GMAT Club World Cup 2022 (DAY 3): Messi and Ronaldo are playing chess

I guess the answer is D

P of winning each game 1/2
and P of 4 games ( 1/2)^4
for 6 games total (1/2)^6
1/2 option C

so atleast six games 3 should win messi and 3 ronaldo

total cases=2^6
favourable case=6(factorial)/3(fac)*3(fac)=20
probability=20/32=5/8 ansE

Since both have equal probability of winning =1/2.

Total 4 games to win

Minimum no of games to get result is 4
And a result will come out in the 7the game ...
Therefore no of games can be 4,5,6,7
Since we need probability of atleast 6 games = 2/4 = 1/2
Answer C

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Given: Messi and Ronaldo are playing chess till one of them wins a total of four games.
Asked: If each of them is equally likely to win any game and no game ends in a draw, what is the probability that at least six games will be played?

The probability that 4 games will be played = The probability of Messi winning 4 games + The probability of Ronaldo winning 4 games = 2(1/2)^4 = 1/8
The probability that 5 games will be played = The probability of Messi winning 4 games and losing 1 + The probability of Ronaldo winning 4 games and losing 1 = 2*5C1*(1/2)^5 = 5/16
The probability that at least 6 games will be played = 1 - (The probability that 4 games will be played+The probability that 5 games will be played) = 1 - (1/8 + 5/16) = 1- 7/16 = 9/16

IMO D

In order to have both the players going into the sixth game, one of the players must win exactly three out of the first 5 games played which can be in any order- and may win or not win the 6th game ( note the keyword- "Atleast" )

Hence:
P= probability that one team wins only 3 matches in the first 5 / total number of possible combinations of wins
p= 5C3 * ([1][/2])^5 = 5/8

The probability of at least 6 games is the complement of the probability of 4 & 5 games.

Probability of 4 games = \(2 * (\frac{1}{2})^4 = \frac{1}{8}\)

Probability of 5 games = \(2 * (4_C_1) * (\frac{1}{2})^5 = 8 * (\frac{1}{2})^5 = \frac{1}{4}\)

Probability of 4 & 5 games = \(\frac{1}{8} + \frac{1}{4} = \frac{3}{8}\)

Hence, probability of at least 6 games = \(1 - \frac{3}{8} = \frac{5}{8}\)

Hence, the answer is E.

Number of times a player wins game consecutively 4 times
MMMM =\( (\frac{1}{2})^4 = \frac{1}{16}\)
RRRR = \( (\frac{1}{2})^4 = \frac{1}{16}\)

Number of times a player is not able to have consecutive wins
MRMMM = \(\frac{5!}{4!} * (\frac{1}{2})^5 = \frac{5}{32}\)
RMRRR = \(\frac{5!}{4!} * (\frac{1}{2})^5 = \frac{5}{32}\)

Probability (Total number of times when the game gets over before 6th round)
\(\frac{1}{16}+\frac{1}{16}+\frac{5}{32}+\frac{5}{32} = \frac{7}{16}\)

Probability (Game played at least 6 times)
\(1-\frac{7}{16} = \frac{9}{16}\)

Messi and Ronaldo are playing chess till one of them wins a total of four games. If each of them is equally likely to win any game and no game ends in a draw, what is the probability that at least six games will be played?

(A) 3/8
(B) 7/16
(C) 1/2
(D) 9/16
(E) 5/8


Games possible (to be payed) = 4,5,6,7 and so on.

Probability (atleast 6 games are payed) = 1 - probabilty ( 4 games OR 5 games)

P (4 games) = 2 * (0.5)^4 = 2/16 = 1/8 (Note. Multiplication by 2 to mean that either Messi or Ronaldo can be the winner)
P (4 games) = 4 * 2 * (0.5)^5 = 4/16 = 1/4 (Note. Multiplication by 4 since one game which is won by overall loser can be won either 1st , or 2nd, or 3rd or 4th game.It can't be the 5Th game; otherwise the match would have been finished in 4 games). Also, Multiplication by 2 to mean that either Messi or Ronaldo can be the winner )

So, Probability (atleast 6 games are payed) = 1 - (1/8 + 1/4)
= 1 - (3/8)
= 5/8

(E) is the CORRECT answer

Hope this helps..

Correct answer: Choice D

Probability of at least 6 games = 1 - (probability of 4 games + probability of 5 games)

Probability of messi winning 4 games = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
Probability of ronaldo winning 4 games = 1/2 * 1/2 * 1/2 * 1/2 = 1/16 (cause anything messi can do, ronaldo can do the same )

so probability of 4 games = 1/16 + 1/16 = 1/8

Probability of 5 games . Lets take the case of Messi winning 4 games and ronaldo 1 =
MMMMR
= 1/2 * 1/2 * 1/2 * 1/2 * 1/2 (5!/4!*1!)
= 5/32
the vice versa of this is RRRRM = 5/32
So total probability of 5 games = 5/32 + 5/32 = 5 /16

Now probabily of 4 games + probability of 5 games = 1/8 + 5/ 16 = 7/16

so the probability of atleast 6 games = ( 1 - 7/16) = 9/16 = choice D

Since a player needs to win 4 games, at least 4 need to be played.
Hence,
P( At least 6 games played ) = 1 - P( Exactly 4 games played ) - P( Exactly 5 games played ) \(\ldots (1)\)

Now we shall apply binomial theorem,
P( Exactly 4 games played ) = \(2 \times \binom{4}{4} (\frac{1}{2})^4 = \frac{1}{8} \ldots (2) \)
(This is the case when only 4 games are played and one of the players wins all of them. Note that either of them can win hence the factor of 2.)

