Aptitude Discussion

Q. |
$A$ and $B$ throws one dice for a stake of Rs 11, which is to be won by the player who first throws a six. The game ends when the stake is won by $A$ or $B$. If $A$ has the first throw, then what are their respective expectations? |

✖ A. |
5 and 6 |

✔ B. |
6 and 5 |

✖ C. |
11 and 10 |

✖ D. |
10 and 11 |

**Solution:**

Option(**B**) is correct

Expectation of winning $A$:

\(\left[\left(\dfrac{1}{6}\right)+\left(\dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{1}{6} \right) +....\right]\times 11\)

\(\left[\dfrac{\dfrac{1}{6}}{1-\dfrac{25}{36}}\right]\times 11=6\)

Similarly we get the expectation of $B$ as 5.

**Edit:** For an laternative solution, check comment by **Jas Sahib.**

**Edit 2:** For yet another alternative solution, check comment by **Sravan Reddy.**

**Sravan Reddy**

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**Jas Sahib**

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Probability of a player winning = 1/6

Probability of a player losing = 5/6

Let W denote a 6 turning up for a player

and L denote 1, 2, 3, 4 or 5 turning up on the dice for a player

If A starts the game then he can win when - W or LLW or LLLLW and so on

Therefore probability of a winning = 1/6 + 5/6 X 5/6 X 1/6 + 5/6 X 5/6 X 5/6 X 5/6 X 1/6 +........till infinity (this is a GP)

Sum = (1/6) / (1-25/36) = 6/11

Therefore stake of A = 6/11 X 11 = Rs.6

Stake of B = 11-6 = Rs. 5

Hello Jas Sahib.

Why the 2nd chance is taken LLW after LW

**Shiv**

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Couldnt understand the concept!

Hope you also got it Shiv.

explained to ruchi.

The answer should be same if you do it for 1st two cases or you continue it for infinity. The expectations don't change.

Probability of A winning in first turn $= \dfrac{1}{6}$

Probability of B winning in the second turn = Losing probability by A * Winning probability by B $= \left(\dfrac{5}{6}\right) \times \left(\dfrac{1}{6}\right) = \dfrac{5}{12}$

So ratio of these 2 should give the answer - $(1/6) : (5/12) = 6:5$