Partisan Force Pools

[Courtenay]

Are the contents of the partisan force pools given anywhere, by year? I know that the later in the war you go, the more likely it is you will get a good partisan, but I would like to know what is possible and the probabilities. I just looked in the player's manuals and did not find anything. Thank you.

[michaelbaldur]

ORIGINAL: Courtenay

Are the contents of the partisan force pools given anywhere, by year? I know that the later in the war you go, the more likely it is you will get a good partisan, but I would like to know what is possible and the probabilities. I just looked in the player's manuals and did not find anything. Thank you.

steve have taken the partisans and given then likelyhood by year and factors.

so when you pick a partisan, there are simply a dice roll. and you get a random.

there is no forcepool, there are unlimited partisans

[michaelbaldur]

and there a a max number of partisans/ by countries.

[Courtenay]

Yes, but there is a distribution of what partisans are available. What is it? What are the probabilities of getting each sort of partisan. Yes, the force pool is unlimited. There is still a force pool.

[Shannon V. OKeets]

ORIGINAL: Courtenay

Yes, but there is a distribution of what partisans are available. What is it? What are the probabilities of getting each sort of partisan. Yes, the force pool is unlimited. There is still a force pool.

The partisan values do not depend on the partisan's country.

The code below is based on the board game WIF countersheets.

===

// ****************************************************************************
// TSubCountry.PickPartisanValues.
// ****************************************************************************
case Game.Date.Year of
1935: TotalUnits := 1;

1936: TotalUnits := 2;

1937: TotalUnits := 3;

1938: TotalUnits := 5;

1939: TotalUnits := 7;

1940: TotalUnits := 9;

1941: TotalUnits := 11;

1942: TotalUnits := 13;

else TotalUnits := 15;
end;

case RollX(rsPickPartisan, 0, TotalUnits - 1) of
0: SetValues(0, 1); // (Combat Factors, Movement Points).

1..2: SetValues(0, 2);

3..4: SetValues(0, 3);

5: SetValues(1, 1);

6..7: SetValues(1, 2);

8: SetValues(1, 3);

9: SetValues(2, 1);

10: SetValues(2, 2);

11: SetValues(2, 3);

12: SetValues(3, 2);

13: SetValues(3, 3);

14: SetValues(4, 2);
end;

[Courtenay]

Thank you.

[Courtenay]

Here is the same information in tabular form.

[Orm]

Thank you for the picture.

[AxelNL]

I am trying to get a feel for the chance of a Russian partisan when USSR forces are in full retreat. Given the German garrison value without any ZOC of USSR troops that chance would be 0, right?

[Centuur]

ORIGINAL: AxelNL

I am trying to get a feel for the chance of a Russian partisan when USSR forces are in full retreat. Given the German garrison value without any ZOC of USSR troops that chance would be 0, right?

F5 is your friend here. That form gives you all the informations on the chance a partisan appears that you need...

[rkr1958]

After owning and playing this game for over 6-years I still don't know the exact details of how partisan generation works.

Maybe someone might help me work through the following two examples?

Example #1. End of turn. Jan/Feb 1940.

India currently has 1 partisan. So at the end of the turn the partisan roll for India was 18. This led to a garrison roll of 6, which resulted in no partisan and no future rolls for India. Might someone explain what just happened here?

Example #2. End of Turn Nov/Dec 1939.

In the previous turn, again India had 1 partisan on the map. The partisan roll was a 3 followed by a garrison roll of another 3. Then this garrison roll was followed by two more rolls both of which were 3. This spawned a second partisan (a 1-1 strength). Forget for the moment the unlikely streak of four 3's in a row, what in the heck were all these rolls for and what did the mean/do?

Thanks!

[Courtenay]

The algorithm for how partisans work is something I know well. I have learned it at least five times. [:)]

First, there is an important piece of information that is not given on the F5 display: How many times a country appears on the partisan chart. You can figure out this information by working backwards from what the chart shows you, but this is more work than is really strictly necessary.

The first die roll the game makes is whether one rolls to see how the partisans match up against the garrison. This uses 2d10, but 1d10 would have worked just as well. India appears on the partisan table twice. This means that one checks for partisans on a roll of 1-20. Yugoslavia, which appears five times, is checked on a roll of 1-50, while the Netherlands, which appears only once on the chart checks only on a roll of 1-10.

So, after been determined that one checks against the garrison, one checks against the garrison. Amazing, right? The way this done is takes the Base Partisan #, which is fixed for each country, except halved for the first calendar year the country is at war, subtracts the garrison, and adds the number of partisans. In the above table, China would be 20 - 13 (12.5 rounded up) = 7 while India at 6 - 4 + 1 = 3. This number is in the "Partisan #" column of the F5 table. One rolls a single ten sided dice. If the roll is higher than the calculated number, no partisan appears. If the roll is less than equal to the number, a partisan will appear. Thus, in this case, if the second roll is 4 or higher, no partisan will appear for India. If it is 1-3, then one will appear. If the roll is 11-20 lower than the partisan #, then two partisans would appear. Don't let the roll be 21 or more less than the partisan #. (Those numbers are what the rules say. I think there is a fencepost error here, and the break points should be 10 and 20, not 11 and 21, but the rules say what they do.)

