Question

A and B have equal chances of winning a single game. A needs 3 games to win a match and B needs 4 games to win the match. What is the probability that A wins the match?

Answer 1

\[P(A)=P(B)=\frac 12\]
Do you agree with this?

Answer 2

Yup. What i thought next is there can be 6 games at most. Not more than that.
Then i made cases, for 3 games: A wins first 3 matches,
for 4 games: A wins 3 matches and B wins 1 match (3 cases here)
and so on... but its too long for an objective question. anything short?

Answer 3

yes, this is the way, it's simple, it's an infinite geometric series

Answer 4

Can you post here the series?

Answer 5

Are you sure this makes a geometric series.
(1/2)^3 + 3*(1/2)^4 + .. what next?

Answer 6

yes,
\[(\frac {1}{2})^3+(\frac {1}{2})^4+(\frac {1}{2})^5+(\frac {1}{2})^6\]
see the common factor is 1/2, it's an infinite geometric series, which converges.

Answer 7

Well this is a geometric series, yes. Infinite, no. And how did you get this?

Answer 8

Oh yes, I missed this, there could be max 6 games, you're right
\[(\frac {1}{2})^3+(\frac {1}{2})^4+(\frac {1}{2})^5+(\frac {1}{2})^6\]
Yeah, you need to find sum of this, use the geometric series sum to find this

Answer 9

But again there will be 3 cases when they play 4 games. one when B wins first, second, third games.

Answer 10

yes, there can be many permutations, we need to account them too !!

Answer 11

How?

Answer 12

oh there is a probability distribution we have to use it here.
have you studied Poisson's distribution?

Answer 13

Yup.

Answer 14

yes, we need to use that here. let me think

Answer 15

And.. no i havent studied poisson distribution. i mistook it for bernoulli trials.

Answer 16

@shubhamsrg

Answer 17

Have you studied negative binomial distribution?

Answer 18

it can finish in 3 games, or 4 games, or 5 games or 6 games or 7 games
for 3 games -> 0.5 * 0.5 * 0.5 *C(3,0)
for 4 games -> 0.5*0.5*0.5*0.5 * C(4,1)
for 5 games -> 0.5*0.5*0.5*0.5*0.5 *C(5,2)
for 6 games -> 0.5*0.5*0.5*0.5*0.5*0.5 *C(6,3)
for 7 games -> 0.5*0.5*0.5*0.5*0.5*0.5*0.5 *C(7,4)

final ans = sum of all

Answer 19

in C(n,r) ,
n= no. of games
r = no. of games B wins

Answer 20

We wont take the case with 7 games no? We dont want B to win?

Answer 21

omg, sorry
yes, 6 cases!
-_-

Answer 22

Aap bhut intelligent ho maharaaj. :')
Thank you.

Answer 23

What's the answer?

Answer 24

gaali de do par mazak nahi banaya karo -_-

Answer 25

|lol| bbj bc

Answer 26

21/32 @drawar

Answer 27

lol shubham. maine book se dekhke btaya. :/

Answer 28

Okay. Let X be the number of games played until the winner is announced. It's clear that 3<=X<=6. Now P{X=n}=C(n-1,2)*(1/2)^n, sum them up from X=3 to X=6 and you'll get 21/32 :)

Answer 29

P{X=n}=C(n-1,2)*(1/2)^n ? Where did this come from? Negative binomial dist?

Answer 30

Yes, you can just argue it as follow: A wins the match when he wins a total of 3 games. In order for A to win his 3rd game at the nth game played, he must win 2 games in the (n-1) games played before (i.e C(n-1,2))*(1/2)^2, and the rest (n-1-2)=(n-3) games is won by B ((1/2)^(n-3). Of course A wins the nth game, the probability of this happening is 1/2. Multiply all these things together, you'll get C(n-1,2)*(1/2)^n

Answer 31

Hmm. It looks good. But I think i'll have to think about it to be sure about it.
I'll stick with what i know, shubham's method for the time being.
Thank you though. i appreciate.