How to map the 2,598,960 VP hands into 134,459 "unique" hands

1. There are combin(13,5)=1,287 ways all five cards can be in the same suit.
2. There are combin(13,4)*13=9,295 ways four cards can be in the same suit and one singleton in another suit.
3. There are combin(13,3)*combin(13,2)=22,308 ways 3 cards can be in the same suit and 2 in another suit.
4. There are combin(13,3)*combin(13+2-1,2)=26,026 ways 3 cards can be in the same suit and 1 each in two other suits.
5. There are (COMBIN(13,2)*(COMBIN(13,2)+1)/2)*13=40,053 ways 2 cards can be in two separate suits and 1 in a third suit.
6. There are COMBIN(13,2)*COMBIN(13+3-1,3)=35,490 ways 2 cards can be in one suit and 3 singletons in the other 3 suits.

Quote: AceTwo

A completely different approach to evaluate poker hands which I found out many years ago and did the algorithm in Excel VBA is as follows:
Assign each rank a Prime Number from 2 to 41.
A Hand evaluates to a unique number which is the product of all of its card ranks.
A second evaluation is done to each cards suit, ie if all suits are equal then it is flush.

The number of possible hands using this unique number are only 6.175
A look Up table is created using this unique number from lowest to highest (so that binary search can be quickly done)
The table has this Unique Number and the ranking of the hand (and a second ranking for flushes) as follows:
NON FLUSH
4 Kind 156 11-23 13
FH 156 24-36 13
Straight 10 1314-1323

Quote: tringlomane

The method used by "Cactus Kev"?

http://suffe.cool/poker/evaluator.html

Quote: tringlomane

The method used by "Cactus Kev"?

http://suffe.cool/poker/evaluator.html

Quote: camapl

Hey Wiz,

Sorry if this is a high jack... Any chance you are planning to use the techniques in this thread to fashion a more time-efficient analysis of Three-Way Action? (If memory serves, your original code took 80 days!) The reason I ask is that I sometimes play on a version whose pay tables are different from those on your WoO page. By using the hand counts shown on that page (which I believe are sub-optimal for my game), I estimate the return to be only a bit higher than 99%, but I'd love to know the actual return and "optimal-ish" strategy, as this may still be a couponing opportunity. I'd like to know if it's a "good bet" in spite of itself! ;)

Quote: camapl

Yes, 3, not Three. Understood. No worries - I only bumped b/c I realized I put "Wiz", not "Wizard."

Thanks for the response. Hope the trip went well!

Quote: Wizard

Quote: MathExtremist

var list = new Dictionary<long, int>();
List<uint> suits = new List<uint>();

for (var c5 = 4; c5 < 52; ++c5) { 
for (var c4 = 3; c4 < c5; ++c4) {  
for (var c3 = 2; c3 < c4; ++c3) {  
for (var c2 = 1; c2 < c3; ++c2) {  
for (var c1 = 0; c1 < c2; ++c1) {  

    // Set a bit for each card in the bitmask
    ulong mask = 1 << c5 | 1 << c4 | 1 << c3 | 1 << c2 | 1 << c1;
    
    // reused variable - start with a blank slate
    suits.clear();  

    // strip the suits out, add them individually to the container
    suits.Add((uint)((mask >> (13 * 0)) & 0x1fffUL));
    suits.Add((uint)((mask >> (13 * 1)) & 0x1fffUL));
    suits.Add((uint)((mask >> (13 * 2)) & 0x1fffUL));
    suits.Add((uint)((mask >> (13 * 3)) & 0x1fffUL));

    // sort them
    suits.Sort();

    // rebuild the handprint
    ulong key = (ulong)suits[0] | (ulong)suits[1] << (13 * 1) | (ulong)suits[2] << (13 * 2) | (ulong)suits[3] << (13 * 3);
    
    // use the key however you want (e.g., the add-or-increment operation you're already doing.)
    // ...
} } } } }

Thank you. I've been trying to figure out the logic of this but your level is obviously higher than mine.