And,
P( Exactly 5 games played ) = \(2 \times \binom{5}{4}(\frac{1}{2})^4 (\frac{1}{2})^1 = \frac{5}{16} \ldots (3) \)

From (1), (2) and (3),
P( At least 6 games played ) = \(1 - \frac{1}{8} - \frac{5}{16}\)
\(= \frac{16-2-5}{16}\)
\(= \frac{9}{16}\)

\(\Rightarrow\) Option D

Maximum games that can be played so that at least one wins a total of four games = 7

So we have to find games where the total game is 6 or 7
For 6 games,
out of first 5 games, 3 is one by one player and 2 by other. Eg MMMRRM
Probability will be 5C2(1/2)^5

For 7 games
out of first 6, 3 is won by one player and other 3 by other player Eg MMMRRRM/MMMRRRRR
Probability will be 6C2(1/2)^6

Total probability = 5C2(1/2)^5 + 6C2(1/2)^6 = 3/8

Given:
Each of them is equally likely to win any game and no game ends in a draw.

Inference
So we will need atleast 4 games to be played.

Question
what is the probability that at least six games will be played

P(at least six games) = 1 - P (fewer than 6 games)

P (fewer than 6 games) = P(four games) + P(five games)

Probability of four games = \((\frac{1}{2})^4\)

Probability of five games = \((\frac{1}{2})^5 * (\frac{5!}{4!})\)

Total probability = \(\frac{1}{16}+\frac{5}{32} = 7/32\)

Note that this is the probability of each individual, so combined = \(\frac{7}{32}*2 = \frac{7}{16}\)

P(at least six games) = \(1-\frac{7}{16} = \frac{9}{16}\)

Option D

P=1-(prob of the game gets finished in 4 rounds +prob of the game getting finished in 5 rounds )
P(game gets finished in 4 rounds)=(1/2*1/2*1/2*1/2)2
P(game getting finished in 5 rounds)=(1/2*1/2*1/2*1/2*1/2)8
1-(1/8+1/4)
1-3/8
=5/8

[quote="Bunuel"]Messi and Ronaldo are playing chess till one of them wins a total of four games. If each of them is equally likely to win any game and no game ends in a draw, what is the probability that at least six games will be played?

(A) 3/8
(B) 7/16
(C) 1/2
(D) 9/16
(E) 5/8


To calculate this value we will consider 2 scenarios
Total number of games will be 4 if one of the player wins all the games.
probability of each player wining =1/2
Ronaldo winning all the 4 games =(1/2)*(1/2)*(1/2)*(1/2)
Messi winning all the 4 games =(1/2)*(1/2)*(1/2)*(1/2)
P(4 games)=2*(1/2)*(1/2)*(1/2)*(1/2)=1/8
Total Number of games played will be 5 if one of the player wins 4 games out of 5.
Total number of such case would be 5C4 - 4C4( The player cannot win all the 4 consecutive games from the first game.then the match will end in 4 games)
=5-1=4
Messi winning 4 games and loosing 1=(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*4
Ronaldo winning 4 games and loosing 1=(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*4
P(5 games)=2*1/8=1/4
Therefore probability of playing at lease 6 games=1-p(4) -p(5)
=1-1/8-1/4
=5/8

To have at least 6 matches, we should NOT have outcomes that complete the encounters within 4 or 5 matches. We can find the probability of 4 and 5 matches and find its complement to get P(at least 6 matches).

Case -1 (Decision in 4 matches) [Let Messi Win be M and Ronaldo win be R]
Thus all 4 Messi wins
P(M,M,M,M) = 1/16
aslo
all 4 Ronaldo wins
P(R,R,R,R) = 1/16
Therefore P(4 matches)=1/16 + 1/16 = 1/8

Case -2 Decision in 5 matches) [Let Messi Win be M and Ronaldo win be R]
4 Messi win and 1 Ronaldo win
P(R,M,M,M,M)= (5!/4! - 1)(1/2)^4(1/2) = 4 (1/2^5)
[*Subtract 1 for the case in which all firs 4 games are won by Messi. This cannot be since the encounter ends with 4 win

Similarly 4 Ronaldo win and 1 Messi win = 4
P(M,R,R,R,R)= (5!/4! - 1)(1/2)^4(1/2) = 4 (1/2^5)
Therefore P(5 matches)= 4 (1/2^5) + 4 (1/2^5) = 1/4

Thus P(at lease 6 matches) = 1 - P(4 matches) - P(5 matches)= 1- 1/8 - 1/4 = 5/8

ANS E

IMO, Option E is correct.

For the first five games, we can have any arrangement of MMMRR or RRRMM. The permutation will give 10 each for a total of 20.

The total outcome possible for the 5 games is 2*2*2*2*2 = 32

Probability = 20/32 = 5/8.