Table of Contents

Tuesday, 24 April 2018

Weather in Wargames

All battles in wargames are usually played in "neutral weather". It never rains, the field is never muddy, there is no fog, and it's not too hot. However, there are plenty historical examples showing that weather did play an important role on the battlefield. Waterloo is a notorious example.

If weather effects are included in a wargame, there are different things to consider:
  1. Determining the type of weather, and any changes in the weather during the wargaming day;
  2. The effects of the weather on movement, firing, visibility, etc.;
  3. For miniature wargames, how weather can be represented visually. But perhaps this is pushing things a bit too far?
Although waether is often ignored in games, it has always been a part of wargaming. The classic board wargame Tactics II (1958) has a weather table, along with effects on movement and combat. The idea is that every turn one rolls a D6 for the weather. Since one turn is supposed to represent one month, it is seasonal weather rather than day-to-day weather, and a different table is used for each season. The chart below is the actual weather table from Tactics II, along with its effects on movement and combat.


Very much the same approach can be found in many miniature wargaming rulesets, in which the weather is rolled for - perhaps depending on season and location - and the resulting weather has an effect on movement, combat etc.

However, suppose we want to include changing weather during the wargaming day? Surely we want a continuous change, i.e. we do not want to have a bright and sunny day on turn 1, a snowstorm on day 2, to turn to cloudy and foggy the turn after.

In the classic book "Practical Wargaming" (1974), Charlie Wesencraft proposed an interesting mechanic to ensure we have this continuous change. The idea is to represent all weather conditions on a linear scale, and roll each turn for a step "up" or "down" on this weather gauge. The actual device could be simply a marker that is put on the weather gauge, or a little pin that is put in drilled holes, etc. The relevant 2 pages from the book are shown below, but the ideas was published a few years earlier in Wargamer's Newsletter, April 1971.


A particular useful variation might be to allow players to move the weather gauge up or down one step (instead of it being a random outcome), depending on some exceptional die roll during the game (e.g. rolling a 12 on a 2D6 during a command roll), much in the same way as we would allow a player to control the length of the game (see our blogpost about The Last Turn in the Game). Thus, a player might want to move the weather to conditions that favour him more on the battlefield for a given scenario or tactical situation.

Apart from a gradual shift each turn, one still has to determine the "starting weather" for the game. A useful tool for this is a "weather die", which has various weather symbols. Since the various outcomes can vary wildly (sun to snow), perhaps a second D6 can be used to indicate variations within the main weather rolled for.



If you don't like rolling for weather at all, but still want to include weather and weather effects during the game, a very useful alternative is to look at historical weather records. These days, many weather records can be consulted online - actual websites may vary from country to country.
Now, suppose we want the weather record for the 18th of June, 1815, near Waterloo. Of course, this pre-dates the actual archived and recorded weather measurements, but we could take the 18th of June in a year for which measurements are available, thus using a plausible weather pattern for our wargame.

Let's say we would like to have the weather at Waterloo, on the 18th of June. We use Brussels as a close enough location, so let's try to find the weather for June 18 in different years. (I used http://www.eurometeo.com/english/ for generating these reports, but there are a number of similar sites available).

This is the weather pattern for June 18, 2012 in Brussels:


And this is the weather pattern for June 18, 2016, again in Brussels:

These seem like interesting weather records, and you can imagine using the sky conditions or the "significant weather" column as the weather pattern for the wargaming day.

So, use weather in your wargames as you see fit! And I am definitely interested in hearing about different weather systems that people have used in their own games ...

Monday, 9 April 2018

Last turn in the game

When playing a scenario, it is often convenient to set a number of turns within which one or both sides need to achieve their objectives. However, how many turns are appropriate for a given scenario? This clearly depends on the ruleset in use, but there are some rules of thumb I have used over the years, and a neat little mechanic to make the number of turns a variable factor in the game.

Number of turns

In my gaming group, we have a running joke that says that a given scenario ends "at the end of the wargaming day". This is a phrase often used by Charles Grant in his famous scenario books, without any further explanation. We always are/were confused whether he meant the end of the wargaming day for the players, or the end of the simulated wargaming day for the armies on the table. Hence, we often refer to it jokingly whenever someone asks about the end of the scenario. It's always "the end of the wargaming day".

