Probability of winning a card game

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Gudjon.Smith

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In a Best of Five card game Player A is playing against Player B. First Player to win 3 sets wins the game. The probability that player A wins a set is 80%.
1)Calculate the probability that player A wins the game.
2)Calculate the probability that player B wins the game.


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The second part of the exercise is easy. It's 1-(probability of A winning the game). The problem is that I'm struggling with the first part of the exercise. First I thought I had to use a simple negative binomial distribution function. But the problem here is that there are 3 different cases I have to consider: Player A winning 3 sets in a row; Player A winning 3 sets and losing 1; Player A winning 3 sets and losing 2. Therefore the negative binomial distribution function doesn't help me to solve this problem. Any suggestions?

Unless I'm overthinking this and all I need is to use the negative binomial distribution after all. I just want to be sure.
 
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Describe the event that player A wins in detail. We will proceed from there after you post your answer.
 
Gudjon.Smith said:
In a Best of Five card game Player A is playing against Player B. First Player to win 3 sets wins the game. The probability that player A wins a set is 80%.
1)Calculate the probability that player A wins the game.
2)Calculate the probability that player B wins the game.
Click to expand...
This is an interesting exercise in counting. The setup is that the game is over in five rounds.
Player \(A\) can win after three hands: Three straight with probability \(\mathcal{P}(A=3)=(0.8)^3\)
Player \(A\) can win after four hands: loosing once in the first three and a win on game four with probability \(\mathcal{P}(A=4)=\dbinom{3}{1}(0.8)^3(0.2)\).
Player \(A\) can win after five hands: loosing twice in the first four and a win on game the fifth game with probability \(\mathcal{P}(A=5)=\dbinom{4}{2}(0.8)^3(0.2)^2\).
 
pka said:
This is an interesting exercise in counting. The setup is that the game is over in five rounds.
Player \(A\) can win after three hands: Three straight with probability \(\mathcal{P}(A=3)=(0.8)^3\)
Player \(A\) can win after four hands: loosing once in the first three and a win on game four with probability \(\mathcal{P}(A=4)=\dbinom{3}{1}(0.8)^3(0.2)\).
Player \(A\) can win after five hands: loosing twice in the first four and a win on game the fifth game with probability \(\mathcal{P}(A=5)=\dbinom{4}{2}(0.8)^3(0.2)^2\).
Click to expand...

Thank you for your answers.
This is how far I came with my answer to the question. The only difference is that for the case in which Player A can win after four hands: loosing once in the first three and a win on game four, I found out that the probability is \(\mathcal{P}(A=4)=\dbinom{3}{2}(0.8)^3(0.2)\) instead of \(\mathcal{P}(A=4)=\dbinom{3}{1}(0.8)^3(0.2)\), so I don't know why my answer is not correct. But this is the part where I'm struggling. I don't know if the exercise is over and all I can do is consider these 3 cases separately or if there is a mathematical formula that allows me to consider these 3 cases together to find 1 final answer to the question.

After thinking for a little while I'm asking myself if I can use the total probability formula to find the final answer. Is my reasoning correct?
 
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Gudjon.Smith said:
Thank you for your answers.
This is how far I came with my answer to the question. The only difference is that for the case in which Player A can win after four hands: loosing once in the first three and a win on game four, I found out that the probability is \(\mathcal{P}(A=4)=\dbinom{3}{2}(0.8)^3(0.2)\) instead of \(\mathcal{P}(A=4)=\dbinom{3}{1}(0.8)^3(0.2)\), so I don't know why my answer is not correct. But this is the part where I'm struggling. I don't know if the exercise is over and all I can do is consider these 3 cases separately or if there is a mathematical formula that allows me to consider these 3 cases together to find 1 final answer to the question.
Click to expand...
Please note that \(\dbinom{3}{1}=\dbinom{3}{2}\).
 
pka said:
Please note that \(\dbinom{3}{1}=\dbinom{3}{2}\).
Click to expand...

Oh yes you're right I'm sorry. By the way I'm still wondering if it's correct to use the total probability formula afterwards or if the exercise is already finished? I have been looking for similar exercises but couldn't find any therefore I don't know if I finished the exercise or if there is more I need to calculate.
 
What do you mean by "the total probability formula "?
 
pka said:
What do you mean by "the total probability formula "?
Click to expand...

Maybe it's called "law of total probability". My probability/statistics course is in french so I might have translated it wrong into english. Sorry.
Because now I know the probabilities of A winning the game for each of the 3 cases calculated above. So by using this law maybe I can calculate the probability of Player A winning the game by considering the 3 cases together. But I don't know if this would be correct.
 
You need to understand what the event A wins really means.
A wins means A wins after 3 games or A wins after 4 games or A wins after 5 games.

So P(A) = P(A wins after 3 games or A wins after 4 games or A wins after 5 games). How can you compute this?
 
Jomo said:
You need to understand what the event A wins really means.
A wins means A wins after 3 games or A wins after 4 games or A wins after 5 games.

So P(A) = P(A wins after 3 games or A wins after 4 games or A wins after 5 games). How can you compute this?
Click to expand...

I would say:

P(A)=P(A wins after 3 games | 3 games are played) x P(3 games are played) + P(A wins after 4 games | 4 games are playes) x P(4 games are played) + P(A wins after 5 games | 5 games are played) x P(5 games are played)

But I don't have the required information to calculate this so this can't be the right way of solving the problem. I've been thinking this through for 2 days now and I can't come up with another way to calculate the probability of A winning the game. I'm out of ideas and the more I think about it the more I get confused.
 
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A wins after 3 games, A wins after 4 games and A wins after 5 games are pairwise mutually exclusive events (lets not get into this one again!).

So P(A wins after 3 games or A wins after 4 games or A wins after 5 games) = P(A wins after 3 games) + P(A wins after 4 games) + P(A wins after 5 games).

So yes, you just add the three probabilities. Knowing how to understand an event is key to doing probability problems.
 
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