Connector-Breaker games on random boards

By now, the Maker-Breaker connectivity game on a complete graph $K_n$ or on a random graph $G\sim G_{n,p}$ is well studied. Recently, London and Pluh\'ar suggested a variant in which Maker always needs to choose her edges in such a way that her graph stays connected. By their results it follows that for this connected version of the game, the threshold bias on $K_n$ and the threshold probability on $G\sim G_{n,p}$ for winning the game drastically differ from the corresponding values for the usual Maker-Breaker version, assuming Maker's bias to be $1$. However, they observed that the threshold biases of both versions played on $K_n$ are still of the same order if instead Maker is allowed to claim two edges in every round. Naturally, this made London and Pluh\'ar ask whether a similar phenomenon can be observed when a $(2:2)$ game is played on $G_{n,p}$. We prove that this is not the case, and determine the threshold probability for winning this game to be of size $n^{-2/3+o(1)}$.


Introduction
A positional game is a perfect information game played by two players on a board X equipped with a family of subsets F Ă 2 X , which represent winning sets. During each round of such a game both players claim previously unclaimed elements of the board. For instance, in the pm : bq Maker-Breaker variant, Maker and Breaker take turns claiming up to m (as Maker) or up to b (as Breaker) such elements. Maker wins the game by claiming all elements of a winning set; Breaker wins otherwise. If m " b " 1, the game is called unbiased. Otherwise, we call the game biased with m and b being the respective biases of Maker and Breaker.
Note that Maker-Breaker games are bias monotone in the sense that claiming more elements of the board never hurts the corresponding player. Given pX, Fq and having Maker's bias m fixed, we thus can find an integer b 0 , called the threshold bias, such that Breaker wins the pm, bq Maker-Breaker game if and only if b ě b 0 holds (except for trivial games, where Maker can win before Breaker's first move).
In our paper, we will consider a variant of such Maker-Breaker games played on a graph G sampled according to the Binomial random graph model G n,p (for short we will write G " G n,p ), where we fix n vertices and each edge appears with probability p independently of all other choices. It is well known that for monontone increasing graph properties F this model always comes with a threshold probability p˚(see e.g. [5]) such that For some properties F there is even a sharp threshold in the sense that if p ě p1`op1qqph olds. One such example will be given in the following paragraph.
Maker-Breaker connectivity game. The Maker-Breaker connectivity game is a game variant played on the edges of a graph G with F consisting of all spanning trees of G. Lehman [20] stated that Maker wins the p1 : 1q Maker-Breaker version of this game as the second player if and only if the graph G contains two edge-disjoint spanning trees. Since the complete graph K n can be decomposed into even more spanning trees, a natural question is to ask what happens when Breaker's power gets increased by making his bias larger. Chvátal and Erdős [6] initiated the study of the p1 : bq variant and they could prove that its threshold bias is bounded from above by p1`op1qqn{ ln n. A matching lower bound was later given by Gebauer and Szabó [14]. Now, if in the pm : bq game on K n Maker and Breaker do not play according to a deterministic strategy but instead they play purely at random, the final graph consisting of Maker's edges will behave similarly to a random graph G " G n,p with p " m{pm`bq. It is well known that the (sharp) threshold probability p˚for G " G n,p being connected, i.e. where G " G n,p turns from almost surely being disconnected to almost surely being connected, satisfies p˚" p1`op1qq ln n{n (see e.g. [4], [18]). Surprisingly, when m " 1, the latter corresponds to b " p1`op1qqn{ ln n and thus perfectly matches the threshold bias mentioned above. In other words, for most values of b, a randomly played p1 : bq Maker-Breaker connectivity game on K n is very likely to end up with the same winner as the corresponding deterministically played game. This phenomenon usually is referred to as probabilistic intuition. There is a wide range of other games fulfilling this property as well, for example the perfect matching game, the Hamiltonicity game [19] and the doubly biased pm : bq connectivity game when Maker's bias satisfies m " opln nq [17]. But there also exist games, where this intuition fails, such as the diameter game [2] and the H-game [3].
A different approach to give Breaker more power is to play unbiased, but to thin the board instead. Stojaković and Szabó [23] initiated the study of Maker-Breaker games played on a random graph G " G n,p , their main question being to find the threshold probability p˚at which an almost sure Breaker's win turns into an almost sure Maker's wins. The existence of such a (not necessarily sharp) threshold is guaranteed by the fact that the property of Maker having Hence, even if Connector's bias gets increased, a much denser random graph is necessary for Connector to have a chance at winning almost surely the connectivity game than in the respective Maker-Breaker variant of this game. Since the proof of our theorem is rather technical and the proofs of the upper and lower bound require different techniques, we split the theorem into two parts. 1.1. Organization of the paper. The main focus of this paper is proving Theorem 1.4 and Theorem 1.5. In Section 2 we will give an overview over all required tools. In the Sections 3 and 4 we will describe Breaker's and Connector's strategy respectively. We will also state some lemmas, from which it will follow that the given strategies succeed almost surely for the respective ranges of the edge probability p. We postpone the proofs of these lemmas to Section 5 (for Breaker's strategy) and Section 6 (for Connector's strategy). Finally, we will give some concluding remarks in Section 7.
1.2. Notation and terminology. The game-theoretic and graph-theoretic notation in our paper is rather standard and most of the times it follows the notation of [16] and [24].
For a positive integer n, we set rns :" tk P N : 1 ď k ď nu. For a graph G " pV, Eq we write V pGq and EpGq for the vertex set and the edge set of G, respectively. If tv, wu is an edge from EpGq, we denote it with vw for short. A vertex w is called a neighbour of v in G if vw P EpGq holds. The neighbourhood of v in G is N G pvq " tw P V pGq : vw P Eu, and with d G pvq " |N G pvq| we denote the degree of v in G. Let subsets A, B Ă V pGq be given. We let N G pv, Aq " N G pvqXA be the neighbourhood of v in A, and we set d G pv, Aq " |N G pv, Aq| to be the degree of v into A. Moreover, we let N G pAq :" Ť vPA N G pvq, e G pAq :" tvw P EpGq : v, w P Au and e G pA, Bq :" tvw P EpGq : v P A, w P Bu.
Let two graphs H and G be given. If V pHq Ă V pGq and EpHq Ă EpGq holds, we call H a subgraph of G, and we write H Ă G for short. We also let GzH " pV pGq, EpGqzEpHqq in this case. If there is a bijection f : V pHq Ñ V pGq such that vw P EpHq holds if and only if CONNECTOR-BREAKER GAMES ON RANDOM BOARDS 5 f pvqf pwq P EpGq holds, the two graphs H and G are called isomorphic (denoted with H -G), and we also say that H is a copy of G in this case.
