I tried this game yesterday with a couple of friends (really interesting, although I did not win, I would definitely recommend it)
here is a small piece of the game:
A player first calls out a number between 2-12,(name this variable x) then rolls two dices, (name the result y) if $y \ge x$, then player gets x number of dollars. If $y<x$, then the player only gets 2 dollars (the lowest possible value)
and if the player has completed a certain task he will also have the following ability:
After rolling two dices, the player can choose either re-roll or replace the number of one dice with the number 4
My question is what should my strategy be before and after I completed the task given that I want to maximize my revenue
Before the task, I am thinking... $$E(12)=12*{1\over36}+2*{35\over36}=2.27$$ $$E(11)=11*{3\over36}+2*{33\over36}=2.75$$ $$E(10)=10*{6\over36}+2*{30\over36}=3.33$$ $$E(9)=9*{10\over36}+2*{26\over36}=3.94$$ $$E(8)=8*{15\over36}+2*{21\over36}=4.5$$ $$E(7)=7*{21\over36}+2*{15\over36}=4.92$$ to be honest I think calling 7 is pretty good, you have like 60% chance of getting it,7 is the most common number for 2 dices, it also has the highest expected value in this case, is it a coincidence? $$E(6)=6*{26\over36}+2*{10\over36}=4.89$$ $$E(5)=5*{30\over36}+2*{6\over36}=4.5$$
After the task, its hard for me to calculate $$E(7)=7*{32\over36}+{4\over36}*(EE(7))=6.768$$ I feel like this is correct, $32\over36$ is the probability of getting at least 1 dice $\ge$ 3 when rolling two dices. $EE(7)=7*{21\over36}+2*{15\over36}=4.92$ which is the expectation for 7 when I don't have the ability to replace/re-roll
and if I was right about my calculation E(8)=7.125 and it should be the biggest expectation? This is pretty neat. I thought even though I completed the task, the expectation distribution should be the same as the previous (7 is still the most likely number? 90+?)
now what should my strategy be? completing the task is going to cost me some time...well it raised my lower bound by a lot(previous 2, now 5:4+anything) but if I just call out 8 every time... I am getting 1+ compare to calling 7 every time(even though the expected value is differ by 2.2, still not much) How should I estimate the cost of getting the task done?
can someone check my answers? and answer my doubts? great thanks...I need my revenge :)
Your calculation before the task looks correct. It is not automatic that you should call the most probable number, that will depend on the probability distribution. An informal way to find the right number to call is to compare calling $x$ with calling $x-1$. You win some amount on average by calling $x$. If you call $x-1$ you lose $1$ if the roll is $x$ or above and gain $x-3$ if the roll is $x-1$. The net is $(x-3)P(x-1)-P(\ge x)$ If you start with $x=12$ you should keep lowering it until this goes negative.
Assuming your only option after the task is to replace one of the two dice with a $4$, you can compute the chance of each result: $$\begin {array} {c c c c c c c c c c c}2&3&4&5&6&7&8&9&10&11&12\\0&0&0&\frac 1{36}&\frac 3{36}&\frac 5{36}&\frac 7{36}&\frac 8{36}&\frac {9}{36}&\frac 2{36}&\frac 1{36} \end{array}$$ where the chance of $10$ comes from rolls of $61,62,63,64,55,46,36,26,16$ and repeat your calculation. I would guess you should call $8$ or $9$.
If a reroll is both dice, you should reroll any number less than the number you call. You can then compute the chance of success for each number. For example, if you call $9$, your chance of success is $1-(1-\frac {10}{36})^2$ You can compute the chance of success for each number you might call and the expected payoff for each call.