# Betting ended after nth round.Find the sum of money NOT WON?

Rahul and Vijay are playing a game with 12-sided die,where both of them lay bets on outcomes of roll of die.They start betting Rs 5 each on first round of the game and the amount bet in each subsequent round is thrice the amount bet in previous round.The game ends when Rahul wins a round .Rahul wins only when die shows a 10,11 or 12.Vijay wins for all other outcomes.If the game ended after the nth round. Which of the following CANNOT BE SUM OF MONEY won by Rahul and Vijay respectively?

• Do you seriously think we'll take the time to answer your question if you don't even take the time for the most rudimentary check whether you've copied it properly? – joriki Jun 25 '16 at 16:25
• @joriki I thought the "owing" was just bad English, but you are right - it is the end of "following"! :) – almagest Jun 25 '16 at 16:39
• @almagest: Tolerance for bad English is essential on a site like this. I have the experience of learning a non-Latin script, so I'm aware how much more difficult it can be to write in another than one's native script. But there are things that one can spot even in a foreign script if one puts in the minimal effort of comparing a copied text to the original. I think "of the following" being replaced by "owing" is one of them. (Also a considerable number of articles are missing that are present in the original.) – joriki Jun 25 '16 at 16:45
• @Yes, that was stupid of me. I just thought I was expected to struggle to read the amounts. I did not pause to wonder why he had missed them out having transcribed most of the rest. – almagest Jun 25 '16 at 16:48
• Well ,Sir I had no intention to paste question on stackflow for any kind of reputation.I was just taking mock test and there i found this question and that prompted me to ask it from stackflow intelligentia.I beg sorry for inconvenience.I will surely take care of it in future. :) – Navneet Kang Jun 25 '16 at 16:51

After $n+1$ rounds we have R $5\cdot3^n$ and V $5(1+3+\dots+3^{n-1})=\frac{5}{2}(3^n-1)$. So R always wins more than twice as much as V.
So it is easy to rule out the 3rd alternative. With a little more effort we can see that in the first alternative R's win corresponds to $n=7$, so V's should be 5465 which it is. In the second alternative R's win corresponds to $n=6$, so V's should be 1820 which it is. In the last alternative R's win corresponds to $n=5$, so V's should be 605 which it is.