# someone pls help, question!

• July 7th 2013, 02:18 PM
whackg
someone pls help, question!
hey guys, just a question if anyone can help or provide some suggestions to approach
Let's say there is a team who you expect to win 75% of its games in a given 82-game sports regular season (and the probability of winning each game = 75%). If you wanted to determine the probability that the team will never lose consecutive games at any point during this 82-game season, what is an approach (or a few approaches) you may use to solve the problem?
• July 7th 2013, 02:55 PM
Plato
Re: someone pls help, question!
Quote:

Originally Posted by whackg
hey guys, just a question if anyone can help or provide some suggestions to approach
Let's say there is a team who you expect to win 75% of its games in a given 82-game sports regular season (and the probability of winning each game = 75%). If you wanted to determine the probability that the team will never lose consecutive games at any point during this 82-game season, what is an approach (or a few approaches) you may use to solve the problem?

I am not sure that I really fully understand the setup here.
Suppose you have either a win or an loss from each of the 82 games with $\mathcal{P}(W)=0.75~\&~\mathcal{P}(L)=0.25$.

You can think of the outcome space as a string of 82 $W's~\&~L's$.
In to have no two consecutive $L's$ we must have at least $41~W's$.
So let $K$ equal the number of wins then $41\le K\le 82$ then $\mathcal{P}(K)=\binom{K+1}{82-K}(0.75)^K(0.25)^{82-K}$
The is the probability of $K$ wins with no two consecutive $L's$.

So to find the probability of a season with no two consecutive loses add those for $K=41\text{ to }82.$.
• July 7th 2013, 03:31 PM
whackg
Re: someone pls help, question!
thanks for the quick response. I'm not sure if I understand your answer correctly, but that seems like there could still be possibility of consecutive losses. Just because the team wins atleast 41 out of 82 does not mean there was no consecutive losses during the stretch. For example, team could have started out season with 10 straight losses but finish the year 72-0 just as an example. Or am I looking at it the wrong way?
• July 7th 2013, 04:55 PM
Plato
Re: someone pls help, question!
Quote:

Originally Posted by whackg
thanks for the quick response. I'm not sure if I understand your answer correctly, but that seems like there could still be possibility of consecutive losses. Just because the team wins atleast 41 out of 82 does not mean there was no consecutive losses during the stretch. For example, team could have started out season with 10 straight losses but finish the year 72-0 just as an example. Or am I looking at it the wrong way?

Now it is quite possible that one of us does not understand the other.
I might tell you that I taught probability theory for over thirty years.

I carefully explained in my first reply how I understood the question.
Did you read it carefully? It does not seem that you did!

If your team wins only 41 wins out of 82 games then there are only two ways that can be done with no consecutive losses.
That probability is $2(0.75)^{41}(0.25)^2$.

If your team wins exactly 75 wins out of 82 games then there are $\binom{76}{7}$ ways that can be done with no consecutive losses.
That probability is $\binom{76}{7}(0.75)^{75}(0.25)^{7}$.
• July 7th 2013, 06:24 PM
Plato
Re: someone pls help, question!
Quote:

Originally Posted by Plato
Now it is quite possible that one of us does not understand the other.
I might tell you that I taught probability theory for over thirty years.

I carefully explained in my first reply how I understood the question.
Did you read it carefully? It does not seem that you did!

If your team wins only 41 wins out of 82 games then there are only two ways that can be done with no consecutive losses.
That probability is $2(0.75)^{41}(0.25)^2$.

If your team wins exactly 75 wins out of 82 games then there are $\binom{76}{7}$ ways that can be done with no consecutive losses.
That probability is $\binom{76}{7}(0.75)^{75}(0.25)^{7}$.

Edit: It should read That probability is $2(0.75)^{41}(0.25)^{41}$.
• July 8th 2013, 09:57 AM
krv
Re: someone pls help, question!
Quote:

Originally Posted by Plato
Now it is quite possible that one of us does not understand the other.
I might tell you that I taught probability theory for over thirty years.

I carefully explained in my first reply how I understood the question.
Did you read it carefully? It does not seem that you did!

If your team wins only 41 wins out of 82 games then there are only two ways that can be done with no consecutive losses.
That probability is $2(0.75)^{41}(0.25)^2$.

If your team wins exactly 75 wins out of 82 games then there are $\binom{76}{7}$ ways that can be done with no consecutive losses.
That probability is $\binom{76}{7}(0.75)^{75}(0.25)^{7}$.

Correct me if I'm wrong, but if a team wins only 41 out of 82 games, it would be $\binom{42}{41}(0.75)^{41}(0.25)^{41}$ according to the formula $\binom{K+1}{82-K}(0.75)^{K}(0.25)^{82-K}$, which means there would be 42 different ways without any consecutive losses, which I think is the wrong number of ways. You are right saying there are only two ways that a team can win 41 games with no consecutive losses, however the equation you put up earlier doesn't support that.
• July 8th 2013, 11:17 AM
Plato
Re: someone pls help, question!
Quote:

Originally Posted by krv
Correct me if I'm wrong, but if a team wins only 41 out of 82 games, it would be $\binom{42}{41}(0.75)^{41}(0.25)^{41}$ according to the formula $\binom{K+1}{82-K}(0.75)^{K}(0.25)^{82-K}$, which means there would be 42 different ways without any consecutive losses, which I think is the wrong number of ways. You are right saying there are only two ways that a team can win 41 games with no consecutive losses, however the equation you put up earlier doesn't support that.

Actually you are right. There are 42 ways to have 41 wins and 41 losses.

Lets simplify the question to five wins and five losses, with no consecutive losses.

Look at the string $\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,}$. Now we have six places to put five $L's$ or $\binom{6}{5}=6$.

I was far to tired to have been working on that last night. Thank you for seeing that that.
• July 8th 2013, 12:40 PM
krv
Re: someone pls help, question!
Quote:

Originally Posted by Plato
Actually you are right. There are 42 ways to have 41 wins and 41 losses.

Lets simplify the question to five wins and five losses, with no consecutive losses.

Look at the string $\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,} W\underline {\,\,\,\,\,\,}$. Now we have six places to put five $L's$ or $\binom{6}{5}=6$.

I was far to tired to have been working on that last night. Thank you for seeing that that.

Ahh ok, that makes much more sense now. This formula is for computing at least 2 consecutive losses correct?