Let's do the math: Witcher/Crones

[Tir_na_Lia]

Let's do the math: Witcher/Crones

I have been wondering about the precise percentage to end with a dead card in hand while playing Witchers or Crones.
Since their use fills 3 valuable Silver slots, I made some math about how useful they are.
I'm writing my calculations here so that people skilled in math can correct me if I made a mistake.

Question: How much is the percentage chance to get an opening draw of 0,1,2 or 3 Witcher cards in a deck of 25?

The deck is composed by 25 cards, but we start drawing 10 of them, so the chance to draw a single particular card is 10/25.
Of course, it changes a little when we want to get 3 particular cards all together.

Let's start with the chance to get all 3 witchers, assuming we 'fix' Eskel on draw slot 1.
We have 10 chances out of 25 to draw him, which means that when we want to draw the next one, Lambert, we haven't 10 draws anymore, they became 9 out of 24.
Next is Vesemir, where we have the previous two draws 'reserved' for his pupils, se he gets 8 out of 23 chances to be drawn and to claim he's too old for that shit.
Since calculating multiple chances is a multiplication, we have:

(10/25) * (9/24) * (8/23) = 720 / 13800 (5.217%)

However, all 3 witchers are interchangeble and need no particular draw order, so the example above happens if we keep 'fixed' Eskel, followed by Lambert and Vesemir.
But no matter how we permute their order, as long as the result is getting 3 witchers out of 3,so we multiply the 'Eskel fixed"'s chance by 3 and get:

3 * [(10/25) * (9/24) * (8/23)] = 2160 / 13800 (15.65%)


Let's check now the other scenarios, for example the chance to get any 2 witchers out of 3.

We fix again the reliable Eskel in slot 1, resulting in the easy 10/25.
Next is Lambert, with the old 9/24.
Now it gets a little different, because for the next 8 draws Vesemir will have NOT to be chosen.
Which means that in the 3rd draw, we have 1 Vesemir out of 23 cards, adding a 22/23 multiplication.
Then we have 21/22, 20/21 and so on until we get 15/16.

To resume, we get this:

(10/25) * (9/24) * (22/23) * (21/22) * (20/21) * (19/20) * (18/19) * (17/18) * (16/17) * (15/16) =
(10/25) * (9/24) * [after simplifying] 15/23 = 1350 / 13800 (9,782%)

Again, we have to multiply by 3, because keeping Eskel fixed is not essential, so we get:

3 * [(10/25) * (9/24) * (15/23)] = 4050 / 13800 (29.35%)

But what if we wanted just a single witcher?
We start with the usual 10/25, but from the second draw to the tenth we have 2 undesired cards, so we get:

(10/25) * (22/24) * (21/23) * (20/22) * (19/21) * (18/20) * (17/19) * (16/18) * (15/17) * (14/16) =
(10/25) * [after simplifying] (15*14)/(24*23) = 2100 / 13800 (15.217%)

Again, we multiply by 3:

3 * [(10/25) * (15/24) * (14/23)] = 6300 / 13800 (45.65%)

And if we wanted no witcher at all?
We can simply sum the chances of all scenarios above put together where we considered the presence of a witcher and subtract it from the total:

13800 - (2160+4050+6300) = 1290 (9.35%)


Now, let's verify.

Chance Percentage

Case 1: Three Witchers = 2160 15.65 %
Case 2: Two Witchers = 4050 29.35 %
Case 3: One Witcher = 6300 45.65 %
Case 4: Zero Witchers = 1290 9.35 %
TOTAL = 13800 100.00 %

It seems to work ^_^.

I'm trying now to calculate the exact chance once the mulligan has been used, but I'd like first to get some feedback to be sure I'm not wrong.

 

[frbfree]

I don't know about math. I was a history major. But I do know that yesterday, on two occasions, different Monster players managed to play Crones on me in the 3rd round with only one in their hand. So what's the percentage of only having 1 Crone after the Mulligan and 3 draws?

 

[arz0]

I don't know about math. I was a history major. But I do know that yesterday, on two occasions, different Monster players managed to play Crones on me in the 3rd round with only one in their hand. So what's the percentage of only having 1 Crone after the Mulligan and 3 draws?

well, that is dependent on the cards you play, and what cards are left in your deck. That is not something you can calculate unless you make assumptions about the match.

