Suppose we play a 1 vs 1 game. I roll a 6 sided fair die, if I get a 1 I win, if not you roll. If you get a 6 you win, otherwise I go again…. What are my chances of winning and what are yours?
[Math] Dice roll game probability
diceprobability
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$a)$ Your answer is correct but you can get it easier. Out of the 36 cases, 6 cases are a tie, 15 G wins, 15 W wins (by symmetry). So the answer is $$\frac{15}{36}=\frac{5}{12}.$$
$b)$ The first roll is a tie with probability $\frac{1}{6}$. After that both have the same chance so the answer is $$\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}.$$ Another way of seeing this is to compute the probability that G wins in round $i$. We must have a tie in rounds $1,2,\dotsc,i-1$ and G winning in round $i$: $$\left(\frac16\right)^{i-1}\frac5{12}.$$ Now we sum this up for $2\le i\le \infty$ to get the probability that G wins but not in round 1: $$\sum_{i=2}^\infty \left(\frac16\right)^{i-1}\frac5{12}=\frac1{12}$$
$c)$ Suppose G gets a $2$. The probability of a tie is then $\frac{1}{6}$ and after that W wins with probability $\frac{1}{2}$. So the probability that W wins and the first round is a tie is $\frac{1}{12}$. The probability that W wins without a tie is $\frac{4}{6}=\frac{2}{3}$. So in total the probability of W winning is $$\frac{1}{12}+\frac{8}{12}=\frac{9}{12}=\frac{3}{4}.$$ Note that here we are looking for a conditional probability given that G gets a 2. This is already given as something that has happened. So we do not need to consider its probability in our calculations.
Let $P$ be the probability that the person about to roll wins the game.
Let $p$ be the probability that a particular roll is a winning result.
Therefore $P= p+ (1-p)(1-P)$, because either the person about to roll immediately obtains a winning result, or they don't and then their opponent is about to roll next. With a little arithmentic rearrangement :
$$P= \frac{1}{2-p}$$
This is also the probability that whoever starts the game wins the game.
On average, you get 6 every $6$ rolls. So on average, the $6$th person to roll will win. So you going first doesn't change the probability (three rolls each), which must therefore be $1/2$.
Sure, the average count for rolls until winning is $6$, but this sais nothing about the probability for who wins. The probability for obtaining this expected value is only about $0.067$, and it is but one outcome in the event for you not winning.
You want the probability that the count for rolls until winning is an odd number.$$P~{=\sum_{k=0}^\infty \mathsf P(X=2k+1)\\=\sum_{k=0}^\infty (1-p)^{2k}p\\ = p+(1-p)^2p+(1-p)^4p+\cdots}\qquad\text{Where }X\sim\mathcal {Geo}_1(p)$$
Best Answer
$\underline{A\; way\; without\; using\; an\; infinite\; geometric\; series}$
Either you win on the first roll, or you don't, and the turn passes to your opponent,who is now exactly in the same position you were at the start of the game.
Thus if P(you eventually win) $=p,\quad p + \frac56p = 1 \to p = \frac6{11} $