Mixing
Probability of football player scoring 2 goals, 3+ goals or anytime given his prob to score if goal is being...
$begingroup$
Assuming that I have a football match, and each team has 11 players. I know for any given goal the probability of scoring for each player (for simplicity these probs remain constant during the game).
What I want to know is how to calculate the probability for each player (all 22 players) of scoring 2 goals, 3 + goals or score anytime.
My initial idea was to use binomial distribution, but I don't know the number of trials.
It is my understanding that:
Pi(0): is the probability that player i scores 0 goals
Pi(1): ---------------------------------------1 goal
Pi(2): ---------------------------------------2 goals
1 - Pi(0): the probability of player i scoring any time (at least one goal)
1 - Pi(0) - Pi(1) - Pi(2): probability of player i scoring 3 or more goals
So any idea how to calculate this?
I can have the probabilities of a goal being scored by each player (p1, p2, ... p22) and have them sum up to 1, or I can have the probability of match ending up 0-0 (no one scored) and have those probs sum up to 1 (p1+p2+...+p22 + p(0-0))
What else do I need to know?
If I knew the probability of having 1 goal for home team, 2 goals, 3 goals etc and then for away team respectively how would that help?
thanks in advance
probability probability-theory probability-distributions binomial-distribution
asked Jan 3 at 15:17
TomTom
1035
$endgroup$
add a comment |
$begingroup$
Assuming that I have a football match, and each team has 11 players. I know for any given goal the probability of scoring for each player (for simplicity these probs remain constant during the game).
What I want to know is how to calculate the probability for each player (all 22 players) of scoring 2 goals, 3 + goals or score anytime.
My initial idea was to use binomial distribution, but I don't know the number of trials.
It is my understanding that:
Pi(0): is the probability that player i scores 0 goals
Pi(1): ---------------------------------------1 goal
Pi(2): ---------------------------------------2 goals
1 - Pi(0): the probability of player i scoring any time (at least one goal)
1 - Pi(0) - Pi(1) - Pi(2): probability of player i scoring 3 or more goals
So any idea how to calculate this?
I can have the probabilities of a goal being scored by each player (p1, p2, ... p22) and have them sum up to 1, or I can have the probability of match ending up 0-0 (no one scored) and have those probs sum up to 1 (p1+p2+...+p22 + p(0-0))
What else do I need to know?
If I knew the probability of having 1 goal for home team, 2 goals, 3 goals etc and then for away team respectively how would that help?
thanks in advance
probability probability-theory probability-distributions binomial-distribution
asked Jan 3 at 15:17
TomTom
1035
$endgroup$
add a comment |
$begingroup$
Assuming that I have a football match, and each team has 11 players. I know for any given goal the probability of scoring for each player (for simplicity these probs remain constant during the game).
What I want to know is how to calculate the probability for each player (all 22 players) of scoring 2 goals, 3 + goals or score anytime.
My initial idea was to use binomial distribution, but I don't know the number of trials.
It is my understanding that:
Pi(0): is the probability that player i scores 0 goals
Pi(1): ---------------------------------------1 goal
Pi(2): ---------------------------------------2 goals
1 - Pi(0): the probability of player i scoring any time (at least one goal)
1 - Pi(0) - Pi(1) - Pi(2): probability of player i scoring 3 or more goals
So any idea how to calculate this?
I can have the probabilities of a goal being scored by each player (p1, p2, ... p22) and have them sum up to 1, or I can have the probability of match ending up 0-0 (no one scored) and have those probs sum up to 1 (p1+p2+...+p22 + p(0-0))
What else do I need to know?
If I knew the probability of having 1 goal for home team, 2 goals, 3 goals etc and then for away team respectively how would that help?
thanks in advance
probability probability-theory probability-distributions binomial-distribution
asked Jan 3 at 15:17
TomTom
1035
$endgroup$
Assuming that I have a football match, and each team has 11 players. I know for any given goal the probability of scoring for each player (for simplicity these probs remain constant during the game).
What I want to know is how to calculate the probability for each player (all 22 players) of scoring 2 goals, 3 + goals or score anytime.
My initial idea was to use binomial distribution, but I don't know the number of trials.
It is my understanding that:
Pi(0): is the probability that player i scores 0 goals
Pi(1): ---------------------------------------1 goal
Pi(2): ---------------------------------------2 goals
1 - Pi(0): the probability of player i scoring any time (at least one goal)
1 - Pi(0) - Pi(1) - Pi(2): probability of player i scoring 3 or more goals
So any idea how to calculate this?
