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[Guide] Forge Point Calculations for Great Building Contests

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DeletedUser4240

[Author’s note: This research paper was written to be understandable by anyone with a high-school level knowledge of algebra. If you find it difficult to follow, I suggest it is because the problem itself is inherently somewhat confusing and not because of the symbol pushing that is going on. I encourage you to persist in reading the document through to the end. I think you find it rewarding if you do. Remember, recreational mathematics is fun!]

A tactic that is sometimes used by experienced contestants in the Great Building contests to win a given prize is to contribute a block of forge points so large that too few forge points will remain available (at the Building’s present level) to allow all the other contenders for that prize to beat the contestant’s score. When this happens the contestant is guaranteed to eventually win the given prize or better. We will explain the idea of prize contenders and the contention set in a little bit.

It is important to understand that locking in a given prize does not prevent you from eventually winning a bigger one.

This paper takes an adaptive approach in which the score needed to win a given prize is calculated as the contest develops. The paper does not address more general questions such as the best timing of forge point contributions or the choice of the best prize.

We need a few basic definitions to start with. Let goal stand for the total number of forge points that are required to raise the Great Building to its next level and let open be the number of forge points that remain to be contributed before that happens. Plainly,
open = goal - total forge points contributed so far

Another way to define open is that it is the number of forge points available for contestants to add to their scores before the Great Building levels.

The subject of this article is a contestant we call player Y. Player Y is basically you. Player Y’s current score in the contest is called y and the score that he needs to guarantee that he wins the prize he is after or better is called y’.

The contention set.

For every prize there is a corresponding contention set. Let p be the scoring position of the prize being sought: 1, 2, 3, etc. Initially, player Y and the p contestants other than player Y who have the highest scores in the contest comprise the contention set for prize p. The members of this set are known as prize contenders, or simply as contenders.

Player Y might not think of all the other contenders as his opponents; some of them may be friends or fellow guild members. So, we will just call these contenders the “other contenders”. Player Y has only to beat one of the other contenders to win the relevant prize or better.

When p contestants cannot be found to serve as contenders (as will be the case at the very beginning of the contest, for example), we will imagine that there are anonymous dummy contenders and use a value of zero for their scores.

Here are three facts to remember:
Player Y is always a member of the contention set.
There is no reason to assume anything about how y is related to the scores of the other contenders.
The owner of the Great Building is not a contestant and thus can never be a prize contender.

The first thing to understand about the contention set is that its members can and often will change as the contest progresses as will their scores. The number of contenders is called n and it too can change. Initially, the contention set will have p+1 members, even if some of them are dummies, and so n = p+1 will be true. But that relationship might not remain true as the contest progresses. (An example of this appears near the end of the paper.)

For every contention set there is a score for player Y, called y’, that, assuming he can reach it, is certain to beat the score of at least one of the contenders by the end of the contest. y’ depends upon the current average score of the contention set, so it is dynamic; it can change with time.

It is easy to misread the previous paragraph and think we claimed that we would beat one of the members of the present contention set. All the members of the present contention set may be replaced by the end of the contest, but one of those will be beaten for the chosen prize if y’ is reached.

We ignore all but the current top scores in selecting the contention set because other contestants’ scores, being lower, would be even easier to beat.

(The identities of the contestants to whom the scores belong are really irrelevant to our calculations. But, for technical mathematical reasons, the contention set cannot be formally defined as a set of scores.)

The 2 contender problem and the 1[SUP]st[/SUP] prize formula.

There is only one other contender in the contest for 1st prize, the top scorer who is not player Y. Let us call his current score z. Any score for player Y that satisfies the condition
y’ > (open + y + z)/2

will eventually win him the 1st prize. Since there is no reason to spend more forge points than is necessary, the smallest whole number value for y’ that wins is considered the ideal solution.
--------------------------------------------------------------------------------------------
Derivation.

[Notice that the prime character is consistently used to distinguish “after” values from “before” values in discussing the changes that occur after player Y raises his score to y’.]