 

[Esmer]

I don't know about math. I was a history major. But I do know that yesterday, on two occasions, different Monster players managed to play Crones on me in the 3rd round with only one in their hand. So what's the percentage of only having 1 Crone after the Mulligan and 3 draws?

Not that low, actually, since monsters don't have many draw mechanics. So we can safely assume they start with 10 cards in their hand (one of which is a crone after a mulligan) and draw another 5 (2 cards Avallach + 2 cards round 2 + 1 card round 3).

The chances to draw another crone are 2/15 + 2/14 + 2/13 + 2/12 + 2/11 ~ 0,78
So it's 22% chance not to get another crone after drawing 5 cards.

If they don't use Avallach, the chance of drawing a crone is only
2/15 + 2/14 + 2/13 ~ 0,43
So it's 57% chance not to get another crone after drawing 3 cards.

Another realistic option is not to have a crone in your starting hand and trying to draw it later.

 

[frbfree]

well, that is dependent on the cards you play, and what cards are left in your deck. That is not something you can calculate unless you make assumptions about the match.

Not that low, actually, since monsters don't have many draw mechanics. So we can safely assume they start with 10 cards in their hand (one of which is a crone after a mulligan) and draw another 5 (2 cards Avallach + 2 cards round 2 + 1 card round 3).

The chances to draw another crone are 2/15 + 2/14 + 2/13 + 2/12 + 2/11 ~ 0,78
So it's 22% chance not to get another crone after drawing 5 cards.

If they don't use Avallach, the chance of drawing a crone is only
2/15 + 2/14 + 2/13 ~ 0,43
So it's 57% chance not to get another crone after drawing 3 cards.

Another realistic option is not to have a crone in your starting hand and trying to draw it later.

My question was purely rhetorical. But, the fact that people on these forums take me seriously is the #1 reason I keep coming back. Excellent work!

 

[4RM3D]

Glaring through your post, it seems you've forgotten the fact you don't have to mulligan once you got one Witcher/Crone.

 

[Tir_na_Lia]

Glaring through your post, it seems you've forgotten the fact you don't have to mulligan once you got one Witcher/Crone.

Witchers and Crones are Silver cards, and every mulligan redraw has the potential of turning a Bronze card to a Gold or Silver.

Unless you start with all your 4 Gold cards in hand, missing the chance to get that improvement would be worse that ending with a Witcher dead card.

 

[zantclick]

There's one other statistic that wasn't listed here:
Chance that I'll facepalm when I have no Witchers Round 1 and draw 2 of them at the start of Round 2 - 100%

 

[Tir_na_Lia]

WARNING: long post

Now, let's wonder about the chance to get the optimal hand when starting the game.
To avoid giving the enemy CA, the optimal number of witchers is one in the hand, the rest in the deck (Case 3: 45.652%).
The results above cannot be considered definitive, because the mulligan has yet to happen.
Let's consider the 4 separate case to think which choices to do with the mulligan.

We suppose that the aim is to avoid to remain with witcher cards dead in our hand, giving them priority over other discardable non-Gold cards.
Also, since witchers (and Crones) are Silver cards, we will always discard 3 times since it's more profitable to aim for a yet undrawn Gold one.
(in case you start with all your 4 Gold cards, you do need to read this at all ^_^)

Case 1: Three Witchers

This is straightforward, and ofc the optimal draw before the mulligan.
2 discards out of 3 will remove 2 witchers (who will not be able to return during the mulligan),and reach the ideal condition while still be able to improve your hand.
It means that the 15.65% of case one ends 100% of the times with the ideal hand.


For this and the next case we'll use a "reverse strategy" calculation.

Since you always have more possible discards than undesired witchers, the only event where you can end with a dead card is when you draw it as a 3rd and cannot redraw later.


Case 2: Two Witchers

With the first discard you remove one of the witchers.
If during the second draw the 3rd witcher shows up, you discard it and there is no chance to remain with a dead card on the 3rd draw.

