Probability in a football game

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Fahim Shahriar
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Probability in a football game

Unread post by Fahim Shahriar » Fri Dec 21, 2012 1:29 am

There is match between $A$ and $B$.
Formation of team $A$ $(3-4-3)$ and Team $B$ $(2-4-4)$.{That means team $B$ has 2 strikers,4 mid-fielders and 4 defenders.}

In this game, only the strikers and the mid-fielders can score goals. But the mid-fielders can't score more than $1$ goal.

Now, Team $A$ will score 4 goals and Team $B$ will score 3 goals.

Player of which team has better probability of making a HAT-TRICK ?
Name: Fahim Shahriar Shakkhor
Notre Dame College

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Fahim Shahriar
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Re: Probability in a football game

Unread post by Fahim Shahriar » Fri Dec 21, 2012 1:35 am

This is a self-made problem. I got the idea of making such problem while playing FIFA game. Adding some conditions, I tried to make it an interesting one.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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kfoozminus
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Re: Probability in a football game

Unread post by kfoozminus » Sun Dec 23, 2012 1:25 am

Team-$A$
$P(A)=\frac{1}{9}$
$P(B)=\frac{1}{12}$
i'm not writing my full solution 'cause i'm so confused about that.

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Fahim Shahriar
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Re: Probability in a football game

Unread post by Fahim Shahriar » Mon Dec 24, 2012 1:42 am

My solution is nothing but a huge calculation.

For Team $A$, when the 3 strikers scored all the goals,there are the possibilities how they can score goals.
[number of possibilities has been marked in () and hat-trick as {H}]

4-0-0 »(3) ways {H}
3-1-0 »(6) {H}
2-2-0 »(3)
2-1-1 »(3)

Now when the midfielders score only 1 goal, strikers score 3 goals as follows.
3-0-0 »3 {H}, midfielders 4 ways. (12 ways)
2-1-0 »6 (24)
1-1-1 »1 (4)

Finally, I got A»21/104 & B»1/16
Name: Fahim Shahriar Shakkhor
Notre Dame College

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kfoozminus
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Re: Probability in a football game

Unread post by kfoozminus » Mon Dec 24, 2012 11:53 am

find my fault...
Team $A$::
among $4$ goals, $1$ goal can be scored in $7$ ways($3$ strikers and $4$ midfielders)
and each of other $3$ goals can be scored in $3$ ways(only $3$ strikers)
so they can score $4$ goals in $7 \cdot 3^3 = 189 $ ways

strikers can make hat-trick,
$*$when all goals are scored by $1$ striker - in $3$ ways
$*$when $3$ goals by each striker and left $1$ goal by any other strikers or mid-fielders - in $3 \cdot 6 = 18$ ways

$P(A)=\frac{21}{189}=\frac{1}{9}$

Team $B$::
among $3$ goals, $1$ goal can be scored in $6$ ways($2$ strikers and $4$ mid-fielders)
and each of other $2$ goals can be scored in $2$ ways(only $2$ strikers)
so they can score $3$ goals in $6 \cdot 2^2 = 24$ ways

strikers can make hat-trick,
$*$when all goals are scored by $1$ striker - in $2$ ways

$P(B)=\frac{2}{24}=\frac{1}{12}$

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Fahim Shahriar
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Re: Probability in a football game

Unread post by Fahim Shahriar » Mon Dec 24, 2012 2:03 pm

There is a little misunderstanding.
I meant each of mid-fielders can't score more than 1 goal.
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Notre Dame College

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kfoozminus
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Re: Probability in a football game

Unread post by kfoozminus » Mon Dec 24, 2012 3:55 pm

Fahim Shahriar wrote:There is a little misunderstanding.
I meant each of mid-fielders can't score more than 1 goal.
i did consider this condition!

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nafistiham
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Re: Probability in a football game

Unread post by nafistiham » Fri Dec 28, 2012 9:10 pm

kfoozminus wrote:find my fault...
Team $A$::
among $4$ goals, $1$ goal can be scored in $7$ ways($3$ strikers and $4$ midfielders)
and each of other $3$ goals can be scored in $3$ ways(only $3$ strikers)

so they can score $4$ goals in $7 \cdot 3^3 = 189 $ ways

strikers can make hat-trick,
$*$when all goals are scored by $1$ striker - in $3$ ways
$*$when $3$ goals by each striker and left $1$ goal by any other strikers or mid-fielders - in $3 \cdot 6 = 18$ ways

$P(A)=\frac{21}{189}=\frac{1}{9}$

Team $B$::
among $3$ goals, $1$ goal can be scored in $6$ ways($2$ strikers and $4$ mid-fielders)
and each of other $2$ goals can be scored in $2$ ways(only $2$ strikers)

so they can score $3$ goals in $6 \cdot 2^2 = 24$ ways

strikers can make hat-trick,
$*$when all goals are scored by $1$ striker - in $2$ ways

$P(B)=\frac{2}{24}=\frac{1}{12}$
Why is that ?
what if each mid fielder scores $1$ goal ?
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kfoozminus
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Re: Probability in a football game

Unread post by kfoozminus » Fri Dec 28, 2012 9:32 pm

thanks for finding the bug... this was so annoying me...
hey, maybe you can look into this too viewtopic.php?f=28&t=2216
i'm not sure about the answer, and looks like nobody's interested in football!

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Fahim Shahriar
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Re: Probability in a football game

Unread post by Fahim Shahriar » Sat Dec 29, 2012 12:09 am

I couldn't post my full solution because my mobile supports limited texts.
And Alas! My computer is now in hospital for treatment.
.
.
You can say it a joke. :D
Name: Fahim Shahriar Shakkhor
Notre Dame College

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