Two teams, A and B, are playing a series of games

[coolusername]

My attempt

I used negative binomial to solve the problem, however I'm left with a polynomial that is difficult to solve? Is there any other way to approach this problem?

I used the inequality because I'm trying to find the range of p. Since the probability of winning the series for team A is written above, the probability should be more than 0.5 since that would mean it would be at an advantage. The sum of the probabilities that I've written in the image shows the probability that team A can win in 4 games , 5 games, 6 games or 7 games.

Is there a way to solve this polynomial?

 

[Ray Vickson]

View attachment 237408

My attempt

View attachment 237409

I used negative binomial to solve the problem, however I'm left with a polynomial that is difficult to solve? Is there any other way to approach this problem?

I used the inequality because I'm trying to find the range of p. Since the probability of winning the series for team A is written above, the probability should be more than 0.5 since that would mean it would be at an advantage. The sum of the probabilities that I've written in the image shows the probability that team A can win in 4 games , 5 games, 6 games or 7 games.

Is there a way to solve this polynomial?

I won't look at images of handwritten work, so I cannot comment on the details of your solution. However, I don't know why you want to use the negative binomial distribution. If I were doing the question that is not the distribution I would use.

 

[coolusername]

I won't look at images of handwritten work, so I cannot comment on the details of your solution. However, I don't know why you want to use the negative binomial distribution. If I were doing the question that is not the distribution I would use.

Since we're trying to find the kth success(4th game won) on nth trial(can be 4,5,6 or 7th game), why wouldn't the negative binomial distribution work for this problem? What distribution should I use?

 

[Ray Vickson]

Since we're trying to find the kth success(4th game won) on nth trial(can be 4,5,6 or 7th game), why wouldn't the negative binomial distribution work for this problem? What distribution should I use?

You are asking about the probability that team A wins at least 4 of the 7 games.

 

[coolusername]

You are asking about the probability that team A wins at least 4 of the 7 games.

A team can't win more than 4 games in a best of 7 series. The series would have stopped at when a team first reaches to 4 wins. Why would setting up the probability that Team A wins at least 4 games when at most you can win is 4 in a best of 4 of 7 game series?

 

[Ray Vickson]

A team can't win more than 4 games in a best of 7 series. The series would have stopped at when a team first reaches to 4 wins. Why would setting up the probability that Team A wins at least 4 games when at most you can win is 4 in a best of 4 of 7 game series?

My interpretation was that they played 7 games, but yours is that they stop when one team has won 4. You get the same tornament-win-probability for A in both cases, because instead of stopping at 4 won games the two teams can just continue to play some useless games until they have played a total of 7.

 

[PeroK]

View attachment 237408

My attempt

View attachment 237409

I used negative binomial to solve the problem, however I'm left with a polynomial that is difficult to solve? Is there any other way to approach this problem?

I used the inequality because I'm trying to find the range of p. Since the probability of winning the series for team A is written above, the probability should be more than 0.5 since that would mean it would be at an advantage. The sum of the probabilities that I've written in the image shows the probability that team A can win in 4 games , 5 games, 6 games or 7 games.

Is there a way to solve this polynomial?

Why do you need to do any calculations at all?

 

[coolusername]

Why do you need to do any calculations at all?

I don't quite understand your point. It's telling me to find the range of p.

 

[PeroK]

I don't quite understand your point. It's telling me to find the range of p.

Suppose $p=0.55$. Would team $A$ have the advantage in a single game? What about in a best of 3? Best of 5? Best of 7? Best of 101 games? Best of $2n+1$ games, for any $n$?

 

[coolusername]

Suppose $p=0.55$. Would team $A$ have the advantage in a single game? What about in a best of 3? Best of 5? Best of 7? Best of 101 games? Best of $2n+1$ games, for any $n$?

I can’t just plug in any p value and see if it works. Is there a way to prove this algebtaically or with an inequality? It would make sense that the probability of p would be more than 0.5 so that team A can have an advantage but I can’t just blatantly say that without proof..

