AI for MWiF - France

[Shannon V. OKeets]

The US entering the war a turn earlier is important, but the effect on their build points over the years is even more important. It is not only the big hit of having their build points jump on the turn they take various US Entry Options, but there is another big hit each year on the anniversary of passing the war appropriations bill.

So depriving France or the Commonwealth of a few resources/production points early in the war has to be balanced against increasing the US build points in the out years.

[Extraneous]

ORIGINAL: paulderynck

20% does not equal impossible. AAMOF if you do it both turns in 1939 the chance for the USA to get one extra chit is 36%. The 1939 chit pool is much better to draw from for the USA than the 39-40 pool. You might be happy with Italy but your axis partners will be less than impressed.

Since "Search & Seizure" only occurs during the production phase by SCS and submarines that stay at sea how are you figuring 36%?

[Shannon V. OKeets]

ORIGINAL: Extraneous

ORIGINAL: paulderynck

20% does not equal impossible. AAMOF if you do it both turns in 1939 the chance for the USA to get one extra chit is 36%. The 1939 chit pool is much better to draw from for the USA than the 39-40 pool. You might be happy with Italy but your axis partners will be less than impressed.

Since "Search & Seizure" only occurs during the production phase by SCS and submarines that stay at sea how are you figuring 36%?

First turn: 80% no chit, 20% chit.
Second turn: 64% no chits, 4% 2 chits, 32% 1 chit.

[Extraneous]

With the number of dice rolls per turn you cannot lump 2 turns together.

My formulae have always been for multiple actions of the same type in a single turn by adding the percents together.

Example: 2x "Search & Seizure" in one turn would be 40%

I feel this is very over generous. What is your formula?

[Shannon V. OKeets]

ORIGINAL: Extraneous

With the number of dice rolls per turn you cannot lump 2 turns together.

My formulae have always been for multiple actions of the same type in a single turn by adding the percents together.

Example: 2x "Search & Seizure" in one turn would be 40%

I feel this is very over generous. What is your formula?

I taught statistics at the college level. The way to combine the probabilities of a sequence of independent events is what I used. It can be shown rather simply using a tree diagram. At the top are the different outcomes of the first die roll. Beneath each of those outcomes insert the full set of possibilities of the second die roll, and so on down the page, building a tree structure.

In this case there were two possible outcomes for each event. So the top level has 2 outcomes, the second level has 4, the third level, 8 and so on increasing by powers of 2.

If there were more possible outcomes (e.g., in land, naval, or air combats), say 5, then the first level would have 5, the second 25, and the third 125. This is what makes calculating the likelihood of success for a groups of attacks on a front line so difficult.

Chess is considered very difficult because the branching logic for possible outcomes (i.e., moves) is so large. But really there are ~20 per side per turn. That makes looking a couple of turns ahead 20**4 = 160,000 possible outcomes/positions. WIF is much worse, because there are easily 100 units that can move and attack in a turn (air + naval + land). Depending on where the units stop their movement, the attacks can vary enormously. Throw in weather and action choices, and the possibilities surely increase to the thousands per side per impulse. 4 impulses would then yield, say, 2000**4 = 16,000,000,000,000.

Perfect strategies in WIF are beyond the capacity of man and machine.

[Shannon V. OKeets]

ORIGINAL: Extraneous

With the number of dice rolls per turn you cannot lump 2 turns together.

My formulae have always been for multiple actions of the same type in a single turn by adding the percents together.

Example: 2x "Search & Seizure" in one turn would be 40%

I feel this is very over generous. What is your formula?

I taught statistics at the college level. The way to combine the probabilities of a sequence of independent events is what I used. It can be shown rather simply using a tree diagram. At the top are the different outcomes of the first die roll. Beneath each of those outcomes insert the full set of possibilities of the second die roll, and so on down the page, building a tree structure.

In this case there were two possible outcomes for each event. So the top level has 2 outcomes, the second level has 4, the third level, 8 and so on increasing by powers of 2.

If there were more possible outcomes (e.g., in land, naval, or air combats), say 5, then the first level would have 5, the second 25, and the third 125. This is what makes calculating the likelihood of success for a groups of attacks on a front line so difficult.

