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Bob is playing a dice game in which he rolls a fair dice multiple times. If he rolls $n$ times under the number $4$ (not necessarily consecutively rolls) he loses the game, and consequently, if he manages to roll $4$ or more $n$ times he wins.

How could we calculate the chance that Bob loses? It is simple to determine the outcome of a single dice roll, and that, in the worst case scenario, he has to roll $2\cdot n-1$ times in order for the game to complete but I am not sure how to approach this problem?

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    $\begingroup$ The game lasts between $n$ and $2n-1$ rolls. Fixing $n \leq i \leq 2n-1$, whats the probability that Bob loses "because of" the ith roll? $\endgroup$
    – Stockfish
    Jan 11, 2019 at 12:32
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    $\begingroup$ It is probably possible to use a symmetry argument to show that the probability must be $0.5$. $\endgroup$ Jan 11, 2019 at 12:35

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Let's call $W_n$ the event of winning in the $n-th$ roll. One can easily see that $P(W_n) \neq 0 \; \forall n \in {n,2n-1}$ and $P(W_n) = 0$ otherwise. So, the probability of winning at any roll (event $W$) will be:

$$ P(W) = \sum_{i=n}^{i=2n-1}P(W_i)$$

Now, we have to find $P(W_i)$. First, let's denote $A_i$ the event of having 4 or more in the $i-th$ roll, and $B_i$ the complementary. Both probabilities are $P(A_i) = P(B_j) = p = 1/2 \; \forall i,j$. Now we can write:

$$ P(W_i) = {i-1 \choose n-1} p^n p^{i-n} = {i-1 \choose n-1} \frac{1}{2^i}$$

Where the last event is forced to be $A_i$, so we have just $n-1$ elements to arrange in $i-1$. Finally, we have:

$$ \boxed{ P(W) = \sum_{i=n}^{2n-1} {i-1 \choose n-1} \frac{1}{2^i}}$$

Which can be shown to be equal to $1/2$

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