Deck size & Probability - A case for (slightly) bigger deck

[MistahBoweh]

So you have 4 SoV in a 29 card deck, your opponent has 4 wbt in a 39 card deck. You will draw more shrieks than they will get wbt, statistically speaking.

[MistahBoweh]

Let's do some simulation exercises to check out the impact of increasing deck size on getting the ideal hand.

As the maths can get quite confusing, i going to put out the maths for everyone to check my steps. If i got the maths wrong, pls correct me (and i will get to learn from my mistakes )

Assumptions:-
1) The ideal hand is T1 Kris, T2 Puwen, T3 Jasmine/ Aldon, T4 JD, T5 Raven.
2) No additional card draws from bad santa or blood frenzy (since we just want to check for ideal hand).
3) First player
4) The deck has 4 copies of the cards in the deck.
5) 1 card is sac every turn.
6) The ideal hand need not be drawn on starting hand. It may be draw through card draws.
7) Each turn probability is calculated independently, i.e. at each turn, i assume that the remaining cards are re-drawn. This actually is more lenient bec at each turn, you sorta "reshuffle" your hand and get a chance to re-draw. This assumption is the most controversial, but it remains here purely bec I can't think of a less cumbersome way of calculating the probability. I didn't think it was accurate just to calculate the odds of getting 1x copy of the ideal hand on the starting hand; bec you always have a chance to draw the card you need next turn. So i resorted to doing it in this manner

i'll appreciate it if someone can help me with the last assumption.

Maths for 39 card deck
1) T1 parameters. Hand size = 6, Target card = 4, Total Deck size = 39. Prob of getting at least 0 Kris = 49.7%.
So prob of getting at least 1 kris = 100-49.7% = 50.2%

2) T2 parameters. Hand size = 5, Target card = 4, Total Deck size = 37. Prob of getting at least 0 Puwen = 54.4%.
So prob of getting at least 1 Puwen = 100-54.4% = 45.5%

(note hand size is reduced bec every turn you have a net loss of 1 card (you draw 1 card, sac 1 card and play a card. Deck size is reduced bec of assumption 7; i have to reduced the deck size each turn, bec Kris was played and 1 card was sac the previous turn.)

3) T3 parameters. Hand size = 4, Target card = 8, Total Deck size = 35. Prob of getting at least 0 Jasmine/Aldon = 33.5%.
So prob of getting at least 1 Jasmine/Aldon = 100-33.5% = 66.4%

4) T4 parameters. Hand size = 3, Target card = 4, Total Deck size = 33. Prob of getting at least 0 JD = 66.9%.
So prob of getting at least 1 JD = 100-66.9% = 33.0%

5) T5 parameters. Hand size = 2, Target card = 4, Total Deck size = 31. Prob of getting at least 0 Raven = 75.4%.
So prob of getting at least 1 Raven = 100-75.4% = 24.5%

So estimated prob of getting the ideal hand (be it from starting hand or from draw) = 1.23%

-------------
now lets repeat the same steps, but with a 41 deck size.

Maths for 41 card deck
1) T1 parameters. Hand size = 6, Target card = 4, Total Deck size = 41. Prob of getting at least 0 Kris = 51.7%.
So prob of getting at least 1 kris = 100-51.7% = 48.2%

2) T2 parameters. Hand size = 5, Target card = 4, Total Deck size = 39. Prob of getting at least 0 Puwen = 56.8%.
So prob of getting at least 1 Puwen = 100-56.8% = 43.6%

(note hand size is reduced bec every turn you have a net loss of 1 card (you draw 1 card, sac 1 card and play a card. Deck size is reduced bec of assumption 7; i have to reduced the deck size each turn, bec Kris was played and 1 card was sac the previous turn.)

3) T3 parameters. Hand size = 4, Target card = 8, Total Deck size = 37. Prob of getting at least 0 Jasmine/Aldon = 35.9%.
So prob of getting at least 1 Jasmine/Aldon = 100-35.9% = 64.1%

4) T4 parameters. Hand size = 3, Target card = 4, Total Deck size = 35. Prob of getting at least 0 JD = 68.8%.
So prob of getting at least 1 JD = 100-68.8% = 31.3%

5) T5 parameters. Hand size = 2, Target card = 4, Total Deck size = 33. Prob of getting at least 0 Raven = 76.8%.
So prob of getting at least 1 Raven = 100-76.8% = 23.1%

So estimated prob of getting the ideal hand (be it from starting hand or from draw) = 0.97%

TL;DR
The ideal hand % for 39-card deck is 1.23%
The ideal hand % for 41-card deck is 0.97%.

