I don't quite get the difference between the probability that the grunters won at the 6th game vs. the probability that the grunters won given there are 6 games. Could you please explain further?

Think of it this way. Let n be the number of ways that the grunters can win in 6 games. The games can end after 4, 5, 6 or 7 games. Let N be the total number of ways the games can end. P(grunters win in 6 games) = n/N. Now suppose you are told that the games after 6 games. Let M be the number of ways the games can end in 6 games. Then P(grunters win in 6 given that games end in 6) = n/M. So, once you are told the games end in 6 games, you are dealing with a reduced sample space.

A simple example would be the roll of a dice. You know that the probability of rolling a 6 would be $\frac{1}{6}$. Suppose you were given that the dice rolled an even number. Then you know that the number rolled is one of {2, 4, 6}. So, the $P(dice\, rolls\, 6\, |\, dice\, rolled\, even) = \frac{1}{3}$

So if the question was "What is the probability the Grunters will win in exactly 6 games", then you are calculating $$\frac{\text{# ways for Grunters to win in 6 matches}}{\text{# ways for game to end in 4, 5, 6, or 7 matches}}$$? I get how the numerator of this fraction is determined by $\binom{5}{3}$, but I don't see how $\frac{1}{2^6}$ counts the number of ways for the game to end in 4,5,6, or 7 matches.

That's not what I meant. If the question asked you "What is the probability that the Grunters win in 6 games", then your original calculation is correct. But, the question was "If the series lasted 6 games, what is the probability that the Grunters win?". In this case, you are told that the series has ended in 6 games. So, ways the series can end in 6 games = ways in which Grunters win in 6 + ways in which Screamers win in 6. That is the numerator of the "fraction" in my answer. Have you been taught conditional probability yet?

I know a little bit of conditional probability. I was indeed able to understand the difference between the two problems problems -- My only remaining question is how my original calculation that answers the problem "probability the grunters win in 6 games calculate the total number of ways the games can end in 4, 5, 6, or 7 ways. I got how the number of ways for grunters to win in 6 matches is determined by $\binom{5}{3}$ but I don't see how $2^6$ is the total number of ways for the game to end in 4, 5, 6, or 7 matches.

OK, I think I see why you are confused. Look again at my answer. The denominator in my "fraction" says P(Grunters win in 6 OR Screamers win in 6). So, we are looking at the total number of ways for the series to end in 6 games. This is because the question is asking you: "What is the probability of the Grunters winning in 6 if you already know that the series ends in 6?". This is a different question than "What is the probability that the Grunters win in 6 games?" SE does not like extended discussions in comments. If you post this question, I can explain further.

hi

Hi

I am answering the question What is the probability the grunters win in 6 games

Not the condition one

So if you are calculating the probaiblity they win in 6 games

OK, so first what is the number of ways Grunters can win in 6?

5C3

times 0.5^6

That gets you the probability, right?

which was your original answer in your post, right?

No, it gets you the number of ways for the Grunters to win in 6

But to calculate the probability don't you need the #ways for grunters to win in 6 divided by the number of total ways

When you multiply by (1/2)^6 doesn't that give you the probability?

OK, lets start againg

Let 1 means Grunters win and a 0 mean they lose

ok

So, for the Grunters to win in 6, we must have something like ?????1

yes

where ? can be 0 or 1

The grunters need to win 4 games out of the 6

But we need 3 of the ? to be 1's

yeah

so 5C3

yup

but for any one of the 5C3 choices, how many ways can you permute them?

Doesn't 5C3 itself count the number of combinations?

Whoops, yes

Now, how many ways for the games to end in 4?

One way

No, 2, 0000, 1111

oh, sorry, right

Now do the same for 5, 6 and 7 games

then add the number of ways for 4, 5, 6 and 7 games

This will be the denominator in your "fraction"

That was 2^6

How did you get 2^6?

I mean when you do 5C3 * (0.5)^4 * (0.5)^2 you get the denominator as 2^6

You calculate the probability of each of each arrangements is (0.5)^6 and there are 5C3 of them

Ok, for 4 games, there are 2 ways. You undestand that right?

yes

Ok, let's try 5 games

How many ways?

It should be 2 * 5C4, right?

01111, 11101, 11011, 10111

Why multiply by 2

one for grunters and one for secreamers

Because either team can win?

ah ok

yes

so now do the same for 6 and 7

Then for 6 team it is 2*6C5

yes

no, 6 C4

and 7C4

got it

OK, now?

yes

wait

?

don't you have to exclude possibilities like 11110

yes

So shouldn't the 5 game case be 2*4C3

but I was trying to clarify the concept

You can fix the last game so either the screamers or grunters win and then from the remaining four you choose 3 other games that they win

I know you now how to work out the number of ways. You were getting confused with 2^6 which had nothing to do with it.

ok

Where does the 2^6 come from

When I calculate $\frac{\binom{5}{3}}{2 \binom{4}{3}+2 \binom{5}{3}+2 \binom{6}{3}}$ I got 5/34 not 5/32 as expected

Each of the 6 games has probability of 0.5, right

yes

So, binomial formula says 5C3 (0.5)^6

yes

right

now let's say p = prob of winning = 0.6 and q = 1 - p

ok

Then the binomial formula would give 5C3 p^4q^2

yes

but since in this case p=q thats why you get 5C3 p^6

yeah

OK, now?

So this gives you the probability of the grunters winning in 6 games

but don't you divide by the number of total outcomes

yes, and you don't know whether the series ended in 4, 5, 6 or 7 games

When I tried to calculate it as 5C3/(2 + 2*4C3 + 2*5C3 + 2*6C3) I got 1/7, not 5/32

No, you either use the binomial formula or you use the number of ways.

you don't use both

Yeah, I tried to use number of ways

I got 1/7

Let G6 be number of ways for Grunters to win in 6

Let S4, S5, S6, S7 be number of ways for series to end in 4, 5 ,6 or 7 games.

G6/(S4 + S5+S6+S7)

The prob = G6/(S4 + S5 + S6 + S7)

That is what I tried and I got the expression 5C3/(2 + 2*4C3 + 2*5C3 + 2*6C3)

But this does not give the right answer

This should be equal to the answer you got using the binomial formula

It isn't thogh

*though

Why is it 2*4C3?

That is the 5 game case

You have 5 games. You fix the final game as either grunters or screamers

Then from the remaining 4 games you choose 3 of them

Let me work out the answer

k

Strange, using binomial I get 10/64

but using counting I get 1/7

Yeah, that is what I got too

Perhaps I'll post that as another question on stackexchange

ok

Ok, I understood everything else. Thanks!

I'd like to find out why too

bye

bye