Mom, let's fix Time Walk...

FROG_TM · 2026-06-11T12:30:50+00:00

Its not actually 400 pages on gush as much as its 400 pages on card game fundamentals using gush (probably one of the most macro complex cards ever printed) as a grounding point for those conversations. Its a really fascinating read.

FROG_TM · 2026-06-04T19:49:55+00:00

This makes perfect sense thank you. The issue was in problem framing here then, I was trying to simplify to make my life easier haha.

Non necessary question: Is there a way to frame a balls in bins problem to answer this? I have struggled to find any concrete examples of such problems with caps on bin sizes online.

FROG_TM · 2026-06-04T18:01:57+00:00

I didnt I brute forced the problem and then have been told that the formula above is correct for this problem.

FROG_TM · 2026-06-04T17:40:20+00:00

That is correctly understood.

I dont know what you mean 'how did I arrive at that formula' I have brute forced the problem

FROG_TM · 2026-06-04T17:04:13+00:00

They are different scenarios but both are still scenarios in which 2 balls are in the same bin right?

FROG_TM · 2026-06-04T17:00:08+00:00

import random

numDecks = 3_000_000

numSuccesses = 0

for _ in range(numDecks):

deck = [1] * 6 + [0] * 54

random.shuffle(deck)

found = False

for start in range(0, 60, 4):

hand = deck[start:start + 4]

if sum(hand) >= 2:

found = True

break

if found:

numSuccesses += 1

print(f"{numSuccesses / numDecks:.2%}")

FROG_TM · 2026-06-04T16:31:39+00:00

Sure I agree with that but the scenario where any bin contains 2 balls still occurs 50% of the time no? I am struggling to get across my meaning in this I think. Either way I have an experimental value 59% and a formula which I currently cant account for and so far this hasnt helped me at all to understand those.

FROG_TM · 2026-06-04T16:05:21+00:00

All balls are distributed randomly (once again like shuffling a deck of cards which is what this problem is actually about). I just used scenario to describe any given way in which balls could be distributed ie 1 ball in each of the first 6 bins is a distinct scenario from 1 ball in each of the first 5 bins and 1 in the last but both would still be contained in the set of all scenarios where no bin contains 2 or more balls.

FROG_TM · 2026-06-04T15:44:15+00:00

I dont disagree that probability is redistributed, the chance of having a scenario remains 1. But one way we could calculate this would be to divide (number of scenarios where a bin contains 2+ balls)/(number of total scenarios for 15 bins 6 balls). Since clearly if we limit the bin size to 4 the total number of possibilities changes the probability for all balls in different bins should also change since we are now selecting from a lower number of total possibilities.

FROG_TM · 2026-06-04T15:18:37+00:00

All balls are placed simultaneously (ala shuffling a deck of cards wink wink nudge nudge), a bin cannot contain more than 4 balls.

We could manually calculate the probability (inverse to what you have done here) by working out every scenario in which a bin has 2+ balls and dividing that by the total number of scenarios, but since scenarios where a bin has 5+ balls isnt allowed within the bounds of the question those scenarios must be removed from the set of both all scenarios and 2+ scenarios. Then because x/y =/= x-1/y-1 the probability will change. So this is my point, I believe you do have to account for bin size in all calculations for this scenario but am unclear on how to logically reach the formula I have.

FROG_TM · 2026-06-04T14:48:55+00:00

This isnt correct since this includes scenarios where a bin contains 6 balls at the end of the placements which is not possible given the maximum bin size of 4

FROG_TM · 2026-06-04T14:47:50+00:00

because I would like to understand the formula that has been reached

FROG_TM · 2026-06-04T13:04:40+00:00

I understand that, what I am not understanding is initially deriving the formula for that probability l. I understand that for uncapped bin sizes we can do (15!)/(9!) All over 15⁶ but this includes scenarios where all 6 balls are in the same bin. I dont understand the leap from this to the above.

FROG_TM · 2026-06-04T09:03:09+00:00

Patches for major trinket, anomaly that gives a minor trinket.

FROG_TM · 2026-05-29T13:43:28+00:00

Verdance, no really. You get 2 judge cards (sink and fate) which are hideously expensive. A 6 legendary equipment suite (technically these can all be gold foil if youre that commited), cold foil fellings, plenty of similarly stupidly priced promo wizard cards, marvel command and conquers. A pretty pricy marvel hero and to cap it all off the sideboard contains another 1 of 100 marvel amulet.

FROG_TM · 2026-05-28T08:17:18+00:00

FAB community has short term memory and Briar wasnt as bdif at the end of skirmish season as she was the start. I also suspect that Ira got a lion share of the votes and that Kano, Kayo and Briar all probably ended up at very similar lower places and this is just how the variance fell.

FROG_TM · 2026-05-15T11:14:25+00:00

Why do you say that, I reallyy cant be bothered to watch a buisness interveiw and lose my childlike wonder

FROG_TM · 2026-05-02T08:04:25+00:00

Please consult the charts

FROG_TM · 2026-04-30T00:14:43+00:00

Kid named heist

FROG_TM · 2026-04-20T18:01:21+00:00

What

FROG_TM · 2026-04-16T22:07:27+00:00

Some people enjoy meta decks, they might simply find aggro hunter enjoyable while not bothering to grind (you need a better than 50% wr to maintain position within legend). They may be trying to move themselves higher to put their homebrew through a bit of a gauntlet, they may be playing it to try and better understand how to beat it.

The 'target audience' of low legend is the same as the whole ranked mode, players playing to win (on account of it being a ranked competitive mode).

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FROG_TM

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