July 25th, 2012 at 1:32:45 PM
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Hi,

So my friend and I are developing a game that uses RPS mechanics at its core. I'm not a fan of draws, and so we were deciding what should happen in the event of a draw. A team member suggested on a draw, the game will ask you to pick your next 3 moves (any combo of R,P,S) and compare it with your opponents next 3 moves. The person with more wins, will break the draw and win.

So after all that, we need some math formula to prove this is probably not the best method to breaking ties .

I created a matrix table to log out the possible outcomes. I came to 27 possible outcomes, with 7 of those tying for 26% chance of ending up back where you started.

So my questions are:

is this correct?

And if it is, is there a formula we can use so that we do not have to use a table like this?

Can anyone think of a better way to break ties that doesn't result in more ties?

So my friend and I are developing a game that uses RPS mechanics at its core. I'm not a fan of draws, and so we were deciding what should happen in the event of a draw. A team member suggested on a draw, the game will ask you to pick your next 3 moves (any combo of R,P,S) and compare it with your opponents next 3 moves. The person with more wins, will break the draw and win.

So after all that, we need some math formula to prove this is probably not the best method to breaking ties .

I created a matrix table to log out the possible outcomes. I came to 27 possible outcomes, with 7 of those tying for 26% chance of ending up back where you started.

So my questions are:

is this correct?

And if it is, is there a formula we can use so that we do not have to use a table like this?

Can anyone think of a better way to break ties that doesn't result in more ties?

July 25th, 2012 at 2:16:41 PM
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If you want to know whats "best" you have to give a definition of the precise "costs" which should be optimized.

If choosing each move has a constant cost, then simply repeat the RPS one by one, untill it doesn't come out as a draw.

Or use a different system that simply doesn't draw.

For example two player each choose "A" or "B". First player wins if both choices are the same. Second player wins of they are different.

You can also combine both systems, i.e. choose from 6 elements "red rock", "red paper", "red scissor" (and corresponding blue ones).

Apply the standard RPS rules independent of color first. If this is a draw, first player wins when both elements have the same color.

If choosing each move has a constant cost, then simply repeat the RPS one by one, untill it doesn't come out as a draw.

Or use a different system that simply doesn't draw.

For example two player each choose "A" or "B". First player wins if both choices are the same. Second player wins of they are different.

You can also combine both systems, i.e. choose from 6 elements "red rock", "red paper", "red scissor" (and corresponding blue ones).

Apply the standard RPS rules independent of color first. If this is a draw, first player wins when both elements have the same color.

July 25th, 2012 at 2:28:47 PM
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Quote:SnoobertCan anyone think of a better way to break ties that doesn't result in more ties?

You could play rock-paper-scissors-lizard-spock. There will be a tie only 1 in 5 games, on average. I think the concept would work for any odd number of symbols. However, one must also consider the simplicity of the game.

If we're restricted to three symbols you could simply do a skins game. For those unfamiliar with the term, for every consecutive tie the stakes of the following game go up by one unit. You could think of it as the first resolved game also resolving all preceding ties since that last resolved game.

It's not whether you win or lose; it's whether or not you had a good bet.

July 25th, 2012 at 2:30:35 PM
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You'll get fewer draws if you do Rock, Paper, Scissors, Spock, Lizard.

Edit: Curses!! Foiled by the wizard again!

Edit: Curses!! Foiled by the wizard again!

Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez

July 25th, 2012 at 3:25:05 PM
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Quote:SnoobertCan anyone think of a better way to break ties that doesn't result in more ties?

First to enter their original selection wins ties. Timing in ten-thousandths of a second should do it. On the once in a lifetime occassion that there is still a tie, the last digit of the time will determine the winner, 0;2;4;6;8; = player 2, 1;3;5;7;9 = player 1. This should encourage players not to stall in hopes of an opponent "flash".

Simplicity is the ultimate sophistication - Leonardo da Vinci

July 25th, 2012 at 4:50:35 PM
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Thanks for the suggestions. Does anyone know how to calculate tie probability if we were to increase the number of games? Is there a special formula? In the OP we listed it at 3 games, but what if we made 5 or 6 etc

Thanks so much for your replies.

Thanks so much for your replies.

July 26th, 2012 at 1:06:37 PM
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Cool Math Post Alert:

This is an amusingly tricky question to answer. After some messing around in Excel, it turns out the probability of tying after n games of RPS can be found from the trinomial expansion, whose coefficients are illustrated on Pascal's Pyramid.

EDIT: I lost part of my post. Darn. It was awesome. Summary:

The sum of the columns in level n gives the number of ways to tie after n games.

You can find this in excel by adding up the previous 3 nearby numbers. I get 8953 for n=10, out of 59049 outcomes, probability 16.5%

Or, if you are into these really cool math artifacts, by doing 111^n, or 10101^n, or 1001001^n, etc.

10101^5 = 105153045514530150501

See how the numbers repeat on either side? Those are the sums of the columns, the number of ways of having any number of points after 5 games. The wiki page shows how to divide it into:

1 5 15 30 45 51 45 30 15 5 1

There are 51 ways to tie after 5 games.

Cool.

This is an amusingly tricky question to answer. After some messing around in Excel, it turns out the probability of tying after n games of RPS can be found from the trinomial expansion, whose coefficients are illustrated on Pascal's Pyramid.

EDIT: I lost part of my post. Darn. It was awesome. Summary:

The sum of the columns in level n gives the number of ways to tie after n games.

You can find this in excel by adding up the previous 3 nearby numbers. I get 8953 for n=10, out of 59049 outcomes, probability 16.5%

Or, if you are into these really cool math artifacts, by doing 111^n, or 10101^n, or 1001001^n, etc.

10101^5 = 105153045514530150501

See how the numbers repeat on either side? Those are the sums of the columns, the number of ways of having any number of points after 5 games. The wiki page shows how to divide it into:

1 5 15 30 45 51 45 30 15 5 1

There are 51 ways to tie after 5 games.

Cool.

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

July 26th, 2012 at 1:51:56 PM
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Really really cool! Thanks dwheatly for this amazing trick. You are a certified genius and a big time stud.

July 26th, 2012 at 2:46:34 PM
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Quote:SnoobertCan anyone think of a better way to break ties that doesn't result in more ties?

How about the "classic" Odd & Even? One player is odd, and the other is even; both "throw" 1 or 2 fingers ("pencil or scissors").

(I say "classic" because it appears in a late-1930s Popeye cartoon.)