Should I throw the dice again if I have rolled 4? My math skills are very basic so it might be a stupid question, I had a discussion with my brother in law and now we have a 'math problem'.
We were playing a game with dices and he threw 4. The challenge was to throw the highest number, you can stop or throw again once, you cant see what the opponent has thrown, you both reveal after finishing. He said if you have 4 you have a 50% change the next time to throw the same or higher. 
(4,5,6) vs (1,2,3) 
and should throw again. 
But I said that I won't throw again on 4 because you already threw 4 and you are not likely to throw 4, 2 times in a row and therefore I would stop at 4. Am I right or is he and do you have a 50 % chance on throwing 4 or higher again? 
Game rules


*

*You throw a normal dice 1/6

*You can choose to throw again or keep the current value

*Your opponent cant see your value, you cant see his.

*The one with the highest value wins.


Short version:
If I threw 4, how big is the chance I throw 4 or more the next time I throw the dice, and should i take that chance?.
Thoughts
Average of 1 dice is 3.5, if i throw 4 im above Average and am more likely to throw below average next time. 

If you and your brother-in-law both have the same strategy, then the game is even. So the only question is, what happens if you re-throw on a 4 and your brother-in-law doesn't? Here is a table of the probability of each outcome:
                   1      2      3      4      5      6
You               1/9    1/9    1/9    1/9    5/18   5/18
Brother-in-law    1/12   1/12   1/12   1/4    1/4    1/4

The probability of a tie is $3 \cdot \dfrac{1}{9} \cdot \dfrac{1}{12} + \dfrac{1}{9} \cdot \dfrac{1}{4} + 2 \cdot \dfrac{5}{18} \cdot \dfrac{1}{4} = \dfrac{7}{36}$  
The probability that you win is $\dfrac{1}{12} \cdot \dfrac{8}{9} + \dfrac{1}{12} \cdot \dfrac{7}{9} + \dfrac{1}{12} \cdot \dfrac{6}{9} + \dfrac{1}{4} \cdot \dfrac{5}{9} + \dfrac{1}{4} \cdot \dfrac{5}{18} = \dfrac{29}{72}$
The probability that you lose is $\dfrac{1}{9} \cdot \dfrac{11}{12} + \dfrac{1}{9} \cdot \dfrac{10}{12} + \dfrac{1}{9} \cdot \dfrac{9}{12} + \dfrac{1}{9} \cdot \dfrac{1}{2} + \dfrac{5}{18} \cdot \dfrac{1}{4} = \dfrac{29}{72}$
So it makes no difference!
