Probability that A wins the game

Question

Question

A and B are 2 players, each throwing 2 dice on his turn, with A starting first and players taking turns alternately. A wins if he throws 7 and B wins if he throws 8. What is the probability that A wins?

My Approach

Let $P,Q$ be the probability that $A$ and $B$ wins respectively.

Getting sum of $7$ on $2$ dice $=\frac{1}{6}$

Getting sum of $8$ on $2$ dice $=\frac{5}{36}$

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$A$ will win

• if it gets the sum $7$ on its first throw
• if A fails on its first turn then B should also fail on its throw and then A should win .

I can write above in equation as-:

$$P=\frac{1}{6} + (1-\frac{5}{36})\times \frac{5}{6}$$

$$P=\frac{192}{216}$$

but answer is given as $\frac{30}{61}$

Where am I wrong?

Answer 1

Well, the answer that you have posted cannot make sense because obviously A has a higher chance of winning. Here is my solution to the problem that was asked. Here is A's winning pathways. Either: A wins on the first round, A wins on the third round, A wins on the fifth round, etc. (A wins on odd rounds) So we can now find the probability of each of these outcomes and sum them. The probability that A wins on the first round is $\frac{1}{6}$. The probability that A wins on the third round is $\frac{5}{6} \times \frac{31}{36} \times \frac{1}{6}$. (A has to miss the first round, then B has to miss, then A has to hit. We can do this in general and see that the probability that A wins on the $nth$ odd round would be $(\frac{5}{6})^n \times (\frac{31}{36})^n \times \frac{1}{6}$. So then we simply need to sum the series. $\sum_{n=0}^\infty ((\frac{5}{6})^n \times (\frac{31}{36})^n \times \frac{1}{6})$, which comes out to $\frac{36}{61}$(maybe you misread the answer?), which makes more sense.

Answer 2

Without summing a series:

There are $1\,296$ possible pairs of rolls; of these, $216$ ($\frac {1}{6} \times 1\,296$) have player A rolling a 7, and $180$ have player B rolling an 8, and $30$ have both - this 30 is counted among both the $216$ and the $180$, so $186$ times A will win without B's potential roll succeeding, and $150$ times B will win, and $30$ ($\frac {1}{6} \times \frac {5}{36} \times 1\,296$) times B could have won but A happened to win first. This gives $366$ total ways the game could end each round, and of those $216$ give A the victory, and otherwise we start from scratch. So A's chances of winning are $\frac{216}{366}=\frac {36}{61}$.

Answer 3

$\underline{Another\;way}$

If both fail in their first chance, we are back to square $1$,
so it is sufficient to consider one round.

P(A wins on first chance) $=\dfrac16 = \dfrac{36}{216},\;\;\;$ P(B wins on her first chance)$=\dfrac56\cdot\dfrac5{36} = \dfrac{25}{216}$

Odds in favor of A $= 36:25$

P(A wins) $=\dfrac{36}{36+25} = \dfrac{36}{61}$

Answer 4

P(Game finishes on $1^{st}$ turn) = $\frac{1}{6}$

P(Game finishes on $2^{nd}$ turn) = $\frac{5}{6}\frac{5}{36}$

P(Game finishes on $3^{rd}$ turn) = $\frac{5}{6}\frac{31}{36}\frac{1}{6}$

P(Game finishes on $4^{th}$ turn) = $\frac{5}{6}\frac{31}{36}\frac{5}{6}\frac{5}{36}$

Clearly, $A$ would win if game ends on odd term

Thus, P(A wins) = Summation of probability of odd terms

The odd terms follow a GP with $r=\frac{(5)(31)}{(6)(36)}$ and $a=\frac{1}{6}$

summation of GP = $a\frac{1-r^n}{1-r}$

since $r < 1$ and $n=\infty$, $1-r^n=1$

and hence sum of gp = $\frac{36}{61}$

What you missed was the fact that game can go on forever with neither of them winning. You only took in account the $1^{st}$ and $3^{rd}$ throw rather than all odd ones