February 4th, 2013 at 12:33:22 PM
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I have been working on a board game and am researching the best way to use dice to resolve challenges. The game utilizes d10 dice numbered 1 to 5 twice on each die, so just call them d5 dice for probability purposes.

Certain challenges in the game involve opponents seeking one or the other of specific pairs of numbers, whcih they take turns in matches to until only one of them gets his or her respective number. Numbers can be obtained on either die or by adding them together. The opposing numbers used in matches are 1 and 10, 2 and 9, 3 and 8, 4 and 7, and 5 and 6. So a 2 is obtainable as a 2 on either die or by a 1 on both dies added together.

I have worked out the probabilities when each player gets one toss of the dice per match, and also when the high number seeker (like a player seekingan 8 in a 3 vs 8 challenge)gets two tosses per match. With either rule, the low number seeker always has better odds, which is fine becausevthe challenger gets to pick the more favorable number as part of how the game works. (BTW, one reason for experimenting with letting the higher-number seeker get two tosses is that it bring the odds closer to together for the two opponents, and also slight decrease the number of matches needed to resolve a challenge.)

Now here is what I want to calculate. I would like to see what happens if each player gets one toss of the dice per match, except if a player obtains their opponent's number then he or she can toss the dice again. So some kind of a conditional pr dependent kind of calculation is needed.

As an example, let's say we have two players, one seeking a 2 and the other a 9. A number can be obtained on one die or by adding them together. The probability of getting a 2 = 10/25 (2 or 2 with 1, 2, 3, 4 or 5 on the other die, and 1+1) and for 9 = 2/25 (4+5 or 5+4). But if the player seeking a 2 obtains 9, he gets to throw again, and yet again if a 9 is again obtained, and similarly if the player seeking 9 gets 2 then that player gets another try.

Taking the above exmple, let's player A is seeking 2 and tosses the dice and gets 9 (5+4), tosses again because he got his opponent's number and gets nether 2 nor 9, ending his try for that match. Now player B, who is seeking 9, tosses the dice and gets 2 (2 on both dies), tosses again and gets 2 (1 +1), then tosses again gets neither 2 nor 9. Now another match and player A tosses and gets neither 2 nor 9, then player B tosses and gets his 9 to win the challenge. How does one mathematically represent this kind of setup in terms of the probability for each player obtaining their respective number, taking into account the proability of obtaining the opponent's number and getting another toss?

Certain challenges in the game involve opponents seeking one or the other of specific pairs of numbers, whcih they take turns in matches to until only one of them gets his or her respective number. Numbers can be obtained on either die or by adding them together. The opposing numbers used in matches are 1 and 10, 2 and 9, 3 and 8, 4 and 7, and 5 and 6. So a 2 is obtainable as a 2 on either die or by a 1 on both dies added together.

I have worked out the probabilities when each player gets one toss of the dice per match, and also when the high number seeker (like a player seekingan 8 in a 3 vs 8 challenge)gets two tosses per match. With either rule, the low number seeker always has better odds, which is fine becausevthe challenger gets to pick the more favorable number as part of how the game works. (BTW, one reason for experimenting with letting the higher-number seeker get two tosses is that it bring the odds closer to together for the two opponents, and also slight decrease the number of matches needed to resolve a challenge.)

Now here is what I want to calculate. I would like to see what happens if each player gets one toss of the dice per match, except if a player obtains their opponent's number then he or she can toss the dice again. So some kind of a conditional pr dependent kind of calculation is needed.

As an example, let's say we have two players, one seeking a 2 and the other a 9. A number can be obtained on one die or by adding them together. The probability of getting a 2 = 10/25 (2 or 2 with 1, 2, 3, 4 or 5 on the other die, and 1+1) and for 9 = 2/25 (4+5 or 5+4). But if the player seeking a 2 obtains 9, he gets to throw again, and yet again if a 9 is again obtained, and similarly if the player seeking 9 gets 2 then that player gets another try.

