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!