However, that does not mean the length of a scenario cannot be set in advance. Clearly, this depends on a number of factors:
  1. How far do troops move per turn, versus the projected total movement distance?
    A typical example is an attack/defence scenario. Suppose the attacker starts at his own baseline, and needs to take a number of objectives (hills, towns, ...) roughly situated at the defender's baseline. Let us further assume that the table is 120cm in width, and that an average move is 15cm (6") per turn. This then translates to 8 turns needed to cross the width of the table. Thus, if you put the scenario length to shorter than 8 turns, the defender will have a hard time - or an impossible time - of reaching the objective.
    However, troops will not move at full speed in a straight line. You need to give some allowance for the attacker to manoeuvre across and around the battlefield. Hence, I often multiply the minimum number of turns needed by 1.5, resulting in 12 turns for the scenario above.
  2. Can troops move and fire, or only move during the turn?
    The above result of 12 turns assumes that units can move every single turn. But suppose your ruleset only allows movement or fire. Sure we want the attacker to fire at the defender as well, and any turn in which firing happens, movement will not happen for a firing unit. In the case of a static defender, the defender does not need to move and can fire every turn. In a typical attack-defence scenario using a move-or-fire ruleset, we might want to give the attacker at least the same firepower as the defender. The attacker therefore needs twice the amounts of units, half of which will/can  fire during a turn, while the other half is moving. Thus, if we want to give every unit the opportunity to reach the objective, we need to double the number of turns, resulting in 24 turns.
  3. Do we have an activation mechanism?
    Some rulesets do not allow every unit to do something. A system using cards or command rolls might restrict the number of units that can be activated during the turn. Suppose any given unit will have a 50% probability of getting activated. That means we have to double the number of turns, in order for our unit to reach the objective. Doubling our 24 turns from above, we now have a 48-turn long scenario.
The calculations above are only here to illustrate the process. If you use different assumptions, or put different restrictions on how many units need to reach the objective or move over a certain total distance, the multipliers will vary, and you will end up with a different total number of turns.

Variable number of turns

Computing how many turns we need is one thing, but should that imply the number of turns is this fixed number?

Most gamers are probably familiar with the last-turn-all-out-attack syndrome. If a player knows it's the last turn, he will risk anything to reach the objectives, because he knows there's no risk for the enemy to take advantage of it if all goes wrong, since there are no more turns left.

Some boardgames therefore use a mechanic to determine the last turn in a random manner. Starting at some point during the game, a random number (die roll, card drawn, ...) determines whether this was the last turn or not. Many different variants exist. There is no reason why something similar cannot be used in a miniature wargame.

E.g. suppose we have set the number fo turns at 20. We want a spread of plus or minus 2 turns for the game to end. The game could end at turn 18, or it could go on till turn 22. We can easily design a mechanic that gives a 20% increment each turn to end the game:
  • Turn 18: 20% chance the game ends (e.g. roll a D10, game ends on a roll of 1-2)
  • Turn 19: 40% chance the game ends (roll a D10, game ends on a roll of 1-4)
  • ...
  • Turn 22:: game ends with 100% probability.
Whatever mechanic you use, be sure that the game does end!

Players control the number of turns

In my games, I often use a mechanic(*)  that allows players to take a certain amount of control over the length of the game. Suppose the scenario is set to last for 20 turns. During the game, whenever a players rolls an exceptional result during some procedure (e.g. a double-6 on a 2D6 for a command roll, or a random event, ... ), he has the opportunity to increase or decrease the number of turns by 1.

Thus, if a player feels he still needs time to reach his objectives, he will often add 1 to the number of remaining turns. If he feels he needs to speed things up, he might subtract a turn. Or he might choose to leave things as they are.

We have used this mechanic in various games with different rulesets, and it often provides a healthy dose of suspense. In some games, it doesn't do much, but in other games, the count has gone up or down a few times. And towards the end of the game, the situation can become really tense and provide a lot of amusement and unexpected outcomes. If the event of altering the turn count happens infrequent enough, players do not have the impression it's an uncontrollable random device, but rather a little extra resource they can use to their benefit.

I even have a special large D20 specifically for this gaming mechanic. The D20 indicates the remaining number of turns (thus, every turn, the die is reduced by 1), but when the special roll or special event occurs, the player can dramatically and with the right amount of pathos turn the die up or down by 1.

My oversized D20, next to a regular D20. Some 42mm toy soldiers for size comparison as well. Winston Churchill noddingly approves.
(*) I encountered this particular mechanic in the first issue of Battlegames magazine (2006), where it was attributed to the ruleset Pieces of Eight.

Sunday, 4 February 2018

Drawing cards from a deck

Drawing cards from a deck is a popular mechanic for unit activation in wargames. In its most simple form, a standard card deck is used. Whenever a red card is drawn, a "red" unit can be activated, and if a black card is drawn, a "black" unit is activated. Variations on this procedure are straightforward: keeping a number of cards in hand and selecting one; cards that can activate groups of units; cards that stipulate specific actions (firing, movement); etc.