Can you please go over more slowly, and in as simple English as possible, what these lines do:


ulong mask = 1 << c5 | 1 << c4 | 1 << c3 | 1 << c2 | 1 << c1;

and

suits.Add((uint)((mask >> (13 * 3)) & 0x1fffUL));

and

suits.Add((uint)((mask >> (13 * 0)) & 0x1fffUL));

If the code is different in C++, could I trouble you to translate it.

Thank you!!!

map<long, int> uniqueMap;
vector<int> suits;

for (int c5 = 4; c5 < 52; ++c5) {
for (int c4 = 3; c4 < c5; ++c4) {
for (int c3 = 2; c3 < c4; ++c3) {
for (int c2 = 1; c2 < c3; ++c2) {
for (int c1 = 0; c1 < c2; ++c1) {
  // Set a bit for each card in the bitmask
  long mask = 1 << c5 | 1 << c4 | 1 << c3 | 1 << c2 | 1 << c1;

  // reused variable - start with a blank slate
  suits.clear();

  // strip the suits out, add them individually to the container
  suits.push_back((int)((mask >> (13 * 0)) & 0x1fffUL));
  suits.push_back((int)((mask >> (13 * 1)) & 0x1fffUL));
  suits.push_back((int)((mask >> (13 * 2)) & 0x1fffUL));
  suits.push_back((int)((mask >> (13 * 3)) & 0x1fffUL));

  // sort them
  sort(suits.begin(), suits.end());

  // rebuild the handprint
  long key = (long)suits[0] | (long)suits[1] << (13 * 1) | (long)suits[2] << (13 * 2) | (long)suits[3] << (13 * 3);
  if (uniqueMap.find(key)!=uniqueMap.end()) {
    uniqueMap[key]++;
  } else {
    uniqueMap.insert(make_pair(key, 1));
  }

}}}}}

Quote: ThatDonGuy

I seem to recall somebody asking a while ago if there was an "easy" way to map the 2,598,960 different possible 5-card hands into the 134,459 "unique" 5-card hands (the ones that took suits into account, but didn't depend on "specific" suits -

Quote: mustangsally

this is a beginning for me.
Combin (52,5) = 2,598,960 possible unique poker hands
7,462 = possible distinct poker hands

data, list and thought process from here:
http://suffe.cool/poker/evaluator.html

Hand Value	Unique	Distinct
Straight Flush	40	10
Four of a Kind	624	156
Full Houses	3744	156
Flush	5108	1277
Straight	10200	10
Three of a Kind	54912	858
Two Pair	123552	858
One Pair	1098240	2860
High Card	1302540	1277
TOTAL	2598960	7462

why do so many say there are 134,459 "unique" 5-card hands
(even S.N. Ethier comes up with that value in his book. I also found out he does not teach any more!)

(I can find this almost everywhere, but the above (that there are only 7,462 where suits do not matter except where they do matter) is found only one place?)

I looked at the text file (he linked to) in Excel and it all looks great and adds up to exactly C(52,5)
I do not see what I may have missed.
and if I did not miss anything
Yahoo!
Sally

Quote: mustangsally

this is a beginning for me.
Combin (52,5) = 2,598,960 possible unique poker hands
7,462 = possible distinct poker hands

data, list and thought process from here:
http://suffe.cool/poker/evaluator.html

Hand Value	Unique	Distinct
Straight Flush	40	10
Four of a Kind	624	156
Full Houses	3744	156
Flush	5108	1277
Straight	10200	10
Three of a Kind	54912	858
Two Pair	123552	858
One Pair	1098240	2860
High Card	1302540	1277
TOTAL	2598960	7462

why do so many say there are 134,459 "unique" 5-card hands
(even S.N. Ethier comes up with that value in his book. I also found out he does not teach any more!)

(I can find this almost everywhere, but the above (that there are only 7,462 where suits do not matter except where they do matter) is found only one place?)

I looked at the text file (he linked to) in Excel and it all looks great and adds up to exactly C(52,5)
I do not see what I may have missed.
and if I did not miss anything
Yahoo!
Sally