Azul Weekly Puzzle #004

[Effgee]

Hi again!
First of all thanks to euklid314 for last week's reply, very well thought through and detailed!
For this week, not so much a puzzle as a intuition test. Here's the context:

I'm in a game, it's extremely and we're wrapping up round 4. I'm slightly ahead point-wise and have the first player tile so a slight advantage for the next and final round. However, my opponent has 3 yellows on their last row, and if they get 2 more, they will not only complete a column but also get the 10 point bonus for all yellows on heir board, and that is an automatic defeat. I've kept track of tiles and know there will be 4 yellows in the final round, randomly displayed on the 5 factories. If I'm starting player, what is the probability I will be able to prevent my opponent from getting 2 yellows? (e.g. if two factories have 2 yellows each, then there's nothing I can do to prevent the worst from happening)

[euklid314]

Your opponent will get 2 yellows in exactly 2 cases (if yellow is all-important):

I) The yellows are distributed 2-2-0-0-0.
II) The yellows are distributed 1-1-1-1-0.

Before calculating, I would guess that cases I) and II) together have a probability of 50%. The remaining cases, that are good for you, are

III) 2-1-1-0-0 (seems the likeliest distribution of all)
IV) 3-1-0-0-0 (feels unlikely)
V) 4-0-0-0-0 (feels impossible

[admitted]

Your opponent will get 2 yellows in exactly 2 cases (if yellow is all-important):

I) The yellows are distributed 2-2-0-0-0.
II) The yellows are distributed 1-1-1-1-0.

Before calculating, I would guess that cases I) and II) together have a probability of 50%. The remaining cases, that are good for you, are

III) 2-1-1-0-0 (seems the likeliest distribution of all)
IV) 3-1-0-0-0 (feels unlikely)
V) 4-0-0-0-0 (feels impossible

I get 328/969 for I or II, which is about 33.849%.
(10*6*4*3*4*3+5*4*4*4*4*4!)*16!/20! =328/969

[euklid314]

I also did the calculation, but I) to V) did not add up to 100%. I must have made sime mistake amd was too lazy to search for it.

Can you verify that III)-V) have 66.151% or can you explain your given formula?

[Edit: I think I can verify your approach. Seems correct. ]

[Propaganda_Panda]

My school days are long over, so I will trust you guys on the exact math

But even though the percentage is clearly below 50%, these are the kind of situations which I always try to avoid. Being forced into moves is never pleasant, much so in the last round.

[Effgee]

Glad this sparked some old probability formulas!
The math is actually pretty tricky because of: (1) permutations, (2) reduced spaces on factories (what I mean here is that once a tile has been placed on a factory, the likelihood of the next tile being placed there is reduced as there are now three spaces left as opposed to 4 on the other factories).
Let's look at the simple cases. I basically assume all factories empty, and randomly place 4 red tiles one by one. Once those are placed, the remaining 16 can go wherever space is free.
1) 4-0-0-0-0: First tile goes anywhere. Second tile must go to the same factory: 3 favorable spaces out of 19 total available. Third tile must also go to the same factory: 2 favorable spaces out of 18 total available. Same for last tile: 1 favorable space out of 17 total available. That's 6 over 5814, 0.1% probability.
2) 1-1-1-1: First tile goes anywhere. Second tile must go to any of the other 4 factories: 16 favorable spaces out of 19 total available. Third tile must go to yet another factory: 12 favorable spaces out of 18 total available. Same for last tile: 8 favorable space out of 17 total available. That's 1536 over 5814, 26.4% probability.

The other cases are harder because of permutations...

So let's tackle from a simulation perspective, much easier to implement and we'll be able to compare with above edge cases to confirm it worked.

out_4_plot.png (20.65 KiB) Viewed 1358 times

Wow, we indeed get 0.1% for the 4-0-0-0-0 case and off by just 0.1% for the 1-1-1-1-0 case.

Back to the original puzzle, I win if I can get 2 out of the 4 yellow tiles, so basically have the following favorable situations:

1) 4-0-0-0-0 (0.1% chance)
2) 3-1-0-0-0 (6.6% chance)
3) 2-1-1 (59.6% chance)

For a total very close to 2/3. So I thought I was pretty lucky being able to prevent opponent from getting 2 yellows, but odds were clearly in my favor.
Small note: I won the game

I have simulations for 2, 3, 4, 5.... tiles, let me know if you'd like to see those!