Thus, the chance of something going WRONG is the remaining witcher NOT showing on the 2nd, but showing on the 3rd

Let's fix Eskel and Lambert in our hand, to see the chance of Vesemir showing up:

1st draw - Remove Lambert
2nd draw - Vesemir NOT showing up
3rd draw - Vesemir shows up

1st draw is 100%, the second is 13/14, the 3rd is 1/13.
So we have:

(13/14) * (1/13) = 1/14 (7.14%)

We kept fixed Eskel and Lambert, but we could do the same with Eskel+Vesemir or Lambert+Vesemir, so we multiply by 3:

3*1/14 = 3/14 (21.43%)


Case 3: One Witcher

You already have one witcher in your hand, so in the 3rd draw the chance of getting a witcher is 2/13, provided it did not also arrived on the 2nd and you dismissed him.
In that case is 1/13, better to show an example:

Eskel is in your hand, Lambert or Vesemir will arrive on the 3rd draw.

E = Eskel
L = Lambert
V = Vesemir
X = non-witcher card


X-X-L 3.1
X-X-V 3.1
X-V-L 3.2.1
X-L-V 3.2.1
V-X-L 3.2.2
L-X-V 3.2.2

Case 3.1 - If only L or V show up in the 3rd draw, the chance is:

(2/13) * (12/14) * (13/15) = 312 / 2730 (11.43%)


Case 3.2 - L or V shows in the 3rd draw, the other shows in the 1st or second.
It changes a little depending on the wicther coming out in the 1st or 2nd draw:

3.2.1
L and V are interchangeable, it happens double the time so we must double the chances.

2 * [(1/13) * (1/14) * (14/15)] = 28 / 2730 (1.03%)

3.2.2
L and V are interchangeable here, too.

2 * [(1/13) * (13/14) * (1/15)] = 26 / 2730 (0.95%)


...Nooooooow, since as always we kept Eskel fixed but we could have keep fixed any of the three, we multiply the above partials by 3 and get:

3.1 = 312*3/2730 = 936 / 2730 (34.29%)
3.1.1 = 28*3/2730 = 84 / 2730 (3.08%)
3.1.1 = 26*3/2730 = 78 / 2730 (2.86%)

TOTAL = 1098 / 2730 (40.22%)


Case 4: No Witchers

This is similar to Case 3, but with the difference that since we do not start with a witcher, we are fine if one of them shows up the 3rd time.
The problem is only if one shows up in the 1st or 2nd, and then one in the 3rd as well.

Let's fix Eskel in the 3rd draw (he'll be undesired this time!), and we get the following:

E = Eskel
L = Lambert
V = Vesemir
X = non-witcher card

X-L-E 4.1
X-V-E 4.1
L-X-E 4.2
V-X-E 4.2
L-V-E 4.3

4.1
L and V are interchangeable, it happens double the time so we must double the chances.

2 * [(1/13) * (1/14) * (14/15)] = 28 / 2730 (1.03%)

4.2
L and V are interchangeable here, too.

2 * [(1/13) * (13/14) * (1/15)] = 26 / 2730 (0.95%)

4.3
Congrats, you got the worst possible mulligan!

(1/13) * (1/14) * (1/15) = 1 / 2730 (0.037%)

...As always, since as always we kept Eskel fixed but we could have keep fixed any of the three, we multiply the above partials by 3 and get:

4.1 = 28*3/2730 = 84 / 2730 (3.08%)
4.2 = 26*3/2730 = 78 / 2730 (2.86%)
4.3 = 1*3/2730 = 3 / 2730 (0.11%)

TOTAL = 165 / 2730 (6.04%)


To summarize, the following are the chances that something will go WRONG with the mulligan:

Case 1: Three Witchers = 0 %
Case 2: Two Witchers = 21.43 %
Case 3: One Witcher = 40.22 %
Case 4: Zero Witchers = 6.04 %


Now it's time to draw some useful conclusion, combining the chances of having the witchers in your hand with those of the mulligan going wrong.

Question: If my 25-cards deck has 3 witchers, how many chances to I have to end with a dead card?

Case 1 happens 15.65% of the time, but it's always safe.