Taking the above exmple, let's player A is seeking 2 and tosses the dice and gets 9 (5+4), tosses again because he got his opponent's number and gets nether 2 nor 9, ending his try for that match. Now player B, who is seeking 9, tosses the dice and gets 2 (2 on both dies), tosses again and gets 2 (1 +1), then tosses again gets neither 2 nor 9. Now another match and player A tosses and gets neither 2 nor 9, then player B tosses and gets his 9 to win the challenge. How does one mathematically represent this kind of setup in terms of the probability for each player obtaining their respective number, taking into account the proability of obtaining the opponent's number and getting another toss?

February 4th, 2013 at 12:41:10 PM
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well i have no idea

February 4th, 2013 at 3:17:45 PM
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Not really understanding your rules on how one player is seeking a 2. is it from the sum of both dice or can the player choose one number from one die??Quote:morphixAs an example, let's say we have two players, one seeking a 2 and the other a 9.

A number can be obtained on one die or by adding them together.

The probability of getting a 2 = 3/25 (2, 2, 1+1) and a 9 = 2/25 (4+5, 5+4).

So you are really making this challenging.

The 3/25 to roll a 2 I do not see that, unless I do not understand what exactly the rules are.

I agree with the 8/100 for the 9. That is easy since there is no way one die could roll a 9 with 1-5 markings.

One can also just write out the polynomial for 2d10 1-5 and solve it too.

(x+x+x^2+x^2+x^3+x^3+x^4+x^4+x^5+x^5)^2 =

4x

^{2}+8x

^{3}+12x

^{4}+16x

^{5}+20x

^{6}+16x

^{7}+12x

^{8}+8x

^{9}+4x

^{10}

That was fun. Have not done that in a long time.

added:

(this just shows a math formula on the number of ways each total can roll from the above 2 dice.

4x

^{2}means there are 4 ways a total of 2 can roll: the x

^{2}and so on.

There are many online calculators that can expand polynomials

Now for the 2.

4/100 is for the sum of 2 from your 2 dice.

But now they can also get a 2 from each die.

That sounds like a large advantage there for Mr. #2.

Now we can add to the 4/100 for the sum just with each die getting a 2 because a 2 can come from the sum OR an individual die.

1/5 + 1/5 but we need to subtract their union because we are over-counting when the roll is a 2,2

(1/5 + 1/5) - (1/5 * 1/5)

(5/25 + 5/25) - (1/25)

(10/25) - (1/25) = 9/25

added: 9/25 = 36/100 meaning there are 36 outcomes from the 100 total that have at least one 2 in it.

9/25 + 4/100 = 40/100 for rolling a sum of 2 or a 2 on one of the two dice.

we add the probabilities because the events are mutually exclusive, meaning both can not happen at the same time. Right?? a 1,1 can roll but not contain a 2

40/100 (#2) vs 8/100 (#9)

(added: I do not see where you got 3/25 for the sum of 2 OR a 2 being rolled on one die)

I still do not get what you are after with your example I just re-read

Maybe I am just having a slow day today.

IMO, why not just keep it simple and have one total vs another.

The multiple rolls thing would be so much easier to see

Good Luck

winsome johnny (not Win some johnny)

February 4th, 2013 at 5:34:02 PM
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I have edited the post to make it clearer. It would take a lengthy explanation to get across why the challenges are lopsided to favor the challenger, but it relates to various things that make sense and are fun within the game. Right now I am just trying to weigh the plusses and minuses of various ways to structure these challenges. If my revised post is still unclear let me know what is unclear and I will do my best to explain.

February 4th, 2013 at 7:23:02 PM
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I agree with 7craps on getting a 2. I calculated 10/25, which, of course, equals 40/100.

I heart Crystal Math.

February 4th, 2013 at 8:09:46 PM
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The OP has another thread that also talks about this game.

I may be wrong here but

I think he needs to really learn how to calculate the probabilities of the sums of any dice say 2d10 with 1-5 faces,

the one he is currently using, he may already know this

or

2d10 with 0-9 faces,

2d10 with 1-10 faces or

even crazy faces like

2d10 with faces 1,1,1,1,2,2,2,2,3,3 on one and 0,0,0,0,0,1,2,3,4,5 on the other.

or even a d4 and a d6

or

a d6 and a d10

the combos are almost endless.

