Dice question, 2 pair of competing dice

In a game where 2 people are each rolling a pair of dice and who ever gets the higher roll wins with re rolling ties, are the odds for each person 50% on any given roll?

Also, if one person gets to add +1 or +2 or +3 to the result what is the percent he would win over the other?

Finally if one person could roll 3 dice keeping the 2 highest as his result, what is the percent chance he will win vs a person who can only roll 2?

Hmmm, well there are 5 numbers below 7 and 5 numbers above.
The game

The probability of rolling a number below 7 is 15/36 or 41.67%. We could term these as "low numbers".
The probability of rolling a number above 7 is 15/36 or 41.67%. We could term these as "high numbers".

Including the 7 the probability of rolling a 7 or below or a 7 or above is 21/36 or 58.3%. Since ties don't matter and the play moves on to a subsequent roll, we don't even have to factor ties in (unless we were calculating a house edge or another long term return percentage).

Assuming that each person rolls their pair of dice, at the same time, they each being independent events occurring simultaneously, the probability of one person winning is exactly 50% (making the above math superfluous).

Now. If we separated the rolls, where Person A rolls his dice for a result, and after seeing said result, Person B rolls his dice. You could calculate the probability of Person B's roll being higher than Person A's. Example: Person A Rolls a 4. Now we can say there are 26 combinations that could beat a 4. 26/36= 72.22%.

But seeing as the game is played in hands or sets consisting of simultaneous, or near simultaneous rolling, in that the other person is contractually obligated to roll their dice, the probability holds at 50%.

Advantage of adding +1,+2, or +3

If Person B gets to add a digit to their result. His odds increase. Let's say they always get to add +1. That means they can never roll a 2. Their spectrum of numbers ranges from 3 to 13 effectively. This means if Person A rolls a 2, no matter what, they will lose. Person A has only 11 possible numbers he can win on, or tie (null). Person B has one number which he wins 100% of the time (6+6 + bonus 1 for 13). Person B has 12 possible numbers for a win or tie. Leaving some math out, this gives Person B an extra 8.3% advantage. Making his probability of winning effectively 58.3% (For adding +1) Notice how this is the same as the probability for rolling a number of 7 or above, or rolling a 7 or below.

Three Dice Are Always Better Than Two

Let's say Person B gets to add a dice, and use the top two method. This gives Person B an effective combination sum of 42 or 41 depending on how you look at it. (If he rolls a 1,1,2 he only is going to have 3. A 1,1,1, he still has 2). If Person B rolls 4,4,6. His safeguard turns his 8 into a 10. Almost like a +2 effect from the advantage of adding example. Now rolling a 1,1,6, turns his 2 into a 7, a spread of 5 at the most, and that only occurs one time out of 216 combinations. The lowest spread would be well 0 at 6,6,6, again a 1/216. This could still result in a tie as well. So each Person still has 12 winning or tie numbers. Averaging the spread at 2.5 gives Person B an effective winning probability of approximately 70.8% (an additional 12.5% edge from the above example).

Good Night!
YoDR11

I'd like to remind all members that
we're not here to do homework assignments.

I realize this may seem like fishing for homework answers since it was my first post, but its not. It is actually a non-gambling game I am working on.

I am trying to figure out the formulas so I can give bonuses to rolls at certain points in the game with out one bonus being more overpowered than the others. Or at least knowing at what point a bonus of +1 is worth more than at another point.

The general rules of the game are that you roll 2 dice and your opponent rolls 2 dice. The higher result wins, ties are a null result and are re-rolled. You can earn bonuses of +1 or have penalties of -1. You can earn multiple of these and apply them as a strategy to give you a better chance to beat your opponent on a given roll. The above poster is correct when he assumed that the rolls are at the same time.

There are also times where you can roll 2 dice and have to get 7 or better on the result (with bonuses and penalties available). This roll is not against an opponent, you just have to get the number or above. I want to know when it is better to roll your dice against another player with bonuses or just roll against the 7.

The real problem I run into is that a +1 (or +2 or +3 etc..) bonus is more likely to give you a win when you or your opponent roll certain numbers because there are more ways to make them. Eg +1 when you roll a 6 is more likely to get you the win than if you roll a 3.

I was hoping that it could be as easy as +1 always gives a 8.33% bonus but as I dug into dice probabilities I found it was way more complicated. I'm fine with complicated, I just want to wrap my head around the formulas so I can know how to handle it in the game and keep it as balanced as possible.

Thanks to the above poster, he got me to where I found that +1 gives an average of 8.33% and +2 gives a 2nd 8.33% advantage. After that it grows in relation to the number of ways you can make the number. For instance +3 gives an 11.11% extra chance to win and +4 gives an additional 13.89%

Now, my math may be off and my brain hurts trying to make sure I am putting everything in the right spots in my spreadsheet, because its been so long since I have done any probabilities studies. It is very possible that all the numbers I have posted here are completely wrong because I am missing part of the relationship between 2 competing pairs of dice, or some other reason I am not aware of.

Although my comment about homework stands, now that you've explained your reasons for the question, I agree that your question is not homework. I hope you accept my appology and understand how easy it was to make that assumption based upon the rather brief query in the original post, combined with the fact that it was your first post.