The number of forge points that player Y needs to add to his score to raise it from y to y’ is given by the expression y’- y. After he adds these forge points they will no longer be available to the other contender. The number of forge points available to him then will be
open’ = open - (y’ - y)

Let z' be the highest score that the other contender can possibly obtain after player Y has made his addition. To achieve his highest possible score, the other contender will add to his current score all the forge points that are still available in the level. It is impossible for him to get his score any higher than this. Therefore,
(z'- z) = open', and thus by substitution​
(z’- z) = open - (y’- y)

Adding y’- y to the left- and right-hand sides of this last equation we arrive at a very important equation,
(y’- y) + (z'- z) = open

This describes a state of the contest in which all the forge points that were initially open have been used up by the two contenders. We call this equation the "base equation” and we will see different forms of it for each value of n that we consider.

Rearranging terms, we get another form of the base equation.
y’ + z’ = open + y + z

Finally, we impose the constraint y' > z' and also demand that the values on both sides of the inequality be whole numbers. (The symbol '>' means 'greater than'.) Later, we will consider the case of a tie score, y' = z', and the circumstances in which it will actually win the prize.

The same operations that preserve equations also preserve inequations; adding equal amounts to both sides is one of them. Adding y’ to both sides of y' > z' gives
2y’ > y’ + z’

By substituting in the right-hand side of the earlier equation, we get
2y’ > open + y + z

The 1st prize formula follows immediately.
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Example.

The example is a level 2 Great Building at 60/150.
goal = 150, open = 150-60 = 90, y = 10

The other contender is in 1st place with a score of 23.
y’ > (90 + 10 + 23)/2

The formula evaluates to y' > 61.5 and the lowest whole number value for y’ that satisfies this is 62. So,
y'= 62, open' = 90 - (62-10) = 38, and z' = 23+38 = 61, which loses to 62 as expected.​

But even something so simple can have a wrinkle. Consider the case in which the other contender has a score of 24 instead of 23. In this case
y’ > (90 + 10 + 24)/2

The formula now evaluates to y' > 62 and the lowest acceptable value for y’ that satisfies the inequality is 63.
y' = 63, open’ = 90 - (63-10) = 37, and z' = 24+37 = 61, which loses to 63, also as expected.​

But suppose instead we had used 62 for y’.
y' = 62, open' = 38, but now z' = 24+38 = 62, which results in a tie.​

Replacing the relation, y’ > z’, by the less restrictive greater-than or equal-to relation, y’ >= z’, permits a tie to occur. A tie will win for player Y if and only if he is ahead of the other contender at the time he reaches the score of y’. This is because in a tie the contestant in the lower place remains in the lower place. Player Y is in the lower place in this case, and unless he can come up with 52 forge points immediately, it is not clear how a tie would work out for him.

It may occur to an insightful reader that the 2 contender formula can be used to beat the score of any contender in any position by using his score for z and that it can therefore be used to lock in any prize. This is entirely true. However, a less costly formula exists for locking in the other prizes as we shall now see.

The 3 contender problem.

The contest to lock in 2nd prize always starts with player Y going again the two top scoring contestants. They are the other contenders for this prize. Let us call their current scores z[SUB]1[/SUB] and z[SUB]2[/SUB]. Any score that satisfies
y’ > (open + y + z[SUB]1[/SUB] + z[SUB]2[/SUB])/3

will guarantee victory to player Y. As before, we regard the smallest whole number that does this as the ideal solution.
--------------------------------------------------------------------------------------------
Derivation.
[The steps in the solution of the three contender problem closely parallel those of the two contender problem. Readers are advised to compare these steps as the derivation progresses.]

