Meepo/Strategy

General Info
is a mobility utility blue hero whose main advantage is the presence of many clones and the presence of periodic active damage abilities (Poof) that allow these clones to move around lines and deal AoE damage. Meepo initially starts out alone, however his signature card Divided We Stand allows him to summon clones, enabling his abilities.

Clones are both independent in the choice of items on them and dependent on the death of one of the clones. If one clone dies, then all Meepos die on all lanes.

The reward for killing a Meepo is 5 * the number of clones.

Meepo has synergy with teleportation items and abilities.

Vulnerabilities
Meepo on the battlefield should try to avoid any damage: whether physical or magical. With a small amount of health, even in the presence of armor, it is a vulnerable target for piercing damage attacks and spells, whether on one target or AoE.

Cleave damage can be considered as a dangerous modifier, which allows unit to inflict damage on the Meepo even if the direction of the main attack is not directed at it.

Synergy with spells

 * Positioning cards such as Cunning Plan and Juke can change an unsuccessful randomly chosen position to a more acceptable one.
 * Buffing cards such as Rumusque Blessing increase survival by increasing overall health or armor.
 * Combat tricks such as Divine Intervention can prevent Meepo and his clones from being removed in a subsequent lane in a critical turn.

Synergy with improvements cards
Lane improvements to armor like Verdant Refuge allow you to reduce or avoid incoming attacks and spells with Normal Damage type.

Probability of draw a card
Having a Meepo in the deck requires the player to use 3 copies of his signature ability in this deck.

Since the size of the deck may vary, it is necessary to estimate the probability of receiving a Divided We Stand to a particular turn depending on the total size of the deck.

In general, a task can be reduced to the following: there is some set of cards of interest to the player (say, a player wants to get one copy of a card for a certain turn).

Below is a table showing the probability of receiving a one card from the interest set for a particular turn depending on the size of the deck (it is assumed that until that turn the one card from the interest set was not got from the deck).

These probabilities are given for a set of one to five cards (сlick on the image to open the data table).











In accordance with the above data, the probability of getting Divided We Stand (it is a 3 card set) up to 2 turns for deck with 40 cards is 44.78 %. Thus, it can be stated that since this spell is key to winning, if there is a Meepo in the team, it is desirable to have as few cards as possible in the deck. For a varied deck with 60 cards, the chance of getting a Devided We Stand is 46.16% on turn 4.

General definitions
Due to the vulnerability of the hero, the player must make decisions on placing the Meepo extremely carefully.

For better clarity, we will present the lane as the following notation

Lane notation:=

, where xi is any unit of the top player on this lane (i = 1, ..., n where n is the number of top player units in this lane) and  yj is any unit of the bottom player on this lane   ( j = 1, ..., k where k is the number of bottom player units in this lane), the sign " - " means the absence of a unit on one of m possible positions. We also keep in mind the ordinal numbers of positions from 1 to m: so the leftmost position has the index 1, and the rightmost one has the m. The maximum number of units that fit within the screen without having to move is 8 so if the lane is less than 8 units on one side, we use notation with 8 positions (or notation with 7 positions as will be shown later):
 * -||-||x1||-||x2||''x3||-||-
 * -||-||y1||y2||-||-||-||-
 * }
 * }



Since in an Artifact, the location of units on a lane must be visually symmetrical, we will also consider the concepts of lane width. The width of the lane can be considered the distance between the leftmost and rightmost units on that lane (take into account the positions on both sides of the lane). Formally, the leftmost and rightmost positions on a lane can be associated with indexes, and the difference between the indexes to which 1 is added serves as the lane width. In the above case, the lane length is 4 (6-3+1). With an odd lane width, it is necessary to take into account the fact that in the lane notation, for convenience, 7 positions are displayed: so the leftmost position has the index 1, and the rightmost one has the 7. For example, on top side of the lane player have unit on position with index 3 and on the bottom side of the lane player have unit on position with index 5 and the lane width is 3 (5-3+1):




 * -||-||x1||align="center"| -||x2||-||-
 * -||-||y1||y2||align="center"| -||-||-
 * }
 * }

If we mentally draw an axis through the center of the board on the lane, then two situations can come out:
 * axis will cross the card of the unit or free slot
 * the axis will pass between the card/slot without intersecting it.

