Error in probability calculations

Quote: ThatDonGuy

I do it slightly differently. First, I calculate the hand values of (a) all 2,598,960 5-card hands, (b) all 270,725 4-card hands, (c) all 22,100 3-card hands, (d) all 1326 2-card hands, (e) all 52 1-card "hands", and (f) the one "zero-card hand," which is the sum of all of the 5-card hands.
Now, for a given hand, I start by assuming the best play is "hold all five cards," with value = the hand's value.
For each card, I calculate the "discard that one card" value by taking the value of the 4-card hand of the 4 cards held, subtracting the 5-card hand value of the original hand (since you can't redraw the discard), and dividing the result by 47 (since there are only 47 ways to draw 1 card).
For each pair of cards, I calculate the "discard those two cards" values by starting with the value of the 3-card hand with the 3 held cards, subtracting each of the 4-card values of those three cards plus one of the discarded cards, and adding back the 5-card hand value of the original 5 cards (this works because of the Inclusion-Exclusion Principle), then dividing by 1081 (as there are 1081 ways to draw 2 cards from 47).
Repeat this for each set of 3 cards held, then each set of 4, and finally, for drawing all 5 cards.
Whichever of these 32 plays returns the highest value is the best play.
link to original post

void storeCount::tCountUpdate(int n, int t, int dt) {
  int row, s;
  for (row = 0; row < HOLD_LIM; row++) {
    s = sierpinski[dt];
    if (s == 2) {
      tCountStore[t] += n;
    } else if (s == 1) {
      tCountStore[t] -= n;
    }
  }
}

  for (dt = 0; dt < HOLD5; dt++) {
    nh = numHeld[dt];
    switch (nh) {
    case 4:
      k = hand.key4(dt);
      for (t = 0; t < maxType; t++) {
        n = hand.array1[t][k];
        if (n) store.tCountUpdate(n, t, dt);
      }
      break;
    case 3:
      k = hand.key3(dt);
      for (t = 0; t < maxType; t++) {
        n = hand.array2[t][k];
        if (n) store.tCountUpdate(n, t, dt);
      }
      break;
    case 2:
      k = hand.key2(dt);
      for (t = 0; t < maxType; t++) {
        n = hand.array3[t][k];
        if (n) store.tCountUpdate(n, t, dt);
      }
      break;
    case 1:
      k = hand.key1(dt);
      for (t = 0; t < maxType; t++) {
        n = hand.array4[t][k];
        if (n) store.tCountUpdate(n, t, dt);
      }
      break;
    case 0:
      for (t = 0; t < maxType; t++) {
        n = hand.array5[t];
        if (n) store.tCountUpdate(n, t, dt);
      }
      break;
    default:
      printf("%2d line %d\n", dt, __LINE__); fflush(stdout);
      exit(0);
      break;
    }
  }

Quote: Mental

I believe that you, I, VPGenius, and the Wiz are all effectively doing the same thing. I just explicitly use the Sierpinski triangle to keep track of the Inclusion-Exclusion math.

Quote: Mental

How do you perform the Inclusion–Exclusion math and keep everything straight?
link to original post

PlayValue = 
(
  Hands2[Hand2(C1, C5)] 
  - Hands3[Hand3(C1, C2, C5)] - Hands3[Hand3(C1, C3, C5)] - Hands3[Hand3(C1, C4, C5)]
  + Hands4[Hand4(C1, C2, C3, C5)] + Hands4[Hand4(C1, C2, C4, C5)] + Hands4[Hand4(C1, C3, C4, C5)] 
  - Hands5[Hand5(C1, C2, C3, C4, C5)]
) / 16215.0;

Quote: ThatDonGuy

Quote: Mental

How do you perform the Inclusion–Exclusion math and keep everything straight?
link to original post

"The long way" - with 32 separate calculations.
For example, to calculate the expected return of discarding the first and fifth cards:

PlayValue = 
(
  Hands2[Hand2(C1, C5)] 
  - Hands3[Hand3(C1, C2, C5)] - Hands3[Hand3(C1, C3, C5)] - Hands3[Hand3(C1, C4, C5)]
  + Hands4[Hand4(C1, C2, C3, C5)] + Hands4[Hand4(C1, C2, C4, C5)] + Hands4[Hand4(C1, C3, C4, C5)] 
  - Hands5[Hand5(C1, C2, C3, C4, C5)]
) / 16215.0;

Here, PlayValue is a double containing the ER for this particular play
C1, C2, C3, C4, and C5 are integers representing the five cards, from 0 to 51 (0 to 12 are A-K of spades, 13-25 are hearts, and so on)
Hand2(C1, C5) converts the pair (C1, C5) to a single number - for example, (0, 1) is 0, (0, 2) is 1, ..., (0, 51) is 50, (1, 2) is 51, (1, 3) is 52, ..., (1, 52) is 99, (2, 3) is 100, and so on
Hands2[n] contains the sum of the values of all of the hands that contain the 2 cards represented by n
Hand3, Hand4, Hand5, Hands3[ ], Hands4[ ], and Hands5[ ] work the same way, but with sets of 3, 4, and 5 cards.
16,215 is the number of sets of 3 cards that are in the 47 cards remaining in the deck
link to original post

Quote: Mental

For the OP or anyone trying to program a VP analyzer, I cannot emphasize how useful it is to understand what this calculation does and how it speeds up a VP analyzer. It takes just 0.22 milliseconds for me to calculate all this information and cache it. Then, I use the cached information over and over again to analyze each of the 152,646 hands. Then it takes 1.4 seconds for me to analyze jacks or better and create a full strategy.
link to original post

Quote: ThatDonGuy

Quote: Mental

For the OP or anyone trying to program a VP analyzer, I cannot emphasize how useful it is to understand what this calculation does and how it speeds up a VP analyzer. It takes just 0.22 milliseconds for me to calculate all this information and cache it. Then, I use the cached information over and over again to analyze each of the 152,646 hands. Then it takes 1.4 seconds for me to analyze jacks or better and create a full strategy.
link to original post

152,646 hands? Shouldn't that be 134,459?
link to original post