Probability with a Converging Infinite Sum

Question

At a fairground, one big prize is being given away by a random chance game, such as spinning a wheel. A finite number of players line up in a queue and take turns with the game. If the person at the front of the queue wins the game, they get the prize. If not, they go to the back of the queue, and the next person in the queue gets a try. This repeats until one person in the queue wins the game of chance and gets the prize.

If chances of winning the game of chance are 'P', the total number of people in the queue is 'S', and your initial position in the queue is 'n', what is the formula for your chances of winning the prize?

I apologise ahead of time for not knowing the proper mathematical terms for things, but this is a question that me and my friends have been trying to debate for ages now, and none of us are particularly gifted at maths, so I thought i'd ask some smarter people the question. So far, I have determined that the formula is some kind of converging infinite sum, but we know not what the formula would be, nor what it may converge to for a given set of input variables.

Answer 1

Let $X$ be a random variable whose distribution is geometric with parameter $p$, that is, $$\mathbb P(X=k)=(1-p)^{k-1}p,\quad k\geqslant 1.$$ This random variable tells us the number of rounds needed for the prize to be won by one of the players. The player at the position $1$ wins if $X=1$, or $X=S+1$, or $X=2S+1$,... and so one, that is, if $X\in\bigcup_{j=0}^{+\infty}\left\{jS+1\right\}$. This probability can be computed: \begin{align} \mathbb P\left(X\in\bigcup_{j=0}^{+\infty}\left\{jS+1\right\}\right)&=\sum_{j=0}^{+\infty} \mathbb P(X=jS+1) \mbox{ since the events }\{jS+1\}\mbox{ are pairwise disjoint}\\ &=\sum_{j=0}^{+\infty}(1-p)^{jS}p\\ &=p\frac 1{1-(1-p)^S}\quad \mbox{using }\sum_{j=0}^{+\infty}r^j=\frac 1{1-r}\mbox{ for }\lvert r\rvert<1. \end{align} Similarly, if $n\in\{1,\dots,S\}$, \begin{align} \mathbb P\left(X\in\bigcup_{j=0}^{+\infty}\left\{jS+n\right\}\right)&=\sum_{j=0}^{+\infty} \mathbb P(X=jS+n) \mbox{ since the events }\{jS+n\}\mbox{ are pairwise disjoint}\\ &=\sum_{j=0}^{+\infty}(1-p)^{jS+n-1}p\\ &=p(1-p)^{n-1}\frac 1{1-(1-p)^S}\quad \mbox{using }\sum_{j=0}^{+\infty}r^j=\frac 1{1-r}\mbox{ for }\lvert r\rvert<1. \end{align}

Answer 2

Let us try to avoid an infinite series

Let the probability of a person in line succeeding when her trial comes be $p$, failure $(1-p) = q$

Let $A$ = probability that person $N$ in line ultimately succeeds, then either she succeeds in the first round, or with everyone failing the round, we are back to start

Thus $A = pq^\left(N-1\right) + q^S\cdot A$

which yields $A = \boxed{\dfrac {pq^\left(N-1\right)}{1-q^S}}$

Answer 3

If chances of winning the game of chance are 'P', the total number of people in the queue is 'S', and your initial position in the queue is 'n', what is the formula for your chances of winning the prize?

Alternate approach:

You can combine recursion + infinite series.
Let $~Q = 1 - P.$

The probability of the 1st person winning is

$$P + Q^{S}P + Q^{2S}P + Q^{3S}P + \cdots = \frac{P}{1 - Q^S}. \tag1 $$

For the $~n$-th person to win, they must first (re recursion) become the first person, which is accomplished by having the first $~n-1~$ people fail.

Therefore, using (1) above, the probability of the $~n$-th person winning is

$$Q^{n-1} \times \frac{P}{1 - Q^S}.$$