GMP#3: Round 1 Mulligans
Table of Contents
With the initial hand draws investigated in GM&P1 it is time to analyze Round 1 mulligan phase arising right after. Due to plenty of topics which I looked into, the R1 probabilities investigation will be splitted into many chapters, so that articles are more readable and ideas are more clear. So here we start with the chapter one of a serie dedicated to Round 1. If you are interested in improving Gwent gameplay with some math i refer you to “Finding Cards – Probability Cheatsheet”. Otherwise – just enjoy your time .
The field of the study will be exactly R1 mulligans, which means that the basic assumption throughout article would be: the initial hand is already known. In other words, we step into the shoes of a Gwent player taking mulligan decisions, rather than a deckbuilder trying to optimize R1 strategy.
In R1 mulligan phase, blue-coin player (going first) may use 3 mulligans, while his red-coin opponent (going second) only 2. For more insight into the general topic of mulligans I refer you to the preceeding GM&P2 article. Let’s move on to see what exactly happens.
1. Returnable Cards and Irreversibility
A naive view on final 10 cards R1 hand would be that you have access to 13(12) cards, so that your R1 hand would be any combination of 13(12) cards on 10 slots. It is not true and the reason is that the last card drawn in mulligan is irremovable.
If you use N mulligans, then ‘1N’ card is guaranteed in your R1 hand. Obviously only a fraction of combinations would contain ‘1N’ amongst elements.
Consequently 13th(12th) card in the deck is a special one. With the common full mulligan use, these cards are the only ones guaranteed in R1 hand, no matter their quality.
We see the practical impact of this fact everyday: “To risk a brick with the last mulligan or not to risk?”.
2. Space of Predetermined Hands after R1 Mulligans (Deterministic)
As we know from GM&P1, when drawing the initial hand a player is put in front of one of 3268760 scenarios. Incoming mulligan phase will let you to explore a chunk of this space. How big is this chunk depending on the number of mulligans used?
Let’s put this question in other words: how many different hands could you create with N mulligans used, starting from the initial one? Assumption: the order of cards in the deck is already set, all cards are distinguishable, the order of cards in hand is irrelevant.
I arrived to the final answer in two independent ways and I invite you to try your own probs skills as well before moving on. I’ll present a good line of reasoning in the section below.
2.1 Method of the Fixed Card
Let’s move back to Fig.3. The last card drawn during mulligan phase would be always in hand, but the remaining 9 could be varied with no limits. Then any combination of cards (1-11) on red and (1-12) on blue could reside on 9 slots.
Therefore, we simply get:
N(red) = Comb(11,9) = 55
N(blue) = Comb(12,9) = 220
, where N is the number of distinct hands which could be achieved.
2.2 Comparison with the Initial Hands Space
Using 0/1/2/3 mulligans leads to disjoint spaces of 1/10/55/220 different hands. It is easy to see why these are disjoint – after each mulligan new card is guaranteed in hand.
Summing these numbers up we get 286 destinations player may land in on blue coin. As we already know, the exploration is not free– each mulligan adds and removes combinations at the same time. That number should be compared with the total number of initial hands: 3268760.
The deterministic space explorable with R1 mulligans is just 8.75*10-5 ~ 1/10000 of the initial hand space.
3. Space of Possible Hands after R1 Mulligans (Non-deterministic)
Deterministic view on Hand Space after R1 Mulligans told us a lot about player’s impact on the hand shape. We dealt there with a ordered deck, where every card drawn in mulligan phase was strictly determined.
That’s reality, but that’s not what a player sees when taking mulligan decisions. Exact order is unknown to him, so he tries to imagine a set of all possible topdecks instead.
3.1. Topdecks Set
This object is rather simple, the only difference with respect to set of n cards chosen from N is that the order matters in the general case. In other words, (A,B,C) topdecks are different than (C,B,A), because various number of mulligans might be chosen. However, when all mulligans are used, the order in topdecks may not be relevant (we neglect order for FullTopdecks computed below).
