Five person A,B,C,D,E throw a dice in the order specified until one of them throws an ace, find their relative chances of winning, supposing the throws to continue till an ace appears.

Answer

To solve the problem of finding the relative chances of winning for persons A, B, C, D, and E when they throw a dice until one of them throws an ace (1), we can follow these steps:

### Step 1: Understand the Probability of Throwing an Ace
The probability of throwing an ace (1) on a single throw of a dice is:
\[ P(\text{Ace}) = \frac{1}{6} \]
The probability of not throwing an ace is:
\[ P(\text{Not Ace}) = 1 - P(\text{Ace}) = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 2: Determine the Winning Probability for Each Person
Each person throws the dice in the order A, B, C, D, E. We will calculate the probability of each person winning.

#### Probability of A Winning
- A can win on the first throw:
\[ P(A \text{ wins}) = P(\text{Ace on A's throw}) = \frac{1}{6} \]
- If A does not win, the probability of this happening is \(\frac{5}{6}\). Then, B, C, D, and E must also not win:
\[ P(\text{A does not win}) = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \left(\frac{5}{6}\right)^4 \]
- If A does not win in the first round, the game resets, and A has another chance to win. Thus, we can express the probability of A winning as:
\[ P(A \text{ wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^4 P(A \text{ wins}) \]
Let \(P(A \text{ wins}) = p_A\):
\[ p_A = \frac{1}{6} + \left(\frac{5}{6}\right)^4 p_A \]
Rearranging gives:
\[ p_A - \left(\frac{5}{6}\right)^4 p_A = \frac{1}{6} \]
\[ p_A \left(1 - \left(\frac{5}{6}\right)^4\right) = \frac{1}{6} \]
\[ p_A = \frac{\frac{1}{6}}{1 - \left(\frac{5}{6}\right)^4} \]

#### Probability of B Winning
For B to win, A must not win, and then B must win:
\[ P(B \text{ wins}) = \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^5 P(B \text{ wins}) \]
Let \(p_B = P(B \text{ wins})\):
\[ p_B = \frac{5}{36} + \left(\frac{5}{6}\right)^5 p_B \]
Rearranging gives:
\[ p_B - \left(\frac{5}{6}\right)^5 p_B = \frac{5}{36} \]
\[ p_B \left(1 - \left(\frac{5}{6}\right)^5\right) = \frac{5}{36} \]
\[ p_B = \frac{\frac{5}{36}}{1 - \left(\frac{5}{6}\right)^5} \]

#### Probability of C, D, and E Winning
Following the same logic:
- For C:
\[ p_C = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^6 p_C \]
- For D:
\[ p_D = \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^7 p_D \]
- For E:
\[ p_E = \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^8 p_E \]

### Step 3: Calculate the Relative Chances
To find the relative chances, we need to express \(p_A, p_B, p_C, p_D, p_E\) in terms of a common factor.

### Final Step: Find the Ratios
The relative chances of winning can be expressed as:
\[ \text{Relative Chances} = p_A : p_B : p_C : p_D : p_E \]