Probability-5

Probability Level 3

Two person A A and B B throw a (fair) die ( six-faced cube with face no. 1 1 to 6 6 ) alternatively, starting with A A . The first person to get an outcome different from the previous one by the opponent wins. The probability that B B wins is x y \frac{x}{y} . Then x + y x+y is

Note: gcd ⁡ ( x , y ) \gcd ( x,y ) is 1 1

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The answer is 13.

Akshay Sharma by Akshay Sharma , 21, India

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1 solution

Sam Bealing
Apr 18, 2016

Let the probability B B wins at the point where it is B B 's go be p p . (This includes B B winning in a later turn).

Clearly A A cannot win on their first turn so we now consider it being B B 's turn.

The probability B B wins from this point is p p (by definition) but we can also calculate it:

The probability B B wins on this turn is 5 6 \frac{5}{6} . (Rolling different to A A )

The probability B B doesn't win and then A A doesn't win is 1 6 × 1 6 = 1 36 \frac{1}{6} \times \frac{1}{6}=\frac{1}{36} (probability of them both rolling the same as the person before).

The probability B B wins from this stage is p p (by definition as we can consider this having a infinite number of turns).

Thus we have by equating the two expressions:

p = 5 6 + 1 36 p ⇒ 35 36 p = 5 6 ⇒ p = 6 7 p=\frac{5}{6}+\frac{1}{36} p \Rightarrow \frac{35}{36}p=\frac{5}{6} \Rightarrow p=\frac{6}{7}

x = 6 , y = 7 ⇒ x + y = 13 x=6,y=7 \Rightarrow x+y=\boxed{13}

I don't think my explanations are great here so I wonder if someone could explain it a bit better.

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Moderator note:

Nice approach using the iterated expectation.

Why not do it this way?Let us assume that B wins on the (k+1)th game.Hence,both A as well as B must get the same number (on their respective dice) for the first k games.On the (k+1)th game,B must get a different number but A must again get the same number so that B wins the game.Find the probability for B to win in terms of k (which is not at all difficult) and then take the summation (you will get an infinite geometric progression ) as k goes from 1 to infinity.You will get the same answer.

Indraneel Mukhopadhyaya - 5 years, 1 month ago

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I know, I was just trying to avoid using an infinite geometric series directly though it's the same idea really.

Sam Bealing - 5 years, 1 month ago

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