I can have the probabilities of a goal being scored by each player (p1, p2, ... p22) and have them sum up to 1, or I can have the probability of match ending up 0-0 (no one scored) and have those probs sum up to 1 (p1+p2+...+p22 + p(0-0))
What else do I need to know?
If I knew the probability of having 1 goal for home team, 2 goals, 3 goals etc and then for away team respectively how would that help?
thanks in advance
probability probability-theory probability-distributions binomial-distribution
probability probability-theory probability-distributions binomial-distribution
asked Jan 3 at 15:17
TomTom
1035
asked Jan 3 at 15:17
TomTom
1035
asked Jan 3 at 15:17
TomTom
1035
asked Jan 3 at 15:17
TomTom
1035
asked Jan 3 at 15:17
TomTom
1035
1035
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The number of trials is $1$ for each player. The chance of one goal is the chance one player scores one goal and all the others score zero, so
$$P(1)=sum_ileft(P_i(1)prod_{j neq i}P_j(0)right)$$
The chance of two goals is the chance one player scores two goals and nobody else scores plus the chance that two different players score one goal each and nobody else scores at all
$$P(2)=sum_ileft(P_i(2)prod_{j neq i}P_j(0)right)+sum_{stackrel {i,j}{i neq j}}left(P_i(1)P_j(1)prod_{k neq i,j}P_k(0)right)$$
It gets more complicated as you add more goals.
This assumes that the chance of one player scoring is independent of the others. Maybe if one player scores the defense is poor and his teammates have a higher chance of scoring. Maybe if the team gives up a goal they play harder, or get discouraged, or something.
The above seems to be going the opposite direction from what you wanted. If you know the chance that each goal is scored by player $i$, call it $p_i$, and the chance of $n$ goals being scored, which is the $P(n)$ above we are trying to calculate the chance player $i$ scored $k$ goals, which is the above $P_i(k)$. If just one goal is scored, we have $P_i(1)=p_i$. In the spirit of generating functions we can use the fact that $sum_ip_i=1$ and raise that the power of $n$, the number of goals that are scored. For $2$ the chance that player $i$ scored them both is $p_i^2$. The chance that players $i$ and $j$ each scored one is $2p_ip_j$. The chance that player $i$ scored exactly one goal is then $$sum_{stackrel {j}{jneq i}}2p_ip_j=2p_i(1-p_i)={2 choose 1}p_i^1(1-p_i)^{(2-1)}$$
The number of trials is just the number of goals scored by the side. If $n$ goals are scored, the chance player $i$ has scored exactly $k$ of them is $${n choose k}p_i^k(1-p_i)^{n-k}$$
answered Jan 3 at 15:25
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
thanks for replying. I need the probability for player i to score 1 goal, not 1 goal in total. This is why I said I don't know the number of trials. I have the prob of success for player i, and I need to calculate the prob of having 1 success , 2 successes etc
$endgroup$
– Tom
Jan 3 at 15:45
$begingroup$
So, if i want the Pi(0) i shloud get something like: P(0)* Pi(0)+P(1)*Pi(0)+....+P(n)*Pi(0) ? And following that rational: for i player scoring 1 goal: P(1)* Pi(1) + P(2)*Pi(1)+...+P(n)*Pi(1) ?
$endgroup$
– Tom
Jan 3 at 17:36
1
$begingroup$
If you want the probability the player scores at least one goal, compute the chance he does not score any and subtract from $1$. You would have to compute the chance he does not score any given the team scores $1,2,3,ldots$ and add them up.
$endgroup$
– Ross Millikan
Jan 3 at 18:05
add a comment |
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1 Answer
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$begingroup$
The number of trials is $1$ for each player. The chance of one goal is the chance one player scores one goal and all the others score zero, so
$$P(1)=sum_ileft(P_i(1)prod_{j neq i}P_j(0)right)$$
The chance of two goals is the chance one player scores two goals and nobody else scores plus the chance that two different players score one goal each and nobody else scores at all
$$P(2)=sum_ileft(P_i(2)prod_{j neq i}P_j(0)right)+sum_{stackrel {i,j}{i neq j}}left(P_i(1)P_j(1)prod_{k neq i,j}P_k(0)right)$$
It gets more complicated as you add more goals.