So in first case we have lane with odd width and in second we have lane with even width.





Intuitive example
It is known that a unit on the lane can attack both a directly opposing unit and its neighbors. Thus, to minimize the possibility of an attack from three sides, one of the best placements can be placing a unit on a more complete dense side of the lane. The dense side of the lane is the one on which there are no gaps between player's units. And the side of the lane where the player has more units is considered more complete. For example, consider the following location of units on the lane:




 * -||-||x1||x2||-||-||-
 * -||-||y1||y2||y3||-||-
 * }
 * }

In this scenario if top player need to deploy Meepo on this lane, then the only possible position for him would be with the index 5. And then the hero is threatened with an attack from y3. Both players have dense sides of lanes, but the bottom player has a more complete lane with greater lane width. If he tries to deploy a Meepo on this lane, then he may take positions with indexes 2 or 6. On the first position Meepo has no threats, and on the second there are no threats.

The counterweight of the dense side of the lane is a sparse side of the lane in which there are unfilled spaces.

Deployment rules
In general, the player must be well aware of all deployment cases on the lane.

Slot is a space intended for unit placement no matter if there is a specific unit on that space or not. Speaking of slots arranged against each other on the lane, we mean a pair of slots or box.

Where are three main types of boxes which we conditionally call:
 * 1) blocked
 * 2) unblocked
 * 3) empty

In the first case another unit should be located opposite the unit in question and in the box both slots are occupied. In the second case, there is free space opposite the unit and in the box one slot in empty. And in third case there are no units located in the box and both slots are empty. Empty box is formed after both units destroy each other in the combat phase and it is removed immediately after that.

Deployment slot is a slot in which it is possible to place a unit during the deployment phase. So at the time of the deployment phase there are no empty boxes on the lane.

To properly understand the deployment slot, you need to take into account the current lane's box types (or else the lane side occupancy), lane width and its change associated with the deploy of new units on the battlefield.

The deployment position of the unit on the lane is determined based on the following:
 * 1) The total lane width is calculated based on the current lane width before placing the units, as well as count of units deployed from each side on that lane.
 * Calculated values ​​for each side of the lane:
 * For the top side, the value of the possible change in the lane width (we call this value TS) is calculated: ( the number of units to be deployed on the top side of the lane ) + ( the current number of units on that side ).
 * For a bottom side, a value of the same is calculated (we call it BS): ( the number of units to be deployed on the bottom side of the lane ) + ( the current number of units on the bottom side of the lane ).
 * The values ​​are compared (TS and BS) and the larger of them is taken into account. If this large value is greater than the lane width, then it replaces the value of the lane width, and if not, the lane width remains unchanged.
 * 1)  Units deployed on each side are randomly assigned to deployment slots on this side of the lane.
 * 2) *If the lane width according to paragraph 1 has not changed, then the available deployment slots are within the indexes that define the lane width.
 * 3) * If the width of the lane changes, then additional deployment slots are created for placement in an amount equal to the difference between the new and the old lane width.
 * Depending on the number of newly created slots for deployment, determined by a change in the lane width (slots for deployment in the interval of the previous lane width are not considered new for deployment) they can be added both to the left and to the right of the existing units on the lane. Such new deployment slots cause a lane shift in one direction or another.
 * For example, when increasing the lane width by 2, 3 cases may arise: 2 deployment slots are added to the right of the existing units on the lane and the lane itself is visually shifted to the left by 2, 2 slots are added to the left of the existing units and the lane is shifted to the right by 2, 1 slot is added to the left and one on the right, the parity of the lane does not change and it does not visually shift.
 * When the lane width increases by 3, 4 cases can occur: adding 3 slots on the right causes a shift of 3 to the left, adding 3 slots on the left will cause a line to shift 3 to the right, adding 1 slot to the left and 2 on the right will shift to the left by 1, adding 1 slot on the right and 2 on the left will cause a right shift of 1. Units can be placed in this case both in the newly added slots for deployment and in the previously available for deployment slots.
 * In general case, when n new deployment slots appear, the the total number of divisions to the left and right is equal to n + 1 for (see  Details ). Each of these divisions leads to a shift of the lane to one of the sides in n + 1 cases, except for the case when the number of new deployment slots is even (in this case, the number of shifts  is n because the division of n/2 "right" and n/2 "left" does not lead to the shift of the lane)
 * This is because the number of variants is described for the case with n equals the number of multisets of length n made up of two ("right and left") elements (counting_multisets). Since this number is (n + 2 - 1)! / n! = n + 1 we get the above result
 * 1) The lane is actually shifts  according to the rules described above. Since there are no empty boxes on the lane to the deployment phase when the lane is shifted, all the boxes on it remain in place, the distances between the units already on the lane do not change. If a lane is shifted, its notation (and visuality because notation reflects the picture on the screen) may change from even to odd or vice versa depending on the initial and final lane width.