 

[scottdave]

Suppose $p=0.55$. Would team $A$ have the advantage in a single game? What about in a best of 3? Best of 5? Best of 7? Best of 101 games? Best of $2n+1$ games, for any $n$?

I agree. Maybe I'm missing something. Wouldn't anything greater than 50% be "an advantage"? That's how the casinos run.

 

[PeroK]

I can’t just plug in any p value and see if it works. Is there a way to prove this algebtaically or with an inequality? It would make sense that the probability of p would be more than 0.5 so that team A can have an advantage but I can’t just blatantly say that without proof..

Possibly you could in this case simply state that $p = 0.5$ is an equal game and $p > 0.5$ then team A has the advantage.

But, if you did want to prove this algebraically, then I would suggest using induction.

 

[PeroK]

Possibly you could in this case simply state that $p = 0.5$ is an equal game and $p > 0.5$ then team A has the advantage.

But, if you did want to prove this algebraically, then I would suggest using induction.

Actually, a better idea might be to line up all the possible results - either for $n = 7$ or for a general $n$ - and show that team A is more likely to win by any particular score. I.e. that team A is more likely to win $4-0$ than team B winning $4-0$ etc.

 

[Ray Vickson]

I don't quite understand your point. It's telling me to find the range of p.

Start at $p = 1/2$ exactly when---probabilistically speaking---neither team has an advantage. Noting that one of the two teams must win the tournament, what is the probability that A wins? What is the probability that B wins?

Now increase $p$ to a value $> 1/2.$ What now would be A's probability of winning the tournament, compared with the $p = 1/2$ case? Would it be less? the same? more?

If you (or your instructor) really does insist on an algebraic demonstration, I suggest you say that
$$P_A \equiv P \{ \text{A wins} \} = P(X=4)+P(X=5)+P(X=6)+P(X=7)\hspace{4ex}(1)$$
where $X \sim \text{Binomial}(7,p)$. (You were skeptical of this, but look back at the last sentence in Post #6.) Expanding (1) gives a 7th degree polynomial for $P_A.$

In order to compare $p > 1/2$ with $p = 1/2$, it is very helpful to substitute $p = 1/2 + x$ into the polynomial $P_A(p)$ and expand it all out. That takes a few minutes by hand and an eye-blink's worth of time using a computer algebra package. The results are very revealing.

 

[StoneTemplePython]

A team can't win more than 4 games in a best of 7 series. The series would have stopped at when a team first reaches to 4 wins. Why would setting up the probability that Team A wins at least 4 games when at most you can win is 4 in a best of 4 of 7 game series?

stopping at 4 wins by a team is actually optional. If you aren't convinced by the suggestions so far, take a look at this thread from mid last year.

https://www.physicsforums.com/threa...g-10-times-in-12-matches.950124/#post-6016733

With a little insight you'll see it's the same as this one, except your problem is easier. In general having people play dummy games is a useful technique to learn though.

- - - -
The easier route has already been suggested but you want a more formal inequality, which is understandable. Here's an approach: use your current formulation where the game stops as soon as 4 wins are achieved by some team.

Now find a simple case you can easily solve directly (i.e. p = 0.5, by symmetry is 50:50 between A and B winning the match). The goal is to build your entire solution off of this simple case.

Now consider some inhomogenous case that is strictly better for team A but very comparable to the symmetric case (suggestion: p = 0.5 for games 1-4, and 5-6 if played, but $p = (0.5 +\delta)$ for game 7 if played, where $\delta \in (0, \frac{1}{2}])$. You should easily be able to decompose this into (states or sample paths or) events and show the probability of $A$ winning the match is strictly better than the case where it is $0.5$ for all games.

Now show that the probability of $A$ winning the match is at least this good when $p = (0.5 +\delta)$ for all matches -- either by a dominance argument all at once, or by iterating and nesting: e.g. show the case of $\frac{1}{2}$ for games 1-5 but 6 and 7 if played are $(\frac{1}{2} + \delta)$ and then consider the splits of $1-4, 5-7$, then splits of $1-3, 4-7$ and so on. It's more work to do theses nested inequalities but it may be more convincing to you.