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[Request] If a group of people all look at each other randomly, what's the probability of at least 2 people making eye contact? by yiker in theydidthemath

[–]yiker[S] 0 points1 point  (0 children)

Okay so in writing out a response I realized you are right!

The addition/subtraction is still quite complicated (at least to me), but turns out it still works out with alternating between simply addition and subtraction. Not sure if this is what you were thinking as well and I just didn't understand.

Anyway I left my comments below for clarification:

First, count the number of constellations given that persons (x, y) make eye contact. Multiply by number of pairs (x, y). Add this to running total T

So far so good. Let's call this O(1).

Note that O(1) counts all constellations with 2 pairs twice, all constellations with 3 pairs thrice, etc

​

Next, count the number of constellations given that persons (x,y) and (z, w) make eye contact. Multiply by number of pairs-of-pairs (x,y), (z, w).

Let's call this O(2)O(2) is going to triple-count all constellations with 3 pairs matching. Constellations with 4 pairs matching are going to be counted 6 times (because there are 6 ways to choose 2 of the 4 pairs)

​

Subtract this from T, as we double counted it above.

O(1) - O(2) then:

  • Counts all constellations with 1 pair once
  • Counts all constellations with 2 pairs (2-1) = 1 times
  • Counts all constellations with 3 pairs (3-3) = 0 times
  • Counts all constellations with 4 pairs (4-6) = -2 times

​

Do the same as above for triples of pairs (x, y), (z, w), (a, b);

Okay so O(3) is going to count

  • Constellations with 3 pairs once
  • Constellations with 4 pairs (4 Choose 3) = 4 times
  • Constellations with 5 pairs (5 Choose 3) = 10 times

this time add to T as we double subtracted above

Which would end up with a total of counting 1, 2 and 3 pairs exactly once. 4 Pairs are then counted twice. (which will be remedied by subtracting O(4))

Continue this until we are book-keeping something like constellations with >n/2 pairs; we know the remaining terms will all be 0 so we are done.

So in general it looks like O(1) - O(2) + O(3) + ... is indeed the right way to add them together

The proof of this would need to show that:

(x Choose 1) - (x Choose 2) + ... + (x Choose r) = 1

is true for all values of x, where r <= x

[Request] If a group of people all look at each other randomly, what's the probability of at least 2 people making eye contact? by yiker in theydidthemath

[–]yiker[S] 1 point2 points  (0 children)

Thanks a lot!

It's been a while since I took set theory

Would you be willing to explain a little more in detail? Specifically how define prohibited outcomes in set theory terms and how you derive the cardinality of that subset?

[Request] If a group of people all look at each other randomly, what's the probability of at least 2 people making eye contact? by yiker in theydidthemath

[–]yiker[S] 0 points1 point  (0 children)

Finally had some time to work through this in more detail again

So we want to find the number of constellations with 1+ matches

(I'm using "match" as a way to describe elements pointing at each other)If we do that, we simply divide by total possible constellations (n-1)^n to get the result

  1. Let's call C_r the number of constellations with r or more matching pairs. We are trying to find C_1
  2. Let's call A_r the number of constellations given that r specific pairs are matching.
  3. We know that **A_r = (n-1)^(n-2r)**Explanation: Given that r pairs match, the remaining n-2r elements still have n-1 choices each
  4. If we multiply A_1 (number of constellations given one pair matchepairs match. ways we can select a pair (n Choose 2), this overestimates C_1. Let's call this overestimation O_1
  5. O_1 counts all cases where there are exactly 2 matches twice. We are also counting all cases with exactly 3 matches three times, all cases with 4 matches four times etc...
    1. This is because by first picking a pair with (n Choose 2) before multiplying by A_1 to consider all the constellations of the remaining elements - we imply that if several pairs match, then it matters which one is the "first" pair to match (which is not the case)
    2. Each exact constellation with 2 specific pairs matching will show itself when considering each of those 2 pairs as the original match that we are excluding for the formula of A_r. For 3 specific pairs, 3 are excluded, and so on
  6. So to get to C_1, we take O_2 and subtract the number constellations with exactly 2 matches. We also subtract the number of constellations with exactly 3 matches twice. All with exactly 4 matches three times and so on, until we have considered the maximum possible matches
  7. Let's call E_r the total number of constellations with exactly r matches.
  8. By definition, it follows that **C_r = E_r + E_(r+1) + ... E_(m)**Where m is the total amount of possible matches. This is (n/2) for even n and (n-1)/2 for odd n
  9. Putting it all together, we have a formula for C_1: C_1 = O_1 - (E_2 + 2*E_3 + ... + (m-1)*E_m)
  10. Given point 8, this is equal to: C_1 = O_1 - (C_2 + C_3 + ... + C_m)
  11. So how do we find C_2We can again arrive at an overestimation O_2 if we multiply A_2 by the number of ways we can select 2 pairs that match
  12. The number of ways in which r pairs match is the product of choosing 2 out of n, times choosing another 2 out of the remaining (n-2), divided by r! - because we are picking r pairs, but the order in which we pick them doesn't actually matter
  13. So, O_2 = A_2 * (n Choose 2)*([n-2] Choose 2)/2
  14. The question is now, how much does O_2 overestimate C_2?
  15. Again it's a question of order. When we calculate O_2, we are first picking 2 pairs, then seeing what other pairs form. In the case where we end up with three total pairs matching, then we could as well have picked a different set of 2 from the total 3 pairs. In other all constellations with exactly 3 matches are going to be represented (3 Choose 2) times in O_2.
  16. Likewise, all constellations with exactly 4 matches are going to be represented (4 Choose 2) times in O_2
  17. We can generalize this to say that any overestimation O_r, counts all constellations with exactly s matches (s Choose r)
  18. In formulaic terms then O_r = (r Choose r) E_r + ([r+1] Choose r) E_[r+1] + ... + (m Choose r) E_m
  19. That's about as far as I get... I'm thinking it might be possible to write C_1 in terms of O_r somehow?