After player Y has added to his own score the other two contenders can use whatever forge points remain to increase their own scores. We make no assumptions about how the remaining points will be divided between them, but, in the worst case situation for player Y, they will certainly consume all the forge points that are left open. This fact leads to another statement of the base equation:
(y'- y) + (z[SUB]1[/SUB]'- z[SUB]1) + [/SUB](z[SUB]2[/SUB]'- z[SUB]2[/SUB]) = open

The logic used to solve this case is a little tricky. As we said, to lock in the 2nd prize, player Y needs to guarantee that he will eventually beat the score of at least one of the other contenders. Moreover, he will accomplish this with the least forge point expenditure if he settles for just tieing the remaining contender. (We must assume that any tie will result in defeat for player Y.) Since the enumeration of the contenders is wholly undefined at this point, we shall simply name the score that gets beaten: z[SUB]1[/SUB]. Now we have these constraints on any solution:
y’ > z[SUB]1[/SUB]’ and y’ = z[SUB]2[/SUB]’

Add 2y’ to both sides of the inequality and then substitute z[SUB]2[/SUB]’ for just one y’ on the right-hand side.
3y’ > y’ + y’ + z[SUB]1[/SUB]’
3y’ > y’ + z[SUB]1[/SUB]’ + z[SUB]2[/SUB]’

Rearranging the terms in the base equation and then applying the previous inequality to the left side gives the following result:
y’ + z[SUB]1[/SUB]' + z[SUB]2[/SUB]' = open + y + z1 + z2
3y’ > open + y + z1 + z2

The formula follows immediately from this.
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Example.
We return to the earlier example: a level 2 Great Building marked as 60/150
goal = 150, open = 90, y = 10

The top two scores are: 23 and 18. The players in 1st and 2nd place are the two other contenders for 2nd prize. Player Y has a score of 10.
y’ > (90 + 10 + 23 + 18)/3

This evaluates to y' > 47 exactly, so the lowest acceptable value for y’ that satisfies the lower bound is 48. We have
y'= 48, open' = 90 - (48-10) = 52, z[SUB]1[/SUB]'+z[SUB]2[/SUB]’ = 23+18+52 = 93

It is clear that both contenders cannot tie player Y’s score of 48. Doing so would require a total score of 96, which means an extra 3 forge points would need to be open that do not exist. A score of y’= 47, on the other hand, would raise z[SUB]1[/SUB]'+z[SUB]2[/SUB]’ to 94 and make it possible for both contenders to tie with player Y. It is presumed that this would defeat him.

This example illustrates the savings one gets by using the 3 contender formula instead of the 2 contender formula. Applying the 2 contender formula to contestant 2 also produces a score that guarantees player Y to win the 2nd prize or better, but what is the cost?
y’ > (90 + 18)/2, which evaluates to y’ > 54 and y’ = 55.​

The cost is 7 extra forge points! Even if it is possible for player Y to win with a tie, 54 forge points is considerably more than 48.

The n contender problem and the pth prize.

By now we have seen enough to justify our jumping wildly ahead and offering the general formula for the critical score when there are n contenders. We mention in passing that, because there are only five prizes to be awarded, the highest practical value for n is 6.

It is time we gave a name to the expression we have been seeing again and again that provides the lower bound for y'. We shall call it ylo. For n contenders its value is
ylo = (open + y + z[SUB]1[/SUB] + z[SUB]2[/SUB] + … + z[SUB]n-1[/SUB]) / n

Any whole number score greater than ylo guarantees player Y a win against n-1 other contenders. As always, we only use the least possible solution.

We will hastily sketch the derivation of this formula, since the basic ideas are already known to the reader. Then we will provide a number of examples that show how the projected values of ylo can change as a contest develops.

Before doing this, however, we have a couple of remarks to make.

1. Prior to any drop-outs the contention set for the pth prize consists of player Y and the p highest scoring contestants, excluding Y. Therefore, in this case n = p+1.

