If i have three rolls of a dice and for the first and second you choose to either keep your roll or keep going, what is the best strategy to getting the best roll and what is the expected value?

On the first roll, we keep a result of 5 or 6, and roll again if it is 4 or lower
On the second roll, we keep any result equal or greater that 4 and we roll again for 1,2 or 3
We keep anything we obtained from the third row

The expected value is 14/3 = 4.666

Step-by-step explanation:

When you throw a roll the expected value is 3.5 (the average of 1,2,3,4,5 and 6). This means that any roll with value 3 or below is worth changing, because the expected value will transform into 3.5 from the 3 or less you had before.

This means that in your second roll you will keep the result as long as it is 4 or more and you will roll once more if it is 3 or less, hence, you can

get a 4, 5 and 6 on your second roll and keep it, so that your value is that number
get 1, 2 or 3 on your second roll and roll again, with an expected value of 3.5.

Rolling a third time has a probability of 3/6 = 0.5, obtaining a 4 has a probability of 1/6, same for 5 and 6. Thus, the expected value from the second roll onwards is

1/2 * 3.5 + 1/6 * 4 + 1/6 * 5 + 1/6 * 6 = 4.25

This means that, after the first roll, we can keep that value or throw again with an expected value of 4.25 (using the strategy depicted above). We can roll again if we got a result of 4 or below expecting a higher value or keep a result of 5 or 6.

The strategy can be ssummarized as follows

On the first roll, we keep a result of 5 or 6, and roll again if it is 4 or lower
On the second roll, we keep any result equal or greater that 4 and we roll again for 1,2 or 3
We keep anything we obtained from the third row

The probability of getting 4 or below is 4/6 = 2/3, and the expected value for the second roll onwards is 4.25, hence the expected value of the strategy we pick is

1/6 * 5 + 1/6*6 + 2/3* 4.25 = 4.666 = 14/3