What is the probability that a person wins the game on the first roll?

Question

I can't seem to find an answer for the title/question above and so I thought I'd ask.

The aim of the game is to get a higher number than your competitor using only one dice. If two people (For Example Jack and Jill) roll the same dice. What is the probability that Jack will beat Jill on his first roll?

I'm not sure if I'm thinking correctly when I do the below

1/6 + 2/6 + 3/6 + 4/6 + 5/6 = 15/30 = 1/2

Is that the right answer or have I got it completely wrong?

Thanks in advance

Answer 1

Let's say Jill rolls first. If she'd roll a 1, Jack would have a $\tfrac{5}{6}$ chance to beat Jill's roll. If she rolls a 2, Jack would have a $\tfrac{4}{6}$ chance to beat her, etc. However, when Jill rolls a 6, Jack could not beat her (only tie), so there is a $\tfrac{0}{6}$ chance of winning for Jack.

So, the calculation should be: $\tfrac{0}{6^2}$ + $\tfrac{1}{6^2}$ + $\tfrac{2}{6^2}$ + $\tfrac{3}{6^2}$ + $\tfrac{4}{6^2}$ + $\tfrac{5}{6^2}$ = $\tfrac{15}{6^2}$ = $\tfrac{5}{12}$.

Answer 2

You can think of the rolls as ordered pairs (Jack,Jill). So, a $(4,2)$ means that Jack rolled a $4$ and Jill rolled a $2$. There are $6$ possibilities for the first slot and $6$ possibilities for the second position, resulting in $36$ possible outcomes.

Whenever the two numbers are the same (which has $6$ occurrences), Jack and Jill tie. This leaves $36-6=30$ outcomes where the numbers are not the same. Whenever the two numbers are different, one of Jack and Jill wins. In half of these situations Jack wins and in the other half, Jill wins (you can see this by pairing up outcome $(a,b)$ with outcome $(b,a)$, one of which Jill wins and one of which Jack wins).

Therefore, there are $30/2=15$ outcomes where Jack wins. Since there are $36$ possible outcomes, the probability that Jack wins on the first try is $\frac{15}{36}=\frac{5}{12}$.

Answer 3

There is an easier solution that uses a basic symmetry argument. Note that there are three possible outcomes: $A)$ Jack $>$ Jill; $B)$ Jack $=$ Jill; and $C)$ Jill $>$ Jack. Note that the probabilities of outcome $A$ and $C$ are identical - no one has an advantage over the other, since the dice are fair. We also know that the sum of the probabilities of $A,$ $B,$ and $C$ is $1.$ We can find $B$ by inspection - it is simply $\frac{6}{36} = \frac{1}{6}.$ So the probability of event $A$ is just $\frac{1 - \frac{1}{6}}{2} = \boxed{\frac{5}{12}}.$