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Do you have the skills to make it to the top? Subscribe to our weekly challenges. Try your best to solve the problem, share your solution, and see how others tackled the same problem. We share our answer too.
Weekly Challenge
Do you have the skills to make it to the top? Subscribe to our weekly challenges. Try your best to solve the problem, share your solution, and see how others tackled the same problem. We share our answer too.
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Challenge #168: Dice Game - Born to Solve

Meteoroid
 
Meteor

It's not an elegant solution, but it gets the job done... 

Meteoroid
It’s my first one. I’ll get better 😊
Meteoroid

Fun challenge! Looks I came up with the same workflow as @EstherB47, but without a 1000 eyes and ears on me! :) Good one Esther! 

 

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Challenge 168.PNG

 

Meteor
Spoiler
I leveraged one formula to take care of calculating all the P&Cs and summarised to identify max probability score. 
challenge168_viveknshah.JPG
Aurora

Congrats to @EstherB47 , and excellent challenge @patrick_digan 

 

I have to echo @NicoleJohnson's sentiment here.  Sorry I missed it, but there was just so much to see at Inspire!

 

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Dan

Pulsar

@danilang I am so glad to have met you in person at Inspire this year!!! Putting faces to community handles was a wonderful highlight of the week.

Aurora

I totally agree @EstherB47 

 

Learning new things is fun, but meeting new people is more enriching.

 

Dan

Asteroid

Hello @T_Willins  I understand that when the 3 dices are thrown it is an independent event, but listing all the possible combinations would mean that there would be 36 combinations where two dices have the same value; for example: 6-6-1 or 1-1-6. Since those combinations belong to the universe of possible combinations, it means that when taking the highest, mean, and minimum of those combinations, a combination would be repeated.

 

I may be wrong, but I think 36 events need to be excluded and filtered out.

 

Thanks for taking the time to reply.

Here is my solution

 

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Solution.png