A path P with V pP q " tv 1 , v 2 , . . . , v k u and EpP q " tv i v i`1 : 1 ď i ď k´1u will be represented by its sequence of vertices, e.g. P " pv 1 , v 2 , . . . , v k q. Its length is its number of edges.
Assume that some Connector-Breaker game, played on the edge set of some graph G, is in progress. At any moment during the game, let C be the graph consisting of Connector's edges and let B be the graph consisting of Breaker's edges. For short, also set V C " V pCq, E C " EpCq and E B " EpBq. If an edge belongs to B Y C, we call it claimed; otherwise it is called free.
Given a distribution D and a random variable X, we write X " D for X being sampled according to the distribution D. With Binpn, pq we denote the binomial distribution with parameters n and p. Moreover, with G n,p we denote the Erdős-Renyi random graph model on n vertices and with edge probability p. If X is a random variable, we let EpXq denote its expectation. If E is an event, we let PpEq denote its probability. A sequence of events E n is said to hold asymptotically almost surely (a.a.s.) if PpE n q Ñ 1 for n Ñ 8.
Our main results are asymptotic. Whenever necessary, we will assume n to be large enough. We will not optimize constants, and whenever these are not crucial, we will omit rounding signs.

Preliminaries
2.1. Maker-Breaker Box game. A simple, yet very useful positional game is the following one, introduced by Chvátal and Erdős [6], which usually is helpful to describe strategies that aim to bound the degrees in the opponent's graph. The game Boxpp, 1; a 1 , . . . , a n q is played on a hypergraph pX, Hq, with H " tF 1 , . . . , F n u consisting of n pairwise disjoint hyperedges (called boxes), satisfying |F i | " a i for every i P rns. In every round, BoxMaker claims at most p elements from X that have not been claimed before, while BoxBreaker solely claims one such element.
If, throughout the game, BoxMaker succeeds in claiming all the elements of a box F i , she is declared the winner of the game. Otherwise, i.e. when BoxBreaker succeeds in claiming at least one element in each box, BoxBreaker wins. The following lemma is a well-known criterion for BoxBreaker to have a winning strategy in the Box game (see e.g. [6], [16]). Lemma 2.1. Let a i " m for every i P rns and assume that m ą ppln n`1q, then BoxBreaker wins the game Boxpp, 1; a 1 , . . . , a n q. A winning strategy S is the following one: in every round, BoxBreaker claims an element which belongs to a box that he does not have an element from and which, among all such boxes, contains the largest number of Maker's elements.
In fact, the first sentence in the above lemma is Theorem 3.4.1 in [16], while the mentioned strategy is contained in its proof. As an immediate corollary of the above lemma we obtain the following: 2.2. Probabilistic tools and basic properties of G n,p . In this section we present a few bounds on large deviations of random variables that will be used to identify typical edge distributions in a random graph G " G n,p . Most of the time, we will use the following inequalities due to Chernoff (see e.g. [1], [18]).
Moreover, we will make use of the well-known Markov inequality (see e.g. [18]).
Lemma 2.5. Let X ě 0 be a random variable. For every t ě 0 it holds that P pX ě tq ď EpXq t .
As a first application of Chernoff's inequalities we will prove a few simple bounds on degrees that are very likely to hold in a random graph G " G n,p . Lemma 2.6. Let ε ą 0, p " n´2 {3´ε and let G " G n,p . Then with probability at least Lemma 2.3 we deduce that P pd G pvq ą 2npq ď exp`´1 4 n 1{3´ε˘. Taking a union bound over all possible vertices v, the claim follows.
Proof. Let A be a fixed set of size n 2{3 . Generating G " G n,p yields that for every vertex v P V pGqzA, we have d G pv, Aq " Binp|A|, pq and thus Epd G pv, Aqq " n ε . By Lemma 2.3 we deduce that P`d G pv, Aq ď 1 2 n ε˘ă exp`´1 8 n ε˘. Taking a union bound over all possible v, the claim follows.
3. Breaker's strategy 3.1. Defining bad vertices. For p " n´2 {3´ε we aim to give a Breaker's strategy that a.a.s. isolates a given vertex x from Connector's graph when a p2 : 2q game is played on G " G n,p . In order to do so, we first define iteratively a set B x of vertices that are bad with respect to the aim of isolating x. If x is carefully chosen (which we will manage later) then Breaker has a strategy to make sure that Maker in her move either does not even reach B x , or in case she reaches B x then Breaker can immediately destroy all potential threads. More details will be given later.
Algorithm 1 decribes how B x is constructed.
Algorithm 1: Bad vertex set B x for given vertex x Input : graph G and vertex x P V pGq The following lemma will be crucial for Breaker's strategy.
We postpone the proof of the above lemma to Section 5 and recommend to read Breaker's strategy first.
3.2. The strategy. In the following we prove Theorem 1.4. Let Connector and Breaker play a p2 : 2q game on G " G n,p . We will show that, under the condition that the property described in 8 DENNIS CLEMENS, LAURIN KIRSCH, AND YANNICK MOGGE Lemma 3.1 holds, Breaker has a strategy that isolates a vertex from Connector's graph. Let V r C denote the set of vertices that are covered by Connector's edges at the end of round r. Immediately after Connector's first move, we have |V 1 C | " 3 and thus, by the property from Lemma 3.1 (applied with M " V 1 C ), we find a vertex x such that Algorithm 1 produces a set B x of vertices and a sequence pB x 1 , . . . , B x rx q of disjoint subsets of B x such that the Properties (B 1)-(B 4) hold with M " V 1 C . Notice that, at this point x R V 1 C Y N G pV 1 C q holds, according to (B 4) and since N G pxq Ă B x . In order to simplify notation, let B x 0 :" txu and set B x ăi :" Breaker's strategy is to make sure that for each round r, immediately after his move the following property holds for every free edge vw: Let us observe first that Breaker keeps x isolated in Connector's graph, if he is indeed able to maintain (Q 1) for every free edge after each of his moves. Assume this is not the case, i.e. there is some round r in which Connector reaches vertex x. Then immediately after Breaker's pr´1q st move, we have that (Q 1) holds for every free edge and still x R V r´1 C . From this it follows that immediately before Connector's r th move there cannot be a free edge xw with w P V r´1 C . Indeed, otherwise we would need x P pN G pV r´1 C qzV r´1 C q X B x 0 and by (Q 1) we would get w P B x ă0 " ∅, a contradiction. Thus, in order to reach x during round r, Connector would need do claim a path pw, v, xq of length 2, starting with some vertex w P V r´1 C and ending in x. It then follows that v P pN G pV r´1 C qzV r´1 C q X B x 1 . However, using (Q 1) for the free edge wv at the end of round r´1, this would give w P B x ă1 " B x 0 and thus x " w, a contradiction. Hence, we know that Connector cannot reach x as long as Breaker restores (Q 1) for every free edge. It thus remains to verify that Breaker can indeed do so. We proceed by induction.