Chess is considered very difficult because the branching logic for possible outcomes (i.e., moves) is so large. But really there are ~20 per side per turn. That makes looking a couple of turns ahead 20**4 = 160,000 possible outcomes/positions. WIF is much worse, because there are easily 100 units that can move and attack in a turn (air + naval + land). Depending on where the units stop their movement, the attacks can vary enormously. Throw in weather and action choices, and the possibilities surely increase to the thousands per side per impulse. 4 impulses would then yield, say, 2000**4 = 16,000,000,000,000.

Perfect strategies in WIF are beyond the capacity of man and machine.

[Extraneous]

I have to say chess moves and random numbers are not the same thing.

While playing chess there are a finite number of first moves (20) for each side.

With random numbers there are considerably more possibilities.

On the first turn:

You have you have an unknown number of dice rolls followed by a 10-sided die roll.

On the second turn:

You have an unknown number of additional dice rolls followed by a 10-sided die roll.


The only thing the 10-sided die rolls have in common is they occur during the same stage of different turns.

While I add percentages together if they occur back to back for the same issue this also is unlikely.

You cannot get an increase in percentages from turn to turn otherwise there is no such thing as a random number.

[Shannon V. OKeets]

ORIGINAL: Extraneous

I have to say chess moves and random numbers are not the same thing.

While playing chess there are a finite number of first moves (20) for each side.

With random numbers there are considerably more possibilities.

On the first turn:

You have you have an unknown number of dice rolls followed by a 10-sided die roll.

On the second turn:

You have an unknown number of additional dice rolls followed by a 10-sided die roll.


The only thing the 10-sided die rolls have in common is they occur during the same stage of different turns.

While I add percentages together if they occur back to back for the same issue this also is unlikely.

You cannot get an increase in percentages from turn to turn otherwise there is no such thing as a random number.

Never visit Las Vegas.

[Extraneous]

ORIGINAL: Shannon V. OKeets
Never visit Las Vegas.

Your advice is too late.

I've been to Las Vegas several times along with Atlantic City and worked at the Hard Rock casino here.


Rhetorical Question: Steve when you go to Las Vegas do you have a system you play that never fails?

Of course you do.

ORIGINAL: Shannon V. OKeets
I taught statistics at the college level. The way to combine the probabilities of a sequence of independent events is what I used. It can be shown rather simply using a tree diagram. At the top are the different outcomes of the first die roll. Beneath each of those outcomes insert the full set of possibilities of the second die roll, and so on down the page, building a tree structure.

In this case there were two possible outcomes for each event. So the top level has 2 outcomes, the second level has 4, the third level, 8 and so on increasing by powers of 2.

If there were more possible outcomes (e.g., in land, naval, or air combats), say 5, then the first level would have 5, the second 25, and the third 125. This is what makes calculating the likelihood of success for a groups of attacks on a front line so difficult.

Using your sequence of independent events please explain:

How does 80% in two rolls become 64%?

While 20% in two rolls becomes 36%,

Working with dice combinations two 10-sided dice give you 100 possible combinations.

Of that 100 possible combinations only three dice combinations will be a 2 or less.

The first die roll is listed horizontally.
The second die roll is listed vertically.


...1234567890
1 xx
2 x
3
4
5
6
7
8
9
0

[Cad908]

ORIGINAL: Extraneous


Using your sequence of independent events please explain:

How does 80% in two rolls become 64%?

While 20% in two rolls becomes 36%,

Working with dice combinations two 10-sided dice give you 100 possible combinations.

Of that 100 possible combinations only three dice combinations will be a 2 or less.

The first die roll is listed horizontally.
The second die roll is listed vertically.


...1234567890
1 xx
2 x
3
4
5
6
7
8
9
0

Oh for heavens sake. Lets try restating your question with this:

An NBA player gets fouled and is awarded 2 Free Throws;
That player's Free Throw (FT) percentage is 80%, which means he makes 8 and misses 2 out of every 10 foul shots;

What is the probability he will:

(1) Make both FT's?
(2) Make one & Miss one FT?
(3) Miss both FT's?

