Little math problems October 2019

[pier4r]

Imagine you have just saw the "royal game of Ur" teaser on youtube.





Nice game, easy to reproduce. I think it is more enjoyable than backgammon where there is more computation to do if one is competitive, and it is not exactly equal to https://en.wikipedia.org/wiki/Mensch_%C3...Dich_nicht or its better sequel (Sequel to Mesch ärgere dich nicht) dog or Tock (https://en.wikipedia.org/wiki/Tock).

Back to the game of Ur. In the teaser there where practically 4 dice, each behaves like a coin.

Nowadays we normally use dice with 6 sides, flipping coins is unpractical and only **cough** nerds **cough** have other type of dice or even calculators (and who on earth would download apps to simulate dice!).

Thus you scavenged some of your popular games (risiko, monopoly, settlers of catan, cluedo and what not) and you found an handful (you decide how many) of 6 sided dice, how you are going to simulate the "flipping of 4 coins"? Please explain it to your fellow players. Of course the solution should be practical.

[jpcuzzourt]

(10-06-2019 07:07 PM)pier4r Wrote:  ...
Back to the game of Ur. In the teaser there where practically 4 dice, each behaves like a coin.

Nowadays we normally use dice with 6 sides, flipping coins is unpractical and only **cough** nerds **cough** have other type of dice or even calculators (and who on earth would download apps to simulate dice!).

Thus you scavenged some of your popular games (risiko, monopoly, settlers of catan, cluedo and what not) and you found an handful (you decide how many) of 6 sided dice, how you are going to simulate the "flipping of 4 coins"? Please explain it to your fellow players. Of course the solution should be practical.

Well, I'll jump in with a couple of naive suggestions. First, I should say I may not be adequately prepared since I haven't watched the Finkel video yet, so these may be too naive.

1. Use 4 dice, and count certain rolls as [heads], and others as [tails].
e.g. 1,2,3 = [heads], 4,5,6 =[tails], or
even=[heads], odd=[tails]

2. Use a single die and roll it 4 successive times with rules as above to determine
[heads] or [tails]

I'm curious as to what I must be missing, as this seems like a trivial problem, so I'll go watch the video now, because I always enjoy Finkel.

[pier4r]

In the video they have "original" dice. I posed the question inspired from the video. I didn't work it out yet but somewhen I'll try to work if I can map a 6 sided dice to the 4 coin probabilities in a easy way (maybe with a piece of paper with some guidelines). This because if you roll more than one dice at once and you want to minimize the mapping, then you need to deal with different probabilities. Of course the immediate solutions you gave are perfectly viable as well.

[jpcuzzourt]

(10-06-2019 08:11 PM)pier4r Wrote:  In the video they have "original" dice. I posed the question inspired from the video. I didn't work it out yet but somewhen I'll try to work if I can map a 6 sided dice to the 4 coin probabilities in a easy way (maybe with a piece of paper with some guidelines). This because if you roll more than one dice at once and you want to minimize the mapping, then you need to deal with different probabilities. Of course the immediate solutions you gave are perfectly viable as well.

The probabilities of 0,1,2,3,4 are 1/16, 1/4, 3/8, 1/4, 1/16 respectively.
I haven't devised a mapping that is easy to explain and more practical than what I suggested earlier. Throwing 4 dice and counting the number of [EVEN] rolls as your move, e.g. is straightforward and practical.
I'll consider it a little more and post if I devise a simple way of producing those probabilities using less dice that doesn't require players to do base 6 arithmetic (which is impractical with most of my friends and family.)
I do look forward to seeing if you or others come up with some clever ways of reproducing those odds simply with fewer throws or simpler rules.

[Thomas Okken]

I don't know how to prove it rigorously, but it seems to me like you can't do better than four dice rolls. Because of the way the probabilities are distributed, whatever scheme you come up with must have a multiple of 16 possible outcomes, and since a die roll has 6 = 2 * 3 possible outcomes, getting the required factor of 2^4 requires 4 dice rolls.

That is, unless you allow re-rolls; then you could do something like this: roll two dice and add the numbers, and then say 3 -> 0; 2, 4, 5 -> 1; 6, 7, 12 -> 2; 8, 10 -> 3; 11 -> 4; 9 -> try again. That means you end up needing, on average, 2 1/4 dice rolls, but of course there is no upper limit to the number of rolls you might end up needing on any given turn.

