2 die roll vs 1 die roll + 7: Player A rolls a 6 sided die twice...

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whatthewhat

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My math knowledge is not great, I made this account looking for a solution to a question I have that I don't know how to work out. It's for a strategy game I play, any help towards the solution is greatly appreciated.

My question is: Player A rolls a 6 sided die twice and adds the results to form his total. Player B rolls a 6 sided die once, and adds his result to 7 to form his total.

Statistically, how often will players B have a higher total than player A?
 
whatthewhat said:
My math knowledge is not great, I made this account looking for a solution to a question I have that I don't know how to work out. It's for a strategy game I play, any help towards the solution is greatly appreciated.

My question is: Player A rolls a 6 sided die twice and adds the results to form his total. Player B rolls a 6 sided die once, and adds his result to 7 to form his total.

Statistically, how often will players B have a higher total than player A?
Click to expand...

You can just build a table.

Player A rolls a 2, Player B can roll anything.
Player A rolls a 3, or 4, or 5, or 7, Player B can roll anything.
Player A Rolls an 8, Player B rolls an 8, they tie, otherwise Player B is over.
You can continue on like this.
 
whatthewhat said:
My math knowledge is not great, I made this account looking for a solution to a question I have that I don't know how to work out. It's for a strategy game I play, any help towards the solution is greatly appreciated.

My question is: Player A rolls a 6 sided die twice and adds the results to form his total. Player B rolls a 6 sided die once, and adds his result to 7 to form his total.

Statistically, how often will players B have a higher total than player A?
Click to expand...

I take you at your word that this is not homework. Please read reply #2 carefully. I have chosen to focus on B's outcomes because you ask for probability that Bs score is greater than that of A.
\(\displaystyle \begin{array}{*{20}{r}} {{\text{Value of B}}}&\|& 8&9&{10}&{11}&{12}&{13}\\ \hline {{\text{Ways A wins}}}&\|& {10}&6&3&1&0&0\\ {{\text{Ways B wins}}}&\|& {21}&{26}&{30}&{33}&{35}&{36}\\ {{\text{Ways to tie}}}&\|& 5&4&3&2&1&0 \end{array}\)
In technical terms we consider 216 ordered triples. For example: \(\displaystyle (4,5;11)\) The first two or A's adds 9, b's add to 11. B wins.
The triple \(\displaystyle (5,5;9)\) means A wins, and \(\displaystyle (4,5;9)\) is a tie.

Using the table can you figure out the probability wins \(\displaystyle \mathscr{P}(B\text{ wins})=\frac{?}{216}\)
 
1,000,000 simulations:
B wins ~838,000 times
A wins ~92,500 times
Leaves ~69,500 ties

S'that close Pka?
 
Seems as though some have assumed that B > A means B WINS. I don't see it in the problem statement.
 
pka said:
I assume having a higher score is a win. It is dice and not golf.
Click to expand...

Since the question didn't mention winning, this seems reasonable to me -- it isn't really an assumption, just a definition. In probability we commonly use the word "success" to mean whatever we are asking for the probability of, even if it's dying from a disease. I take "win" here in the same sense -- we want the probability that B>A, so that is what we are thinking of as "winning" (success), within the question.
 
Dr.Peterson said:
Since the question didn't mention winning, this seems reasonable to me -- it isn't really an assumption, just a definition. In probability we commonly use the word "success" to mean whatever we are asking for the probability of, even if it's dying from a disease. I take "win" here in the same sense -- we want the probability that B>A, so that is what we are thinking of as "winning" (success), within the question.
Click to expand...
The Woolsey professor of New Testament at Yale would say that "professors use two-bit words where plain English would be much better." Surely 'a win' is plain English for better score, higher than, more than. Lets not loose site of the original post: "how many times is B's total more​ than A's total.
 
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