Answer:

First FT: Make 80% (Outcome A) Miss 20% (Outcome B)
Second FT: Make 80% (Outcome C) Miss 20% (Outcome D)

Therefore:

(1) Outcome A * Outcome C = .8 * .8 = .64 or 64%
(2) Outcome A * Outcome D + Outcome B * Outcome C = (.8 * .2) + (.8 * .2) = .16 + .16 = .32 or 32%
(3) Outcome B * Outcome D = .2 * .2 = .04 or 4 %

So, from the Axis perspective 64% of the time ZERO entry markers will be added, 32% ONE marker will be added, and 4% of the time TWO markers will be added

Stated in another way, on average, the Allies will get 32 + 8 = 40 markers added on 100 draws of 2 search and seizures, which works out to 20% + 20%.

[michaelbaldur]

Of that 100 possible combinations only three dice combinations will be a 2 or less.

what ... only 1 combination will be 2 or less. both ones ....

[Shannon V. OKeets]

ORIGINAL: Extraneous

ORIGINAL: Shannon V. OKeets
Never visit Las Vegas.

Your advice is too late.

I've been to Las Vegas several times along with Atlantic City and worked at the Hard Rock casino here.


Rhetorical Question: Steve when you go to Las Vegas do you have a system you play that never fails?

Of course you do.

ORIGINAL: Shannon V. OKeets
I taught statistics at the college level. The way to combine the probabilities of a sequence of independent events is what I used. It can be shown rather simply using a tree diagram. At the top are the different outcomes of the first die roll. Beneath each of those outcomes insert the full set of possibilities of the second die roll, and so on down the page, building a tree structure.

In this case there were two possible outcomes for each event. So the top level has 2 outcomes, the second level has 4, the third level, 8 and so on increasing by powers of 2.

If there were more possible outcomes (e.g., in land, naval, or air combats), say 5, then the first level would have 5, the second 25, and the third 125. This is what makes calculating the likelihood of success for a groups of attacks on a front line so difficult.

Using your sequence of independent events please explain:

How does 80% in two rolls become 64%?

While 20% in two rolls becomes 36%,

Working with dice combinations two 10-sided dice give you 100 possible combinations.

Of that 100 possible combinations only three dice combinations will be a 2 or less.

The first die roll is listed horizontally.
The second die roll is listed vertically.


...1234567890
1 xx
2 x
3
4
5
6
7
8
9
0

My 'system', which I learned in college, is that it is always best to be the house. I don't gamble against the house - it is like giving money away. For instance, in Las Vegas the casinos will tell you the odds of you winning at each game of chance; it's never 100+%. Poker among friends is a different story, but I will only play very small stakes even there. While I am an excellent chess player, my skill at poker isn't worth risking my hard earned money.

---

Here's how the odds for the land combat tables are explained in the Players Manual.

[Extraneous]

ORIGINAL: Shannon V. OKeets
An NBA player gets fouled and is awarded 2 Free Throws;
That player's Free Throw (FT) percentage is 80%, which means he makes 8 and misses 2 out of every 10 foul shots;

What is the probability he will:

(1) Make both FT's?
(2) Make one & Miss one FT?
(3) Miss both FT's?

Answer:

First FT: Make 80% (Outcome A) Miss 20% (Outcome B)
Second FT: Make 80% (Outcome C) Miss 20% (Outcome D)

Therefore:

(1) Outcome A * Outcome C = .8 * .8 = .64 or 64%
(2) Outcome A * Outcome D + Outcome B * Outcome C = (.8 * .2) + (.8 * .2) = .16 + .16 = .32 or 32%
(3) Outcome B * Outcome D = .2 * .2 = .04 or 4 %

So, from the Axis perspective 64% of the time ZERO entry markers will be added, 32% ONE marker will be added, and 4% of the time TWO markers will be added

Stated in another way, on average, the Allies will get 32 + 8 = 40 markers added on 100 draws of 2 search and seizures, which works out to 20% + 20%.

First FT: Make 80% (Outcome A) Miss 20% (Outcome B)
Second FT: Make 80% (Outcome C) Miss 20% (Outcome D)

Therefore:

(1) Outcome A * Outcome C = .8 * .8 = .64 or 64%
(2) (Outcome A * Outcome D) + (Outcome B * Outcome C) = (.8 * .2) + (.8 * .2) = .16 + .16 = .32 or 32%
(3) Outcome B * Outcome D = .2 * .2 = .04 or 4%
[/quote]

Rolling one 6-sided die to roll a one the odds are 6 to 1 (16.6%).

