Challenge #168: Dice Game - Born to Solve

With another Alteryx Inspire behind us, we wanted to share a workflow from a session in the Community Track: 'Born to Solve - Think Like an Alteryx Champion'

In this session, @patrick_digan presented a analytical challenge to @estherb47 sight-unseen! The intent of the session was to show more than solving a problem, but show attendees how some of the top challenge leaders approach and think about problems. Would you like to try it yourself? See below!

THE CREW

Challenge

Suppose you have 3 ordinary dice (6 sides with the numbers 1-6). You roll all 3 dice and compute the score like so: Multiply the largest number by the second largest number. Then add the remaining number. Given this unique method of scoring, which is the most likely score?

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Accepted answers

All comments

NicoleJohnson

I was SO sad to miss this session with Weekly Challenge powerhouses @patrick_digan and @estherb47 ... Esther, I heard you did an excellent job, and it was super cool to see someone "think out loud" about how to tackle a challenge!

Spoiler

So happy to have a Weekly Challenge to start off my post-Inspire Monday... I'm suffering severe withdrawals already

Cheers!

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Kenda

Spoiler

Finished my first workflow using 2019.2!

Thanks all for a fun session at Inspire! I sort of did the "cheating" way and took advantage of the fact that there are only three dice, so I used the Summarize tool to get the max, min, and median for each roll. If there were more than 3 dice, this wouldn't have worked.

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Verakso

Must have been a brilliant session, and there is many ways to solve challenges in Alteryx, so here is mine.

It works, but is not very dynamic, and extended challenge would have been to do this with n numbers of dices...

Spoiler

I'll did it all in the Formula tool

Still Climbing

/Verakso

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estherb47

Cheating slightly on completing so quickly, as I already did the solve

So, included two more approaches, including one to make it completely dynamic and work 3 or more dice. Take a look, and please check out the challenge deck in the Inspire Buzz.

Taking the stage with @patrick_digan and @JoeM was such a treat!!

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bdaniels

All this talk about working with more dice made me wanna try solving for that too

Spoiler

Works for 3+ dice
Screen Shot 2019-06-17 at 11.04.21 AM.png

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neilgallen

is being at the session considered cheating? That said, I would have approached it the same way as @estherb47 did at Inspire.

Well done to @JoeM, @estherb47 and @patrick_digan as well. Great session.

Spoiler

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pjdit

My Solution.. It was fun watching this at Inspire and seeing how someone else thinks through the process.

Just glad I didn't have to blow up the big Dice before I could solve it.. That looked hard!

Spoiler

challenge_168_PD_Solution.yxmd

DataG8

Some wise people once said to always visualize your data, so I listen. Interesting that a score of 36 is not possible, someone more mathematician can probably explain. I wanted to use this in some fancy long run Monte Carlo simulation but maybe when we go to 5 dice.

Spoiler

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RolandSchubert

My solution - I decided to use a very simple approach

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jbrowne02

I attended this session at the conference so I had a head start but here is my solution.

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Ladarthure

My solution, funny one

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JORGE4900

I could not arrive at the same conclusion on the output; here is why.

I calculate there are (6*6*6) = 216 possible permutations; however, there are only 180 where Dice1 is greater than Dice2 or Dice 2 is greater than Dice 1, which is the same as Dice1 != Dice2. The mode of those 180 calculations is 7 and happens 12 times.

Where did I go wrong?

Spoiler

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pasccout

Fun challenge!

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phottovy

Got "Inspire"d by the conference and here is my first submission for a weekly challenge.

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JohnJPS

Fun stuff. I had a brain seizure along the way and found the expected value first.

Spoiler

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Ckaralis

Here is my solution.

I tried to keep in simple...

challenge_168_CKaralis_Solution.yxmd

Jbrooks

Fun challenge

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unknown

This was my first weekly challenge after hearing about it at Inspire - so fun!

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Greg_Murray

Spoiler

My solution.

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cplewis90

This is such a great idea to get the thoughts of how someone solves a problem! I am so sad I missed this! Excellent work @patrick_digan and @estherb47! Definitely a fun challenge!

Spoiler

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Claje

This was an interesting one!