This assumes that the chance of one player scoring is independent of the others. Maybe if one player scores the defense is poor and his teammates have a higher chance of scoring. Maybe if the team gives up a goal they play harder, or get discouraged, or something.
The above seems to be going the opposite direction from what you wanted. If you know the chance that each goal is scored by player $i$, call it $p_i$, and the chance of $n$ goals being scored, which is the $P(n)$ above we are trying to calculate the chance player $i$ scored $k$ goals, which is the above $P_i(k)$. If just one goal is scored, we have $P_i(1)=p_i$. In the spirit of generating functions we can use the fact that $sum_ip_i=1$ and raise that the power of $n$, the number of goals that are scored. For $2$ the chance that player $i$ scored them both is $p_i^2$. The chance that players $i$ and $j$ each scored one is $2p_ip_j$. The chance that player $i$ scored exactly one goal is then $$sum_{stackrel {j}{jneq i}}2p_ip_j=2p_i(1-p_i)={2 choose 1}p_i^1(1-p_i)^{(2-1)}$$
The number of trials is just the number of goals scored by the side. If $n$ goals are scored, the chance player $i$ has scored exactly $k$ of them is $${n choose k}p_i^k(1-p_i)^{n-k}$$
answered Jan 3 at 15:25
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
thanks for replying. I need the probability for player i to score 1 goal, not 1 goal in total. This is why I said I don't know the number of trials. I have the prob of success for player i, and I need to calculate the prob of having 1 success , 2 successes etc
$endgroup$
– Tom
Jan 3 at 15:45
$begingroup$
So, if i want the Pi(0) i shloud get something like: P(0)* Pi(0)+P(1)*Pi(0)+....+P(n)*Pi(0) ? And following that rational: for i player scoring 1 goal: P(1)* Pi(1) + P(2)*Pi(1)+...+P(n)*Pi(1) ?
$endgroup$
– Tom
Jan 3 at 17:36
1
$begingroup$
If you want the probability the player scores at least one goal, compute the chance he does not score any and subtract from $1$. You would have to compute the chance he does not score any given the team scores $1,2,3,ldots$ and add them up.
$endgroup$
– Ross Millikan
Jan 3 at 18:05
add a comment |
$begingroup$
The number of trials is $1$ for each player. The chance of one goal is the chance one player scores one goal and all the others score zero, so
$$P(1)=sum_ileft(P_i(1)prod_{j neq i}P_j(0)right)$$
The chance of two goals is the chance one player scores two goals and nobody else scores plus the chance that two different players score one goal each and nobody else scores at all
$$P(2)=sum_ileft(P_i(2)prod_{j neq i}P_j(0)right)+sum_{stackrel {i,j}{i neq j}}left(P_i(1)P_j(1)prod_{k neq i,j}P_k(0)right)$$
It gets more complicated as you add more goals.
This assumes that the chance of one player scoring is independent of the others. Maybe if one player scores the defense is poor and his teammates have a higher chance of scoring. Maybe if the team gives up a goal they play harder, or get discouraged, or something.
The above seems to be going the opposite direction from what you wanted. If you know the chance that each goal is scored by player $i$, call it $p_i$, and the chance of $n$ goals being scored, which is the $P(n)$ above we are trying to calculate the chance player $i$ scored $k$ goals, which is the above $P_i(k)$. If just one goal is scored, we have $P_i(1)=p_i$. In the spirit of generating functions we can use the fact that $sum_ip_i=1$ and raise that the power of $n$, the number of goals that are scored. For $2$ the chance that player $i$ scored them both is $p_i^2$. The chance that players $i$ and $j$ each scored one is $2p_ip_j$. The chance that player $i$ scored exactly one goal is then $$sum_{stackrel {j}{jneq i}}2p_ip_j=2p_i(1-p_i)={2 choose 1}p_i^1(1-p_i)^{(2-1)}$$
The number of trials is just the number of goals scored by the side. If $n$ goals are scored, the chance player $i$ has scored exactly $k$ of them is $${n choose k}p_i^k(1-p_i)^{n-k}$$
answered Jan 3 at 15:25
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
thanks for replying. I need the probability for player i to score 1 goal, not 1 goal in total. This is why I said I don't know the number of trials. I have the prob of success for player i, and I need to calculate the prob of having 1 success , 2 successes etc
$endgroup$
– Tom
Jan 3 at 15:45
$begingroup$
So, if i want the Pi(0) i shloud get something like: P(0)* Pi(0)+P(1)*Pi(0)+....+P(n)*Pi(0) ? And following that rational: for i player scoring 1 goal: P(1)* Pi(1) + P(2)*Pi(1)+...+P(n)*Pi(1) ?