For example, consider the following location of units on the lane before deployment:


 * -||-||x1||align="center"| -||x2||-||-
 * -||-||align="center"| -||y1||align="center"| -||-||-
 * }
 * }

Suppose both players deploy two units on this lane: x3, x4, y2, y3. The lane width is 3 before deployment. On the upper lane, we get the value of TS equal to 4 (2 units on the lane + 2 units will deploy). On the bottom lane, the value BS is 3 (1 unit currently on the lane + 2 units will deploy). Comparing them, we find that TS is larger then BS and TS is larger than the current lane width. That is, the lane will shift as its length has changed from 3 to 4 units (the notation also changes from odd to even).

We will not consider all possible options for the placement of units x3, x4, y2, y3 between the deployment slots, but suppose that units is placed from left to right and along the left border. There are only two possibilities to shift: one unit to the right or left.



! colspan="11"| 1<= ! colspan="12"| before shift ! colspan="11"| => 1
 * -|| x1 || x3 || x2 || x4 ||-|| || || || || ||
 * -||-||x1||align="center"| -||x2||-||-|| || || || || ||
 * -||x3||x1||x4||x2||-|| || || || || ||
 * -||y2||y1||y3||align="center"| -||-|| || || || || ||
 * -||-||align="center"| -||y1||align="center"| -||-||-|| || || || || ||
 * -||y2||y3||y1||align="center"| -||-|| || || || || ||
 * }
 * -||y2||y3||y1||align="center"| -||-|| || || || || ||
 * }

Probability of taking damage
For a better assessment of the possibility of taking damage, player should know the probabilities of attacked units and taking damage from units in standard situations.

By default, a unit has the following probabilities of attacking, if in front of it all slots are empty or with the above terminology it is in the unblocking box (for more details see Pathing):



This is looks as follows: one of four possible Pathing Cards (the card "forward" has a probability of twice as much as each of the side ("left" and "right") cards) is placed before the unit (see  Details)



These probabilities change in the following situations:

Next, we calculate the probability of being attacked by two units at the same time. Since events are independent we have the following probability:

This is followed by the case when a unit can be attacked by either one unit, or another, or two of them simultaneously. Calculate the sum of probabilities of events.

Now when we can calculate the probability of receiving damage on each of the positions, consider the following examples.

1. This lane is proposed to deploy Meepo. Consider the total opportunity to receive lethal damage on lane (here it is assumed only the main damage already available to the units and cases are not considered when it is necessary to avoid deploying to the lane on which the opponent has a black hero or a red hero for example). The probability of being deployed in a particular slot and getting lethal damage P(LD^S) is equal to the product of the probability of being deployed on a certain slot P(S) and getting lethal damage, provided that it was deployed on this slot P(LD|S)   (see conditional probability).

The total probability of receiving lethal damage during deployment is equal to the sum of the probabilities of receiving lethal damage in each of the positions: 33.33%*25%+33.33%*43.75%+100%*33.33% = 56.25%

2. Consider another example when creeps deploys on the lane with Meepo.

The number of deployment variations in this case is based on permutations of multisets. In our case we have 3 types of objects: Meepo in the amount of 1, 2 creeps and 1 empty slot. Of these, you can make (3! / (1! * 2! * 1!)) =12 variations. The probability of being deployed in a particular slot and getting lethal damage P(LD^S) is equal to the product of the probability of being deployed on a certain case (units deployed on concrete slot) P(S) and getting lethal damage on that case P(LD|S). We depict in the figures all possible cases of placement and the probability of receiving lethal damage in each of them (see  Details )

The total probability of receiving lethal damage during deployment is equal to the sum of the probabilities of receiving lethal damage in each of the positions: 1/48+1/12+1/48+1/12+1/12+1/12+1/12+1/12+1/12+1/12 = 18/24 = 75%.

