"Simple" Dice Probability

Evilbob

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Sep 3, 2012
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Problem:
Person A and Person B each possess a d20 die (a die with 20 sides from 1-20). Person B believes her dice are extra lucky and challenges Person A to a "dice-off".


Person B is so convinced of her luck that in each Event, she sets up the following arrangements:
-In each Event, Person B will roll twice and pick the highest number from those rolls.
-In each Event, Person A will roll thrice, pick the highest number from those rolls and add 1/4 of Person B's highest number, rounded up to the nearest integer, to his own number.
-Whoever has the highest total number at the end of each Event "wins" that Event. If both numbers are equal, it is considered a "tie".


Assuming that Person B is not as lucky as she thinks she is (and isn't cheating in any way), for any given Event, what is the probability that:
a) Person A will "win"?
b) Person B will "win"?
c) The Event will result in a "tie"?


d) If Person A only added 1/5 of Person B's highest number instead of 1/4: how will the probability change for a), b), and c), respectively?
---



I secretly suspect our professor is either a D&D nerd or a sadist. (or maybe even both...)


I considered and tried listing the possibilities to see if I could see a pattern, but it quickly got out of control, and I'm pretty sure there's an easier way to solve this problem; I'm not sure how to approach it or really to start; it seems a bit daunting.


It seems like there's almost too much going on here at the same time. (three rolls+highest fraction of the two rolls??? vs. two rolls???)

Please let me know if there's a way to actually solve this problem like a sane person instead of brute-forcing cracking it by listing every single possibility...
 
Sometimes to wrap your brain around a problem like this it helps to do a small similar problem first. Assume the dice are actually d3. For person B we have:

11 12 13 ---> 1 2 3
21 22 23 ---> 2 2 3
31 32 33 ---> 3 3 3

So for person B, P(1)=1/9, P(2)=3/9, P(3) = 5/9 which is clearly P(t)=(2t-1)/9. 1<=t<=3

For person A, we have:

111 112 113 121 122 123 131 132 133 ---> 1 2 3 2 2 3 3 3 3
211 212 213 221 222 223 231 232 233 ---> 2 2 3 2 2 3 3 3 3
311 312 313 321 322 323 331 332 333 ---> 3 3 3 3 3 3 3 3 3

So for person A, P(1)=1/27, P(2)=7/27, P(3)=19/27 (numerator is +6, +12 etc...)
This appears to be P(n)=[3(n^2)-3n+1]/27

I don't have time to continue right now, but thinking ahead you will have to find which combos result in ties (when B rolls multiples of 4) which gives me the gut feeling like this isn't so simple to derive a nice neat formula from but I could be wrong. Brute force it with a spreadsheet program like Excel is always a fairly straight forward option. Don't be crazy and try to brute force it by hand when computers can do it :p
 
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