$endgroup$
– Tom
Jan 3 at 17:36
1
$begingroup$
If you want the probability the player scores at least one goal, compute the chance he does not score any and subtract from $1$. You would have to compute the chance he does not score any given the team scores $1,2,3,ldots$ and add them up.
$endgroup$
– Ross Millikan
Jan 3 at 18:05
add a comment |
$begingroup$
The number of trials is $1$ for each player. The chance of one goal is the chance one player scores one goal and all the others score zero, so
$$P(1)=sum_ileft(P_i(1)prod_{j neq i}P_j(0)right)$$
The chance of two goals is the chance one player scores two goals and nobody else scores plus the chance that two different players score one goal each and nobody else scores at all
$$P(2)=sum_ileft(P_i(2)prod_{j neq i}P_j(0)right)+sum_{stackrel {i,j}{i neq j}}left(P_i(1)P_j(1)prod_{k neq i,j}P_k(0)right)$$
It gets more complicated as you add more goals.
This assumes that the chance of one player scoring is independent of the others. Maybe if one player scores the defense is poor and his teammates have a higher chance of scoring. Maybe if the team gives up a goal they play harder, or get discouraged, or something.
The above seems to be going the opposite direction from what you wanted. If you know the chance that each goal is scored by player $i$, call it $p_i$, and the chance of $n$ goals being scored, which is the $P(n)$ above we are trying to calculate the chance player $i$ scored $k$ goals, which is the above $P_i(k)$. If just one goal is scored, we have $P_i(1)=p_i$. In the spirit of generating functions we can use the fact that $sum_ip_i=1$ and raise that the power of $n$, the number of goals that are scored. For $2$ the chance that player $i$ scored them both is $p_i^2$. The chance that players $i$ and $j$ each scored one is $2p_ip_j$. The chance that player $i$ scored exactly one goal is then $$sum_{stackrel {j}{jneq i}}2p_ip_j=2p_i(1-p_i)={2 choose 1}p_i^1(1-p_i)^{(2-1)}$$
The number of trials is just the number of goals scored by the side. If $n$ goals are scored, the chance player $i$ has scored exactly $k$ of them is $${n choose k}p_i^k(1-p_i)^{n-k}$$
answered Jan 3 at 15:25
Ross MillikanRoss Millikan
293k23197371
$endgroup$
The number of trials is $1$ for each player. The chance of one goal is the chance one player scores one goal and all the others score zero, so
$$P(1)=sum_ileft(P_i(1)prod_{j neq i}P_j(0)right)$$
The chance of two goals is the chance one player scores two goals and nobody else scores plus the chance that two different players score one goal each and nobody else scores at all
$$P(2)=sum_ileft(P_i(2)prod_{j neq i}P_j(0)right)+sum_{stackrel {i,j}{i neq j}}left(P_i(1)P_j(1)prod_{k neq i,j}P_k(0)right)$$
It gets more complicated as you add more goals.
This assumes that the chance of one player scoring is independent of the others. Maybe if one player scores the defense is poor and his teammates have a higher chance of scoring. Maybe if the team gives up a goal they play harder, or get discouraged, or something.