For round 1, observe that immediately after Connector's first move, there is no edge between V 1 C and B x Y txu, according to Property (B 4) (with M " V 1 C ). Thus, Property (Q 1) holds at the end of round 1 for every free edge, independent of what Breaker's first move is, as there does not exist any edge vw as described in that property.
Let us assume then, that (Q 1) is satisfied immediately after Breaker's pr´1q st move for every free edge, and let us explain how Breaker restores (Q 1) in the next round. Without loss of generality we may assume that in round r Connector reaches exactly two new vertices, say w 1 and w 2 , i.e. V r C " V r´1 C Y tw 1 , w 2 u. If after Connector's r th move, there exist at most two free edges that fail to satisfy Property (Q 1) (with V C " V r C ), then Breaker claims these edges and by this easily restores that (Q 1) holds for every free edge at the end of round r. So, assume for a contradiction that immediately after Connector's r th move there are at least three free edges that do not satisfy (Q 1). All of these edges need to be incident to w 1 or w 2 , as before Connector's move the Property (Q 1) was true for every free edge (where V C " V r´1 C ). Without loss of generality let w 2 be incident to at least two of these edges, say w 2 v 1 and w 2 v 2 . As these edges fail to hold (Q 1) after Connector's r th Consider first the case that in round r Connector reaches w 2 by claiming a free edge yw 2 with and, since (Q 1) was true for yw 2 has three neighbours in B x ďk (namely v 1 , v 2 and y), a contradiction to (B 2). Consider then the case that in round r Connector does not reach w 2 as in the first case. That is, in round r Connector claims a path py, w 1 , w 2 q with y P V r´1 C and w 1 P N G pV r´1 C qzV r´1 C . We know that w 2 P B k has exactly two neighbours in B x ďk according to Property (B 2), and these neighbours need to be v 1 and v 2 . It follows that the third edge, which does not satisfy (Q 1) immediately before Breaker's r th move, cannot be incident to w 2 and thus needs to be of the form Then v 3 , w 2 P B x are two neighbours of w 1 and hence Algorithm 1 must have added w 1 to B x at some point, say w 1 P B x t . Since again w 2 P B x k has exactly two neighbours in B x ďk and these are v 1 and v 2 , we must have w 1 R B x ďk , i.e. t ą k . But now, by induction, Property (Q 1) was true for the free edge yw 1 at the end of round r´1, and thus y P B x ăt . Moreover, as we assumed v 3 w 1 to be an edge not satisfying (Q 1) after Connector's r th move, we have w 1 R B x ăi 3 and thus i 3 ď t. Hence, we obtain that the three neighbours and y P B x ăt . This again leads to a contradiction with (B 2). which Connector a.a.s. can reach every vertex of G " G n,p . In order to do so, we will first describe a few useful structures, that are typically contained in G even after deleting a few edges and which will help Connector later on to reach any fixed vertex within a small number of rounds.
Recall that E B denotes the set of Breaker's edges at any moment during a Connector-Breaker game, while V C denotes the set of vertices incident to Connector's edges. Moreover, denote with T k the full binary tree with k levels.
T is the orientation where the edges are oriented from the root to the leaves, then for Lemma 4.2 (Base strategy). Assume a p2 : 2q Connector-Breaker game on some graph H is in progress with Connector being the next player to make a move. Let x P V pHqzV C and let k ě 2 be any integer. Moreover, assume that H contains a binary tree T -T k which is good with respect to px, Hq and such that its root r belongs to V C already. Then Connector has a strategy S x to reach x (i.e. to add x to V C ) within at most k rounds.
Proof. We prove the statement by induction on k. For k " 2, by assumption we are given a tree T -T 2 the leaves of which are adjacent with x in HzB, according to Definition 4.1. If one of the leaves belongs to V C , then Connector can take the edge between that leaf and x.
Otherwise, according to (2) we obtain EpT q X E B " ∅. Then, since the root r of T belongs to V C by assumption, Connector can claim one edge between r and a leaf of T , and for her second edge she can claim the edge between that leaf and x. Thus, she reaches x within 1 round.
Let k ą 2 then. Let T -T k be a tree as described in the assumption of the lemma. Denote the root of T with r, let r 1 and r 2 be the neighbours of r in T , and let r 1,1 , r 1,2 and r 2,1 , r 2,2 be the respective children of r 1 and r 2 in T . Each of the vertices r i,j is the root of a subtree T i,j -T k´2 the leaves of which are adjacent with x in HzB. Let for every 1 ď i, j ď 2, and observe that the four sets E i,j are pairwise disjoint. For the first round, Connector makes sure that r 1 and r 2 are added to V C if they do not belong to V C already. This is possible since for every i P r2s we have that r i P V C already before that round or r i r R E B according to (2) in Definition 4.1. After Breaker's following move we know that there are at least two sets E i,j with 1 ď i, j, ď 2 that Breaker did not touch in his move. Taking the union of two such sets, say E i 1 ,j 1 and E i 2 ,j 2 , while identifying r 1 with r 2 if i 1 ‰ i 2 , we obtain a binary induction Connector needs at most k´1 further rounds for reaching x.
Connector's main strategy will be split into different stages. Depending on the number of rounds played so far, she will use similar but different structures that help to increase V C until every vertex is reached. These structures are given by the following lemmas while the proofs of the lemmas will be given in Section 6.

Lemma 4.3 (Good structures for Stage I).
For every constant δ ą 0 there exists an integer k 1 P N such that the following holds. Let G " G n,p with p " n´2 {3`δ , then with probability at least 1´n´1 the following is true for every r, x P V pGq: Let B be any subgraph of G with epBq ď n 1{3 ln n, then the graph GzB contains a copy T of T k 1 such that r is the root of T , x R V pT q and every leaf of T is adjacent to x in GzB.