Shooting craps
To roll snake eyes (a combination of 1 and 1) the odds are 36 to 1 (2.7%).

...123456
1 x
2
3
4
5
6

When shooting craps house odds are weighted in the house favor at 31 to 1.


First craps roll: no snake eyes 97.3% (Outcome A) Makes snake eyes 2.7% (Outcome B)
Second craps roll: no snake eyes 97.3% (Outcome C) Makes snake eyes 2.7% (Outcome D)

(1) (Outcome A * Outcome C) * 100 or (.973 * .973) * 100
or 94.6% to not roll snake eyes on your second roll.

(2) ((Outcome A * Outcome D) + (Outcome D * Outcome C)) * 100
Or ((.973 * .027) + (.973 * .027)) * 100
Or (2.6% + 2.6%) = 5.2% to roll snake eyes on your second roll. Even if you pass the dice go to dinner and come back.

(3) ((Outcome B * Outcome D) * 100) or ((.027 + .027) * 100) = 5.4%

When shooting craps play the odds not the statistics.

[brian brian]

The Allies are The House in World in Flames. Excessive betting against them on the first two turns is poor strategy.

[Cad908]

ORIGINAL: Extraneous

ORIGINAL: Shannon V. OKeets
An NBA player gets fouled and is awarded 2 Free Throws;
That player's Free Throw (FT) percentage is 80%, which means he makes 8 and misses 2 out of every 10 foul shots;

What is the probability he will:

(1) Make both FT's?
(2) Make one & Miss one FT?
(3) Miss both FT's?

Answer:

First FT: Make 80% (Outcome A) Miss 20% (Outcome B)
Second FT: Make 80% (Outcome C) Miss 20% (Outcome D)

Therefore:

(1) Outcome A * Outcome C = .8 * .8 = .64 or 64%
(2) Outcome A * Outcome D + Outcome B * Outcome C = (.8 * .2) + (.8 * .2) = .16 + .16 = .32 or 32%
(3) Outcome B * Outcome D = .2 * .2 = .04 or 4 %

So, from the Axis perspective 64% of the time ZERO entry markers will be added, 32% ONE marker will be added, and 4% of the time TWO markers will be added

Stated in another way, on average, the Allies will get 32 + 8 = 40 markers added on 100 draws of 2 search and seizures, which works out to 20% + 20%.

First FT: Make 80% (Outcome A) Miss 20% (Outcome B)
Second FT: Make 80% (Outcome C) Miss 20% (Outcome D)

Therefore:

(1) Outcome A * Outcome C = .8 * .8 = .64 or 64%
(2) (Outcome A * Outcome D) + (Outcome B * Outcome C) = (80% * 20%) + (80% * 20%) = 160% + 160% = 320%
(3) Outcome B * Outcome D = 20% * 20% = .04 or 4 %

Really, 320%? This probability distribution should add up to 100%, or 64 + 32 + 4 = 100.

In my world, 80% * 20% = .8 * .2 = .16 or 16%, not 160%.

If you are having difficulties with this example, you should really not be offering AI advice. I think brian's rule in the post above is a good one to follow.

[Extraneous]

ORIGINAL: Cad908

Really, 320%? This probability distribution should add up to 100%, or 64 + 32 + 4 = 100.

In my world, 80% * 20% = .8 * .2 = .16 or 16%, not 160%.

If you are having difficulties with this example, you should really not be offering AI advice. I think brian's rule in the post above is a good one to follow.

Excuse me for trying to understand a poorly written formula. Is it better now?


It only took me three posts to get them to post the formula in the first place. Then I had to figure it out.


Did you note my changes to the formula to arrive at a percentage (1) (Outcome A * Outcome C) * 100?

That is how Steve should have posted the formula.


Have you ever programmed a computer professionally?

I'm glad you agree with Brian others have posted just the opposite. To me 20% is a good bet.


Since you just started posting here today "Welcome to the MWiF forum".

Control Alternate Delete 906.

[Cad908]

ORIGINAL: Extraneous

ORIGINAL: Cad908

Really, 320%? This probability distribution should add up to 100%, or 64 + 32 + 4 = 100.

