# Determing the losing probability of a die game by rolling $n$ times under a value.

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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• 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? – Stockfish Jan 11 at 12:32
• It is probably possible to use a symmetry argument to show that the probability must be $0.5$. – Riccardo Orlando Jan 11 at 12:35

## 1 Answer

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 $$1/2$$