Problem:
Satvik proposes a game of dice to Agnishom.
He says, "There are $2$ standard six-sided dice and the aim is to have the highest total of two dice. First you will throw $2$ dice. If you are happy with your total, that's great!
If you aren't happy with the total you have thrown, you can roll both dice again. However if you roll a second time, then whatever total you get, you must keep.
Now I will roll both dice. I have to accept the total I throw the first time, however if we both have rolled the same total then I win."
Given optimal strategy by Agnishom, what is the probability for Agnishom winning this game?
Approach:
There are a total of 36 outcomes of rolling two dice. The probabilities for the rolls are
2, 12 = 1/36
3,11 -- 2/36
4, 10 -- 3/36
5, 10 -- 4/36
6, 8 -- 5/36
7 --- 6/36
Formulating Optimal Strategy - - -
If Agnishom rolls a 6 and keeps it, he can win only if Satvik rolls a 2, 3, 4, 5 with a 10/36 = 0.27 chance. Remember Satvik also wins if he rolls a 6. Thus its obvious that any roll 6 and below, Agnishom must reroll.
If Agnishom rolls a 8, he wins if Satvik rolls a 2, 3, 4, 5, 6, 7 with 21/36 = 0.58 chance. Thus its obvious that any roll of 8 or higher Agnishom must keep.
Now lets see what is optimal strategy if Agnishom rolls a 7. Agnishom wins if Satvik rolls a 2, 3, 4, 5, 6 with 15/36 = 0.4166 chance. Can we improve that by re-rolling ?
If you roll a second time, then you have to take the sum of the probabilities of each roll multiplied by the probability of winning with that roll. This gives --
2 : 1/36 * 0 = 0
3 : 2/36 * 1/36 = 2/1296
4: 3/36 * 3/36 = 9/1296
5: 4/36 * 6/36 = 24/1296
6: 5/36 * 10/36 = 50/1296
7: 6/36 * 15/36 = 90/1296
8: 5/36 * 21/36 = 105/1296
9: 4/36 * 26/36 = 104/1296
10: 3/36 * 30/36 = 90/1296
11: 2/36 * 33/36 = 66/1296
12: 1/36 * 35/36 = 35/1296
We add all these probabilities to get 575/1296 = 0.443 chance which has improved our chances of winning if we re-roll. So now our strategy is clear.
Any roll 7 or below, Agnishom must re-roll. And keep any roll 8 and above.
Take the probability for each roll for the first throw of the dice multiplied by the probability of winning after that first roll. Any number 8 or higher has the same probabilities calculated above.
That's 105 + 104 + 90 + 66 +35 = 400/1296 = 0. 3086 chance of winning with the first roll.
Any number 7 or lower will be re-rolled, and from above we know that the probability of winning on the second roll is 575/1296. The probability of rolling 7 or lower is 21/36. This gives a probability for 2-7 roll when you re-roll at 21/36 * 575/1296 = 12075/46656 = 0.2588.
Total probability of winning on first and second roll is 0.3086 + 0.2588 = 0.567.
Is there any better way to solve it.Also I would like to ask if my solution is correct or not
Let $A_1, A_2$ denote the result of Agishom's first and second roll, respectively. I assume that the optimal is strategy among the following ones:
With this, let's add one more outcome to Agishom's second roll: $A_2=0\text{ if } A_1\geq x.$
If $A^x$ denotes Agishom's result then $$A^x=\begin{cases}A_1&\text{ if }&A_1\geq x,\\A_2&\text{ if }&A_1<x.\end{cases}$$ Finally, let $S$ denote Satvik's result.
Now, let's calculate the $x$-dependent probability that Agishom wins: $P(A^x>S)$ for the meaningful $x's$.
For $x=2$
$$P(A^2>S)=P(A_1>S\ \cap \ A_1 \geq 2)+P(A_2>S\ \cap \ A_1<2)=$$ $$=\sum_{i=2}^{i=12}P(A_1>i\ \cap \ A_1 \geq 2\ \cap \ S=i)=$$ $$=\sum_{i=2}^{i=12}P(A_1>i\ \cap \ S=i)=\sum_{i=2}^{i=12}P(A_1>i)P(S=i).$$
In general
$$P(A^x>S)=P(A_1>S\ \cap \ A_1 \geq x)+P(A_2>S\ \cap \ A_1<x)=$$ $$=\sum_{i=2}^{i=12}P(A_1>i\ \cap \ A_1 \geq x\ \cap \ S=i)+\sum_{i=2}^{i=12}P(A_2>i\ \cap \ A_1<x\ \cap \ S=i)=$$ $$=P(A_1\geq x)\sum_{i=2}^{x-1}P(S=i)+\sum_{i=x}^{i=12}P(A_1>i)P(S=i)+P(A_1<x)\sum_{i=2}^{i=12}P(A_2>i)P(S=i).$$
Based on the $P(A_1=i)=P(A_2=i)=P(S=i)$ table the probabilities above can be calculated. The answer is the maximum in $x$.
The results of the calculation is shown below. Here the probability that Agishom wins is depicted as a function of $x$:
Tha maximum takes place at $x=8$. So, if Agishom rolls the second time if $A_1<8$ then the probability that she wins is $\approx 0.57.$-