...Comment -
Yes, you can do that too by using expected goals, altough the more simple way obviously would be to use correct score predictions.
However there must be a logical/math way to calculate draw probabilities based only on the winning chances of only one team.
I think the best starting point should be the chance for a draw of two teams with exactly the same chances, say cloning a team to play itself.
The assumption is that the probabilities in such case are equal for all three outcomes, thus 33.3%. However is it really true?
I tend to intuitively think that in such case the probability for a draw is 50%, leaving 25% for a winner. Still, can't find the rationale for this except the statistical data.
The second part would be how much draw ratio changes by moving the winning chances into one direction.Comment -
Simply put, two equally strong teams can hardly beat each other, that's what common sense tells you.
On the other side, the intuition is mostly feminine thing and that's why they're always wrong.
In fact, the stats very vary regarding this matter. In Italian competition the ratio is close to 1/3, e.g. Juventus vs Milan.
But comparing two legends Barcelona & Real Madrid provides completely surprising w-d-l results: roughly 40-20-40%!
The meaning of my last sentence was that I can only use statistical data to (try to) make conclusions, but can't think of the math solution.
Can you?Comment -
Your only relevancy in this forum is related to trolling and spamming in a very similar way as seen in other threads. For a moment I thought you're some kind of software used for keeping threads alive.
My age doesn't allow me to spend too much time on such a BS and peacocks without essence, but I hope administrators could find it to clear this place.
Anyway I would appreciate any meaningful answer to my original question.Comment -
Sports betting, basically, is a number's game based on probabilities.
It does not deal in "it seems to me...", "I tend to intuitively think...", "two equal teams can hardly beat each other..."
or anything else of that sort.
I see you appeal to common sense and yet, you make statement like this:
which make no sense what so ever (same as your original question, btw).However there must be a logical/math way to calculate draw probabilities based only on the winning chances of only one team.
It is not a rocket science, really. If your age allows for knowing any 5th grader, ask him. He'll tell you the same thing.
All in all, it is painfully obvious that you are quite a bit confused as to the nature of that beast you are dealing with.
I tried to help by asking a simple question, that should have moved you from the land of confusion to the land of arithmetic (order) and got slammed with constantly recycled.when better arguments are not there, "trolling" crap.
I guess stupidity is a diagnose after all. And in most cases, this sickness is incurable. It is just sad.
Good luck in your sports betting, sir.
P.S.
A+B+C=100
A=30
What is the value of C?
Does this make any kind of sense to anybody?Last edited by hutennis; 04-29-13, 08:27 PM.Comment -
O.T.: /* Usually I have no problem with narrow minded people. The problem arises when they conclude their way of thinking is a definite one and start trying to impose it to the world in an arrogant way. */
Going back to H(ome)/D(raw)/A(way), based only on probability of H in the beginning of the game and hypotheticaly assuming other conditions as being fair, the following can be drawn:
Sure thing (almost
):
# H probability has already incorporated goal difference and time in it
# H=100-(D+A)
# if H>50 then H>A; H>D
Hypothesis, true for statistical data:
# if H > 40 then H>A
# if H > 60 then D>A
# if H < 20 then A>D
.... and so on.
It is true that the relationship between D and A is also closely related to H. E.g. there is no way that having H with probability of 1% can have D=98%. There must be some constant somewhere.Comment -
How can you calculate the chance of H without also being able to calculate A. In order to be able to accurately calculate the probability of H win you must have taken into account the chances of A win, therefore you can calculate D/A as well as H.O.T.: /* Usually I have no problem with narrow minded people. The problem arises when they conclude their way of thinking is a definite one and start trying to impose it to the world in an arrogant way. */
Going back to H(ome)/D(raw)/A(way), based only on probability of H in the beginning of the game and hypotheticaly assuming other conditions as being fair, the following can be drawn:
Sure thing (almost ):
# H probability has already incorporated goal difference and time in it
# H=100-(D+A)
# if H>50 then H>A; H>D
Hypothesis, true for statistical data:
# if H > 40 then H>A
# if H > 60 then D>A
# if H < 20 then A>D
.... and so on.
It is true that the relationship between D and A is also closely related to H. E.g. there is no way that having H with probability of 1% can have D=98%. There must be some constant somewhere.
