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Hello, i have some doubts(?) about Weighting the enumerations..

first of all.. the weight table is used when is calculated the potential and strength of the hand, right?.. then



is this code right? or "total += weightTable[i]" should be "total += 1"

second..
According to Davidson paper:


"The relative value for each hand is increased or decreased to be consistent with every opponent action."

i can't see the way of increase the value with WT[E] = WT[E] * PT[A], if WT[E] is between 0 to 1, and in each street is mutiplied by other value between 0 to 1... how this work?

Thanks

If you believe opponent holds a uniformly distributed hand then every weightTable[i] is the same and could be 1. If you believe opponent is more likely to hold some cards than others then some weightTable[i] are greater than others.


To get the correct probabilities you would just normalise WT. As it is the the relative values of WT are correct.

Yes, I understand it, I do this thinking that is more likely to hold some cards than others, my question is more stupid, for example:
weightTable lenght = 3.. I lost (with 0.50 prob. hold it), won (with 0.75 prob. hold it) and tie (with 0.25 prob. hold it)..
then is..
strength = (0.75+0.25/2) / (0.50+0.75+0.25)
or
strength = (0.75+0.25/2) / (1+1+1)
??


Then for example.. if opp hold 72 preflop and call a raise, WT[72] was equal to 1, becomes 0.1.. in flop 725, opp raise, then.. WT[72] = 0.1 * 0.9 = 0.09 ??

Thanks so much

Consider a toy game where both players can hold only A, K, Q or J.
- Hero holds K.
- Villain's weight table is {0.15, 0.5, 0.25, 0.1}.
- Hero's strength = (0.5/2 + 0.25 + 0.1) / (0.15 + 0.5 + 0.25 + 0.1)

thank you, i still confused with the updated method.. for me is more logic this..

I will continue reading.. thanks

It is hard but i love this, all about AI in general, I hope I can contribute in the future

The following is just Bayes Theorem. https://en.wikipedia.org/wiki/Bayes%27_theorem
probability of villain holding ace given action is raise = probability of action is raise given villain is holding ace * probability of villain holding ace / probability of action is raise

Yes, that code looks right, and "total += 1" looks wrong.



With that algorithm, you are right, WT[x] will only decrease. But the word "relative" is crucial here.

Example: suppose there are only two cards, Ace and King, and initially WT[K] = 0.5, WT[A] = 0.5. And suppose that the opponent calls, and the opponent model says that PT[call] for Ace is 40%, PT[call] for King is 80%, so now WT[A] = 0.5*0.4 = 0.2, WT[K] = 0.5*0.8 = 0.4. The probability that the opponent has ace is 0.2/(0.2+0.4) = 33%, king: 0.4/(0.2+0.4) = 67%. And if we have an ace (supposing ace is better than king), then with your strength computation algorithm, total becomes 0.4+0.2 = 0.6, ahead = WT[K] = 0.4, tied = WT[A] = 0.2 so strength = (ahead+tied/2)/total = (0.4+0.2/2)/0.6 = 83%.

aaaaah, i now understand, especially the meaning of normalise ...
Thank you so much