2. The formula for ylo can also be written this way:
ylo = open/n + (y + z[SUB]1[/SUB] + z[SUB]2[/SUB] + … + z[SUB]n-1[/SUB])/n

The second term is, of course, the average score of all the contenders. We mention this to call attention to the highly symmetric nature of this expression. (It might be even more obvious if y were given the alias name z[SUB]0[/SUB]). All members of the contention set play an equal role in the expression, a fact with an important implication. As long as player Y is one of the top n scorers in the contest (which will become the case as he approaches his target), all contenders will calculate the same value for ylo.Thus, the score that guarantees player Y will win a given prize or better also guarantees any of the other contenders the same thing. If all other contenders achieve the critical score before player Y, he will have to settle for a lesser prize at best.

3. There is another fact of some interest. When y is increased by a given amount, open decreases by the same amount and the value of open+y is left unchanged. This, in turn, implies that ylo is left unchanged. And naturally the same thing is true regarding any of the other contenders; no addition to any contender’s score can affect the value of ylo. Only contributions that reduce open without affecting the average score of the contenders can do that. This means contributions that come from non-contenders, including the owner of the Great Building.

4. At the very outset of a contest, when all scores are zero and open = goal, we have ylo = goal/n. This just says, “Grab more than your equal share of the forge points and you will beat one of the other contenders.”
--------------------------------------------------------------------------------------------
Derivation.
We have seen most of this before, starting with the by now very familiar “base equation”
(y'- y) + (z[SUB]1[/SUB]'- z[SUB]1) + [/SUB](z[SUB]2[/SUB]'- z[SUB]2[/SUB]) + … + (z[SUB]n-1[/SUB]’- z[SUB]n-1[/SUB]) = open

Player Y can win the prize he is seeking or better at the least expense to himself by making sure he beats one contender and at least ties all the others. There is no loss of generality if we enumerate the contention set so that z[SUB]1[/SUB] is the score that is beaten. The constraints on y’ are now
y’ > z[SUB]1[/SUB]’ and y’ = z[SUB]2[/SUB]’, ..., y’ = z[SUB]n-1[/SUB]’

Adding all the equations together, then adding 2y’ to both sides of the resulting equation, and finally substituting z[SUB]1[/SUB]’ for y’ on its right-hand side gives
(n-2)y’ = z[SUB]2[/SUB]' + … + z[SUB]n-1[/SUB]’
(n-2)y’+ 2y’ = y’ + y’ + z[SUB]2[/SUB]' + … + z[SUB]n-1[/SUB]’
ny’ > y’ + z[SUB]1[/SUB]’ + z[SUB]2[/SUB]' + … + z[SUB]n-1[/SUB]’

Rearranging the terms in the base equation and applying the previous inequation,
y’ + z[SUB]1[/SUB]' + z[SUB]2[/SUB]' + … + z[SUB]n-1[/SUB]’ = open + y + z[SUB]1[/SUB] + z[SUB]2[/SUB] + … + z[SUB]n-1[/SUB]
ny’ > open + y + z[SUB]1[/SUB] + z[SUB]2[/SUB] + … + z[SUB]n-1[/SUB]

The n contender formula follows immediately from this. Of future relevance is this inequality:
(y'- y) + (y'- z[SUB]1) + [/SUB](y'- z[SUB]2[/SUB]) + … + (y’- z[SUB]n-1[/SUB]) > open

It is merely a rearrangement of the terms in the result that preceded it. We will refer to it as [A].
--------------------------------------------------------------------------------------------

The History of a Contest.

In this real-life example we will record the progress of a Great Building contest, taking snapshots called frames along the way as the contestants and its owner work to raise the Building to the next level.

The subject is a level 3 Great Building that was marked 33/220 when we first arrived at the construction site. As suggested by the low scores and large number of open forge points, this contest is just starting. All five of the prizes for leveling this building yield a single blueprint and, since blueprints are all we personally care about at this time, we decide to go for locking in the 5th prize.

A Contest Frame.