Lemma 4.4 (Good structures for Stage II).
For every constant δ ą 0 there exists an integer k 2 P N such that the following holds. Let G " G n,p with p " n´2 {3`δ , and let A Ă V pGq be of size n 1{3 , then with probability at least 1´n´1 the following is true for every x P V zA: and such that epBq ď n 2{3 ln n, then G contains a vertex z P N GzB pAq and four vertex disjoint copies T of T k 2 with roots r such that for every P r4s we have: T is the orientation where the edges are oriented from the root to the leaves, then for We postpone the proofs of the above lemmas to Section 6 and recommend to read Connector's strategy first.
4.2. The strategy. In the following we prove Theorem 1.5. Let ε ą 0 be given, and let k 1 and k 2 be integers promised by Lemma 4.3 and Lemma 4.4 (applied with δ " ε), respectively. Set k :" maxtk 1 , k 2 u`2. Before revealing G " G n,p on the vertex set V " rns, we fix an arbitrary set A 1 Ă rns of size n 1{3 and an arbitrary set A 2 Ă rns of size n 2{3 . Then, with probability tending to 1, all the properties from Lemma 4.3, Lemma 4.4 (applied for A " A 1 ) and Lemma 2.7 (applied for A " A 2 ) hold. From now on, let us condition on these. Let Connector and Breaker play a p2 : 2q game on G. In the following we will first describe a strategy for Connector, and afterwards we will show that indeed it constitutes a winning strategy for the connectivity game on G, when we assume all the properties that we conditioned on above to hold. The strategy will be described through the following two stages between which Connector alternates. If at any moment Connector cannot follow the strategy while V ‰ V C still holds, then she forfeits the game. (We will show later that this does not happen).
Strategy description: Fix a vertex r P V to be Connector's start vertex, and set V C " tru before the game starts. As long as V ‰ V C holds, Connector plays as follows, starting with Stage I for her very first move.
Since such a vertex x is always chosen to have maximal Breaker degree among all vertices in V zV C and since such a choice always repeats within at most 2k rounds, we obtain This proves the observation. Observation 4.6. As long as Connector can follow the proposed strategy the following holds: we finally will show that Connector can always follow the proposed strategy. That is, assuming that so far Connector could follow her strategy, we will show that when she fixes her next vertex x according to Stage I or Stage II, she can really add this vertex to V C within at most k rounds.
In order to do so, we will consider three cases.
In this case we have epBq ď n 1{3 ln n according to Observation 4.6. Thus, by the property from Lemma 4.3 we can find a copy T of T k 1 in GzB such that r is the root of T , such that x R V pT q and such that every leaf of T is adjacent to x in GzB. In particular, T is good with respect to px, GzBq. Thus, following the base strategy S x from Lemma 4.2, Connector can reach x within k 1 ď k rounds. and four vertex disjoint copies T of T k 2 with roots r such that for every P r4s we have that zr P EpGzBq and T is good with respect to px, GzBq. In the first round, Connector claims an edge between A 1 and z which is possible as A 1 Ă V C and z P N GzB pA 1 q. Afterwards, consider the pairwise disjoint sets E :" tzr u Y EpT q Y txw : w is a leaf of T u for P r4s. As in the meantime Breaker claims only two edges, there will be at least two of these sets that Breaker does not touch until Connector's next move. Without loss of generality let these be the sets E 1 and E 2 . Then the union tzr 1 , zr 2 u Y EpT 1 q Y EpT 2 q induces a copy of T k 2`1 , which is good with respect to px, GzBq. Therefore, following the base strategy S x from Lemma 4.2, Connector can reach x within at most k 2`1 further rounds. Hence, in total, Connector needs at most k 2`2 ď k rounds in this case.
According to Observation 4.5, we have d B pxq ă ln 2 n before Connector wants to add x to V C . Following the property from Lemma 2.7 (with A " A 2 ) we then conclude that d G px, A 2 q ą n ε{2 ą d B pxq. Therefore, since A 2 Ă V C , Connector immediately can claim an edge leading to x. l

Analysis of Algorithm 1
The aim of this section is to prove Lemma 3.1. For that reason we will prove a slightly more general lemma, Lemma 5.1, from which Lemma 3.1 will follow. For Lemma 5.1 we are going to apply Algorithm 1 to a set A " tx 1 , . . . , x t u of vertices, later choosing one of them carefully to obtain a vertex x as promised by Lemma 3.1. That is, we first fix x 1 and apply Algorithm 1 in order to determine the set B x 1 , then we repeat the algorithm for x 2 and so on. Amongst other properties we will obtain that it is very likely that all the sets B x j are pairwise disjoint and satisfy certain degree conditions. To simplify notation we set B pj,iq :" That is, B pj,iq is the set of all bad vertices that are determined immediately after B x j i is created. In particular, B pt,rx t q " Ť xPA B x is the union of all bad vertices after the algorithm is proceeded for all vertices x j . Moreover, we let apj, iq :" denote the pair coming immediately before pj, iq in lexicographic order, for pj, iq ‰ p1, 1q.
. .  . , x t the following holds for every j P rts and i ďr j :" mintr x j , r1{εsu: k˘" 2, proving (B 2). Let us prove (ii) then. For any k ă j, we have B x k Ă B apj,1q by Definition (5.1) and since Thus, using Property (P 4) we conclude that B x j and B x k are disjoint. Moreover, since B x k Ă B apj,iq we also obtain that N G pB x k q Ă N G pB apj,iq q. Thus, using Property (P 4) again, we get that G does not have any edges between B x j and B x k . As a consequence we have that every vertex v which is adjacent to but not contained in B x j for some j P r7s needs to be element of V zB pt,rx t q . However, according to Property (P 5) and since r x j ď r1{εs holds by Property (P 6), we obtain N 3 pt,rx t q " ∅ for large enough n. This implies that every vertex of V zB pt,rx t q is adjacent to at most two of the sets B x j with j P r7s.
We conclude that at most 3 of the pairwise disjoint sets B x j may contain a vertex of M . If a vertex v P M belongs to some set B x j with j P r7s, then v R B x k Y N G pB x k q for every k ‰ j. If otherwise a vertex v P M belongs to V zB pt,rx t q , then it is adjacent to at most two of the sets B x j .
Hence, there are at most six vertices x P A such that M X pB x Y N G pB x qq ‰ ∅. This proves statement (ii).