The above seems to be going the opposite direction from what you wanted. If you know the chance that each goal is scored by player $i$, call it $p_i$, and the chance of $n$ goals being scored, which is the $P(n)$ above we are trying to calculate the chance player $i$ scored $k$ goals, which is the above $P_i(k)$. If just one goal is scored, we have $P_i(1)=p_i$. In the spirit of generating functions we can use the fact that $sum_ip_i=1$ and raise that the power of $n$, the number of goals that are scored. For $2$ the chance that player $i$ scored them both is $p_i^2$. The chance that players $i$ and $j$ each scored one is $2p_ip_j$. The chance that player $i$ scored exactly one goal is then $$sum_{stackrel {j}{jneq i}}2p_ip_j=2p_i(1-p_i)={2 choose 1}p_i^1(1-p_i)^{(2-1)}$$
The number of trials is just the number of goals scored by the side. If $n$ goals are scored, the chance player $i$ has scored exactly $k$ of them is $${n choose k}p_i^k(1-p_i)^{n-k}$$
answered Jan 3 at 15:25
Ross MillikanRoss Millikan
293k23197371
edited Jan 3 at 17:20
answered Jan 3 at 15:25
Ross MillikanRoss Millikan
293k23197371
answered Jan 3 at 15:25
Ross MillikanRoss Millikan
293k23197371
answered Jan 3 at 15:25
Ross MillikanRoss Millikan
293k23197371
293k23197371
$begingroup$
thanks for replying. I need the probability for player i to score 1 goal, not 1 goal in total. This is why I said I don't know the number of trials. I have the prob of success for player i, and I need to calculate the prob of having 1 success , 2 successes etc
$endgroup$
– Tom
Jan 3 at 15:45
$begingroup$
So, if i want the Pi(0) i shloud get something like: P(0)* Pi(0)+P(1)*Pi(0)+....+P(n)*Pi(0) ? And following that rational: for i player scoring 1 goal: P(1)* Pi(1) + P(2)*Pi(1)+...+P(n)*Pi(1) ?
$endgroup$
– Tom
Jan 3 at 17:36
1
$begingroup$
If you want the probability the player scores at least one goal, compute the chance he does not score any and subtract from $1$. You would have to compute the chance he does not score any given the team scores $1,2,3,ldots$ and add them up.
$endgroup$
– Ross Millikan
Jan 3 at 18:05
add a comment |
$begingroup$
thanks for replying. I need the probability for player i to score 1 goal, not 1 goal in total. This is why I said I don't know the number of trials. I have the prob of success for player i, and I need to calculate the prob of having 1 success , 2 successes etc
$endgroup$
– Tom
Jan 3 at 15:45
$begingroup$
So, if i want the Pi(0) i shloud get something like: P(0)* Pi(0)+P(1)*Pi(0)+....+P(n)*Pi(0) ? And following that rational: for i player scoring 1 goal: P(1)* Pi(1) + P(2)*Pi(1)+...+P(n)*Pi(1) ?
$endgroup$
– Tom
Jan 3 at 17:36
1
$begingroup$
If you want the probability the player scores at least one goal, compute the chance he does not score any and subtract from $1$. You would have to compute the chance he does not score any given the team scores $1,2,3,ldots$ and add them up.
$endgroup$
– Ross Millikan
Jan 3 at 18:05
$begingroup$
thanks for replying. I need the probability for player i to score 1 goal, not 1 goal in total. This is why I said I don't know the number of trials. I have the prob of success for player i, and I need to calculate the prob of having 1 success , 2 successes etc
$endgroup$
– Tom
Jan 3 at 15:45
$begingroup$
thanks for replying. I need the probability for player i to score 1 goal, not 1 goal in total. This is why I said I don't know the number of trials. I have the prob of success for player i, and I need to calculate the prob of having 1 success , 2 successes etc
$endgroup$
– Tom
Jan 3 at 15:45
$begingroup$
So, if i want the Pi(0) i shloud get something like: P(0)* Pi(0)+P(1)*Pi(0)+....+P(n)*Pi(0) ? And following that rational: for i player scoring 1 goal: P(1)* Pi(1) + P(2)*Pi(1)+...+P(n)*Pi(1) ?
$endgroup$
– Tom
Jan 3 at 17:36
$begingroup$
So, if i want the Pi(0) i shloud get something like: P(0)* Pi(0)+P(1)*Pi(0)+....+P(n)*Pi(0) ? And following that rational: for i player scoring 1 goal: P(1)* Pi(1) + P(2)*Pi(1)+...+P(n)*Pi(1) ?
$endgroup$
– Tom
Jan 3 at 17:36
1
1
$begingroup$
If you want the probability the player scores at least one goal, compute the chance he does not score any and subtract from $1$. You would have to compute the chance he does not score any given the team scores $1,2,3,ldots$ and add them up.
$endgroup$
– Ross Millikan
Jan 3 at 18:05
$begingroup$
If you want the probability the player scores at least one goal, compute the chance he does not score any and subtract from $1$. You would have to compute the chance he does not score any given the team scores $1,2,3,ldots$ and add them up.
$endgroup$
– Ross Millikan
Jan 3 at 18:05
add a comment |
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