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Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2012 based on Probability.

Two players A and B play rock - paper - scissors continuously until player A wins 2 consecutive games. Suppose each player is equally likely to use each hand - sign in every game . What is the expected number of games they will play?

- 12
- 15
- 16
- 20

Probability

Permutation Combination

But try the problem first...

Answer: 12

Source

Suggested Reading

Singapore Mathematical Olympiad

Challenges and Thrills - Pre - College Mathematics

First hint

We can start with a general case :

Two players are playing a series of games of Rock - Paper - scissors. There are a total of **K** games played. Player 1 has a sequence of moves denoted by string **A** and similarly player 2 has string **B**. If any player reaches the end of their string, they move back to the start of the string. The task is to count the number of games won by each of the player when exactly **K** games are being played.

To start with this particular problem let's set an expectation as k.

If A doesnot win , so the probability will be = \(\frac {no of event}{total event} = \frac {2}{3}\)

so game will be restarted.

Try to find out the rest of the cases..............

Second Hint

Continue from the 1st hint :

Again case 2: If A wins at first and then losses so again the probability will be \(\frac {1}{3} \times \frac {2}{3}\). again new game will start.

Last case : A wins in consecutive two games the probability will be \(\frac {1}{3} \times \frac{1}{3} \)

Do the rest of the sum........

Final Step

The total number of games that they will play :

K = \(\frac {2}{3} \times (K+1) \times \frac {2}{9} (K+2) \times \frac {1}{9} \times 2 \)

Now solve this equation and at the end E will be 12. [Check it yourself]

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