Happy 8th birthday to the Maveryx Community! Take a walk down memory lane in our birthday blog, and don't miss out on the awesome birthday present that all Maveryx Community members get to take advantage of!

Weekly Challenge

Solve the challenge, share your solution and summit the ranks of our Community!

Also available in | Français | Português | Español | 日本語

Want to get involved? We're always looking for ideas and content for Weekly Challenges.


Challenge #73: Plinko Probabilities

8 - Asteroid

Tried to not over complicate this too much!

10 - Fireball

Here is a solution

18 - Pollux
18 - Pollux

I modeled the board and then Monte Carlo'd until the pins wore off.


Like @patrick_digan@jamielaird and @Kenda, I found the bounded model produces higher probabilities that token will end up in Bucket 1 or  5 if started in those columns




Because I modeled the board in the data input, I only needed 2 tools in the macro to simulate a run.


8 - Asteroid

My solution.


Ummm wow, this one was exhausting for me.


Glad to know the posted solution looked up a pascal triangle because I spent a good half hour trying to manually count the paths to each space and gave up after a bit.  Definitely not proud of my efficiency on this one.  I walked away and came back to it on and off for 3 days researching different articles on Plinko probability.


So here it is, Didn't bother with the bonus, mostly cause i was burnt out by the time i got to the solution.  Just glad to just check the box on this one and be done.  :)  No offense   


I imagine you could add some interface tools on this approach to adjust the size of triangle you put through the calculation. 



8 - Asteroid

I followed the posted solution for this exercise....

8 - Asteroid

Finally got it (I think...)


You input start column (in letters - A to E) and it gives you the output...


Challenge 73 Result.jpg
15 - Aurora
15 - Aurora

I went with an iterative macro which runs 1000 simulations for each starting position, and then works out the % of iterations which ended in the winning position. Worked with @George_Walker on this one :)

Starting in the second position seems to be the best place - ending in the winning spot ~ 26% of the time. 

The formula updating the position in the macro uses whether a random integer is odd or even to decide whether to go left or right. We also need to appreciate that if a row has 6 blocks, (what I called a T(ight) row) and we are in the outer positions, we can only stay in the outer position. If we are in a row with 5 blocks (what I called a L(oose) row) we can always move left or right. 
Challenge 73.png
11 - Bolide

Hi! Here my solution :)


I developed an app that ask two questions to the user: how many rows and the starting position.

Considering the game as an iteration of groups of two rows (a 4-pin row and a 5-pin row) plus a possible 5-pin row, I calculated the probabilities for a 2-row group and iterated as many times as groups existing; then, applying the posible 5-pin row probabilities.



11 - Bolide

That took a while but eventually got there. First built a macro that builds Pascal's Triangle, then built a batch macro that calculates the chances of reaching each bin. The batch macro can be customised on the number of slots and number of rows.

8 - Asteroid

Here's my solution to challenge 73!

I created an iterative macro that will allow you to decide the number of stages and tell you the probability of landing in any given position from any given space.

I define a stage as two sets of pegs - so every time the pegs line up to match the first row of pegs, that's a new stage.

I'm really happy with this as it took a lot of work to figure out how to iterate properly and figure out a cheeky way of passing multiple fields back in a single field (concat, pass back, seperate)!