[Note: The Forum editor cleverly figures that, since zero is the same thing as nothing, it can omit all zero values from its crudely-drawn tables. Hence the absurd 0. notation.]
contrib
goal
open
y
1
2
3
4
5
FRAME 1
33
220
187
9
23
1
0.
0.
0.
36.7
0.
other
p=5
28
14
36
37
37
37
189 > 187
33
y’ =
37
37
37
37
37
37
n=6

This table, called a contest frame, captures the values that are relevant to us at a given instant in the life of the contest. Here is a description of its elements:

The first column of the frame shows the total number of forge points contributed to the contest so far. It is split into two parts: the total contribution from contenders, and the contribution from all non-contenders, meaning the owner and the other contestants. (It is desirable to track ‘other’ contributions because they are what reduce the value of y’.) The sum of the two parts is at the foot of the column.

goal was defined at the start of this paper. Recall that open = goal - total contrib. The statement p=5 identifies the prize we are seeking. The value in boldface to the right of open is player Y’s score. Next on the line come the top five contestant scores, excluding y. At the end of the line is ylo, the lower bound for y’. The current number of contenders is given in the lower right corner of the frame. Watch for when n changes. (You can easily calculate ylo from the data in the first line; just add the top number in the ‘contrib’ column to open and divide the sum by n.)

The number below player Y’s score is y’- y, the number of forge points he needs to add to reach his target score. Each of the five numbers to its right is the number of forge points that the respective contestant would require to tie with player Y after player Y has reached y’. These numbers are exactly the values of the terms on the left side of the inequality we named [A]. It follows that their sum must be greater than open. The last entry on the line confirms this and thus proves that there are not enough open forge points for all the other contenders to tie with player Y. One of them must be defeated. The relation provides an internal check. If it ever fails to hold true, something is wrong.

The name of the frame appears in the upper right corner. Its sequence number increases whenever time has passed and new contributions have been made. Successive versions of the frame where no time has passed -- for instance, new calculations -- have names that differ only in an additional character.

For those interested in making a spreadsheet to calculate the values of open and ylo, please notice that the data in the top line of the frame, along with the total ‘contrib’, are all that is really needed.

The Contest Develops.

At this point, because only contenders have contributed, the value of ylo is the same as it was at the very outset, when all scores were zero. Specifically, ylo = 220/6 = 36.7. This still true in the next frame.

contrib
goal
open
y
1
2
3
4
5
FRAME 2
53
220
167
13
36
3
1
0.
0.
36.7
0.
other
p=5
24
1
34
36
37
37
169 > 167
53
y’ =
37
37
37
37
37
37
n=6

There is not much to say here. A few forge points have been added. Contestant 1 is being somewhat aggressive.

contrib
goal
open
y
1
2
3
4
5
FRAME 3
79
220
141
25
48
3
2
1
0.
36.7
0.
other
p=5
12
-11
34
35
36
37
143 > 141
79
y’ =
37
37
37
37
37
37
n=6

Still more forge points have been added in frame 3, but there is no sign of contributions from the owner or other contestants. On second look ... ? What is a negative value doing in the “needed to tie” row? That seems to imply that it might be possible for contender 1 to take back some of the forge points he contributed to the Great Building and return them to the open pool. Plainly, that cannot happen.

So what should we do? A negative value in this row tells us that the contender has passed the critical score and is guaranteed to win at least prize p; he does not need to contribute any more forge points to do so. Let us try replacing the negative number with zero.

contrib
goal
open
y
1
2
3
4
5
FRAME 3a
79
220
141
25
48
3
2
1
0.
36.7
0.
other
p=5
12
0.
34
35
36
37
154 > 141
79
y’ =
37
37
37
37
37
n=6

This seems alright because it maintains the vital relation sum > open, but look how much higher than open the sum of the row has now become! If y’ were to decrease by a couple of points, it would lower the sum and still satisfy the greater-than relation. But the formula for ylo will not allow that.

A Contender Drops Out.