In my world, 80% * 20% = .8 * .2 = .16 or 16%, not 160%.

If you are having difficulties with this example, you should really not be offering AI advice. I think brian's rule in the post above is a good one to follow.

Excuse me for trying to understand a poorly written formula. Is it better now?


It only took me three posts to get them to post the formula in the first place. Then I had to figure it out.


Did you note my changes to the formula to arrive at a percentage (1) (Outcome A * Outcome C) * 100?

That is how Steve should have posted the formula.


Have you ever programmed a computer professionally?

I'm glad you agree with Brian others have posted just the opposite. To me 20% is a good bet.


Since you just started posting here today "Welcome to the MWiF forum".

Control Alternate Delete 906.

The "poorly written formula" you describe is a pretty simple probability distribution not unlike flipping a coin twice. The expected outcomes from that experiment are:

25% TWO Heads
25% TWO Tails
50% ONE Head and ONE Tail

because each event (Head or Tail) has a 50% probability. I hope we can agree on that.

I have posted several times in this forum, though in different threads. My profession is corporate finance so I do not write software for a living. Though, for the life of me, I cannot see what that has to do with the probability distribution we have been discussing. Also, for what it is worth, I have been beta testing MWiF during the last 2 1/2 years.

[Tonqeen]

Example: If you cast your bait and have a 10% chance to get a mackerel per cast, what is your probability to get two mackerel in a row?
It is 10% to get the first one (0,1) times the second one 10% (0,1) = 1% (0,01)
IF i got a mackerel in my first cast and holding it in the tail THEN I have 10% (0,1) chance to get a second one...

[Extraneous]

ORIGINAL: Cad908

1) The "poorly written formula" you describe is a pretty simple probability distribution not unlike flipping a coin twice. The expected outcomes from that experiment are:

2) 25% TWO Heads
25% TWO Tails
50% ONE Head and ONE Tail

because each event (Head or Tail) has a 50% probability. I hope we can agree on that.

3)I have posted several times in this forum, though in different threads. My profession is corporate finance so I do not write software for a living. Though, for the life of me, I cannot see what that has to do with the probability distribution we have been discussing. Also, for what it is worth, I have been beta testing MWiF during the last 2 1/2 years.

1) To a programmer you would be considered an "end user" to you the formula is simple and understandable. As a programmer it is a poorly written set of specifications.


Example:

Steve changed 80% to .8 and 20% to .2 unnecessaryly, with no reason, or instructions as to why.

I started multiplying 80 and 20 but didn't divide the result by 100 to get a percent (Outcome A * Outcome C) /100. I then thought this would probably confuse people since it wouldn't look like Steve's formula. The way a formula looks means a lot to people.

So I decided to submit my own example of the formula to make my point. But forgot that I needed to change the part of the document that I had been working on.


2) Statistically yes. But flipping coins is a bad example after the probability study on flipping coins disproved the statistics.

But the situation is not flipping coins we are rolling dice looking for a set of values.


3) Your profile shows only two posts on this forum since October 12, 2012.

Have you ever submitted specifications to your IT people for coding? They would probably explain that your computer has the comparable intelligence to that of a three-year-old human. You have to be specific with your instructions.

Two and a half years in the beta I'm glad to see they got some new people in there.


New business
You have suggested I cease posting advice on AI.

I have tried to get members of the beta to post on what they like about the game.

They have all refused.

One even posted he would rather flame people than post about the game.

Other than the "Best WWII movie?" thread there have been ZERO posts on the game not initiated by me.

I originally started this under a thread to discuss player global strategy. Steve moved this to the AI threads.


Since you are from Finance I would think the production aspect would be of interest to you. Why haven't you posted something in two and a half years?

[Neilster]

80% is 80 divided by 100, which is 0.8. Extraneous, all your above posts indicate no knowledge of probability and a pretty poor grasp of basic mathematics but you lash out and blame others for your ignorance.

Plenty of people have started posts about the game. I've started several but from memory they were some time ago. In short, you appear to have a very strange attitude and have trouble getting along with people.

Just in case you're wondering, I am a commercial programmer and achieved a perfect score for university probability [;)]

Neilster