Case 2 happens 29.35% of the time, and goes wrong in 21.43%
The 21.43% of 29.35 is:
100 : 29.35 = 21.43 : x
x = 6.29%

Case 3 happens 45.65% of the time, and goes wrong in 40.22%
The 40.22% of 45.65 is:
100 : 45.65 = 40.22 : x
x = 18.36 %%

Case 4 happens 9.35% of the time, and goes wrong in 6.04%
The 6.04% of 9.35 is:
100 : 9.35 = 6.04 : x
x = 0.56%


Chance Goes wrong on Result

Case 1: Three Witchers = 15.65 % 0% = 0%
Case 2: Two Witchers = 29.35 % 21.43% = 6.29%
Case 3: One Witcher = 45.65 % 40.22% = 18.36%
Case 4: Zero Witchers = 9.35 % 6.04% = 0.56%

TOTAL = 6.29+18.36+0.56 = 25.21%

Conclusion: You get a dead card 25.21% of the draws, more or less one out of four.

Of course, it's true for every kind of triplet, like Witchers, Crones, Wild Hunt Raiders and so on.

 

[Superpueppi]

Hi,

interesting topic, I was thinking about this today as well, but from the more limited perspective of the 3rd mulligan, which was the most relevant point to improving my play.

I think however that your math isn't fully correct, if I remember correctly these are all cases of probability that are resolved through forumlas with faculty (if that the correct translation, the math operator is !).

Aside from it there are a few aspects, you did not consider, but make the whole thing more complex.
1) the chance to just draw a specific card in the opening hand isnt 1/25, but since the 10 draws of the opening hand are actually consecutive draws, the probability gets higher. I.e. in the draw of your first card in the opening hand it's 1/25, but in the draw for the second card it's already 1/24 and so on.
2) blacklisting is making the remaining card pools smaller, increasing the chance for 2nd crone. E.g. in a consume deck with 3 ghouls, your opening hand has 1 witcher and 1 ghoul - if you mulligan the ghoul, the chance for another witcher in that specific draw is 2/13, because the blacklisting is eliminating two more cards from the potential pool.

Unfortunately I am too rusted in my math skills to calculate all this properly, just wanted to give the input.

To still add some benefit to the topic:
I ran a few scenarios for my consume deck to understand how big the chances are to end up with a 2nd crone in the 3rd mulligan (i.e. stuck with "dead" card as a penalty for being greedy).
All the other stuff doesn't matter much to me from a play perspective, bc typically I use the first two mulligan draws in any case to optimise my opening hand unless it's perfect

The chances ranged from best case 15% (1 crone in hand, no blacklisting - then the 3rd mulligan is 2/13) to worst case 20% (blacklisting one duplicate card in all 3 mulligan draws, last mulligan draw is then 2/10).
Theoretically it could go up to 29%, if you are blacklisting two duplicates in all 3 Mulligan cards, leading to a 2/7 probability to draw the second crone. This is however a rare case, because your deck would include at least 9 cards you don't want in the opening hand (aka crappy deck).

So the question to ask is, how much you can gain in that last mulligan vs the 15-20% risk of the 2nd crone. E.g. if you don't have any gold cards or key cards to beat the specific opponent in hand, might be more worthwhile than if you are just fishing for a more optimal hand.
"Dead" card might be overstated though and maybe the negative feeling to be stuck with a second crone is worse than it actually is. From a play perspective you loose the power of the last card you mulliganed + tempo from the crone combo, but you don't actually give card advantage away or something.

And I think that the chances for 2nd crone go up for avallach or 2nd turn draws, as these are then a 2-consecutive draw senario out of 15 cards with 2 "bad" cards - again would need the proper forumla but I think I read something in the 20-25% range in a related article.
So in any case enough incentive to strongly recommend to use them in the first round.

 

[Tir_na_Lia]

Thank you for the reply and feedback, I was waiting for someone to check my calculations .

It's been ages since I studied maths, so I'm pretty sure I made some mistakes.

1) It might be true, I remember there were two different formulaes regarding consecutive draws in this kind of scenario, with the chance of drawing rising every time.
But in some problems I remember the teacher taught us to solve them by "fixing" a single slot with the relevant card, and proceeding by exclusion; later, multiply the "fixed" scenario according to the interchangebility.
But as I said it has been a lot of time, I should probably check the book if I want to find the appropriate formula.

2) Blacklisting during the mulligan is something I did not take into consideration at all (I learnt of that after writing the post), so the chances should be adjusted as you wrote.

I am glad to read that you too give importance to playing the Crones on first turn, to avoid further draws or Avallac'h messing with your hand.
(btw, is there anyone playing Monsters who does NOT use Avallac'h ?)

