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Challenge #73: Plinko Probabilities

8 - Asteroid

Tried to not over complicate this too much!

10 - Fireball

Here is a solution

19 - Altair
19 - Altair

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)!