NicolleLouis
commented
6 months ago Hey, thanks for your question.
I have a first questions: where is your 13 factor coming from in the situation computation? A proposition: 3e9 is indeed doable territorry (even if the current code is probably a bit too slow for that to be realistic) but double discard scenario is not imaginable for this scale.
So maybe: this computation could give you the best discard possible:
- Given a hand, generate the 219 potential dicards D.
- For each D, generate all the possible card draws (between 1086008 and 48).
- For each of this draw, compute a hand score representing the closeness to a straight (must not follow this kind of line, but can be a score close to the StraightEngineV3 which will be way faster).
- Compute the average of those scores.
- Return the best discard.
It's not perfect since you have to rely on a non exhaustive computation in the middle (because of multiple discard and the total number of cases with multiple decisions), but you will probably get quite a good result already
Agoraaa
Hi, I've tried to write an algorithm to find the best discard for getting straight before but failed. I want to ask if there is a solution to the problem I'll say, also you might be interested in implementing it in this repo.
I want to find the best discard for getting straight in a 8 hand size & 2 discard scenario. For first draw I just draw 8 cards. Then there'll be 219 possible discard combinations. Then I'll draw (at most) ${52-8 \choose 5} =1086008$ possible card combination for each hand. After that I'll have 1 discard remaining. Thankfully I don't need to iterate through every possible draw for every possible discard because I can count the outs and find the straight probability. For example, if I were calculating flush draw chance for spades with 3 spades in hand can be calculated with hypergeometric distribution with parameters $k=2,n=5,K=10,N=46$. Similar thing can be done with calculating Ace high, King high etc. straight chances independently. This is where my problem occurs, I can't simply add all straight chances to calculate the total straight chance because events are not independent. It's way easier to get a Q-high straight if you have a K-high one. I don't know how to approximate this value in a $O(1)$ manner. But if I could, calculating the best straight discard would require evaluating $219\cdot1086008\cdot13=3\cdot10^9$ possible situations which would take ~30 seconds in a single thread without optimizations. So it is in the doable territory. So my question is, is there an $O(1)$ formula for calculating straight chance?