Topdecks(blue) = 15!/12! = 2730
Topdecks(red) = 15!/13! = 210
And when all mulligans are used…
FullTopdecks(blue) = 455
FullTopdecks(red) = 105
3.2. Non-deterministic Hand Space after R1 Mulligans
The lack of determinism introduces some variety in possible topdecks and the exact description of this variety was done in the section above. The question is how this variety transfers into the space of possible final hands if the initial one is given.
Very important thing to notice in non-deterministic approach is that space of hands after each mulligan is no longer disjoint! For example, when mulliganing out a card from one and the same slot, the new card X may appear on 1st, 2nd or 3rd draw, leading to exact same hand everytime.
Therefore, we have to look at the problem in a new way in order to avoid duplicates.
In the picture above we go by the number of new cards (N) in the hand, rather than each subsequent mulligan. The sets with 0/1/2/3 new cards are then obviously disjoint, which makes the math easier.
For each N there would be Comb(15,N) ways of choosing new cards and then Comb(10,N) ways of choosing slots in which new cards reside. Attention: 3 new cards are not identical with blue coin, 2 with red etc; instead sum up to N should be used.
When taking R1 mulligans you could think of 59476 (4876) possible starting hands without knowing deck order. It makes for 1.82% of the total hand space on blue coin and only 0.15% on red coin. Let’s conclude with two final graphs: comparison of non-deterministic space with the total space as well as deterministic vs non-deterministic world (on blue coin).
Non-deterministic space is considerably bigger, but still only a small fraction of the total space of R1 hands. The size of the green(red) ball with respect to yellow says how much you could remodel your hand. Apparently, sometimes you cannot escape too far from the doom…
The proportion between Deterministic and Non-Deterministic spaces is in a sense the amount of bias we have to avoid when watching games of Gwent. Streamer trying to obtain the best R1 hand is the big circle, retrospective chat knowing the outcome is the small circle.
4. R1 Mulligans - Probabilities
In the preceeding ‘General Info’ section we learned about spaces associated with the possible hands after mulligans (starting from a given initial hand) as well as the space of possible topdecks. Let’s move to a more practical investigation of mulligan probabilities.
We are in the mulligan phase right after drawing the initial hand. What are the odds of finding crucial cards not present in the hand yet?
4.1. Finding Cards - Probability Cheatsheet
Hypergeometic probabilities are the ultimate answer to the most frequently asked questions about chances of drawing k cards from N, where K of N belongs to the packet of interest. All the basic probs important for practical play are collected in the cheatsheet below (you could also find it as Google sheet here).
The heading refers to mulligan/set of mulligans. 1st means the very first mulligan and so on. 1st∩2nd means two first mulligans and similarily 2nd∩3rd means two last. Columns 1st∩2nd and 1st∩2nd∩3rd are simply full mulligans on red/blue coin, painted with respective colors.
‘To find’ column specifies the exact event considered. ‘1 of 1’ means simply finding a single card, or in other words, finding 1 card out of 1 card package in the deck. Wherever package size is greater than number of cards seeked, ‘>=’ symbol is used. It mean accessing at least k out of K cards; i.e. >=2 of 4 means drawing at least 2 cards from 4 cards package. The reason to use ‘>=’ rather than ‘=’ is that ‘>=’ has way higher relevance in practice.
Finally last 3 rows describes the probability of finding a card mulled out in R1. Sometimes you need to put your golds unplayable in R1 back into the deck in order to contest the round. The calculation is rather straightforward: 100%-(% missing gold in R2)*(% missing gold in R3) in the case of total chances to find the card back.
Of course, all values refer to 25 cards deck without thinning!
4.2. Bricking in R1 Mulligans
The only reason to not take mulligans (especially the last mulligan) is the risk of worsening the hand. Sometimes a perfect initial hand is drawn and there is no reason to fix it. More often though, there is too high risk of drawing a brick – an unplayable or very weak card, usually lowering hand quality by 4 points or more with respect to bronze card mulled out. Some examples of bricks in Gwent v9.0 are:
- one card from of a thinning pair with second already in hand – e.g. Wild Hunt Riders, Archespores, Dun Banners
- cards summoned on the board from the deck – e.g. The Flying Redanian, Madoc, Roach, Knickers…
- cards with a dead effect – e.g. Mage Assassin, Tuirseach Skirmisher/Morkvarg without discard, Necromancy with no possible targets…
- cards being exclusive targets for tutors – e.g. Bear Witcher Adepts as only Portal targets in SK Witchers
4.2.1. Understanding the Problem
When it comes to math, finding a brick in a single mulligan is no different to finding any card from a K cards package. Taking decision on whether to take the last mulligan or not, one can simply use ‘1 of K’ cells from the cheatsheet, where ‘K’ is the number of bricks in the deck.