I can't see how you can only calculate only the H and not the other two.Comment -
H is a known variable. Any arbitrary number > 1.
D+A is obviously 100-H. Given to H e.g. an odd of 1.50 or 66.6% probability, D+A combined have the odd of 3.00 or 33.33%.In order to be able to accurately calculate the probability of H win you must have taken into account the chances of A win, therefore you can calculate D/A as well as H.
I can't see how you can only calculate only the H and not the other two.
Individually in this case D can be e.g. 4.5 (22.22%) and A 9.0 (11.11%), but NOT any possible number such as D=90 (1.1%); A=3.1(32.2%).
The ratio of D and A is closely related in some way to the value of H, which makes the problem more plausible and solvable.
There is some magic number that determines the relationship among the three values.Comment -
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Uhh... not sure why you would intuitively think this.
Let's do the math then. Let's pretend 0, 1, 2, 3, 4, and 5 goals are possibly scored in games. What is the probability that in each game the result is a draw, assuming the distribution of scoring is random between the two teams?
0 Goals: 100% draw
1 Goal: 0% draw
2 goals: 50% draw
AA
AB
BA
BB
3 goals: 0% draw
4 goals: 6/16 draw.
AAAA
AAAB
AABA
AABB
ABAA
ABAB
ABBA
ABBB
BAAA
BAAB
BABA
BABB
BBAA
BBAB
BBBA
BBBB
5 goals: 0% draw
Then, to determine how likely a draw is overall, you weight each %. I don't have any data on me, but I hope you see the point. if 0/1/2/3/4/5 goals are each 16.7% (which I made up from my ass), the math would be
100(.167) + 0(.167) + 50(.167) + 0(.167) + 6/16(.167) + 0(.167) = 16.7 + 0 + 8.35 + 0 + 6.3 + 0 = 31.35% chance of draw.Comment -
First of all, thanks for jumping in.
Let's do the math then. Let's pretend 0, 1, 2, 3, 4, and 5 goals are possibly scored in games.This is another number of goals approach, which is no necessarily wrong, but as I pointed earlier, it is already incorporated in odds (time+goals difference).if 0/1/2/3/4/5 goals are each 16.7% (which I made up from my ass), the math would be
100(.167) + 0(.167) + 50(.167) + 0(.167) + 6/16(.167) + 0(.167) = 16.7 + 0 + 8.35 + 0 + 6.3 + 0 = 31.35% chance of draw.
Don't take it in a wrong way: five goals per game is wrong and the distribution of .167 is even more wrong.
Please don't let me delute the original question and further elaborate this as I will accept your final conclusion of 33.3% (or if you insist 31.35%) for a draw between the teams of a equal strength on a neutral field.
My next question is how much would be the draw skewed if one team has 66.66% chance of winning?Comment -
Why would it be "skewed" anyway? Relative to what?
Since when odds for individual game are governed by some magical "constant" and not by careful calculations based on all known factors specific to this particular game?
Today there were 6 soccer games where favorite had no vig odds of 1.54 (65%)
No vig Draw odds where different in every game:
Which one of those draws is "skewed"?5.09 4.81 4.71 4.59 4.53 4.52
Which one would you like to argue with?
I have a suggestion.
Why don't you back every draw above average (4.71), lay every draw below average, declare it a "handicapping model" and make a lot of money fast?Last edited by hutennis; 05-01-13, 07:14 PM.Comment -
My magic number for the proportion of draws for the EPL is 0.301*.936^r where r is the win ratio of the favourite to the underdog. This is based on a regression of the no-vig odds for EPL games. It is very difficult to come up with a more theoretical formula as Poisson underestimates draws. So for Hutennis's list above this formula would give draw odds of 4.51 so either my formula is too conservative or the games in question came from a lower scoring league than the EPL.Comment -
you can try and add the "no draw" streak to the calculation, as when a team has a long streak with no draws the % of a draw occurring increases. -anyone has any thought about this?Comment -
wow as i know soccer is so simple, 50% chance of winning, its like left or right...Comment -
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mostly because of stats.
Seriously, look up the past years in EPL, all mid table teams play between 7 - 12 Draws a season, annually!
French league has even more draws, and so is argentina... so yes, when a team played for 15 games and got 2 draws, that's the sign for me that more draws are up ahead.Comment
