Mastering Probability: Solving Football Penalty Shoot-Outs with Ease

  • Thread starter phalanx123
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In summary: Therefore, the probability that eventually one team or the other will win is given by:P(\text{one team or the other wins}) = \sum_{k=0}^\infty \left(\begin{array}{cc}10+2k+1\\k+1\end{array}\right) p^{k+1}(1-p)^{9+k} = 1This is because the sum is
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phalanx123
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Ok here is the question

If a football match ends in a draw, there may be a ‘penalty shoot-out’. In one
version of a penalty shoot-out, each team initially takes 5 shots at the goal. If one teambscores more times than the other, then that team wins.

(i) Two teams take part in a penalty shoot-out. The probability that any player scores from a single shot is p, where 0 < p < 1. Prove carefully that the probability, [tex]\alpha[/tex], that neither side has won at the end of the initial 10-shot period is given by
[tex]\alpha=\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n}[/tex]

In the case p = 1 2 , evaluate and find the probability that the first team to shoot wins at the end of the initial period.

(ii) If neither team has won at the end of the initial period, the teams take another shot each. If one team scores and the other does not, then the team that scores wins; if neither team wins, then the teams take another shot each and the process continues until one of the teams wins. Find the probability, for 0 < p < 1, that a total of 10 + 2k + 2 shots (k >= 0) are required to win the game. Verify that the probability that, eventually, one
team or the other will win is 1.


I got fine with the first part of the question,but stuck on part (ii). Here is my attempt to the solution

I started by argueing that
the propability that 12 shots are needed is [tex]2p(1-p)\alpha[/tex]
the propability that 14 shots are needed is [tex]2ppp(1-p)+2(1-p)(1-p)p(1-p)\alpha=2p(1-p)(p^2+(1-p)^2)\alpha[/tex]
the propability that 16 shots are needed is [tex]2ppppp(1-p)+2(1-p)(1-p)(1-p)(1-p)p(1-p)\alpha=2p(1-p)(p^4+(1-p)^4)\alpha[/tex]
and so on

so the probability p of 10+2k+2 shots are needed is
[tex]p=\left\{\begin{array}{cc}2p(1-p)\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n},&\mbox{ if } k=0\\2p(1-p)(p^{2i}+(1-p)^{2i})\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n}, & \mbox{ if } k>0\end{array}\right[/tex]
where i>=1

I am not sure about this answer as it doesn't look right to me and I don't know how to verify that the probability that eventually one team or the other will win is 1 either:frown: . Could somebody help me please? Thanks a million:redface:
 
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  • #2


Hello,

Thank you for your question. Let me first address your attempt at the solution for part (ii). Your reasoning is correct, but there are a few small errors in your calculations. First, the probability that 12 shots are needed is actually 2p(1-p)^2 \alpha, not 2p(1-p)\alpha. Similarly, the probability that 14 shots are needed is 2p^2(1-p)^3 \alpha, and the probability that 16 shots are needed is 2p^3(1-p)^4 \alpha.

Overall, the probability that 10+2k+2 shots are needed is given by:

p = \left\{\begin{array}{cc}2p(1-p)^2\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n},&\mbox{ if } k=0\\2p^i(1-p)^{i+2}\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n}, & \mbox{ if } k>0\end{array}\right

Note that the power of p in the second term should be i, not 2i. Also, the sum should be multiplied by (1-p)^2, not (1-p).

To verify that the probability that eventually one team or the other will win is 1, we can use the following reasoning:

If 10+2k+2 shots are needed to win, then the game must have gone through 10+2k+1 shots without a winner. This means that both teams must have scored the same number of times (k+1) out of the first 10+2k+1 shots. The probability of this happening is given by:

P(\text{both teams score k+1 times out of 10+2k+1 shots}) = \left(\begin{array}{cc}10+2k+1\\k+1\end{array}\right) p^{k+1}(1-p)^{9+k}

This is because there are \left(\begin{array}{cc}
 

What is probability?

Probability is the likelihood or chance of an event occurring. It is often expressed as a number between 0 and 1, with 0 representing impossibility and 1 representing certainty.

How is probability used in football penalty shoot-outs?

In football penalty shoot-outs, probability is used to predict the outcome of each penalty kick and determine which team has a higher chance of winning. It is also used to strategize and make decisions on which players should take the penalties.

What factors affect the probability of scoring a penalty kick?

The probability of scoring a penalty kick is affected by various factors such as the skill level and strategy of the player, the goalkeeper's performance, the pressure and stress of the situation, and the weather conditions.

Can probability be manipulated to increase the chances of winning a penalty shoot-out?

Yes, probability can be manipulated by using statistical analysis and data to determine the best strategy for taking penalty kicks. This can increase the chances of winning a penalty shoot-out.

Are there any other applications of probability in sports?

Yes, probability is commonly used in sports to predict the outcomes of games and tournaments, evaluate player performance, and make strategic decisions. It is also used in sports betting and fantasy sports.

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