What to do? Since contender 1 has already locked in at least 5th prize, he is no longer in actual contention for it. For this reason he must be removed from its contention set. When this happens, it is known as contender drop-out.

contrib
goal
open
y
1
2
3
4
5
FRAME 3b
31
220
141
25
48
3
2
1
0.
34.4
48
other
p=5
12
34
35
36
37
144 > 141
79
y’ =
35
35
35
35
35
n=5

We want to go on tracking the top scores, so we have just dimmed contestant 1’s score to show that he no longer participates as a contender. His score has been shifted to ‘other’ in the ‘contrib’ column. The sum of the contenders’ scores is now 31 and the value of ylo is now
ylo = (141+31)/5 = 34.4, which means the ideal solution has become y’ = 35.​

We call your attention to the change in n. Lowering y’ has brought the sum in the second row down quite a bit, but it is still greater than open, as it must be. We can see from the lower value of y’ that dropping the contender has worked to player Y’s advantage. But would such a drop-out always be advantageous? Removing one of the terms in the numerator of the expression for ylo will certainly decrease its value, but lowering the denominator of that expression, which is n, has the opposite effect. Which effect will win?

Dropping a contender whose score has become greater than y’ always helps player Y, as is shown in [1] at the end of this paper.

We are past 5th prize!
contrib
goal
open
y
1
2
3
4
5
FRAME 4
43
220
114
34
58
5
3
1
0.
31.4
63
other
p=5
-2
27
29
31
32
117 > 114
106
y’ =
32
32
32
32
32
n=5

Hurray! We are now guaranteed to win 5th prize or better! Even as we were piling in forge points trying to reach the last frame’s critical value of 35, contestant 1 was busily adding to his own score and bringing ylo down some more. (Remember, contestant 1 is now a non-contender.) As the result we overshot the critical value by two forge points. The negative value below our score tells us that we are no longer a contender for this prize and we can rest on our laurels if we choose to. But it is more fun to go for the 4th prize now.

A new contention set.
contrib
goal
open
y
1
2
3
4
5
FRAME 4a
101
220
114
34
58
5
3
1
0.
43.0
5
other
p=4
10
-14
39
41
43
119 > 114
106
y’ =
44
44
44
44
44
n=5

To form the new contention set for 4th prize we take the top four not-Y scores, which include that of contestant 1. Contestant 5 is now removed from contention and contestant 1 is (briefly) back in it. Notice that p has been decreased to 4 and we have a new value for ylo. (Incidentally, the ‘other’ contribution of 5 is from the owner.) Right away we see that contestant 1 is also past being a contender for the 4th prize, so he drops right out again.
contrib
goal
open
y
1
2
3
4
5
FRAME 4b
43
220
114
34
58
5
3
1
0.
39.3
63
other
p=4
6
35
37
39
117 > 114
106
y’ =
40
40
40
40
n=4

The notation n=4 shows that the contention set now has only four members. And again we obtain a new value for ylo.

More contributions.
contrib
goal
open
y
1
2
3
4
5
FRAME 5
51
220
85
34
77
9
5
3
0.
34.0
84
other
p=4
1
26
30
32
89 > 85
135
y’ =
35
35
35
35
n=4

More contributions from contestant 1 and the owner reduce y’ to 35. It is good we waited before adding more forge points.

We are past 4th!
contrib
goal
open
y
1
2
3
4
5
FRAME 6
51
220
69
34
87
9
5
3
0.
30.0
100
other
p=4
-3
22
26
28
73 > 69
151
y’ =
31
31
31
31
n=4

Contestant 1 is still eagerly contributing and pushing down ylo. The appearance of a negative value now tells us that we are beyond contention for 4th prize and it is time to starting thinking about 3rd. Contestant 1 is very close to winning 1st place. Using the 2 contender formula with him as the other contender gives
ylo = (69+34+87)/2 = 95.0

Even if we pursue him to the last forge point, if he reaches the score of 95 first, he will win in a tie contest.