Spoiler

I think I definitely overcomplicated the flow. Looking at some other solutions, there are definitely good ways to solve this using a more condensed flow. I like my answer because it should be completely dynamic and it uses the Sample tool to exclude the lowest die, and then uses a join to get the data back. This is certainly a slower solution than many of the others posted.

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Eduardo_Pavan

My solution certainly takes a lot longer but it is more hands on and fun!

Spoiler

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ryangensel

Spoiler

I was actually present during this session at Inspire. Without it, I probably wouldn't have used the "tile" tool in my solution.

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garthn555

Spoiler

My go at this....now to check other solutions !

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Alekh

Tried it with transpose, cross tab and then straight formula too.

Spoiler

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bnelly1987

My route seems a bit different from what I am seeing on here. I've always liked the Cross Tab and Transpose tools.

Spoiler

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garthn555

After I get spoilers working properly

T_Willins

You can actually just logic why the solution cannot be 36. Since the last number (smallest) is added, the product of the first two can be no smaller than 30 and no larger than 35 (last die cannot be greater than 6 or less than 1). Factoring the numbers between 30 & 35 leads to only one number that has 2 paired factors between 1 & 6: 30, with a factor pair of 5 & 6. Given the constraint that the added number must be less than or equal to the lower number the maximum result with a product of 30 is 35 (6*5+5).

T_Willins

I watched this live at #Inspire19, but I wanted to try the approach I was thinking of, which is to use a series of MAX functions. I'm happy with my result.

Spoiler

challenge_168_solution.yxmd

T_Willins

Hi @JORGE4900 . The rules are that the highest die and 2nd highest die get multiplied together, then the 3rd die gets added to the product. The die are independent of each other, so the highest die and 2nd highest die could be the same number. All 216 permutations should be in play to calculate the correct result.

I love Mondays - it's the day the Weekly Challenge comes out!

T_Willins

@NicoleJohnson - I know your withdrawal. I'm waiting for the ferry back from Orcas Island by logging into the Alteryx Community and doing the Weekly Challenge.

BradWerner

Done!

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paulrstats

My first weekly challenge, having been inspired by Inspire 2019

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papalow

Not an elegant solution -- but it works.

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dhtay

I solved it by writing a nested if-else formula.

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mbogusz

Spoiler

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Cal_A

I like the visualisation, it helps give comfort that you understand the result and makes you check your expectations/intuition.

36 is not possible because to get in the ballpark you need to have:

Dice1*Dice2 >= 30 [IE 6x5 or 6x6]

6x6 => cannot get 36 as still have to add dice3 (smallest dice), which will be 1-6.

6x5 => cannot get 36 as dice3 must be <= dice 2 (dice2 is the second largest, or middle dice). Dice 3 cannot be 6, has to be 5,4,3,2,1.

Cal_A

It is a question of interpretation.

In a result of 5, 5, 5 I would say:

- Largest number is 5

- Smallest number is also 5

- Median number is also 5

This is also how the functions in Alteryx and Excel work (e.g. Max(5,5,5) = 5)

To me, it makes sense. Imagine a city has a rainfall over 5 days of 5mm, 10mm, 6mm, 10mm, 5mm. If I asked you what the maximum/largest rainfaill was, you would probably say "10mm", you wouldn't say Null.

piotrzawistowski

My version.

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NickC

🙂

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Kale_Malcom

Spoiler

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TungThanhHo

my solution

challenge_168_Thanh_Solution.yxmd

bpiscoya

Hello Esther, I read your post, kudos!!!. I was wondering whether there is a video link of your solution that you might share.

In my first approach, I treated it as 3 independent events with a random of 0-6 each one to calculate the score. That was my initial approach.

Spoiler

Computernerd

I feel so good that my solution was basically identical to yours. I was wondering how far off I'd be on the "right" solution. :-)

vakachiranjeevi

estherb47

Thanks for your solution, I have learned few new things, however your solution doesn't work in case of dynamic inputs - . Have added 4th dice and it is not working as expected.

flaw I have observed:

1. Static column picking while finding the sum. instead of static use Min,max and median to achieve this.

AHill06

Here is my solution...its not pretty but works!

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rapetter

Done

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