No apology necessary. Completely valid assumption under the circumstances.

Quote: DJTeddyBear

I'd like to remind all members that
we're not here to do homework assignments.

LOL

Quote: Raawr

I realize this may seem like fishing for homework answers since it was my first post, but its not. It is actually a non-gambling game I am working on.

This is what I thought from how you described it.

Quote: Raawr

Thanks to the above poster, he got me to where I found that ...

No problem! Sounds like a cool game.

And BTW A person always has better odds trying to roll a 7 or higher, rather than trying to just roll against a player. (Assuming no bonuses).

YoDR11

Glad this is not homework.
I wonder if the gaming and hospitality students at the UNLV more likely - or less likely - to short-cut their way through....

Quote: DJTeddyBear

I'd like to remind all members that
we're not here to do homework assignments.

Is this how we welcome new members now? Screaming at them in big black letters for simply asking a question?

Welcome to the site Raawr and IGC. Had you been long standing members, I doubt that your treatment would have been that unfriendly. Please stick around. We don't all feel this way.

Quote: Raawr

In a game where 2 people are each rolling a pair of dice and who ever gets the higher roll wins with re rolling ties, are the odds for each person 50% on any given roll?

Absolutely. Each player has 575 ways of out-totalling the other and 146 ways to tie sums, with 40 ways to each player of equalling with a combination which could (if you wanted) be consistently ruled as "beating" the other player's (e.g. a 5&5 over a 4&6 or a 2&4 over a 1&5), and there are 66 ways to tie on exactly the same combination ("two-pairing"), six of which are hardways ("quadrupling"). So, there are either 1,150 or 1,230 declarable win-loss games (depending on how you choose to handle tie-sum W&X-vs.-Y&Z situations) out of 1,296 rolls. That's one sure thing about dice over cards - reciprocity of wagers. No watter who goes first, the odds remain the same for either side of a "doey-don't" proposition.

Quote: Raawr

Also, if one person gets to add +1 or +2 or +3 to the result what is the percent he would win over the other?

Haven't reflected on this one yet. I'm not sure I understand the question.

Quote: Raawr

Finally if one person could roll 3 dice keeping the 2 highest as his result, what is the percent chance he will win vs a person who can only roll 2 dice?

This question sounds similar to a riddle I worked recently: What is the probability that a given dice pair will out-total another dice trio?

Quote: NowTheSerpent

Haven't reflected on this one yet. I'm not sure I understand the question.
This question sounds similar to a riddle I worked recently: What is the probability that a given dice pair will out-total another dice trio?

Just scroll up for the answer.

Edit: On Page 1 of course.

@1BB: thanks for the welcome :)

@ NowTheSerpent: Thanks for the expanded explanation of rolling against an opponent coming out to 50% for each side. I think I will stick with re rolling ties because it becomes too much to remember with tie breakers. The game would bog down at that point and its not really necessary since the odds are the same for each player on a re roll.

The thing I was trying to figure out is how much of an advantage each +1 gives an individual player. That way I can figure out how much it should cost to obtain the bonus.

What I ended up doing is filling out a spreadsheet that had lots of cells I input manually then added and multiplied as needed to get percentages. It turns out that the +1 stacking bonus runs at a diminishing return because, the more bonuses you have the less of a chance you have to be beat.

And if anyone is interested, here is the table of what a +1 through a +10 bonus will get you in expected win percentages.

Bonus	Win Percentage	Percent of Change
None	50%	None
+1	59.83%	9.83%
+2	69.15%	9.31%
+3	77.60%	8.45%
+4	84.86%	7.27%
+5	90.67%	5.80%
+6	94.75%	4.08%
+7	97.41%	2.66%
+8	98.97%	1.57%
+9	99.75%	0.77%
+10	100%	0.25%

Quote: NowTheSerpent

This question sounds similar to a riddle I worked recently: What is the probability that a given dice pair will out-total another dice trio?

Yeah, and mine has an added layer of complexity in that from that trio of dice you can only pick 2 dice for your total so that on one hand you have 36 ways to make 12 results and on the other hand you have 216 ways to make 12 results. AND what is the chances that one total will be more than the other.

My tables show that if you roll 2 dice and your opponent can roll 3 and use the best 2 then you will only win 41.56% of the time and he wins 58.44% of the time. The way I got this is I found out how many rolls you can beat for any given number from 2 to 12 then I divided that by the total number of possible rolls the opponent can make. Averaging all the win percentages gives me the 8.44% difference.

For instance: Player A has 2 dice and rolls a 5. Player B has 3 dice and if he rolls a 2, 3, or 4 he looses. Player B can roll a 2 only once, roll a 3 in 4 different ways, and roll a 4 in 7 different ways. This means Player A can beat Player B 11 times out of the possible 216 rolls. Player A will win 5.09% of the time.

Now lets look at Player B rolling a 5. Player A can roll a 2 only once, roll a 3 in 2 different ways, and roll a 4 only in 3 ways. Player B can win 6 times out of a possible 36 rolls. Player B will win 16.67% of the time.

Here is a table showing how many ways you can roll 2 through 12 with 3 dice

Roll	Possible Ways
2	1
3	3
4	7
5	12
6	19
7	28
8	34
9	36
10	33
11	27
12	16