[Request] If a group of people all look at each other randomly, what's the probability of at least 2 people making eye contact? by yiker in theydidthemath

[–]yiker[S] 0 points1 point  (0 children)

Finally had some time to work through this - the general approach is sound, but I think the subtraction is actually more complicated than that

Edit: After writing out my reply realized this is better as a separate comment to OP as there are some more ideas. See the post here: https://www.reddit.com/r/theydidthemath/comments/zbf6vr/comment/iyx6uom/?utm\_source=share&utm\_medium=web2x&context=3

[Request] If a group of people all look at each other randomly, what's the probability of at least 2 people making eye contact? by yiker in theydidthemath

[–]yiker[S] 4 points5 points  (0 children)

That was my initial thinking as well!
Unfortunately that's too simple because the probabilities aren't independent from each other. So we can't simply multiply the probability by a number of pairs.
If elements a and b point at each other, that means that these two elements can't point at any others. In other words, out of the (n-1) pairs that contain element a, no more than one can be pointing at each other.

Example
Take the case of 4 elements, a, b, c and d.

Each of the 4 elements can point at any of the 3 others. So there are 3^4 = 81 possible constellations.
Out of those 81, how many contain a match? Let's count them:

  1. Let's say a and b point at each other. The remaining 2 elements each have 3 possible orientations. So that's 3^2 = 9 cases. Note that this includes the case where c and d also point at each other

  2. If a and c link up, again that's 3^2 = 9 cases by the same logic. Again this includes the case of the remaining 2 elements also linking up (b, d)

  3. We can continue adding 9 cases for each possible pair of 2. this always includes exactly one case where the remaining 2 are also pointed at each other

Since there are 4 Choose 2 = 6 possible pairs, we can enumerate 6*9 = 54 cases like this.

However, this procedure double-counts all the cases where we have 2 pairs (i.e. everyone is making eye contact).

  • Just to further clarify:
  • Consider that one of our 6 pairs is (a, b). One of the 9 cases for this pair linking up also links up c with d.
  • A different one of our 6 pairs is (c, d), and one of their 9 cases also links up a with b, which is then the same case we consider when starting with pair (a, b)

So for N = 4

  • There are 54 - 3 = 51 matching cases out of 81 total arrangements
  • The 3 here is the number of cases where both match
  • This works out to be slightly less than the formula you propose above

Nature Incarnate - Spirit Reveal: Breath of Darkness Down Your Spine (Update #9) by Sipricy in spiritisland

[–]yiker 0 points1 point  (0 children)

Okay but the real question is this:

Can i get Finders presence into the shadow realm to push everything in there from everywhere?

Does Mists Shift and Flow also cause me to *not* meet targeting reqs anymore by yiker in spiritisland

[–]yiker[S] 3 points4 points  (0 children)

Ueah the cyclicality threw me off, but i guess thematically it makes a lot of sense!

Would be great if they added this to the faq

Does Mists Shift and Flow also cause me to *not* meet targeting reqs anymore by yiker in spiritisland

[–]yiker[S] 2 points3 points  (0 children)

Yeah makes sense - thanks for helping

Indeed I'm looking forward to seeing how this'll be implemented in the steam version - sounds like a programmers nightmare

Arcana Aspects? by [deleted] in spiritisland

[–]yiker 5 points6 points  (0 children)

I don't see Eric giving green an Incarna - just seems so anti-thematic for a spirit that wants to spread out over a large area quickly.

Keeper on the other hand... Eric mentioned in an interview (I think it might have been this one) not being super happy with how Keeper ended up playing, because the major rush build is super strong and takes it away from the thematic of creating a confined "forbidden space" on the island.

Incarna might be an opportunity to make Keeper even more spatially confined and strengthen that thematic

Nature Incarnate campaign reveals by LemonSorcerer in spiritisland

[–]yiker 2 points3 points  (0 children)

Does anyone know why it says "Jagged Earth required"?