Now one has to find out what size partisan is produced. This is where the table I presented up above comes in. However, I am confused by what you reported. I do not know why there are two more rolls, there should have been one. A 3 should have produced a 0-2, not a 1-1. At least it does when I debug. Maybe in the game, it collapses the 1-2 and 3-4 results into 1 and 2, but it really shouldn't. As I said, I have no idea at all what the fourth die roll was.

[Shannon V. OKeets]

ORIGINAL: Courtenay

The algorithm for how partisans work is something I know well. I have learned it at least five times. [:)]

First, there is an important piece of information that is not given on the F5 display: How many times a country appears on the partisan chart. You can figure out this information by working backwards from what the chart shows you, but this is more work than is really strictly necessary.

The first die roll the game makes is whether one rolls to see how the partisans match up against the garrison. This uses 2d10, but 1d10 would have worked just as well. India appears on the partisan table twice. This means that one checks for partisans on a roll of 1-20. Yugoslavia, which appears five times, is checked on a roll of 1-50, while the Netherlands, which appears only once on the chart checks only on a roll of 1-10.

So, after been determined that one checks against the garrison, one checks against the garrison. Amazing, right? The way this done is takes the Base Partisan #, which is fixed for each country, except halved for the first calendar year the country is at war, subtracts the garrison, and adds the number of partisans. In the above table, China would be 20 - 13 (12.5 rounded up) = 7 while India at 6 - 4 + 1 = 3. This number is in the "Partisan #" column of the F5 table. One rolls a single ten sided dice. If the roll is higher than the calculated number, no partisan appears. If the roll is less than equal to the number, a partisan will appear. Thus, in this case, if the second roll is 4 or higher, no partisan will appear for India. If it is 1-3, then one will appear. If the roll is 11-20 lower than the partisan #, then two partisans would appear. Don't let the roll be 21 or more less than the partisan #. (Those numbers are what the rules say. I think there is a fencepost error here, and the break points should be 10 and 20, not 11 and 21, but the rules say what they do.)

Now one has to find out what size partisan is produced. This is where the table I presented up above comes in. However, I am confused by what you reported. I do not know why there are two more rolls, there should have been one. A 3 should have produced a 0-2, not a 1-1. At least it does when I debug. Maybe in the game, it collapses the 1-2 and 3-4 results into 1 and 2, but it really shouldn't. As I said, I have no idea at all what the fourth die roll was.

The program generates at lease two die rolls for the size of the partisan. Only the second one is actually used to determine the size of the partisan to be placed on the map. The first die roll is to create a temporary partisan unit to check if there is room for the unit to be placed on the map. Not infrequently, there is no place to put the partisan that is in an enemy controlled hex in the partisan's home country and also not in the ZOC of an enemy unit. This gets even more important if there are 2 or 3 partisans to place on the map. So the program creates 1, 2, or 3 dummy partisan units and tries to find a place for them on the map. If there is insufficient space, then the number of partisans to arrive is reduced to the space available.

As far as the player is concerned, the extra die rolls for dummy partisan units can be ignored.

And yeah, this was a delight to code. I didn't write these rules; I just wrote the code to implement them.

[rkr1958]

@ Courtenay & Steve - thanks! That made things much clearer. Though I'd be stretching it if I said I was 100% confident that I now understood the process. I think I might?

@ Courtenay - I was missing the key piece, which was the "Partisan Table". I was under the impression that the first 0 - 99 (or is it 1 - 100?), 2D10 roll was checking against the "0 1 2 3 Partisan" probabilities. So I was total confused at the very start. [&:]

[Courtenay]

Than you Steve. Now I know why I could never figure out what the third roll does. Since the result of the roll does nothing, why not always set the roll to one, and not show it to the players? Less confusing and a trifle faster that way.

I just did some partisans yesterday, and the partisan size is as I said it was, based on the fourth roll. I still don't know why Ronnie's roll of a three for the fourth number produced a 1-2 partisan; is should have been a 0-2. Certainly, last night the program produced a 1-2 for me on a 6 and and a 2-1 on a 9, agreeing with my table.

[Courtenay]

ORIGINAL: rkr1958
@ Courtenay - I was missing the key piece, which was the "Partisan Table". I was under the impression that the first 0 - 99 (or is it 1 - 100?), 2D10 roll was checking against the "0 1 2 3 Partisan" probabilities. So I was total confused at the very start. [&:]

I don't know if the first roll is from 0-99 or 1-100. My write up above assumed 1-100. If it is 0-99, then you would have to change my figures appropriately (India 0-19, instead of 1-20, for example.) I just assumed that it was 1-100 because WiF uses 1-10 for a d10, not 0-9, but I don't know how a d100 is interpreted.

[Courtenay]

ORIGINAL: Courtenay

ORIGINAL: rkr1958
@ Courtenay - I was missing the key piece, which was the "Partisan Table". I was under the impression that the first 0 - 99 (or is it 1 - 100?), 2D10 roll was checking against the "0 1 2 3 Partisan" probabilities. So I was total confused at the very start. [&:]

I don't know if the first roll is from 0-99 or 1-100. My write up above assumed 1-100. If it is 0-99, then you would have to change my figures appropriately (India 0-19, instead of 1-20, for example.) I just assumed that it was 1-100 because WiF uses 1-10 for a d10, not 0-9, but I don't know how a d100 is interpreted.

Today I saw a partisan roll of 100, so the partisan rolls are from 1-100.