There several versions of "1 to 31" game, involving one die and two players (say you and your opponent), both gifted mathematicians.

My choice is the version defining as a final goal being the first to get a score

of 31

**or more**.

Here we go:

First: You throw the die and note the outcome.

Then: Your opponent makes a quarter turn* and add this outcome to the earlier sum.

Then: You make a quarter turn and add... and so on...

Until: Whoever gets first the score of 31, or more WINS...

Obviously - there are no draws.

*The quarter turn gets you one of the 4 neighboring faces - never the opposite one!

Design the optimal strategy, - for each of the players and evaluate their chances of victory, provided each makes the best possible moves and that the first number

is a perfectly random digit between 1 and 6.

I am assuming the die is rolled once and all moves are sequential quarter turns of the die, as opposed to a random roll each round as in Charlie's answer.

If the current total is 26 to 30 then the current player can always win immediately.

If the current total is 25 and the die is showing 2, 3, 4, or 5 then the current player can still win immediately. But if the die is showing 1 or 6 then the current player will be forced to leave a winning state for the other player.

If the current total is 24 and the die is showing 2, 3, 4, or 5 then a winning play is to make the quarter turn to 1, putting the opponent in the losing state of having a total of 25 and die showing 1.

This idea can be extended indefinitely to conclude that any state with the die showing 1 or 6 is a losing state if you cannot immediately reach 31 or higher.

So a simple strategy emerges. If you can win immediately then do so. Otherwise if the die is showing 2, 3, 4, or 5 then turn the die to show 1 (or possibly 6 if the total is less than 19). Finally, if the die is showing 1 or 6 and the total is 25 or lower then you are in a losing state.

So then the two players' chances of victory come down to the initial die roll. If it is 1 or 6 then player 1 will win, a probability of 1/3; otherwise player 2 will win, a probability if 2/3.