When such a mechanic is discussed, invariably the argument is used that it is possible "... to draw 5 red cards in a row, while the opponent cannot do a single thing! Thus, this mechanic is flawed! I hate card activations!"

But is this really so? Let's look in some detail at the probabilities involved in drawing cards and try to find out what is the typical length of a run of cards all of the same colour?


A first approximation ...

First, some basic assumptions:
  • We use a single deck of standard playing cards: 26 red cards, 26 black cards.
  • We will simply look at the colour of the cards being drawn (red or black), and not the suit (diamonds, hearts, clubs, spades).
  • We use a discard deck, i.e. drawn cards do not go immediately back into the pool. 
  • We call a seuence of drawn cards, all of the same colour, a run of cards.
Let's simplify our problem, and assume we have a 50% probability of drawing a red card, and 50% probability of drawing a black card whenever we draw a card from the deck. I know this is not correct, because the more red cards we draw, the less red cards remain in the deck, but as I said, we will keep it simple ...

The first card we draw is either red or black. Let us assume the colour is red. But what is the probability the 2nd card is also red? It's rather obvious that in 50% cases the card will be red, and in 50% of the cases, the card will be black.
Thus, we have 2 situations:
  • 1st card is red, 2nd card is black: 50% probability
  • 1st card is red, 2nd card is also red: 50% probability 
The first case stops the run of red cards. So, we can already conclude that there is a 50% probability that a run of cards of the same colour only lasts for 1 card. In 50% of the cases, we will have at least 2 cards of the same colour, but maybe there are more?
By using the same reasoning, we can extend the 2nd case:
  • 1st card is red, 2nd card is black: 50% probability
  • 1st card is red, 2nd card is also red: 50% probability 
    • ... and the third card is black: again 50%, so a cumulative probability of 25%
    • ... and the third card is red: again 50%, so a cumulative probability of 25%
And we can extend this even further:
  • 1st card is red, 2nd card is black: 50% probability
  • 1st card is red, 2nd card is also red:
    • ... and the third card is black: 25%
    • ... and the third card is red:
      • ... and the fourth card is black:  12.5%
      • ... and the fourth card is red: 12.5%
        • ...
So, we see a pattern here. The probability of drawing N red cards in a row, followed by a black card, equals 1/2 to the power of N. Thus:
  • drawing 1 red card, followed by a black one: 0.5^1 = 0.5 = 50%
  • drawing 2 red cards, followed by a black one: 0.5^2 = 0.25 = 25%
  • drawing 3 red cards, followed by a black one: 0.5^3 = 0.125 = 12.5%
  • drawing 4 red cards, followed by a black one: 0.5^4 = 0.0625 = 6.3%
  • drawing 5 red cards, followed by a black one: 0.5^5 = 0.03125 = 3.1%
  • drawing 6 red cards, followed by a black one: 0.5^6 = 0.015625 = 1.6%
  • ...
Note that the probability is exactly halved for adding an additional card to the run.

For the expected length of a run of cards, we add all the various lengths, weighted by their respective probabilities:

1*0.5 + 2*0.25 + 3*0.125 + 4*0.0625 + 5*0.03125 + 6*0.015625= 1.875.

However, we should also add the terms for even longer runs. This comes down to the sum of all terms N/(2^N), for N ranging from 1 to 26. Mathematics learns us that this sum eventually reaches the value 2 when N goes to infinity, which is a reasonable approximation in our case, since the terms for larger values of N are very, very small. Hence, we can say the expected length of a run of cards of the same colours is 2.

But, we approximated the probabilites for drawing a red or black card. So let's look at the same problem in a more accurate manner.

A second approximation ...

Above we assumed the probability of drawing either a red of black card was always 50%. However, this is an upper bound for drawing red cards, since there are fewer red cards left in the deck. On the other hand, the probability for drawing black cards will increase slightly, because there will be relatively more black cards remaining after first drawing one or more cards. In principle, we also would have to take into account how many red and black cards have been drawn so far (in all previous runs), but let's assume we start with a fresh deck, consisting of 26 red and 26 black cards.