Then learn to add the probabilities of different events, both mutually exclusive and joint events.

For many and even me it takes time to learn this, but once the basics are down you should be home free.

Practice helps too.

The math is challenging until you see how to do it.

Then it is so simple it really is fun stuff IMO.

The best info I have seen for this (dice sum probabilities) is a pdf by (plug, it is free)

Matthew M. Conroy

http://www.madandmoonly.com/doctormatt/mathematics/mathematics.htm

The link is Dice Problems

Then find

Appendix C

Dice Sums and Polynomials

(the pdf is a good one to go through, many questions with good solutions)

I think I pointed this out before.

Get good and go wild with your new game!

Good Luck

OP, how did you arrive at 3/25 for the probability of 2??

We really should start there first.

Not putting you down, just trying to help out.

This is fun stuffs.

I may be wrong here but

I think he needs to really learn how to calculate the probabilities of the sums of any dice say 2d10 with 1-5 faces,

the one he is currently using, he may already know this

or

2d10 with 0-9 faces,

2d10 with 1-10 faces or

even crazy faces like

2d10 with faces 1,1,1,1,2,2,2,2,3,3 on one and 0,0,0,0,0,1,2,3,4,5 on the other.

or even a d4 and a d6

or

a d6 and a d10

the combos are almost endless.

Then learn to add the probabilities of different events, both mutually exclusive and joint events.

For many and even me it takes time to learn this, but once the basics are down you should be home free.

Practice helps too.

The math is challenging until you see how to do it.

Then it is so simple it really is fun stuff IMO.

The best info I have seen for this (dice sum probabilities) is a pdf by (plug, it is free)

Matthew M. Conroy

http://www.madandmoonly.com/doctormatt/mathematics/mathematics.htm

The link is Dice Problems

Then find

Appendix C

Dice Sums and Polynomials

(the pdf is a good one to go through, many questions with good solutions)

I think I pointed this out before.

Get good and go wild with your new game!

Good Luck

OP, how did you arrive at 3/25 for the probability of 2??

We really should start there first.

Not putting you down, just trying to help out.

This is fun stuffs.

winsome johnny (not Win some johnny)

February 5th, 2013 at 6:15:37 AM
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On dice numbered 1 to 5 there are 25 possible combinations involving the addition of numbers from each die, as seen on a 5 x 5 table. To that I add the possibility of getting some sought after numbers on either die individually, such as a 1, 2, 3, 4 or a 5. So a 2, for example, can be obtained with a 1 on both dies added together, or by getting a 2 on either die with a 1,2, 3, 4 or 5 on the other die, so 10/25. ( Corrected after comment below.)

February 5th, 2013 at 6:42:35 AM
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Quote:morphixOn dice numbered 1 to 5 there are 25 possible combinations involving the addition of numbers from each die, as seen on a 5 x 5 table. To that I add the possibility of getting some sought after numbers on either die individually, such as a 1, 2, 3, 4 or a 5. So a 2, for example, can be obtained with a 1 on both dies added together, or by getting a 2 on either die, so 3/25, unless there is some other way I should be doing it?

There is some other way (the correct way) to do it:

Pr( 2 on die 1) = 1/5

Pr( 2 on die 2, not on die 1) = 1/5 * 4/5

Pr( 1 on both die) = 1/25

Pr( 2 anyway ) = sum of the above 3 = 10/25

You can also use your 5x5 table, where you should see there is an entire row & column that achieves a 2 (9 spots), plus the 1,1 spot to make 10/25

You need to know how to do this in order to calculate your more complicated probabilities. If you are serious about your board game design, you may want to pay someone to analyze this.

Wisdom is the quality that keeps you out of situations where you would otherwise need it

February 5th, 2013 at 8:17:25 AM
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I go the easy way for a double-check with only simple adding required,

by brute force using all 100 possible outcomes (10*10)

Just count1,6,51,56 are the sums for 2. There are 4 of them.