Thus, the lane given in this example is much more dangerous from the point of view of receiving lethal damage than the one given in example 1.

3. Let us consider the third lane where, on one side, Meepo and creep, and on the other side, creep and Phantom Assassin are deployed. As it is possible to judge from the rules of deployment on the lane, two new slots for deployment are formed on it.

But how to calculate the probability in the case of the emergence of new slots for deployment?

The probability in the general case should be calculated by the formula: the probability of an elementary location on the lane depending on the deployment = the number of possible combinations of new slots for deployment * the number of object locations on the first side of the lane by slots for deployment * the number of possible locations of objects on the second side of the lane on deployment slots.

In our case we have 2 * (3*2)*(2) possible elementary events. And the probability of the elementary units location is 1/24. The probability of taking lethal damage should be considered for each of the elementary cases (see  Details ). In the case when the new deployment slot appears on the left see Details.

In the case when the new deployment slot appears on the right see Details</i>.

So the total probability to take lethal damage after deployment is (1/24)*12 + ((1/24)*(1/4))*4)=13/24=54.17%.

Deployment strategy
In most cases, you should be guided when deploying Meepo (or another hero) the above calculations: minimizing the possibility of receiving lethal damage.

But in a number of game situations, you must also take into account the initiative and color of your opponent’s heroes on the lane to avoid the possibility of receiving this or that damage.

One of the Meepo's strengths is the ability to inflict damage on any of the lanes and before the action phase on lane (also used the term "cross lane"). This has only a number of heroes, spells, improvements. Among others it is possible to name such cards as Heartstopper Aura (It dealt damage before the action phase), Grazing Shot (dealt damage on any lane), Steam Cannon. The damage from the Poof is also applied to the area. The course of the battle rushes from the left to the right lane, but if you place the Meepo with the Blink Dagger on the central or right lane, you can then dealt damage on the left lane by applying the item and then Poof another clones to it (this is the possibility of causing damage to any point on the map).

Consider a number of examples.

Example 1. On the Top (leftmost) lane opponent have Phantom Assassin and now turn 4 (6 mana turn). On turn 2 you used the ability Divided We Stand and both Meepo died on combat phase. On turn 4 they are ready to be deploy again.

Suppose also that the enemy has the initiative on the Top lane. If you place a Meepo on the Top lane, then the opponent can immediately use the Coup de Grace. If, however, they are placed on other lanes under the assumption that the enemy will not have heal to the Phantom Assassin, then it will be possible to use the blink and poof to the first lane and killed Phantom Assassin. Black heroes, as the most dangerous for Meepo, have other abilities, for example, a Gank, which in this situation will nullify our placement efforts, but we consider the general idea of ​​avoiding damage on one of the lanes and then applying it in the opposite direction of the battle on the lanes (right to left).

In general, it does not matter how the Meepos is deployed on the center lane, however, let us assume that the placement happened as follows:

Our actions on the Center lane will consist of the following:
 * We move to the free field of the Top lane (using blink)
 * And then apply Poof on the Meepo, remaining on the Center lane to the Meepo, moved with the help of a blink to the Top lane:
 * So Phantom Assassin dies

Example2. Consider the following case of placement on the lane. As you can see Meepo dies after the combat phase. But Meepo can move to a more advantageous position after an unsuccessful deployment by applying a Poof on himself. After applying the Poof, Meepo disappears from the lane and then appears at the position available for deployment including the one with which it has performed the Poof. Meepo then moves to one of the possible free deployment positions and simultaneously deal damage from the Poof. When moving to the previous position, it will still die, but when redeploying to a new one, it will remain alive.

Example3. In some situations, if one of the Meepo is threatened on the lane, the sequence of Poofs is important for redeploy. That is, the order in which the Poofs are performed is important.

Black
Black heroes and spells are potentially one of the most dangerous threats to Meepo, because black spells can act on any position on the lane, and a number even on any position on another lane. Therefore, the Meepo advantages in mobility are completely non-leveled.

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