Aaron,That's why I thought Anthis meant the pay-off is only the last throw...

Good point So you think that if each roll has expected payoff of 1.5$ then N rolls worth N*1.5$But do you think that the option to quit has no value in any potential scenario of the game?Think this: In a game of 6 rounds, in the first three rolls you have brought three times a big number then the average accumulated pofit at that point in time is in the range of 12$ to 18$ or in more general terms you have accumulated profit higher than 6*1.5$=9$. Wouldnt you exercise your option to exit?RegardsAnthis

QuoteOriginally posted by: AnthisSo you think that if each roll has expected payoff of 1.5$ then N rolls worth N*1.5$. But do you think that the option to quit has no value in any potential scenario of the game? Think this: In a game of 6 rounds, in the first three rolls you have brought three times a big number then the average accumulated pofit at that point in time is in the range of 12$ to 18$ or in more general terms you have accumulated profit higher than 6*1.5$=9$. Wouldnt you exercise your option to exit?If I am risk-neutral, no, because on average I will make $4.50 from the last three rolls.If I am risk-averse then I'm more likely to quit when I'm down a lot. Say I lose $9 on the first three rolls so I'm out $18. I might choose to quit rather than risk losing $27, which I might not be able to afford.

QuoteIf we play many such games each with a large number of rounds (dice throws per game), what would be the average duration of a game? Duration = number of dice throws per game.Yes. Anthis & Aaron are right. As Aaron pointed out, it's a textbook problem - a waiting time (or first arrival) problem. Of course, from the standpoint of the game it doesn't make much sense if we play such that the number of rounds is large - you pay $6 and keep rolling till you hit a 6, and get your $6 back......

Hey guys!Just had a phone interview and was asked to solve the "dice problem" this thread is about....Here is my solution using conditional expectation: where X is a dice outcome and E[X1] = 3.5. The reasoning is the following: if the first outcome is less than the price of n-1 - step game, then we continue to play and have n-1 steps more....price of n-1 game we know, it's E(X_{n-1}). The rest is clear.It gives the same answer as the formula suggested by Aaron....but somehow I don't quite understand how to derive the other formula.Couldn't do it during the phone interview by the way...and got rejected.....

Your formula is more general, it covers any distribution for the gamble. Mine covers only the specific case of six equal-probability outcomes.To get my formula, rewrite your formula conditional on rolling a 1, 2, 3, 4, 5 or 6. One of your probabilities will be 1 and one will be 0 in each case. In place of the conditional expectation, use the roll number itself. Add all six results together and you get my formula.