You can look at the 2 figures and have 2 possible interpretations
a) "Crap, having 2 more cards reduce my % of getting ideal hand by 0.25%. Which equates to 0.79 times less likely to get more ideal hand (0.97/1.23). I'm not going to add that 2 cards in my deck."

b) "Crap, the ideal hand % is already so low at 1.23%. I'm very unlikely to rely on the ideal hand to win, so i might as well add in that 2 additional enrage that i need, but could never find space to include it in."

The stats are shown clearly. You make the decision of whether to add that 2 cards in.

Your math is wrong because hand size stays the same. Sure, you're saccing one card a turn, but that sac will NEVER interrupt your ideal play (ideal was 5 cards, starting hand is 6) so it shouldn't count for the purposes of calculating the odds of having specific cards.

[Atomzed]

Your math is wrong because hand size stays the same. Sure, you're saccing one card a turn, but that sac will NEVER interrupt your ideal play (ideal was 5 cards, starting hand is 6) so it shouldn't count for the purposes of calculating the odds of having specific cards.

Read assumption 7. If you have a good idea of helping out with my assumption, i will gladly listen to it.

[MistahBoweh]

...so? You start with six cards. Assuming you have the ideal play, that number goes up one and then down one each turn. You shouldn't calculate one less card each time, since the number that matters is the number of unassigned cards the player's seen, not the quantity in their hand. The fact that cards are being sacced is irrelevant to the calculation.

I fail to see what this has to do with #7

To clarify:

T1 you have six cards. None of these are known. What are the odds of one being Kris?
T2 You have seen seven cards. One is a Kris. What are the odds of an unknown card being Puwen?
T3 You have seen eight cards. One is Kris, one is Puwen. Of the six unassigned cards, what are the odds that one is Jas?

Etc.

[Ringel]

The ideal hand exercise: Your total draw hits 10 by turn 5. As long as you have the cards you need, you always have a sac. choice.
The probability calculations are quite complex-- I think you need a technique called inclusion exclusion.

K= Kris on turn 1. nK= no Kris by turn 1
P = Puwen on turn 2. nP = no Puwen by turn 2
A = Aldon or Jasmine on turn 3. nA = neither by turn 3
J = Jeweler's Dream on turn 4. nJ = not by turn 4
R = Raven on turn 5. nR = not by turn 5

I'm going to try and calculate this as an exercise (It looks crazy tough) and I expect the probabilities to be pretty small, but I want to point out some other possible interpretations besides the 2 given. (edit: I don't expect the probabilities are going to be so different that they undermine the rest of the conversation)

Possible interpretation: Good decks do not rely on perfect draws. They rely on good enough draws. In order to hit a good enough draw consistently game after game you want your deck as taught as possible.

Possible interpretation: Your measure of perfect cards includes exactly 24 cards. That means you have 15 other cards to take out of the deck to keep your total at 39.

Possible interpretation: Your good deck relies on too many 4 -ofs to be competetive.

[Ringel]

Where I'm running into trouble trying to make an arugment, there really are 3 decks to consider:

Your original deck.
Your new 42 card deck with 2 extra cards.
Your original deck with 2 cards replaced.

And it is the last that is hard to measure. How do you decide how much pain you have taken by substituting the cards instead of just adding them on?
How do you model that? Most or the arguments compare the first two decks, but you really want to compare the last 2.

[Atomzed]

sigh..

Let me gives a better explanation why its insufficient to check for ideal hand, by just looking at the starting hand.

You start with 6 cards.

Assuming you get 1x Kris, 1 Puwen, 1 Jasmine / Aldon, 1 JD 1x Raven, 1x Aeon (which is irrelevant). This is your perfect hand.
Probability wise, it will be P(Kris = 1) x P (Puwen =1) and so on and so forth.
That gives you .39 x .39 x. 41 x . 39 x .39 = 1.04%.