Proof of Lemma 5.1. For the proof of Lemma 5.1 we expose the edges of G " G n,p step by step with respect to the given algorithm, and only during the process we choose the vertices of A randomly. To be more precise, we proceed as follows: We first choose x 1 uniformly at random from V pGq " rns and then apply Algorithm 1 for x 1 . Once, Algorithm 1 has been applied for x j´1 and afterwards B x j´1 is determined, we choose x j uniformly at random from rns and apply Algorithm 1 for x j . While doing this, we always expose only those edges which have not been exposed yet and which are needed to determine the next set B x j i in the algorithm. For example: When applying the algorithm for x 1 , we first expose only the edges incident to x 1 so that we are able to determine B x 1 1 . Once this set is fixed, we expose all edges incident to B x 1 1 that have not been exposed yet, so that we can find B x 1 2 . We then expose all edges incident to B x 1 2 that have not been exposed yet, and so on.
For the analysis of the algorithm, we consider the pairs pj, iq, with j P rts and i P rr j s, in lexicographic order. We consider the following event: for all pairs until and including pj, iq the Properties (P 1)-(P 5) hold, and the Property (P 6) is true for all k ă j . Proof. Observe first that for every j P rts the events E pj,rx j q and E pj,r j q are equivalent. Indeed, by definition E pj,rx j q implies E pj,r j q , since r x j ěr j . Now, let E pj,r j q be given and let us explain why E pj,rx j q follows then. If we assume that the latter does not hold, thenr j ‰ r x j , and by definition ofr j we then haver j " r1{εs ă r x j . Applying (P 2) for pj,r j q, which is given under assumption of E pj,r j q , we obtain B x j r j " ∅. But this means that Algorithm 1, when processed for vertex x j , must have already stopped, i.e. r x j ăr j , a contradiction.
Moreover, by looking at the above argument more carefully we see that whenever one of the events E pj,rx j q and E pj,r j q holds, we must have r x j "r j ď r1{εs.
For every j P rts we now conclude that P´E pj,rx j q¯" P`E pj,r j q˘ďr j ÿ i"1 P´E pj,iqˇEapj,iq¯`P`Eapj,1qp
Applying the above inequality recursively we finally obtain P`E pt,rx t q and r xt "r t˘" P`E pt,rx t q˘ă t¨1 0 ε n´ε 3 ă n´ε 4 as claimed.
It thus remains to prove (5.2). We start with a few observations.

Observation 5.3. If Algorithm 1 adds a vertex v to the set
provided n is large enough. Thus, (i) follows.
For (ii) observe that, according to the algorithm, no vertex from Ť kďi´1 B x j k can be added to B x j i . Moreover, using Property (P 4), no vertex in B apj,1q has a neighbour in B x j i´1 (or in tx j u in the case when i " 1, because of (P 1)), while every vertex being added to B x j i needs to have such a neighbour according to Observation 5.3 (or since B Now, for each j P rts and i P rr j s we will prove (5. recall that under condition of E apj,1q " E pj´1,r j´1 q we also have that r x j´1 "r j´1 (as shown in the proof of Claim 5.2), making sure that (P 6) holds for pj, 1q as well. We discuss each of the Properties (P 1)-(P 5) seperately.
Property (P 1): The statement is trivially true for j " 1. So, let j ą 1 and let us condition on E apj,1q . The vertex x j is chosen uniformly at random from rns after Algorithm 1 has been applied for x 1 , . . . , x j´1 and B apj,1q was determined. Now, conditioned on E apj,1q , we have |B apj,1q Y N G pB apj,1q q| ă n 2{3 due to Observation 5.4. It thus follows that P`(P 1) fails for jˇˇE apj,1q˘ă n´1 3 ă n´ε 3 . So, let i ą 1 from now on and consider the moment immediately after B x j i´1 was determined, i.e. when all remaining edges incident to B x j i´1 get exposed in order to determine B x j i . When we condition on E apj,iq , only vertices from v P V zB apj,iq can be added to B x j i according to Observation 5.4. Moreover, before B x j i´1 was determined, for every vertex v P V zB apj,iq all the edges towards B x j i´1 have not been exposed so far. Now, if a vertex v P V zB apj,iq is added to B Conditioned on E apj,iq , the expected number of vertices in (i) is smaller than n¨|B x j k˘ˇă n 2{3´ε according to Observation 5.4, we get that the expected number of vertices in V zB apj,iq satisfying (ii) is at most n 2{3´ε¨| B x j i´1 |¨p (P 2) ă n 1{3´pi`5qε{3 . Summing up, we get that the (conditional) expected size of B i q has been exposed before. With probability at least 1´n´4 ε{3 we get |B x j i | ă n p1´iεq{3 , according to (5.4). If we condition on the latter, the expectation of e G pB ă n´4 ε{3 . Thus, using Markov's inequality (Lemma 2.5), Property (P 4): Let i " 1. If j " 1 then the statement is trivially true. Otherwise, we know from (5.3) that, under condition of E apj,iq , we have x j R B apj,1q Y N G pB apj,1q q with probability at least 1´n´1 {3 . This implies B x j 1 X B apj,1q " ∅ and thus it remains to check that it is unlikely to have a vertex from N G pB apj,1q qzB apj,1q landing in B x j 1 . Note that before B x j 1 gets determined none of the edges between N G pB apj,1q qzB apj,1q and x j has been exposed so far, when x j R B apj,1q Y N G pB apj,1q q.