For a roll that has at least a 2 on either Die

2: 2,7

10: 11-20

8: 22,27,32,37,42,47,52,57

10: 61-70

6: 72,77,82,87,92,97

So, 2+10+8+10+6 = 36

36+4 = 40

1/100 * 40 = the probability of rolling a sum of 2 or a 2 on at least one die

Good Luck

Why only faces with 1-5?

by brute force using all 100 possible outcomes (10*10)

Just count

# Die 1 Die 2 Sum # Die 1 Die 2 Sum

1 1 1 2 51 1 1 2

2 1 2 3 52 1 2 3

3 1 3 4 53 1 3 4

4 1 4 5 54 1 4 5

5 1 5 6 55 1 5 6

6 1 1 2 56 1 1 2

7 1 2 3 57 1 2 3

8 1 3 4 58 1 3 4

9 1 4 5 59 1 4 5

10 1 5 6 60 1 5 6

11 2 1 3 61 2 1 3

12 2 2 4 62 2 2 4

13 2 3 5 63 2 3 5

14 2 4 6 64 2 4 6

15 2 5 7 65 2 5 7

16 2 1 3 66 2 1 3

17 2 2 4 67 2 2 4

18 2 3 5 68 2 3 5

19 2 4 6 69 2 4 6

20 2 5 7 70 2 5 7

21 3 1 4 71 3 1 4

22 3 2 5 72 3 2 5

23 3 3 6 73 3 3 6

24 3 4 7 74 3 4 7

25 3 5 8 75 3 5 8

26 3 1 4 76 3 1 4

27 3 2 5 77 3 2 5

28 3 3 6 78 3 3 6

29 3 4 7 79 3 4 7

30 3 5 8 80 3 5 8

31 4 1 5 81 4 1 5

32 4 2 6 82 4 2 6

33 4 3 7 83 4 3 7

34 4 4 8 84 4 4 8

35 4 5 9 85 4 5 9

36 4 1 5 86 4 1 5

37 4 2 6 87 4 2 6

38 4 3 7 88 4 3 7

39 4 4 8 89 4 4 8

40 4 5 9 90 4 5 9

41 5 1 6 91 5 1 6

42 5 2 7 92 5 2 7

43 5 3 8 93 5 3 8

44 5 4 9 94 5 4 9

45 5 5 10 95 5 5 10

46 5 1 6 96 5 1 6

47 5 2 7 97 5 2 7

48 5 3 8 98 5 3 8

49 5 4 9 99 5 4 9

50 5 5 10 100 5 5 10

For a roll that has at least a 2 on either Die

2: 2,7

10: 11-20

8: 22,27,32,37,42,47,52,57

10: 61-70

6: 72,77,82,87,92,97

So, 2+10+8+10+6 = 36

36+4 = 40

1/100 * 40 = the probability of rolling a sum of 2 or a 2 on at least one die

Good Luck

Why only faces with 1-5?

winsome johnny (not Win some johnny)

February 5th, 2013 at 7:19:46 PM
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You are correct, my brain failed me when I wrote the post. The probabilities for obtaining the particular numbers with one toss of dice for both high and low number seeker on 1-5 dice

1 @ 9/25 = 36%

10 @ 1/25 = 4%

2 @ 10/25 = 40%

9 @ 2/25 = 8%

3 @ 11/25 = 44%

8 @ 3/25 = 12%

4 @ 12/25 = 48%

7 @ 4/25 = 16%

5 @ 13 /25 = 52%

6 @ 5/25 = 20%

I corrected this in the original post and emphasized the crux of my question at the end so that others can go right to the question without getting sidetracked.

1 @ 9/25 = 36%

10 @ 1/25 = 4%

2 @ 10/25 = 40%

9 @ 2/25 = 8%

3 @ 11/25 = 44%

8 @ 3/25 = 12%

4 @ 12/25 = 48%

7 @ 4/25 = 16%

5 @ 13 /25 = 52%

6 @ 5/25 = 20%

I corrected this in the original post and emphasized the crux of my question at the end so that others can go right to the question without getting sidetracked.