Whatever my deck, I always keep one single triplet of Crones/Witchers/Wild Hunt Warriors and play them asap before later drawing can avoid the positive trimming effect.
Happy to see this strategy is common.
Losing a tempo might not be the end of the world and does not give CA, but I had the impression that too many strategy rely upon "telling the last word" with a Dim Bomb/Ragnarok/Horn and potentially change the outcome of the match, and a dead card reduces your effectiveness in rounds 1-2 when you must make the opponent bleed to get CA.

 

[Superpueppi]

You're welcome - has been ~14 years since I had this in statistics at university and rarely used it since, so I also don't know the proper formula and would have to look it up. This fixing logic is related, but I believe it doesn't properly apply to a situation where the pool drawn from is getting smaller. Anyhow, the %ages probably do not change too drastically just from my gutfeel for numbers - which I think to have after working 14years in market research After all, your questions are rather about low/med/high chance understanding to make an informed gameplay decision, not the latest 2% up or down.
And I guess this would actually get into some hard work with the other factors like blacklisting.

Blacklisting is a key advanced thing to work with and quite crucial in many decks to optimize the chances to get a good starting hand.

Crones in the first round is same as with WH warriors typically a good move to increase chances of better draws successively, however Crones are not like the default first play, as WH warriors typically are, unless you want to set up nithral first (much more potent after upcoming patch).
Crones are one of the key tempo plays for monsters at the moment, so playing them at the right time is also key - e.g. in consume decks, which start out slower you keep them until you need them to speed up to your opponents score.
In some cases it is worth not to play them though in the first turn resp too early - e.g. yencon if they would be the highest cards, as it is setting them up for scorch/igni immediately - or vs weather decks as they can counter them too easily with water hag.
I many telling the last card scenarios duplicate cards actually are not that much of an issue - simply because the last card is the deciding factor because of the huge swing it provides without the other player being able to react. But yes, not so great to play a card, that you would have gotten for free of course.

Crones work surpisingly well in weather decks as well, as they can give you a great and most times unexpected tempo move, that weather is typically lacking (aside from the weather itself of course).

Avallach - yes, autoinclude as one of the few options for MS to gain card advantage.

 

[the.real.kowboy]

Firstly the real point of these cards is to deck thin which I think is getting lost in the discourse regarding when to play.

Secondly the equation doesn't change for figuring it out one variable does, its 2/(unseen cards in the deck) per draw. Only thing that changes is the how many unseen cards there are in the deck.

(1-(2/15)) * (1-(2/14)) = you have a 74.3 chance of not drawing a crone on turn 2. Does get cluster fucky with black listing on mulligans but same basic general equasion.

 

[Colt_McBauch]

TRIGGER WARNING for : Long post with math and criticism which is not meant to be ad hominem




Hi there, Checco515!

I'm kind of proficient in statistics, when reading your first post regarding the chances of drawing X witchers/crones on the initial draw before mulligan, i came upon something that felt wrong.

See, when drawing 10 cards out of 25 cards, basically you are dividing your deck into two stacks, one with 10 cards, the other with 15 cards. Three of these cards are witchers. But, when comparing case 1 and case 4 you will notice that your number suggest, that it is more likely to have all three witchers in your 10 hand cards than it is for them to be in your 15 "remaining deck" cards. Drawing zero witchers must be more likely than drawing all three.

The reason your sum of probabilites is 100% in the end is due to a flaw in your methodology. You just sum up all your cases who have at least one witcher and subtract that from your total. It would have been better to calculate it on its own, because firstly this proves nothing and secondly, if you made any errors this projects on your last figure as well, as it did.

Another thing that didn't match my experience was having a dead card 25% of the time, i would have guessed less.







PART I, your first post:




I tried calculating it with a different, more complex method, by dividing the sum of all desired possibilitiies by the sum of all possible possibilities.

Here are the results:

Case 1 (3 witchers drawn):

All possible possibilities to draw 10 cards out of 25 cards is "25!/(10!*15!)" which equals to 3268760.

All desired possibilities are all cases where I draw three witchers and 7 random cards out of the 22 non-witcher cards. There only one possibility to draw three witchers out of three witchers, therefore: 1 * 22!/(15!*7!) = 170544

These two divided are: 170544/3268760 = 0,05173913

The same number you calculated before multiplying it by 3, indicating this multiplication was invalid.




