Is there an elegant way to deal with the Ace in Bl

Question

My kiddo had a homework assignment to write Blackjack in Java. I helped him a little but for the most part he did it all himself and it actually plays pretty well. He even caught an error I didn't see in how it was calculating hand values. However, there is a snag that he hasn't dealt with and every solution I can think of is really complicated and way beyond what he's going to be able to easily code up with his still rudimentary Java skills.

The Ace. In fact, not just one Ace, there's four of them and you could possibly get all four of them in a single hand. How do you elegantly deal with calculating the value of a hand of cards when there's one or more Aces, each of which might be valued at one or eleven. I feel like there should be a graceful algorithm for it, but I'm not seeing it. Of course, part of it could just be that I'm tired, but maybe you can help.

Answer 1

No matter what every ace should be counted to sum value as 11, then when the total sum has reached over 21 subtract 10 from hand, but the thing is you must make sure you keep a count of how many times you subtract 10 and how many times you add 11( an ace),

add 11 >= subtract 10 -must always be satisfied

alorithm example:

int sum=0;
int ace=0;
int subtract=0;
while(!busted or !stay)
{
  Hitme(value);
  if(value=11)ace++;
  sum+=value;
  if(sum>21) 
  {
      if(ace>=1)
      {
         if(ace>=subtract)
         {
           sum-=10
           subtract++;
         }
         else
         {
            busted;
         }
      }
      else
      {
          busted;
      }
  }
  else
  {
    hit or stay;
    //of course if sum== 21 then force stay

  }
}