Let i ą 1 then. Under assumption of E apj,iq , we have that B x j i´1 X N G pB apj,1q q " ∅ according to (P 4). But then, according to Observation 5.3, no vertex from B apj,1q is added to B It thus remains to check that it is unlikely to have a vertex from N G pB apj,1q qzB apj,1q landing in B x j i . Using that B x j i´1 X B apj,1q " ∅ by (P 4), we note that before B x j i gets determined, no edge between N G pB apj,1q qzB apj,1q and B x j i´1 has been exposed. Now, applying Observation 5.3, a vertex v from N G pB apj,1q qzB apj,1q is added to B x j k q ě 1. Hereby, again using Observation 5.4 as well as (P 2), the (conditional) expected number of vertices in (i) is at most |N G pB apj,iq q|¨|B Property (P 5): Consider first the case when pj, iq " p1, 1q. The bound on |N 0 p1,1q | is trivially true. So, let s ě 1. Immediately after B p1,1q " N G px 1 q is determined, none of the edges between V zB p1,1q and B p1,1q has been exposed. Moreover, according to Lemma 2.6, with probability at least 1´expp´n 1{3´2ε q we have |B p1,1q | ă n p1´εq{3 . Thus, if we condition on that bound, the expected size of N s p1,1q is at most n¨`|B p1,1q |¨p˘s ă n p3´sp1`4εqq{3 . It follows that P´(P 5) fails for p1, 1q¯ď So let pj, iq ‰ p1, 1q from now on. Again, the bound on |N 0 pj,iq | is trivially true. Under condition of E apj,iq we have |B x j i | ă n p1´iεq{3 with probability at least 1´n´4 ε{3 , according to (5.4). Condition on the latter from now on. Given that E apj,iq holds, we additionally geťˇN s apj,iqˇ(  Now, for s P r3s, if a vertex v ends up being in N s pj,iq zN s apj,iq then, by definition, v P V zB pj,iq Ă V zB apj,iq and d G`v , B apj,iq˘" t for some t ă s. But this means that v P N t apj,iq and, in order to be added to N s pj,iq , the vertex v needs to get at least s´t edges towards B x j i (which get exposed only after B x j i has been determined, since v P V zB pj,iq ). We conclude that the (conditional) expected size ofˇˇN s pj,iq zN s apj,iqˇi s at most where the first inequality uses (P 2) and (P 5), and the last inequality uses that i ďr j ď 2ε´1. for every s P r3s, and thus P`(P 5) fails for pj, iqˇˇE apj,iq˘ă n´4 This finishes the proof of Lemma 5.1.
6. Good structures for Connector 6.1. Technical Lemma. Throughout Section 6, we will consider the sequence pα i q iPN given by and note that α 1 " 1 and α i`1 " 2pα i`1 q hold. In order to simplify our argument, in the next lemma we will consider ε to be a real number of the form ε " 1{p9¨2 k´2´3 q with k P N. Note that this yields Given G " G n,p and x P V pGq, with high probability the following lemma will provide us with a suitable subgraph H of G which later (see e.g. Claim 6.2) will turn out to contain a bunch of copies of T k that help to prove Lemma 4.3 and Lemma 4.4.
In order to simplify notation, we set Starting with vertex disjoint sets V pi,j, q Ă V pGq for pi, j, q P I k we will iteratively find wellbehaving subsets M pi,j, q of those which later turn out to be good candidate sets for embedding the vertices of T k , even when some edges of G are not allowed to be used. Hereby, the tuple pi, jq will represent the position of a vertex in the desired tree, while the component is used in order to apply our argument on disjoint subsets of V pGq labeled with distinct indices , so that we will be able to find a few edge-disjoint copies of T k . Lemma 6.1 (Decomposition of G n,p ). Let k ě 3 be an integer, let ε " p9¨2 k´2´3 q´1 and let n P N be large enough. Let G " G n,p with p " n´2 {3`ε . Fix x P V pGq and let be any partition of V pGq such that |V pi,j, q | " n{p2 k`4 q holds. Then with probability at least such that the following is true for every pi, j, q P I k : where we define M p0,j, q :" txu for every j P r2 k s and P r4s.
M (i,j, ) Proof. Fix x P V pGq and let be any partition of V pGq such that |V pi,j, q | " n{p2 k`4 q. We will iteratively construct a subgraph H together with sets M pi,j, q Ă V pi,j, q through Algorithm 2 and we will prove that with probability at least 1´expp´ln 1.5 nq the algorithm succeeds in creating these in such a way that all the Properties (D 1)-(D 6) hold. Again, we will expose the edges of G " G n,p while the algorithm is running. That is, whenever a new set is going to be determined, we will only expose those edges which have not been exposed before and which are needed for the corresponding step in the algorithm.

Algorithm 2:
Good subgraph H for vertex x Input : graph G, vertex x P V pGq, partition of V pGq as described in Lemma 6.1 Output : subgraph H, sets M pi,j, q M p0,j, q :" txu for every j P r2 k s and P r4s; V pHq :" txu and EpHq :" ∅ ; if |M pi,j, q | ě n 1{3`α i ε ln´2 α i n then remove randomly selected vertices from M pi,j, q until |M pi,j, q | " n 1{3`α i ε ln´2 α i n; Let E t be the event that the Properties (D 2) and (D 4) hold for every i ď t (and every j ď 2 k´i and P r4s). In the following we will show that PpE 1 q ď exp´´n 1 3¯a nd P´E tˇEt´1¯ď exp`´2 ln 1.5 nf or every 2 ď t ď k. Observe that, once these two inequalities are proven, we can deduce that PpE k q ě 1´k exp`´2 ln 1.5 n˘ě 1´exp`´ln 1.5 n˘, from which Lemma 6.1 follows.
The event E 1 : Property (D 4) holds trivially when i " 1. For (D 2) observe that a standard Chernoff argument (apply Lemma 2.3 and union bound) yields that, with probability at least 1´expp´n 1{3 q, for every j P r2 k´1 s we have holds, and let k 1 " k`1. We will prove the lemma for p " n´2 {3`ε and notice that by the monotonicity the lemma then follows for p " n´2 {3`δ as well. As before, we set I k " pi, j, q P N 3 : 1 ď i ď k, 1 ď j ď 2 k´i , 1 ď ď 4 ( .

DENNIS CLEMENS, LAURIN KIRSCH, AND YANNICK MOGGE
Let V " rns be the vertex set, and let x, r P V be fixed. Before exposing all the edges of G " G n,p we fix a partition rns " txu Y ď pi,j, qPI k V pi,j, q Y R of the vertex set such that r P R and |V pi,j, q | " n{p2 k`4 q for every pi, j, q P I k . We will show that with probability at least 1´n´3 a random graph G " G n,p is such that for every subgraph B with epBq ď n 1{3 ln n the graph GzB contains a copy T of T k 1 as desired, additionally satisfying that V pT q Ă prnszRq Y tru. Taking a union bound over all choices of r and x, it then follows that the property described in Lemma 4.3 holds with probability at least 1´n´1 .
In order to do so, we first expose the edges of G on rnszR. By Lemma 6.1 we know that with probability at least 1´expp´ln 1.5 nq there exist subsets M pi,j, q Ă V pi,j, q and a subgraph H Ă G on the vertex set Ť pi,j, qPI k M pi,j, q Y txu such that all the Properties (D 1)-(D 6) hold. Let E H be the event that such a graph H with vertex sets M pi,j, q exists. From now on, we will condition on E H to hold. Recall the definition of M in (6.3) and L i in (6.4). We first observe that there must be many copies of T k in H with all leaves being adjacent to x in H. Claim 6.2. For every P r4s and every v P M pk,1, q there exists a tree T v -T k such that Proof. Label the vertices of T k in such a way that the root gets label pk, 1q and for every vertex with label pi, jq and i ą 1 its two children get the labels pi´1, 2j´1q and pi´1, 2jq, respectively.