Case 2 (2 witchers drawn):

All possible possibilities are still "25!/(10!*15!)" which equals to 3268760.

Now all desired possibilities are all cases where we draw two of the three witcher cards and 8 random cards out of the 22 non-witcher cards. There are three possibilities of drawing two witcher cards out of three witcher cards, therefore: 3 * 22!/(14!*8!) = 959 310

Again, lets divide those two numbers: 3268760/959 310 = 0.293478261

Same result you got!




Case 3 (1 witcher drawn):

All possible possibilities are still "25!/(10!*15!)" which equals to 3268760.

This time, all desired possibilities are all cases where we draw one of the three witcher cards and 9 random cards out of the 22 non-witcher cards. There are three possibilities of drawing one witcher cards out of three witcher cards, therefore: 3 * 22!/(13!*9!) = 1492260

Once again, lets divide those two numbers: 3268760/959310 = 0.456521739

Also, the same result you got!




Case 4 (no witcher drawn):

Now I could sum up these three cases and subtract them from the count of all possibilities, but this is lazy and not suitable for proving the calculation right.

So, another method! (In my opinion easier to understand than the other three cases)

The first card we draw has to be a non-witcher card, ergo the chance is 22/25

The second card also has to be a non-witcher card, the chance is now 21/24

The third's card probability to be a non-witcher card is 20/23

And so on.

The final tenth card's probability is 13/16

Let's calculate: (22/25)*(21/24)*(20/23)* [rinse and repeat] *(13/16) = (22!/12!)/(25!/15!) = 0.197826087




Let's summarize:




Case 1 (3 witchers drawn): 5.173913%

Case 2 (2 witchers drawn): 29.3478261%

Case 3 (1 witcher drawn): 45.6521739%

Case 4 (no witcher drawn): 19.7826087%

Sum: 1




NOTE: these figures are rounded, so if you add these it's not exactly 1, but if you add the terms described above, it will equal exactly 1)

As you see, case 2 and 3 were correct in your calculation, case 1 was wrong and case 4 subsequently as well.
















PART II, your second post:




CASE 1




Ok, case 1 is pretty straightforward, you just discard two of your witchers and are done, nothing can go wrong, 100 percent of the time, it works every time.







CASE 2




Case 2 is more difficult.




Scenario 1:

You start with two witchers, discard one, get the third witcher in exchange, discard him as well, get a non-witcher card in return, the third replacement move does not matter at all, since you discard another non-witcher card and only are able to get a non-witcher card for it.



redraw #1: witcher

redraw #2: non-witcher

redraw #3: non-witcher



The chance to get witcher #3 on the first redraw is 1/15, rest doesn't matter




Scenario 2:

You start with two witchers, discard one, get a non-witcher card in exchange, discard a random non-witcher card, get a witcher card in return, which you discard with your third move. The card you can get on your third move can also only be a non-witcher card.



redraw #1: non-witcher

redraw #2: witcher

redraw #3: non-witcher



The chance to get a non-witcher on the first redraw is 14/15. The chance to get the third witcher on the second redraw is 1/14 --> (14/15)*(1/14) = 1/15




Scenario 3: (The only bad scenario in case 2!)

You start with two witchers, discard one, get a non-witcher card in exchange, discard a random non-witcher card, get another non-witcher card in return, which you discard with your third move, BUT this time you are unlucky and draw the third witcher as a dead card you must keep.



redraw #1: non-witcher

redraw #2: non-witcher

redraw #3: witcher



The chance to get a non-witcher on the first redraw is 14/15. The chance to another non-witcher on the second redraw is 13/14. The chance to get the third witcher on redraw number three is 1/13 --> (14/15)*(13/14)*(1/13) = 1/15




Scenario 4:

You start with two witchers, discard one, do your remaining redraws, but witcher number three never appears.



redraw #1: non-witcher

redraw #2: non-witcher

redraw #3: non-witcher



(14/15)*(13/14)*(12/13) = 12/15




Sum of all scenarios = 15/15 = 1




Let's summarize, chance for a dead card after the redraw in case 2 is 1/15 = 0,066666666667







CASE 3

Case 3 is one witcher drawn before mulligan.

Here again we have different scenarios to look at.




