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. 
 A: 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!
A: Going back the the question of should you throw again I would say the answer is no you should not. If you roll a 4 the chance that you win is 50%, the chance that you lose is 33.33% and the chance that you tie is about 16.66%. So if you roll a 4 you don't lose 66.66% of the time. 
A: Answer updated after edit: It does not matter
An analytical solution has already been given by @TonyK. As such I have just added the simulation code to represent this situation:
% Generate first rolls and potential second rolls for each strategy
strategy1= randi(6,2,1000000);
strategy2= randi(6,2,1000000);
% Determine whether to reroll or to keep
idx_keep1 = strategy1(1,:)>=4; %Keep the 4
idx_keep2 = strategy2(1,:)> 4; %Don't keep the 4
% Use the first value if you keep, the second if you reroll
strategy1(2,idx_keep1)= strategy1(1,idx_keep1);
strategy2(2,idx_keep2)= strategy2(1,idx_keep2);
% Check the statistics
h=hist(sign(strategy1(2,:)-strategy2(2,:)))

Results of strategy1 vs strategy 2
If you reroll the 4 and the other player does not, these are the resulting probabilities:


*

*Chance that you win: 40.3%

*Chance that you draw: 19.4%

*Chance that you lose: 40.3%




Answer before edit:

How to win the game
For completeness I will repeat that if you roll again, you have indeed got a 50% chance of throwing at least four. But here is why you should not do it!
Assuming the game is you may roll once or twice, then your opponent rolls once looks at your score and may decide to roll again, and that the one with the highest score wins and it is otherwise a tie:
Opponent roll 1:


*

*There is a 2/6 chance that he will roll above 4

*There is a 1/6 chance that he will roll 4 (and thus should stop as the second roll is expected to be below 4)

*There is a 3/6 chance that he will roll below 4 (and thus should roll again)


Opponent roll 2:


*

*There is a 2/6 chance that he will roll above 4

*There is a 1/6 chance that he will roll 4 

*There is a 3/6 chance that he will roll below 4 


Resulting probabilities
These are the probabilities given that you don't reroll after getting a 4


*

*Chance that your opponent wins: 2/6+3/6*2/6 = 50%

*Chance that your opponent draws: 1/6+3/6*1/6 = 25%

*Chance that your opponent loses: 3/6*3/6 = 25%


Conclusion
If you roll again you are expected to roll lower than what you have, I didn't do the math here but intuition tells me that in the resulting situation (you roll once, your opponent rolls twice) you will face worse odds than you have now (with your above average roll of 4). The simulation below seems to confirm this:

Simulation to calculate odds if you do throw again
r1= randi(6,1,1000000);
r2= randi(6,2,1000000);
r2(2,r2(1,:)>=4)=0;
r2 = max(r2);
h = hist(sign(r1-r2))

Note that it is Matlab. If you want to simulate the odds of beating 4, just replace r1 by 4 in the last line.
The results for if you do throw again are approximately:


*

*Chance that your opponent wins: 2/6+3/6*2/6 = 56%

*Chance that your opponent draws: 1/6+3/6*1/6 = 17%

*Chance that your opponent loses: 3/6*3/6 = 27%

A: You have exactly the same probability of getting a 4 on every single throw. The next dice doesn't care what you threw last. It doesn't look at the other dice, and think "well, 4 is already taken sooo...".
We call this "independent events".
A: 
He said if you have 4 you have a 50% change the next time to throw the same or higher.

This is true, but is misleading. It's also true that you have a 4/6 chance to throw the same or lower.
If you wish to maximize the expected value, this is the strategy to follow:
The expected value each time you roll is $3.5$ (since this is the average of the values, and they have equal probabilities). Since your roll of $4$ is above this, rerolling is expected to lower your score, not raise it. You should reroll $(1,2,3)$ and hold $(4,5,6)$. Note that the rolls are completely independent events, so rolling a 4 the first time does not decrease the probability that you'll roll a 4 the second time.
Whether this actually results in you beating your opponent more than 50% of the time depends on his strategy, which likely depends on what he knows of your strategy.
A: The chance of rolling a four or higher on your next roll is independent of your original roll. The fact you just rolled a four doesn't make it any less likely to roll one again. So your chance of rolling a four or higher is indeed 1/2, since you have three ways of rolling a four or higher and six total outcomes. 3/6=1/2.
So to answer your full question, if you roll again, you have a 1/2 chance of doing the same or better. However your chance of doing the same or worse is the number of ways to roll a four or lower (4), divided by the total outcomes (6). 4/6=2/3 is about 67%, so it would not be better to roll again. You are right to not roll again because you odds of rolling higher or the same are worse than your odds of rolling lower or the same.
A: As André Nicolas commented, whether on not to throw again depends on the precise objective of the game. Vincent also asked whether or not the opponent has the choice of rolling twice makes any difference.
If we assume that all wins give the same payoff, and a loss and a draw are both worthless, then it turns out that it doesn't matter whether the opponent gets to roll twice, you are still better off keeping a four (and re-rolling a three).
If we assume the opponent rolls once, then, if I roll $i$, the probability of him winning is $\frac{6-i}{6}$. If he is allowed to roll twice, then there is a $\frac{6-i}{6}$ chance that he wins on his first roll and likewise on his second roll, thus the probability of the opponent winning is $\frac{6-i}{6}+\frac{i}{6}\cdot\frac{6-i}{6}=\frac{6-i+i(6-1)}{36}=\frac{(6-i)(6+i)}{36}$.
Plotting these curves gives:

The horizontal lines show the expected value, assuming your roll is uniformly distributed, which it would be if you roll again. You can see that if you roll a four, your opponent will have a lower probability of winning than the expectation if you rolled again. So it is better to not roll if you have a four (under the assumptions that loss and draw are worthless and all wins are equal).
A: To answer the short version:
"If I threw 4, how big is the chance I throw 4 or more the next time I throw the dice"
The answer is 50%.
The unspoken questions seems to be "if I don't know what my opponent threw and I threw a 4, should I roll again or 'stick'?"
This boils down to 'what are the chances of improving on a 4?' and the answer to THAT is ⅓ (33.3333%). Since this is less than 50%, you might as well just stick on a 4 (because you have a 33⅓% chance of improving but a 50% chance of getting worse).
A: Interestingly, it seems that, in the limit of n, the Nash optimal strategy for a die with n-faces is to rethrow whenever your first trie is less than $\frac{n}{\phi}$ where $\phi$ is the golden ratio.
You should not rethrow a 4, but if these were 20 sided die, you would want to rethrow anything below 12.