When our first card is a red card, we have a deck left of 51 cards, with 26 black and 25 red cards. Thus, there is a 26/51 probability of drawing a black card as the next card, and a 25/51 probability of drawing a red card. Next, we have 50 cards left, and again, we can express the probabilities of drawing a red or black card by comparing the red or black cards left in the deck to the total cards remaining. By calculating cumulative probabilities for drawing red or black cards from the remaining deck after having drawn a red card first, we have the following series of outcomes:
  • drawing 1 red card, followed by a black one: 26/51 = 0.5098 = 50.98%
  • drawing 2 red cards, followed by a black one: 25/51 * 26/50 = 0.2549 = 25.49%
  • drawing 3 red cards, followed by a black one: 25/51 * 24/50 * 26/49 = 0.1248 = 12.48%
  • drawing 4 red cards, followed by a black one: 25/51 * 24/50 * 23/49 * 26/48 = 0.0598 = 5.98%
  • drawing 5 red cards, followed by a black one: 25/51 * 24/50 * 23/49 * 22/48 * 26/47 = 0.0280 = 2.80%
  • drawing 6 red cards, followed by a black one: 25/51 * 24/50 * 23/49 * 22/48 * 21/47 * 26/46 = 0.0127 = 1.27%
  • ...
So you see that the results are not that far off from the first approximation - extending the run by one card rughly halves the probability. The difference becomes more noticeable when calculating the probabilities for longer card runs, but those probabilities are so small anyway they are of little or no concern when actually playing a game.

We could also compute the expected length of a run using these adjusted probabilities for the first 6 terms:

1*0.5098 + 2*0.2549 + 3*0.1248 + 4*0.0598 + 5*0.0280 + 6*0.0127 = 1.8494

So, our average card run is a little bit less compared to the earlier calculation, but not by that much.

In practice ...

In practice, drawing a black card after a run of red cards means the start of a new run for the other player. Also, the deck has a "memory", which we did not take into account so far. A long run of red cards increases the probability of a long run of black cards, as long as we do not reset the deck. So, there are some correlated effects taking place. So, could we calculate such probabilities? This becomes a much more difficult exercise, but we could simulate it.

Let's try a little experiment. I took a card deck, shuffled the cards thoroughly, and went through the entire deck 12 times, writing down the length of card runs for both colours. Below you see my tallied results, along with the relative frequencies.


Length of run
Counted runs
Relative frequencies
1
113
48%
2
53
23%
3
29
12%
4
15
6%
5
9
4%
6
5
2%
7
6
3%
8
2
1%
9
2
1%
10
0
0%

I counted a total of 234 runs, with an average length of 2.21. You can see that the relative frequencies pretty much follow the previously computed pattern - roughly halving the probability for each additional card in the run. Of coure, with such a low number of runs, you always see discrepancies, especially in the lower frequencies. And there's also a slight other problem. I counted the last run in the deck by itself, not continuing the run over a reshuffled deck. This should favour shorter runs just a little bit more.

Let's simulate some more ...

Many rulesets that use a card drawing emchanism often have a "Joker" card. Whenever this card is drawn, the entire deck is reshuffled. This has the effect that the deck is reset, and so we start again with a no-memory deck of 26 cards for each colour, close approximating the probabilities derived before. Moreover, it handles the ending of one deck and starting a new one rather elegantly.


I didn't try to compute probabilities for including a Joker card, but I wrote a little computer program in Python that simulates exactly this. The program simulated the drawing of cards, including a Joker, and counted how long each run was. I counted runs that extended over the Joker - thus 2 red cards that showed up before the Joker, together with 2 cards after the reshuffling -  as a run of 4 cards.

It took me a short while to write such a program, and I let the simulation run till I reached a total of 100 million runs. After all, the deck never runs out of cards, since the Joker will turn up eventually. Below you see the results.



Length of run
Counted runs
Relative frequencies
1
50035520
50%
2
25490112
25%
3
12728374
13%
4
6220250
6%
5
2978897
3%
6
1398615
1.4%
7
640678
0.6%
8
287923
0.3%
9
126047
0.1%
10
54784
0.05%
11
22918
0.02%
12
9483
0.01%
13
3823
0%
14
1585
0%
15
621
0%
16
220
0%
17
90
0%
18
38
0%
19
10
0%
20
7
0%
21
1
0%
22
3
0%


As you see, the overall trend is again noticeable: runs with a length of 1 occur roughly 50% of the time, and adding a card to the run length means halving the probability. The average run length is 1.96, again very close to 2.

Since so many runs were simulated, it no surprise that extremely-low-probability run lengths such as 21 or 22 also occured a few times.

How to use in wargaming?

As I have said various times before on this blog, a specific rule mechanic doesn't mean much if you do not embed it in a wider ruleset.

Personally, I do like card-based activation mechanisms instead of a more classic IGO-UGO turn sequence. I think it adds some unpredictability, and keeps players involved. Also, it forces you to think a bit more about what the most important action is you want to do right now. On the other hand, I realize it's a more difficult to coordinate actions between different units.

But at least the statistics above show that on average, we can expect card runs that have a length of 2, and that 50% of the runs only consist of 1 card. So, if you do not like card-based activation, never again use the argument that card-based activation is a bad mechanic because runs of 10 cards show up all the time. Find some other argument instead! :-)