# What is the average probability of rolling a dice until a 6 turns up?

• Last Updated : 30 Dec, 2021

Probability is also known as a possibility, which works in the happening of a likely event. The utility is designated from zero to one. In math, Probability has been obvious to approximate how possible events are to occur. Basically, the probability is the scope to which something is to be expected to take place.

### Probability

To understand probability more exactly, let’s understand an example of flipping a coin, the possible outcomes are – head and tail. The possibility of happening any of the likely event is 1/2. As the possibility of occurring any of the likely event is the same so there is an equal possibility of happening any favourable affair, in this case, it is 1/2.

Formula of Probability

P(A) = {Number of affair A} ⁄ {Total number of affair}

DICE

Dice is a small block that has between one and six marks or tints on its boundary and is used in games to give a random integer. Dice are small, tossable blocks with a detectable boundary that can stop in respective figures. They are handed-down to give stand-up to respective figures, often as part of sideboard games, as well as dice games, board games, role-playing games, and games of chance.

A usual die is a block with each of its six sides detectable with a different integers of figures from one to six. When throwable or rolled, the die comes to pause shows a random number from one to six on its higher side, with the happening of each affair being equally likely. Dice may also have concave or unequal shapes and may have faces noticeable with figures or characters instead of pit. Filled dice are drawn to favour some results over others for break out or relaxation.

### What is the average probability of rolling a dice until a 6 turns up?

Solution:

(5/6)n – 1 is the probability that one rolls any other number at least n times until rolling a six. Setting this to 1/2 gives,

n = -1/log2(5/6) + 1

This is the median of the diffusion: the arithmetical value splitting the excessive half of the diffusion from the lower half.

It is not the mean (standard).

### Similar Problems

Question 1: If a fair 6-sided die is thrown three times, what is the probab

y that precisely one 3 occur?

Solution:

Total ways in which a 6-sided die can be thrown three times = 6 × 6 × 6 = 216

To get precisely one 3, there are three ways,

A 3 on the first throw and non 3 on other two throw. This can be done in 1 × 5 × 5 = 25 ways.

The 3 could be on the second or third throw too. So total likely results = 25 × 3 = 75

Required Probability = 75/216 = 25/72

Question 2: If a fair 6-sided die is thrown two times, what is the probability that precisely one 4 occur?

Solution:

Total ways in which a 6-sided die can be thrown two times = 6 × 6 = 36

T

t precisely one 4, there are two ways,

A 4 on the first throw and non 4 on other throw. This can be done in 1 × 5 = 5 ways.

The 4 could be on the second throw too. So total likely results = 5 × 2 = 10

Required Probability = 10/36 = 5/18

Question 3: If you throw a fair six-sided die twice, what’s the probability that you get the same number both times?

Solution:

There are six possible numerals that can be thrown twice. For a precise number, say 1, the chances of throwing it twice is equal to,

[1/61st roll] × [1/62nd roll] = [1/36] (by the rule of the product).

Since this can occur for 2, 3, 4, 5, 6, as well as for 1,

The probability of throwing the same nu

twice is

6 × 1/6 × 1/6 = 1/6

Question 4: If a fair 6-sided die is thrown three times, what is the probability that precisely one 3 is rolled?

Solution:

Total ways in which a 6 sided die can be thrown three times = 6 × 6 × 6 = 216

to get precisely one 3, there are three ways,

A 3 on the first throw and non 3 on other two throws. This can be done in 1 × 5 × 5 = 25 ways.

The 3 could be on the second or third throw too. So total likely events = 25 × 3 = 75

Required probability = 75/216

= 25/72

Question 5: What is the probability to occur 6 to three times in a row when throwing a dice?<

ong>

Solution:

Probability of an affair = (number of favourable event) / (total number of event).

P(B) = {Number of affair B } ⁄ {Total number of affair}.

Probability of occurring 6 = 1/6

Throwing a dice is an free event, it is not dependent on how many times it’s been thrown.

Probability of occurring 6 three times in a row = probability of occurring 6 first time × probability of occurring 6 second time × probability of occurring 6 third time.

Probability of occurring 6 three times in a row  = (1/6) × (1/6) × (1/6) = 1/216.

Hence, the probability of occurring 6 three times in a row is 0.463%.

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