Scenario 1:

We discard a non-witcher card, get a witcher in return, discard the witcher #2, get witcher #3 in return, discard witcher #3 and get a non-witcher card in return

redraw #1: witcher

redraw #2: witcher

redraw #3: non-witcher

The chance for this scenario is: (2/15)*(1/14) = 26/2730 = 0,00952380952




Scenario 2 (a BAD scenario):

We discard a non-witcher card, get a witcher in return, discard the witcher #2, get a non-witcher in return, discard a random non-witcher card and get the third witcher card in return and have to keep it

redraw #1: witcher

redraw #2: non-witcher

redraw #3: witcher

The chance for this scenario is: (2/15)*(13/14)*(1/13) = 26/230 = 0,00952380952




Scenario 3:

We discard a non-witcher card, get a witcher in return, discard the second witcher card, get a non-witcher in return, discard a random non-witcher card and get another non-witcher.

redraw #1: witcher

redraw #2: non-witcher

redraw #3: non-witcher

The chance for this scenario is: (2/15)*(13/14)*(12/13) = 312/2730 = 0,114285714




Scenario 4 (a BAD scenario):

We discard a non-witcher card, get another non-witcher in return, discard a random non-witcher card, get the second witcher in return, discard the second witcher and get the third witcher we have to keep.

redraw #1: non-witcher

redraw #2: witcher

redraw #3: witcher

The chance for this scenario is: (13/15)*(2/14)*(1/13) = 26/2730 = 0,00952380952




Scenario 5:

We discard a non-witcher card, get a non-witcher in return, discard a random non-witcher card, get a witcher in return, discard the witcher and get a non-witcher in return.

redraw #1: non-witcher

redraw #2: witcher

redraw #3: non-witcher

The chance for this scenario is: (13/15)*(2/14)*(12/13) = 312/210 = 0,114285714




Scenario 6 (another BAD scenario):

We discard a non-witcher card, get a non-witcher in return, discard a random non-witcher card, get another non-witcher in return, discard a third non-witcher and get a witcher on the third redraw.

redraw #1: non-witcher

redraw #2: non-witcher

redraw #3: witcher

The chance for this scenario is: (13/15)*(12/14)*(2/13) = 312/2730 = 0,114285714




Scenario 7:

We discard non-witcher cards 3 times and never draw one of the two remaining witchers.

redraw #1: non-witcher

redraw #2: non-witcher

redraw #3: non-witcher

Chances are: (13/15)*(12/14)*(11/13) = 1716/2730 = 0,628571429




All chances added equal 2730/2730 = 1

Chance for a dead card in case 3 is: 26/2730 + 26/2730 + 312/2730 = 364/2730 = 0,1333333333







CASE 4:




In case 4 we start out without a witcher and our tactic will be to keep the first witcher we draw and discard any follow-up witcher.




Scenario 1 (BAD):

We draw a witcher on the first redraw, keep it, discard a non-witcher and get witcher #2 in return, discard witcher #2 and get witcher #3 and have to keep him.

redraw #1: witcher

redraw #2: witcher

redraw #3: witcher

Chances: (3/15)*(2/14)*(1/13) = 6/2730 = 0,0021978022




Scenario 2:

We draw a witcher on the first redraw, keep it, discard a non-witcher and get witcher #2 in return, discard witcher #2 and get a non-witcher.

redraw #1: witcher

redraw #2: witcher

redraw #3: non-witcher

Chances: (3/15)*(2/14)*(12/13) = 72/2730 = 0,0263736264




Scenario 3 (BAD):

We draw a witcher on the first redraw, keep it, discard a non-witcher and a non-witcher in return, discard a random non-witcher and get a witcher and have to keep him.

redraw #1: witcher

redraw #2: non-witcher

redraw #3: witcher

Chances: (3/15)*(12/14)*(2/13) = 72/2730 = 0,0263736264




Scenario 4:

We draw a witcher on the first redraw, keep it, discard a non-witcher and get a non-witcher in return, discard a random non-witcher and get another non-witcher.

redraw #1: witcher

redraw #2: non-witcher

redraw #3: non-witcher

Chances: (3/15)*(12/14)*(11/13) = 396/2730 = 0,145054945




Scenario 5 (BAD):

We draw a non-witcher on the first redraw, discard a random non-witcher and get a witcher in return, keep it, discard another non-witcher and draw a second witcher at the third redraw.

