Fair Value of Game Involving Last Strictly Increasing Sequence Assume there is a game that costs $\$x$ dollars to play, and costs an additional $\$1$ to roll a die each turn. You are allowed to stop the game at any time, and the amount of money you win is equal to the sum of the rolls in the last strictly increasing sequence. 
For example, if you decide to roll the die 6 times and your $6$ rolls are $1, 2, 3, 4, 5, 6$, then you would earn $(1+2+3+4+5+6)-6 = 15$. If you decide to roll the die 3 times and your $3$ rolls are $3, 4, 1$, then you would earn $1 - 3 = -2$. 
What is a fair value $\$x$ (a range is okay) to pay for this game? 
What I've tried so far is analyzing only the expected payout from the last roll. For example, if you roll a $1$ on your first roll, and decide to roll again, your expected "profit" from rolling again is 
$-1 + \frac{1}{6}(0) + \frac{1}{6}(2 + 3 + 4 + 5 + 6) = 7/3$. 
However, I'm not sure how to incorporate the loss involved when your sequence of rolls looks something like $2, 3, 4, 5, 6, 1$. If you had stopped after the $6$, you would have earned $20 - 5 = 15$. However, the last roll of a $1$ brings your earnings to $1 - 6 = -5$. 
 A: You need to start by formulating a strategy of when to quit.  If your last roll was $1$ you can quit and collect $1$ minus the number of rolls or you can roll again.  If you roll again and get $1$ you lose $1$ because you are in the same spot except you have rolled once more.  If you get any higher number you are ahead because you collect at least $3$ and have rolled once more, so you are at least $2$ better off.  The expected value of the strategy roll one more and quit is $\frac 16 \cdot (-1)+\frac 16 \cdot 2 +\frac 16 \cdot 3+\frac 16 \cdot 4+\frac 16 \cdot 5+\frac 16 \cdot 6=\frac {19}6$.  In fact the expected value of rolling once more after you have $1$ is greater because you should roll again after $2$ as well.  
The two inputs to your strategy will be the last roll and the sum of the sequence ending with that roll.  If your last rolls are $1,2,3$ you could quit and take $6$.  If your last rolls are $4,3$ and quit you only get $3$ so the risk is higher in the first case.  The output will be roll or quit.  You know that if your last roll is $6$ you should quit because things can only get worse.  It shouldn't be hard to see that if your last roll is $5$ you should quit as well. The hardest questions are whether you should roll again if you have $1,2,3$ so far and whether you should roll again if you have $4$ alone so far.  You can guess my guesses by the way I asked the question.  Now you can evaluate the Markov chain for each roll and be done.