However, you don't need to get 1x Kris, 1x Puwen, 1x Jasmine/ Aldon, 1x JD , 1x Raven in starting hand to play the ideal hand.

For example, I get 2x Kris, 2x Jasmine, 2x Raven.
T1, i sac kris, play Kris.
T2, i draw Puwen, sac Jasmine, play Puwen
T3, I draw KP, sac KP, play Jasmine
T4, I draw JD, sac Raven, play JD
T5, i draw Aeon, sac Aeon, play Raven.

See that the card drawing actually gives u additional chances to draw the cards needed for ideal hand.

Looking at T2 only, bec i have already played Kris, and sac a card, the chances of me drawing a Puwen:-
a) Hand size = 4 from previous hand + 1 new drawn card
b) Deck size = 37

In order to be more accurate in calculating the probabilities, i need to consider the possibility of drawing the correct card at the right turn. And this is what i did to take that into consideration.

Someone well-versed in stats will say that i need to consider the actual identity of the cards drawn. Say in the example, bec I have drawn 2 jasmine, and 2 raven, the chances of drawing a puwen at T2 will actually increase. But to do this, i will have to consider ALL POSSIBILITIES. Which is something beyond my knowledge.

Hence, i went for the easy route, and assume that each turn is independent of the card drawn the previous round.

I hope i making some sense here.

[Atomzed]

The ideal hand exercise: Your total draw hits 10 by turn 5. As long as you have the cards you need, you always have a sac. choice.
The probability calculations are quite complex-- I think you need a technique called inclusion exclusion.

K= Kris on turn 1. nK= no Kris by turn 1
P = Puwen on turn 2. nP = no Puwen by turn 2
A = Aldon or Jasmine on turn 3. nA = neither by turn 3
J = Jeweler's Dream on turn 4. nJ = not by turn 4
R = Raven on turn 5. nR = not by turn 5

I'm going to try and calculate this as an exercise (It looks crazy tough) and I expect the probabilities to be pretty small, but I want to point out some other possible interpretations besides the 2 given. (edit: I don't expect the probabilities are going to be so different that they undermine the rest of the conversation)

Thank you Ringel, thank god at least 1 person understood why I include assumption 7.

And yes, i agree with you that the maths is crazy... you have to consider the P(X=1) given the previous hand. Conditional probability. And you have to include all the possible scenarios. I am not going to attempt it...

Ringel, let me know if you can get the maths done. I assume you will have to use some modelling techniques...

[Atomzed]

Where I'm running into trouble trying to make an arugment, there really are 3 decks to consider:

Your original deck.
Your new 42 card deck with 2 extra cards.
Your original deck with 2 cards replaced.

And it is the last that is hard to measure. How do you decide how much pain you have taken by substituting the cards instead of just adding them on?
How do you model that? Most or the arguments compare the first two decks, but you really want to compare the last 2.

You are right that there are 3 situations. In the OP, the stats are based more on scenario 2 and scenario 3.

let me mull this through on how best to show the stats...hmmm...

[Ringel]

(Ooops, not for me)

Anyway the rest of the inclusion-exclusion calculations are this: P() stands for probability

P(KPAJR) = 1 - P(nK) - P(nP) - P(nA) - P(nJ) - P(nR)
+P(nK,nP) + P(nK,nA)+P(nK,nJ)+P(nK,nR)+P(nP,nA)+P(nP,nJ)+P(nP, nR)+P(nA,nJ)+P(nA,nR)+P(nJ,nR)
-P(nK,nP,nA) - P(nK,nP,nJ) - P(nK,nP,nR) - P(nK,nA,nJ) - P(nK,nA,nR) - P(nK,nJ,nR) - P(nP,nA,nJ) - P(nP,nA,nR) - P(nP,nJ,nR) - P(nA,nJ,nR)
+P(nK,nP,nA,nJ) + P(nK,nP,nA,nR) + P(nK,nP,nR,nJ)+P(nK,nA,nR,nJ)+P(nP,nA,nR,nJ)
-P(nK,nP,nA,nJ,nR)

I'm afraid there is no easier way to get an exact answer.

Let me emphasize-- I expect both numbers to be small and close together. So I don't think the discussion should hinge on the exact answers.