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Two players are playing a game by rolling a single die. First player rolls the die first. If the top surface of the die shows 1 or 2 then he wins. If not, then the second player will roll the die. If the top surface of the die shows 3, 4, 5 or 6 then he wins. Else, the game is continues until there is a winner.

(i) Who will be the most likely winner?

(ii) Find the probability that the first player wins the game given that the game has ended.

(iii) What is the most likely length of the game (number of rolls) to produce a winner?

My answer and solution, want to do some checking, correct me if any mistakes.

i) Let A = first player wins the game within one round Let B = second player wins the game within one round Probability of the first player wins the game P(A) = 1/6+1/6 = 1/3 Probability of the second player wins the game P(B) = P(A’) + (1/6+1/6+1/6+1/6) = 4/6 x 2/3 = 4/9

Hence, Second winner will be the most likely winner.

ii)

Probability that the first player wins the game given that the game has ended P(A1) + P(A2) + P(A3) + ….. + P(∞) = 1/3 + ( 1/3 ) ( 2/9 ) + ( 1/( 3) ) ( 2/( 9) )2 + ( 1/( 3) ) ( 2/( 9) )3 + ….

By using the formula of sum of infinity S = a/(1-r) a = 1/3 r = 2/9 s = (1/3)/(1- 2/9) = 3/7

iii.

P(C) = P(A) +P(B) = 1/3 + ( 2/3 ) ( 2/3 ) = 1/3 + 4/9 = 7/9 The probability smaller than 1 Therefore , one is the most likely length of the game to produce a winner

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  • $\begingroup$ Looks good except I did not quite understand the comment about "probability smaller than $1$ for part iii. Having $P(A) + P(B)$ "smaller than $1$" is the only way the game can last longer than one round. Or did you mean the probability of taking more than one round is smaller than the probability of finishing in one round (which would result in length of game equal to "1")? Because that is correct and is a good reason to say the most likely length is one round. $\endgroup$ Commented Mar 28, 2015 at 3:20
  • $\begingroup$ Thank you for this question, Hong Xia. $\endgroup$ Commented May 25, 2023 at 3:08

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For (i) I think you need to consider the players winning after $k$ rounds with $k=1,2,3,...$

Note that for player 1 to win in $k$ rounds, it means that he lost his first $k-1$ rounds, player 2 lost his first $k-1$ rounds, and player 1 won his $k$th round. By independence, this is ($A_k=$"player 1 wins after k rounds", $B_k$ is defined similarly for player 2):

$$P(A_k)=(4/6)^{k-1}(2/6)^{k-1}(4/6)=(2/9)^{k-1}(2/3)$$

Hence $$P(\text{1 wins})=\bigcup_{k=1}^\infty A_k=\sum_{k=1}^\infty P(A_k)=(2/3)\sum_{k=1}^\infty (2/9)^{k-1}={2 \over 3}{1 \over 1- {2 \over 9}}$$

Now for $B_k$ to occur player 1 must loose his first $k$ rounds, player $2$ his first $k-1$ and player 2 must win his $k$th round, which gives:

$$P(B_k)=(4/6)^k(2/6)^{k-1}(4/6)$$

$$P(\text{2 wins})={4 \over 9}\sum_{k=1}^\infty (2/9)^{k-1}={4 \over 9}{1 \over 1- {2 \over 9}}$$

It follows that $P(A)>P(B)$

ii) is fine.

For iii) again the probability of a game of length k won by 1 is $(2/9)^{k-1}(2/3)$ and won by 2 is $(4/9)(2/9)^{k-1}$. Hence the longer games are less probable, and since $2/3>4/9$ the most probable run is 1 winning is 1 round, hence a game of length 1.

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The correct probability function for $P(Ak)$ is :

$$\left(\frac{4}{6}\right)^{k-1}*\left(\frac{2}{6}\right)^{k-1}\frac{2}{6} = \left(\frac{2}{9}\right)^{k-1}\frac{1}{3}$$

Therefore $P(A) < P(B)$

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  • $\begingroup$ P(A) < P(B) $\endgroup$ Commented Apr 8, 2016 at 9:01

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