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game 2 - release your prisoners 1. what is the least number of rolls it…

Question

game 2 - release your prisoners

  1. what is the least number of rolls it will take to release all your prisoners?
  2. what is the probability of having the difference in dice be 0?
  3. what are the odds of the difference being 3?
  4. what is the probability of rolling a difference of 4?

Explanation:

Step1: Analyze the first - question

Since no information about how many prisoners and how dice - rolls relate to releasing prisoners is given, assume the best - case scenario. If each roll can release a prisoner and there are 6 prisoners (as a standard die has 6 faces), the least number of rolls to release all prisoners is 6.

Step2: Analyze the second - question

When rolling two dice, the total number of possible outcomes is \(n(S)=6\times6 = 36\). The cases where the difference between the two dice is 0 are \((1,1)\), \((2,2)\), \((3,3)\), \((4,4)\), \((5,5)\), \((6,6)\), so \(n(A)=6\). The probability \(P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\).

Step3: Analyze the third - question

The cases where the difference between the two dice is 3 are \((1,4)\), \((4,1)\), \((2,5)\), \((5,2)\), \((3,6)\), \((6,3)\), so \(n = 6\). The total number of outcomes \(n(S)=36\). The odds in favor of an event \(E\) is given by \(\frac{P(E)}{1 - P(E)}\), where \(P(E)=\frac{n}{n(S)}=\frac{6}{36}=\frac{1}{6}\), and \(1 - P(E)=\frac{5}{6}\), so the odds are \(\frac{1}{5}\).

Step4: Analyze the fourth - question

The cases where the difference between the two dice is 4 are \((1,5)\), \((5,1)\), \((2,6)\), \((6,2)\), so \(n = 4\). The total number of outcomes \(n(S)=36\). The probability \(P=\frac{4}{36}=\frac{1}{9}\).

Answer:

  1. 6
  2. \(\frac{1}{6}\)
  3. \(\frac{1}{5}\)
  4. \(\frac{1}{9}\)