Contestant 1 takes 1st prize.
contrib
goal
open
y
1
2
3
4
5
FRAME 7
48
220
61
34
95
9
5
3
36.3
111
other
p=3
3
28
32
63 > 61
159
y’ =
37
37
37
n=3

When we next visit the Great Building site, contestant 1 has, unsurprisingly, increased his score to 95. He has won 1st prize. We are now focussed on the 3rd prize, which means contestant 4 is no longer in contention. It is evident that contestant 1 is out of contention also since he just won 1st prize.

3rd prize pays an extra 5 FP and more medals than 4th. The cost to win it is 3 FP. That sounds like a good deal, so we take it.

We go past 3rd!
contrib
goal
open
y
1
2
3
4
5
FRAME 8
46
220
44
37
95
9
9
5
3
45.0
130
other
p=2
9
37
46 > 44
176
y’ =
46
46
n=2

With a score of 37 we have locked in 3rd prize or better. This frame shows what is required to win 2nd. It is between contestant 2 and us. (‘Other’ has increased due to getting contestant 3’s forge points, plus additional contributions from the owner.)

Should we make a run for 2nd? It pays an extra 10 FP over 3rd prize, plus a few additional medals. The cost to get it at this point is 9 FP -- profitable, but not very. We will stay loose a bit longer.

The price is right.
contrib
goal
open
y
1
2
3
4
5
FRAME 9
52
220
31
37
95
15
9
5
3
41.5
137
other
p=2
5
27
32 > 31
189
y’ =
42
42
n=2

Contestant 2 contributes 6 more forge points and the owner tosses in 7 more, which push ylo down again. The cost of 2nd prize is now only 5 FP. We will take it and win back more forge points.

Safe at 2nd!
contrib
goal
open
y
1
2
3
4
5
FRAME 9a
57
220
26
42
95
15
9
5
3
41.5
137
other
p=2
27
27 > 26
194
y’ =
42
42
n=2

We add the 5 FP it takes to capture 2nd prize and rest with this frame. Let others finish the contest for us.

The game is over. Postmortem analysis.
contrib
goal
open
y
1
2
3
4
5
FRAME 10
220
220
42
95
20
9
5
3

This shows the final scores in the contest. Almost all the remaining forge points were added by the Building’s owner, who probably realized that contestant contributions were pretty much over. Contestant 2 added 5 FP more near the end, probably to further strengthen his claim on 3rd prize. He won it. Contestants 3 and 4 came in with the 4th and 5th prizes, respectively. Contestant 5 lost out.

Our forge point cost was 42 minus the 15 we won, a net sum of 27 FP, versus 34 FP if we had stopped back at frame 4. By continuing to play we saved 7 FP and won an extra 48 medals for a total of 53 medals. Compare this to contestant 1, who spent 95-25=70 FP and won 106 medals, twice the number that we won, but who got the same number of blueprints that we did.

I hope that you will find this paper a useful guide to your own gameplay.

----------------------------------------------------------------------------------

[1] The argument made here uses the case discussed earlier, but is clearly repeatable in any other case. It compares the ylo that was originally calculated in FRAME 3a, before the drop-out, with the value of ylo that was calculated in FRAME 3b, after the drop-out. We will call the latter value ylo’.
ylo = (141 + (25 + 48 + 3 + 2 + 1 + 0)) / 6
ylo’ = (141 + (25 + 3 + 2 + 1 + 0)) / 5

Multiplying the top equation by 6 and the bottom equation by 5 and subtracting each side of the top equation from the bottom equation gives
5ylo’ - 6ylo = -48

Adding ylo to both sides then gives
5(ylo’ - ylo) = ylo - 48

Using y’ from FRAME 3a, we have ylo < y’ so that ylo - 48 < y’- 48, and since y’- 48 = -11 was the negative number we threw out, it follows that
5(ylo’ - ylo) = ylo - 48 < -11 < 0, and therefore​
ylo’ < ylo

Lowering the lower bound allows y’ to go lower.

 
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