Similarily you could think about chains of bricks – drawing only bricks one after another – as just drawing ‘2 of K’ on red or ‘3 of K’ on blue. Notice that the probability of getting a brick chain is very low in general. Unless you have optimal hand already, there is no reason to hesitate with the first mulligan on blue (‘3 of 5’ is 2.2% only). On red the risk is considerably higher, but still about 4 bricks in the deck (5.7% brick chain risk) and decent hand is the point in which you could consider dropping first mulligan altogether.
That’s what would happen most often in practice, and as you can see, the answers are very simple. More complex questions would be: ‘I start my mulligan phase: what is the chance I will brick if using all the mulligans available (and no bricks is in the hand yet)?’
4.2.2. Full Mulligan Brick Risk
To answer this question we need to make few remarks:
- Last mulligan will be the decisive factor
- Drawing a brick early in the mulligan phase could lower or even eliminate the bricking risk.
- We assume by default that the brick is always mulled out when possible
Let’s start from a model example of a single brick in the deck.
The chance to brick in the last mull would be 7.7% proven that the brick wasn’t drawn earlier (otherwise – 0%). The chance to draw brick earlier is ‘1 of 1’ in 1st∩2nd, which is 13.3%. In 86.7% of cases there will emerge 7.7% risk, so we got finally 6.7%.
The chance to brick in the last mull would be 7.1% proven that the brick wasn’t drawn earlier (otherwise – 0%). The chance to draw brick earlier is ‘1 of 1’ in 1st, which is 6.7%. In 93.3% of cases there will emerge 7.1% risk, so we got finally 6.6%.
The a’priori bricking risks with 1 brick in the deck are practically the same on blue and red coin thanks to factor described in 3). If you carelessly use every mulligan, then 1 brick in the deck will mean ~6.7% bricking risk for you! (I use rounded numbers from cheatsheet here)
As shown above, the possibility of drawing some bricks early is the main complication and flavor for estimating bricking probabilities. On red it happens always in the first mull, on blue in 1st and 2nd.
Chance to draw brick in 1st is always ‘1 of K’. If it happens, then in 2nd mulligan the chance is ‘1 of (K-1)’. Otherwise simply ‘1 of K’.
P = ‘1 of K’(1st) * ‘1 of (K-1)’(2nd) + ‘0 of K’(1st) * ‘1 of K’(2nd)
Drawing (0,1,2) bricks in first mulligans are separate scenarios to consider. Drawing 0 is simply 1 – ‘>=1 of K’(1st∩2nd). Getting 2 bricks in a row is ‘>=2 of K’(1st∩2nd). Getting exactly 1 would be ‘>=1 of K’(1st∩2nd) – ‘>=2 of K’(1st∩2nd)’. The expression would be too long, so I won’t write it below 😉 Let’s see the results.
4.3. Don't think so.
The problem of bricking risk with full mulligans is a very simple one, but in a disguise. The only real question to ask is: ‘What is the chance for a brick at 2nd(red)/3rd(blue) position of the deck?’. The poll result would look way different in such case. Approaching this problem people often raised arguments of greater chances to brick last mulligan on blue (if not finding brick earlier) as well as higher chances to unbrick thanks to drawing the brick as the 1st/2nd card. Those are good points and as we saw in the section above, could lead to correct results with proper math. Anyway, I have to say it, you’ve been fooled in majority! 😉 The chances of bricking are exactly N/15 (using convention from the tweet) and the number of mulligans doesn’t matter! Could be even one or ten mulligans, the result would not change!
Thanks for reading!
This article is a very abstract one (especially in first chapters) but we will get back to matters of great practical relevance very soon. Most of the follow-up article on R1 hand probabilities is ready and would be published asap!