[jpcuzzourt]

(10-06-2019 09:40 PM)Thomas Okken Wrote:  I don't know how to prove it rigorously, but it seems to me like you can't do better than four dice rolls. Because of the way the probabilities are distributed, whatever scheme you come up with must have a multiple of 16 possible outcomes, and since a die roll has 6 = 2 * 3 possible outcomes, getting the required factor of 2^4 requires 4 dice rolls.

That is, unless you allow re-rolls; then you could do something like this: roll two dice and add the numbers, and then say 3 -> 0; 2, 4, 5 -> 1; 6, 7, 12 -> 2; 8, 10 -> 3; 11 -> 4; 9 -> try again. That means you end up needing, on average, 2 1/4 dice rolls, but of course there is no upper limit to the number of rolls you might end up needing on any given turn.

Well, since we require a space of at least 16 possibilities, 2 6-sided dice yields 36 possible states, if taken in order. I've worked out a mapping that allows us to discard 4 of those as invalid, requiring a re-roll if they happen (11.11% probability), and all the others are mapped such that the order doesn't matter, condensing the remaining 32 states to 16. Despite my best efforts at making it logical, it would not be easily explained or learned, I think. Here is one possible mapping. BTW, lets redesignate the '6' as a '0' so that each die displays 0-5.

03,30,24,25 are disallowed = rethrow

01 or 10 = '0'
45 or 54 = '4'
12,21,14,41,15,51,05,50 = '1'
13,31,23,32,34,43,35,53 = '3'
00,11,22,33,44,55,02,20,04,40,05,50 = '2'

You can see that you get a '2' if you roll doubles or a 0 with anything but a 1 (or 3, since 03, 30 are invalid.)
stating the other rules is a bit trickier. Let me simplify the table:

01 = '0'
05,12,14,15 = '1'
13,23,34,35 = '3'
45 = '4'
doubles and 02,04,05 = '2'

That's the best I can do for 'minimal', but the rules are pretty ugly. Someone might devise prettier rules, but as long as I could scrounge 4 dice, I think I'd do it that way. Or throw 1 die 4 times, or 2 dice, twice!
I'll keep my eye on this space for better solutions though.

[Paul Dale]

Using six sided dice to get exact probabilities can only be done as has been suggested (roll four counting high/low or a complex table with rerolls).

More interesting might be an approximation, roll two dice and sum the total:

Code:

outcome    roll     probability     desired probability
   0        3           1/18            1/16
   1        5, 6        1/4             1/4
   2        others      7/18            3/8
   3        8, 9        1/4             1/4
   4        11          1/18            1/16

Pauli

[Albert Chan]

Instead of rolling the dice, put it in the box, such that it only have room to spin.

Shake the box to do a random spin, and do it twice, you got 1 of 16 outcomes.

[jpcuzzourt]

(10-07-2019 12:10 AM)Paul Dale Wrote:  Using six sided dice to get exact probabilities can only be done as has been suggested (roll four counting high/low or a complex table with rerolls).

More interesting might be an approximation, roll two dice and sum the total:

Code:

outcome    roll     probability     desired probability
   0        3           1/18            1/16
   1        5, 6        1/4             1/4
   2        others      7/18            3/8
   3        8, 9        1/4             1/4
   4        11          1/18            1/16

Pauli

That's a very reasonable approximation. I'm sure it'd be acceptable to most engineers.
:-)

[mfleming]

(10-07-2019 01:19 AM)jpcuzzourt Wrote:  That's a very reasonable approximation. I'm sure it'd be acceptable to most engineers.
:-)

+1
Horseshoes and hand grenades and engineers with sliderules...

[jpcuzzourt]

(10-07-2019 04:07 PM)mfleming Wrote:  

(10-07-2019 01:19 AM)jpcuzzourt Wrote:  That's a very reasonable approximation. I'm sure it'd be acceptable to most engineers.
:-)

+1
Horseshoes and hand grenades and engineers with sliderules...

I have quite a collection of those (Sliderules, not horseshoes or hand grenades )