07-01-2017, 01:35 PM
07-01-2017, 06:53 PM
(07-01-2017 01:35 PM)Don Shepherd Wrote: [ -> ]Here is an interesting game.
What would be a good strategy for winning?
Hi Don,
I love these simple games and whether there could be a strategy to winning them. Unless I have made a mistake (which would be very likely), the following are the possible combinations of dice rolls:
There are 36 possible dice rolls. Here's the possible results for each number:
1 would be 18 out of the 36
2 would be 19 out of 36
3 would be 14 out of 36
4 would be 13 out of 36
5 would be 10 out of 36
6 would be 10 out of 36
7 would be 6 out of 36
8 would be 7 out of 36
9 would be 5 out of 36
10 would be 5 out of 36
So if possible, I would try to do 9 and 10 first, then 7 & 8, then 5 & 6, then 3 & 4, and save 1 & 2 towards the end.
I had never heard of this game. These is also a variation on this using Letters on the Dice that you would then try to create words from.
bill
Smithville, NJ
07-01-2017, 07:09 PM
The simple approach: look at the totals you can make with your roll, and choose the "rarest" one.
[attachment=4992]
[attachment=4992]
07-01-2017, 11:56 PM
Thanks Bill and Dave.
I must be doing something wrong, my figures don't seem to jive with either of yours, although the general approach seems right (go for the biggest numbers [9, 10] first since they are the rarest, I think).
I don't consider 36 possible rolls, because 1-3 and 3-1 are the same. I considered these possible rolls:
1-1
1-2
1-3
1-4
1-5
1-6
2-2
2-3
2-4
2-5
2-6
3-3
3-4
3-5
3-6
4-4
4-5
4-6
5-5
5-6
6-6
Considering the add, subtract, multiply, and divide for these combinations, I get the following possibilities:
1 - 12
2 - 9
3 - 7
4 - 7
5 - 5
6 - 6
7 - 3
8 - 4
9 - 3
10 - 3
And I assume division must leave no remainder, so for example a 3-2 roll could only yield 5, 1, or 6.
Don
I must be doing something wrong, my figures don't seem to jive with either of yours, although the general approach seems right (go for the biggest numbers [9, 10] first since they are the rarest, I think).
I don't consider 36 possible rolls, because 1-3 and 3-1 are the same. I considered these possible rolls:
1-1
1-2
1-3
1-4
1-5
1-6
2-2
2-3
2-4
2-5
2-6
3-3
3-4
3-5
3-6
4-4
4-5
4-6
5-5
5-6
6-6
Considering the add, subtract, multiply, and divide for these combinations, I get the following possibilities:
1 - 12
2 - 9
3 - 7
4 - 7
5 - 5
6 - 6
7 - 3
8 - 4
9 - 3
10 - 3
And I assume division must leave no remainder, so for example a 3-2 roll could only yield 5, 1, or 6.
Don
07-02-2017, 12:21 AM
You have to remember that 3, 1 and 1, 3 are considered different outcomes for computing probability.
I just made a table of all possible outcomes in 1-2-3 (6x6x4), extracted all the integer results between 1 and 10, while removing duplicates (I think 1+1=1 and 1/1=1 is the only one), counted the frequencies, and slapped a graph on the results.
I just made a table of all possible outcomes in 1-2-3 (6x6x4), extracted all the integer results between 1 and 10, while removing duplicates (I think 1+1=1 and 1/1=1 is the only one), counted the frequencies, and slapped a graph on the results.
07-02-2017, 12:25 AM
(07-01-2017 11:56 PM)Don Shepherd Wrote: [ -> ]I don't consider 36 possible rolls, because 1-3 and 3-1 are the same. I considered these possible rolls:
If you do this, you will have outcomes with unequal probabilities. E.g. it is twice as likely to roll 1-2 as it is 1-1. You'll have to weight unequal rolls twice that of pairs.
Pauli
07-02-2017, 03:51 AM
OK, considering 36 possible rolls, I get these frequencies:
1-17
2-17
3-14
4-12
5-10
6-11
7-6
8-7
9-5
10-5
This was from paper-and-pencil tallies and me eyeballing and summarizing the results, so I wouldn't swear to it but I think it is correct.
The winning strategy is the same, go for the bigger numbers first.
It would be interesting to write a simulation of, say, 1000 game iterations with 4 players, one of which follows this strategy and the other three just choose a random solution each time, and see the results.
1-17
2-17
3-14
4-12
5-10
6-11
7-6
8-7
9-5
10-5
This was from paper-and-pencil tallies and me eyeballing and summarizing the results, so I wouldn't swear to it but I think it is correct.
The winning strategy is the same, go for the bigger numbers first.
It would be interesting to write a simulation of, say, 1000 game iterations with 4 players, one of which follows this strategy and the other three just choose a random solution each time, and see the results.
07-02-2017, 11:28 AM
(07-02-2017 03:51 AM)Don Shepherd Wrote: [ -> ]The winning strategy is the same, go for the bigger numbers first.
It would be interesting to write a simulation of, say, 1000 game iterations with 4 players, one of which follows this strategy and the other three just choose a random solution each time, and see the results.
A more fun test would be to play the game with a young person. Too bad my grandchildren are too old to want to play. But it would be interesting to see how long it would take for a young person to figure out what the better strategy is and see if they adjust their play over time. Of course, it also would be a great opportunity to discuss probabilities with them and how it can affect their game play.
Bill
Smithville, NJ
07-02-2017, 01:01 PM
(07-02-2017 11:28 AM)Bill (Smithville NJ) Wrote: [ -> ]A more fun test would be to play the game with a young person. Too bad my grandchildren are too old to want to play. But it would be interesting to see how long it would take for a young person to figure out what the better strategy is and see if they adjust their play over time. Of course, it also would be a great opportunity to discuss probabilities with them and how it can affect their game play.
Bill
Smithville, NJ
That's a good idea. My 3-month-old granddaughter is probably too young, however!