Don't roll a six!

Probability Level 4

Matthew plays a game with a die (shaped like a cube, has six numbers, {1,2,3,4,5,6} on it). First, he pays a fee of 10€ to start the game. Then he rolls the die. If he rolls a 6, the game ends. In other cases, he wins 1€ and rolls again. If he can avoid the 6 now, he wins 2€ and goes on, otherwise he wins nothing and the game ends.

Matthew goes on rolling the die until he rolls a 6, and every time he rolls a number other than 6, he wins 1€ more than on the previous roll. What is the probability for Matthew's total earnings to be more than 250€? Give the answer as the value of a with the accuracy of two decimals, when the probability is a %, .

The answer is 1.51.

1 solution

Tarmo Taipale
Jun 15, 2016

First, the probability of rolling a number other than six with a single throw is $\frac{5}{6}$ . The probability of Matthew avoiding the 6 n times in a row is ( $\frac{5}{6}$ ) $^{n}$ . To earn 250€ totally, Matthew must collect 260€ from the rolling, as he also needs to cover the fee of 10€. The total amount of cash collected after n non-six throws is (1+2+3+...+( n -1)+ n ) = $\frac{n(n+1)}{2}$ . We will get an equation

$\frac{n(n+1)}{2}$ = 260

n $^{2}$ +n= 520

n= $\frac{-1+(\sqrt{1+4 \times1 \times 520})}{2}$ (n= $\frac{-1-(\sqrt{1+4 \times1 \times 520})}{2}$ would result in n being negative)

n= $\frac{1}{2}$ + $\frac{\sqrt{2081}}{2}$ , so n is approximately 22.309.

This means Matthew needs 23 rolls to collect more than 260€ with them. The probability of this is ( $\frac{5}{6}$ ) $^{23}$ , which is approximately $\boxed{1.51}$ %.

Wouldn't there be an infinite number of ways to win more than 250€ :

P(23 non-6's) +

P(24 non-6's) +

P(25 non-6's) + ...

= (5/6)^23 + (5/6)^24 + (5/6)^25 + ...

= (5/6)^23 / (1-(5/6))

= 6*(5/6)^23

≈9.06%

Luke Videckis - 4 years, 12 months ago

There is an infinite number of ways to win more than250€, you're right, but when you have thrown 23 non-sixes it's no matter what you throw next (you have already won 250€). But if you want to calculate P(23 or more non-sixes) = P(exactly 23)+ P(exactly 24)+ P(exactly 25)... =[ (5/6)^23] x (1/6) + [(5/6)^24] x (1/6) +[(5/6)^25 x (1/6)]= (1/6)[(5/6)^23 + (5/6)^24 + (5/6)^25 + ...], because you have to throw 23 non-sixes AND then a six, to throw exactly 23 non-sixes, for example. We get
(1/6)[(5/6)^23 + (5/6)^24 + (5/6)^25 + ...] = [(1/6)(5/6)^23] / (1-(5/6))=(1/6) 6 (5/6)^23≈1.51

Tarmo Taipale - 4 years, 12 months ago

I understand now. Thanks for explaining!

Luke Videckis - 4 years, 11 months ago

@Luke Videckis – You're Welcome!

Tarmo Taipale - 4 years, 11 months ago

Don't roll a six!

The answer is 1.51.

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