redraw #1: non-witcher

redraw #2: witcher

redraw #3: witcher

Chances: (12/15)*(3/14)*(2/13) = 72/2730 = 0,0263736264




Scenario 6:

We draw a non-witcher on the first redraw, discard a random non-witcher and get a witcher in return, keep it, discard another non-witcher and luckily don't draw a witcher at the third redraw.

redraw #1: non-witcher

redraw #2: witcher

redraw #3: non-witcher

Chances: (12/15)*(3/14)*(11/13) = 396/2730 = 0,145054945




Scenario 7 (BAD):

We draw a non-witcher on the first redraw, discard a random non-witcher and get another non-witcher in return, discard another non-witcher and luckily draw our first witcher at the third redraw.

redraw #1: non-witcher

redraw #2: non-witcher

redraw #3: witcher

Chances: (12/15)*(11/14)*(3/13) = 396/2730 = 0,145054945




Scenario 8 (SAD):

We redraw three times but never see a witcher's face.

redraw #1: non-witcher

redraw #2: non-witcher

redraw #3: non-witcher

Chances: (12/15)*(11/14)*(10/13) = 1320/2730 = 0,483516484




All chances added equal 2730/2730 = 1

Chance for a dead card in case 4: 6/2730 + 72/2730 + 72/2730 + 396/2730 = 546/2730 = 0,2

Chance for no witcher in hand after mulligan in case 4 = 0,483516484










Summary of summaries:

Now we can calculate our combined chances of having a dead card when using the following discarding tactic:

"Use every redraw you get. If you have more than one witcher, always discard a witcher, if you have one witcher, never discard the witcher"




Case 1 (three witchers before mulligan) occurs in 5.173913%

Chance of having case 1 + dead card after mulligan: 0%




Case 2 (two witchers before mulligan) occurs in 29.3478261%

Chance of having case 2 + dead card after mulligan: 1.95652174%




Case 3 (one witcher before mulligan) occurs in 45.6521739%

Chance of having case 3 + dead card after mulligan: 6.08695652%




Case 4 (no witcher before mulligan) occurs in 19.7826087%

Chance of having case 4 + dead card after mulligan: 3.95652174%

Chance of having case 4 + no witcher after mulligan: 9.56521739%




Total chance of having two witchers if aforementioned tactics are applied: exactly 12%

Total chance of having no witcher after mulligan: 9.56521739%

Total chance of having exactly one witcher: 78.4347826%




One last thing, if you change the tactic to "only discard witchers until you have exactly one witcher, then stop redrawing; if no witchers are present, discard non-witchers until you get one, then stop redrawing" you'll never end up with a dead card, but the chance of not having a witcher at all still remains the same.

 

[Tir_na_Lia]

First of all, thanks for the time you spent to analyze the problem , and in no way I feel offended.
(actually, I'm happy someone wished to delve in this thing)

Drawing 3 Witchers CANNOT be easier than drawing none, if we do the 10 cards/15 cards split, it's certainly as you say.
I wonder why I could not see it at a glance...I'm really too old for this :what2:

The metodology to calculate all possible cases is often very long and boring, though accurate.
That's why I was told to extrapolate the desired case and rule out the rest, provided it is done correctly, as they taught me in high school.
Not much left of that world...

Case 2 and 3 we get the same result, case 4 is surely as you explain, so case 1 must be the remaining percentage.
(I'm lazy, taking the shortcut)

The mulligan works as you wrote, I got a headache just reading it.
(at the beginning I started calculating the mulligan chances doing what you just did, I gave up because I got a painful headache...)

And finally, the puzzle has been solved!
Kudos to you for finding the solution!

So it's 12% the chance to end with a dead card.
It's very useful to know, when assembling a deck.

I'm still adamant about using all mulligan redraws, because sometimes you trade a Bronze for a Gold, and it's a huge advantage.
No reason to stop the mulligan to protect Silver cards if Gold could emerge.
(coincidentally, I can use these odds when I use Bronze Wild Hunt Riders instead of Witchers, in that case it's even more important to redraw)

Thank you very much for your effort!!:cheers:

 

[Anen]

Of course, it changes a little when we want to get 3 particular cards all together.

another mistake. you only need one of them.
i dont have any correction about your stat.

Crones is better because you melee is full with ghoul and nekker and crones is stronger than vesemir and comrades