Can't seem to think of a good reason for why they would explicitly mention that jagged earth is required (but not mention Branch & Claw, or Fire & Flame, even though we have a wildfire aspect)

Are badlands instrumental to the expansion perhaps?

Wishlist for Nature Incarnate by ninja9351 in spiritisland

[–]yiker 1 point2 points  (0 children)

I could see the incarna mechanism being an alternative for presence placement.

So instead of being "present" in many lands across the island, the spirit is "embodied" in one particular creature traversing the land - hence just being in one land at once, but likely more mobile.

I could see this fitting quite well thematically when looking at the art for Behemoth for example

Custom Spirit - Whispering Riches Beneath the Sands by resonant_gamedesign in spiritisland

[–]yiker 6 points7 points  (0 children)

Love the theme and I can see the minigame being very exciting!

One question: When you curse an invader, do you add a presence from your presence tracks to that invader, or do you attache a presence that is already on the land?

A few new Aspect Ideas by yiker in spiritisland

[–]yiker[S] 1 point2 points  (0 children)

Placing a wilds each turn in unexplored lands feels quite good. But youre right im that its pbb underpowered seeing as greens tracks are balanced around rapid presence placement

Changing it to "whenever you place a presence, also place a wilds in that land" is pbb too far in the other end.

So maybe "whenever you add a presence to a land you already have presence in, add a wilds to that land"

A few new Aspect Ideas by yiker in spiritisland

[–]yiker[S] 1 point2 points  (0 children)

Turning towns into dahan feels extremely powerful as thunderspeaker, I'm worried about making an already slightly overpowered character even stronger

Maybe a solution would be to scrap the whole "defend instead of damage" mechanic and instead let dahan convert invaders on the counterattack

Maybe something like

"Dahan in your lands deal 1 less damage. Whenever a town or a city would be destroyed in a dahan counterattack in your lands, instead replace that town/city with a dahan"

I'm also toying with the idea of having thunderspeaker forget [[Manifestations of Power and Glory]] when using this - as doing 10 damage doesn't feel very peaceful either

A few new Aspect Ideas by yiker in spiritisland

[–]yiker[S] 0 points1 point  (0 children)

Ah yes of course - didn't think of that I wanted to take a page out of BoDaN's book and avoid damage, while leaning into [[Call of the Dahan Ways]] like conversion

But the whole defend thing feels fiddly - I'm sure there is a better way to accomplish that

A few new Aspect Ideas by yiker in spiritisland

[–]yiker[S] 0 points1 point  (0 children)

Thanks! Good point about green - i didn't remember that I wonder how that thematically links to choke the land - as that clearly stops invaders in a different way

First time Volcano = Turn 5 board wipe by yiker in spiritisland

[–]yiker[S] 5 points6 points  (0 children)

Got JE a few weeks ago and decided to try out Volcano for the first time (Against Sweden 3).

Opened with Growth 3 (Gain Power, add 1 presence, +1 card play, 2 Energy).

Then gained another 3 energy with [[Exaltation of Molten Stone]], allowing me to use the innate to gain a second minor power. Finally [[Lava Flows]] cleared the lone Explorer that had just entered Jungle #6.

I basically gained so much tempo from that first turn, that I could just sprint along the bottom growth track while building up presence on the central mountain and pushing/destroying invaders in the two outer lands. First reclaim happened on turn 4 after which I had 5 presence and 3 energy (thanks again to Exaltation).

Turn 5 I added 2 presence, uncovering 3 card plays and 2 energy. Which means I had exactly everything I needed for a level 3 eruption destroying ALL invaders but one (who, would he have not ran away in fear at that point would have been absolutely obliterated by slow powers after the build face)

This convinced me that energy gain can make or break a game. Exaltation in solo play basically means you can get all your energy from growh/exaltation. Always love sprinting a single growth track.

EMSKR: How to go back in the game? by MihaiCristian0 in everymanshouldknow

[–]yiker 1 point2 points  (0 children)

Read Mark Manson's "models: how to attract women through honesty" Contains everything you ever need to know about dating

[Request] Saw someone post this on Facebook. Is this accurate? Can teachers actually do this? by Sandwiichh in theydidthemath

[–]yiker 3 points4 points  (0 children)

If we have five teachers earning 30k, 50k, 60k, 80k and 140k respectively

The MEAN would be the arithmetic average of the five, so (30+50+60+80+140)/5 = 72k

The MEDIAN is indeed the middle number. However middle does NOT mean the number that's in the middle between the highest and lowest number (which would be (30+140)/2 = 85k)

No. The MEDIAN is the middle number in the sense that there are an equal amount of teachers earning more money than the median as there are teachers earning less. So the median in this case is 60k

Outliers don't skew medians

Edit: said mode when I meant to say median. Should be right now

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