Probability – Probability of Rolling a Dice 8 Times Before All Numbers Are Shown

combinatoricsprobability

What is the probability of having to roll a (six sided) dice at least 8 times before you get to see all of the numbers at least once?

I don't really have a clue how to work this out.

Edit: If we are trying to find the number of situations where all of the numbers are shown, for seven rolls, a favorable outcome would be one in which there are two numbers the same, for example: 1123456. These numbers can be arranged in 7!/2!5! ways, and there are 6 different numbers that could be repeated.
So the probability of getting all 6 numbers with 7 throws is (6*7!/2!5!)/6^7 = 126/279936.
Is that right?
Then 1 minus this is the probability?

Edit: prob. of not getting all six numbers with seven rolls = 1-prob of getting all six numbers with seven rolls

Six numbers same with seven rolls means one number repeated twice 7!/2! ways of doing this for each repeated number
6*(7!/2!) is number of ways of getting all six numbers with seven rolls. Total number of outcomes 6^7.
Prob of getting all six numbers with seven rolls = (6*(7!/2!))/6^7 = 0.054
Prob of not getting all six numbers with seven rolls (=prob of needing at least 8 rolls to get all numbers) = 1-0.054 = 0.946

Best Answer

Rephrase the question:

What is the probability of not seeing all $6$ values when rolling a die $7$ times?

Find the probability of the complementary event:

What is the probability of seeing all $6$ values when rolling a die $7$ times?

Use the inclusion/exclusion principle in order to count the number of ways:

  • Include the number of ways to roll a die $7$ times and see up to $6$ different values: $\binom66\cdot6^7$
  • Exclude the number of ways to roll a die $7$ times and see up to $5$ different values: $\binom65\cdot5^7$
  • Include the number of ways to roll a die $7$ times and see up to $4$ different values: $\binom64\cdot4^7$
  • Exclude the number of ways to roll a die $7$ times and see up to $3$ different values: $\binom63\cdot3^7$
  • Include the number of ways to roll a die $7$ times and see up to $2$ different values: $\binom62\cdot2^7$
  • Exclude the number of ways to roll a die $7$ times and see up to $1$ different values: $\binom61\cdot1^7$

Divide the result by the total number of ways, which is $6^7$:

$$\frac{\binom66\cdot6^7-\binom65\cdot5^7+\binom64\cdot4^7-\binom63\cdot3^7+\binom62\cdot2^7-\binom61\cdot1^7}{6^7}=\frac{35}{648}$$

Calculate the probability of the original event by subtracting the result from $1$:

$$1-\frac{35}{648}=\frac{613}{648}\approx94.6\%$$

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