Let P r4s and v P M pk,1, q be given. Applying Property (D 3) iteratively we find an embedding of T k into H, such that the root of T k is mapped to v, and such that the vertex with label pi, jq in T k is mapped to a vertex in M pi,j, q Ă M X L i . Let T v denote the resulting copy of T k in H.
Also every leaf of T k has some label p1, jq with 1 ď j ď 2 k´1 and thus the corresponding leaf in T v is contained in M p1,j, q . By Property (D 5) it follows that the latter is adjacent to x in H.
Thus the claim follows.
From now on, for every v P L k " Ť Pr4s M pk,1, q fix a tree T v as described above. Since, for the property that we are aiming for, we need to have control on how many such trees become useless when some edge is removed from H, we define S e :" tv P L k : e P EpT v q Y txw : w is a leaf of T v uu (6.5) for every edge e P EpHq. Under assumption of the event E H , we next deduce that S e does not get too large. Proof. Let e P EpHq, then e is incident to at least one vertex y P M pi,j, q for some pi, j, q P I k , because of Property (D 6). For every vertex v P S e there must exist a path P in T v leading from y to v. According to Property (T 4), this path P needs to be of the form P " py, v i`1 , v i`2 , . . . , v k´1 , vq with v s P L s for every i`1 ď s ď k´1. Following Property (D 4) and Property (D 6), we have d H pv s , L s`1 q ď n pα s`1´αs qε for every i`1 ď s ď k´1 and d H py, L i`1 q ď n pα i`1´αi qε . Thus, the number of all possible such y-v-paths P that belong to some T v with v P L k is bounded from above by where the last inequality holds by (6.2) and since α i ě α 1 " 1. This proves the claim.
We next expose the edges incident to r P R. By a standard Chernoff argument (apply Lemma 2.4 and union bound), we conclude that with probability at least 1´expp´0.5 ln 2 nq it holds that d G pr, S e q ă ln 2 n (6.6) for every e P EpHq, and d G`r , M pk,1, q˘ą 1 2 p|M pk,1, q | (D 2) " n´1 3`p α k`1 qε 2 ln 2α k n p6.2q " n 1 3`ε 2 ln 2α k n (6.7) for every P r4s. Conditioning on (6.6) and (6.7) as well as the event E H , which together hold with probability at least 1´expp´0.5 ln 1.5 nq ě 1´n´3, it remains to prove that for every subgraph B with epBq ď n 1{3 ln n the graph HzB Ă GzB contains a copy T of T k 1 as was described at the beginning of the proof.
So, let B of size epBq ď n 1{3 ln n be given. Set S B :" Ť ePB S e and observe that ă |B|¨ln 2 n ă n 1 3 ln 3 n .
For every P r4s we thus obtain d GzB`r , M pk,1, q˘ě d G`r , M pk,1, q˘´e pBq p6.7q ą n provided n is large enough. Hence, for every P r4s, we find a vertex x P N GzB prq with x P M pk,1, q zS B . By the choice of T x (Claim 6.2), the tree T x is a copy of T k in H with x being its root, such that every leaf of T x is adjacent to x in H Ă G and such that V pT x q Ă M .
Moreover, by the definition of S e in (6.5), and since x R S B , we find that EpT x q Y txw : w is a leaf of T x u Ă EpHqzEpBq .

DENNIS CLEMENS, LAURIN KIRSCH, AND YANNICK MOGGE
Now, set T to be the union of T x 1 , T x 2 and tx 1 r, x 2 ru. Then, since M 1 X M 2 " ∅, we get that T Ă GzB is a copy of T k`1 " T k 1 with all its leaves being adjacent to x in GzB. Moreover, r is the root of T , and we have x R V pT q, since x R M Ą V pT x q.

Finding good structures -Part II.
Proof of Lemma 4.4. Let δ ą 0 be given. Fix k P N such that δ ě ε :" 1{p9¨2 k´2´3 q holds, and let k 2 " k. Again, it is enough to prove the lemma for p " n´2 {3`ε and to use the monotonicity of the desired property. Again, we set Let V " rns be the vertex set, let A Ă V with |A| " n 1{3 be fixed, and let x P V zA. Before exposing the edges of G " G n,p let be any partition of the vertex set satisfying A Ă R and |V pi,j, q | " n{p2 k`4 q for every pi, j, q P I k .
We will show that with probability at least 1´n´2 a random graph G " G n,p is such that for every vertex set M Ă V ztxu and every subgraph B, with epBq ď n 2{3 ln n and d B pvq ď ln 2 n for every v P V zM , the graph GzB contains a vertex z and four binary trees T as described by the lemma, additionally satisfying that z P RzA and V pT q Ă V :" Ť pi,jq V pi,j, q for every P r4s. Then, taking union bound over all choices of x, Lemma 4.4 follows immediately.
In order to do so, we first expose the edges of G on rnszR. By Lemma 6.1 we know that with probability at least 1´expp´ln 1.5 nq there exists a subgraph H Ă G with V pHq Ă rnszR as well as subsets M pi,j, q Ă V pi,j, q satisfying the Properties (D 1)-(D 6). Let E H be the event that such a graph H with vertex sets M pi,j, q exists. From now on, we will condition on E H to hold. Following Claim 6.2 we then know that for every P r4s and every vertex v P M pk,1, q there exists a tree T v -T k in H such that V pT v q Ă V and such that all the leaves of T v are adjacent to x in H.
Proof. For (a) notice that d G pv, M pk,1, q q " Binp|M pk,1, q |, pq and thus Epd G pv, M pk,1, q qq " |M pk,1, q |p Applying Chernoff's inequality (Lemma 2.3) and a union bound over all choice of P r4s and v P R 1 we get Pp(a) failsq ď 4n expp´n 1{3 q .
For (b), observe first that |R 1 | ą n 4 . Applying Chernoff's inequality (Lemmas 2.3 and 2.4) and a union bound, we see that with probability at least 1´expp´0.9 ln 2 nq we have and d G pv, Aq ă ln 2 n for every v P R 1 .
