# combining 2 probabilities

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• November 20th 2010, 06:47 AM
neeeel
combining 2 probabilities
I have 2 probabilities

p1 is the probability that team A will score 0 goals
p2 is the probability that team B will concede 0 goals

how do i combine the 2 and calculate the probability of team A scoring 0 goals into team B?

should the probability of A scoring 0 goals into team B not be the same as the probability of team B conceding 0 goals from team A?

Is it as simple as p1*p2? I thought i might need to use conditional probability to work out the intersection of p1 and p2, but it doesnt seem to look right when I do it

thanks for any help
neil
• November 20th 2010, 11:28 AM
ANDS!
Unknown since we know nothing about Event A (Team A score 0 goals) or Event B (Team B concedes 0 goals - no idea what that means).
• November 20th 2010, 11:36 AM
neeeel
ok i have the average number of goals that team A scores
using the poisson formula = ((lambda^k) *( e^-lambda))/k! you can work out the probability that team A scores 0 goals( where k = 0, lambda is the average number of goals scored)

I have the average number of goals that team B concedes
using the poisson formula I can work out the probability that team B concedes 0 goals.

is there any way of combining the 2 probabilities to work out the probability that team A scores no goals into team B?

not sure how else to explain it
• November 20th 2010, 11:57 AM
neeeel
concede means, to let in, have scored against them
• November 20th 2010, 12:23 PM
ANDS!
I have no idea how sports probabilities work (what assumptions and what not goes into them), and I would imagine there would be some conditional dependence based on who Team A is playing - for example a crappy team playing a crappy team versus a good team playing a better team; so for a real world model - I do not believe you can infer that the two probabilities are independent of each other. If it was a simple model - then yea, just multiply - your probability would be a sort of "general" probability that team A scores 0 against an arbitrary team, and Team B concedes 0 goals against an arbitrary team.

That's my interpretation at the least.
• November 20th 2010, 12:45 PM
Quote:

Originally Posted by neeeel
I have 2 probabilities

p1 is the probability that team A will score 0 goals
p2 is the probability that team B will concede 0 goals

how do i combine the 2 and calculate the probability of team A scoring 0 goals into team B?

should the probability of A scoring 0 goals into team B not be the same as the probability of team B conceding 0 goals from team A?

Yes, the probability of A not scoring against B is the probability of B not conceding against A more or less.
(B could conceiveably concede against themselves ... an "own goal")
However the probability of A not scoring "in a game" is based on their fixture history,
if they failed to score in 30 of 100 on average, the probability of not scoring is 0.3
and the probability of B not conceding is based on their past games also.

However, this does not mean that the probability of A not scoring against B is 0.3..
Statistically speaking, we'd need to examine the previous games between the clubs
and many other current factors.

Is it as simple as p1*p2? I thought i might need to use conditional probability to work out the intersection of p1 and p2, but it doesnt seem to look right when I do it

Not in this case, since they are playing each other.
If they were playing different teams, then the probability of A not scoring and B not conceding is the
product of the probabilities, not taking into account the details of the opponents.

For example, if B have always failed to keep a clean sheet,
and A score in half their games,
would the probability of a 0-0 score be close to 0.5 ?

thanks for any help
neil

There are 4 possibilities

(1) A score and B score
(2) A score and B do not
(3) A do not score but B do
(4) neither score

You could sum the probabilities of (3) and (4), but since some of the data is not available

you need to calculate the probability that A do not score against B
"given that" the probability of B not conceding a goal is $p_2$
• November 20th 2010, 01:29 PM
neeeel
Quote:

you need to calculate the probability that A do not score against B
"given that" the probability of B not conceding a goal is $p_2$

this seems to imply using conditional probability

if p1 = the probability that A do not score
p2 = the probability of B not conceding a goal

P(p1 | p2 ) = P(p1 intersection p2) / p2

in this case though, i just end up with p1 as the answer, since (p1 intersection p2)/p2 = (p1 *p2)/p2 = p1

Im still confused as to how i get the probability that team A do not score into team B
• November 20th 2010, 02:15 PM
Quote:

Originally Posted by neeeel
this seems to imply using conditional probability

if p1 = the probability that A do not score
p2 = the probability of B not conceding a goal

P(p1 | p2 ) = P(p1 intersection p2) / p2

in this case though, i just end up with p1 as the answer, since (p1 intersection p2)/p2 = (p1 *p2)/p2 = p1

Im still confused as to how i get the probability that team A do not score into team B

Do you see how you lost the information about team B ?
You need to dig deeper.
• November 20th 2010, 03:19 PM
neeeel
I havent lost it, I never had it in the first place :S

I know u want me to work it out myself, I do too, but i need a push in the right direction
neil
• November 20th 2010, 03:43 PM
Quote:

Originally Posted by neeeel
I havent lost it, I never had it in the first place :S

I know u want me to work it out myself, I do too, but i need a push in the right direction
neil

What I mean is that the probability $p_1$ is just the ongoing average chance
of Team A not scoring.

That is not the probability of Team A not scoring against Team B.

The solution will involve $p_2$

so the solution tried does not work.
It surely requires $p_2$ to be part of the solution, not cancelled.

The probability of Team A not scoring if Team B do not concede is 1.
The probability of Team B not conceding if Team A does not score is also 1
(own goals ignored).

P(AB) cannot be the multiplication of the probabilities because in this game
Team A not scoring and Team B not conceding are not 2 events.
• November 21st 2010, 03:03 AM
neeeel
Only thing I can come up with is taking the mean, ie $(p_1 + p_2)/2$ but i dont think thats a very good solution
• November 21st 2010, 05:28 AM
Quote:

Originally Posted by neeeel
Only thing I can come up with is taking the mean, ie $(p_1 + p_2)/2$ but i dont think thats a very good solution

The reason the conditional probability solution you tried earlier doesn't work
is.... it applies to A and B playing in different games.
The probability of A not scoring against an arbitrary team
given that B did not concede against another arbitrary team
will end up giving you the probability that A do not score on average
if you use that conditional probability formula.
The events are then independent and multiplying the probabilities is feasible
to give the probability of both of 2 events happening.

In this case there is a complete overlap between A not scoring and B not conceding.
Also, if the probability of B not conceding is very small,
then the probability of A not scoring against B is much lower than
if the probability of B not conceding is very high.
• November 21st 2010, 05:55 AM
neeeel
ok im totally confused now

In an earlier post, you said that multiplying the probabilities was not the solution

now you are saying that it is the solution?

so

probability that A doesnt score into B = probability that A doesnt score * probability that B doesnt concede???
• November 21st 2010, 06:41 AM