For (c) we start by observing that according to a Chernoff and union bound argument (applying Lemma 2.4) the following property holds with probability at least 1´expp´nq: for every subsets X Ă L k and Y Ă R 1 of sizes |X| " n 1´2ε and |Y | " n 2{3`ε{2 we have e G pX, Y q ă n 1`ε{2 . Given that property, assume that (c) fails to hold. Then there exist Q Ă L k of size n 1´2ε and a subset Big 1 Q Ă Big Q with |Big 1 Q | " n 2{3`ε{2 . We then have e G pQ, Big 1 Q q ă n 1`ε{2 under assumption of the property mentioned above; but also e G pQ, Big 1 Q q ą n 1{3`ε{2¨| Big 1 Q | " n 1`ε according to the definition of Big Q in (6.8), a contradiction. Thus, Pp(c) failsq ď expp´nq.
Finally, summing up all the failure probabilities, that were obtained above, the claim follows.
Conditioning on the event E H and on the properties described in Claim 6.4, it remains to prove that for every vertex set M Ă V ztxu and every subgraph B, with epBq ď n 2{3 ln n and d B pvq ď ln 2 n for every v P V zM , the graph GzB contains a vertex z and four binary trees T as described at the beginning of the proof. So, let any such M and B be given. Similarly to the proof of Lemma 4.3 consider the set S e " tv P L k : e P EpT v q Y txw : w is a leaf of T v uu for every edge e P EpHq, and let S X :" Ť ePX S e for every X Ă V . Further, let B i :" te P B X H : e " ab with a P L i´1 zM and b P L i u and B˚:" Similarly to Claim 6.3, we prove the following Claim 6.5. Let E H hold. Then |S B˚| ď n 1´2ε .
Proof. We first bound |S B i | for every i P rks. If v is a vertex in S B i , then this means that the edge set EpT v q Y txw : w is a leaf of T v u needs to contain an edge e " ab P B between a vertex a P L i´1 zM and a vertex b P L i . By assumption on B we have d B paq ď ln 2 n. Moreover, there must exist a path P in T v leading from b to v which is of the form P " pb, v i`1 , v i`2 , . . . , v k´1 , vq with v s P L s for every i`1 ď s ď k´1. Analogously to the proof of Claim 6.3 we know that the number of such paths is at most ś k´1 s"i n pα s`1´αs qε " n pα k´αi qε . Provided n is large enough, we thus conclude that for i ě 2 it holds that |S B i | ă |L i´1 |¨ln 2 n¨n pα k´αi qε " 2 k´i`3 n 1 3`α i´1 ε ln 2 n¨n 2 3´α i ε ă n 1´pα i´αi´1 qε ln 3 n ď n 1´3ε ln 3 n where for the equation we make use of Definition (6.4), Property (D 2), the equation α k ε " 2 3 from (6.2), and where in the last inequality we use that, according to (6.1), α i´αi´1 " 3¨2 i´2 ě 3 holds. Moreover, since L 0 " txu and using (6.1) again, |S B 1 | ă |L 0 |¨ln 2 n¨n pα k´α1 qε " n 2 3´ε ln 2 n .
Hence, |S B˚| ď ř iPrks |S B i | ă n 1´2ε , which finishes the proof of the claim.
Now, under assumption of |S B˚| ď n 1´2ε and the properties in Claim 6.4 it follows that |pN GzB pAq X R 1 qzBig S B˚| ě |N G pAq X R 1 |´epBq´|Big S B˚| provided n is large enough. Thus, there exists a vertex z P pN GzB pAq X R 1 qzBig S Bs uch that d B pzq " 0. In particular, we then obtain that d GzB pz, M pk,1, q q " d G pz, M pk,1, q q (a) ą n 1 3 p1`2εq ą d G pz, S B˚q for every P r4s, where in the last inequality we use that z R Big S B˚. For every P r4s we thus find a vertex r P M pk,1, q zS B˚s uch that zr P EpGzBq. The latter already ensures that Property (S 2) holds. By the choice of T :" T r (Claim 6.2), we know that T is a copy of T k in H such that V pT q Ă M and such that all the leaves of T are adjacent to x in H. Since r R S Bẘ e also know that the set EpT q Y txw : w is a leaf of T u does not contain any edges from B i for i P rks. Thus, if there is an edge e " ab P B that belongs to EpT q with a P L i´1 and b P L i for some 2 ď i ď k, then a P M must hold, according to the definition of B i , B˚and S B˚. This yields Property (S 3). Analogously, if there were edges from B incident to x and to a leaf of T , then x P M would need to hold. However, we have x R M by assumption, and thus Property (S 4) follows. Finally, (S 1) holds, since V pT q Ă M by Property (T 2), and x R M .

Concluding remarks
Adding more constraints. Another variant of Maker-Breaker games are Walker-Breaker games (see e.g. [7], [9] and [13]) which put even more constraints on the edges that Maker may choose from in every round. Here, Maker is only allowed to claim her edges according to a walk. That is, in each round she must claim a free edge or she must walk along one of her already claimed edges, such that this edge is incident to her current position in the graph. It is quite natural to ask what happens in the connectivity game on G " G n,p when the game is played in the Walker-Breaker setting. This is work in progress already. Moreover, one may consider the variant in which Breaker also needs to play as a Walker, as suggested in [9] and [12]. We have not considered this variant, but we would be interested in how it behaves compared to the usual Maker-Breaker game and the Connector-Breaker game, respectively.
Considering different graph properties. In our paper we consider the Connector-Breaker game on G " G n,p in which Connector aims for a spanning tree. By combining our argument with the randomized strategy given by Ferber, Krivelevich and Naves [11], we can even show that n´2 {3`op1q is the size of the threshold probability when Connector aims for a Hamilton cycle. For a clearer presentation in this paper, we however skip the full argument here. A proof will appear in a follow-up paper. Furthermore, it would be interesting to consider other graph properties and to study the relation between the Connector-Breaker game, the Walker-Breaker game and their Maker-Breaker analogue. For example, consider the H-game where Maker (or Connector/Walker) wins if she claims all the edges of a copy of a given (constant size) graph H. Following [3] and the approach given in [7] it turns our that for all the three types of games the threshold bias for a p1 : bq game played on K n is of the same order, namely Θpn 1{m 2 pHq q, with m 2 pHq being the maximum 2-density of H. This is in contrast to the connectivity game discussed in this paper. We wonder whether in the unbiased H-game on G " G n,p it also holds that the threshold probabilities for winning either variant are of the same order.