# S1 Probability question

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#2

so you have a normal dice:

https://cdn.shopify.com/s/files/1/17...g?v=1501628890

what is the chance it lands on a 6 ?

https://cdn.shopify.com/s/files/1/17...g?v=1501628890

what is the chance it lands on a 6 ?

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(Original post by

so you have a normal dice:

https://cdn.shopify.com/s/files/1/17...g?v=1501628890

what is the chance it lands on a 6 ?

**the bear**)so you have a normal dice:

https://cdn.shopify.com/s/files/1/17...g?v=1501628890

what is the chance it lands on a 6 ?

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#4

(Original post by

1/6. The answer says something specifically about it being a cubical dice and I have no idea what difference it makes. The answer says a cubical die has 18 sides.. so yeah I'm really confused.

**dont know it**)1/6. The answer says something specifically about it being a cubical dice and I have no idea what difference it makes. The answer says a cubical die has 18 sides.. so yeah I'm really confused.

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#5

(Original post by

The answer says a cubical die has 18 sides..

**dont know it**)The answer says a cubical die has 18 sides..

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(Original post by

a cubical die is just a normal 6 sided cube.

**the bear**)a cubical die is just a normal 6 sided cube.

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#9

(Original post by

Oh I see. I still don't understand why it's not 1/6 though .

**dont know it**)Oh I see. I still don't understand why it's not 1/6 though .

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it IS 1/6 if you do it correctly. they are supposing that some doofus thinks a dice has 18 sides.

**the bear**)it IS 1/6 if you do it correctly. they are supposing that some doofus thinks a dice has 18 sides.

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#11

It can be simplified greatly by considering what events must occur in each of the 4 cases.

For example, for Chris to win on his first roll, we need the first two people to NOT win, and for the third person TO win. The probability of any roll resulting in a six is 1/6, so the probability of a win is 1/6 and the probability of a loss is 5/6. so the probability that the first roll is a loss is 5/6, the second roll is a loss is also 5/6, and the third roll is a win is 1/6. Multiply these to get 25/216, which is your answer for the first part.

Similarly for the next three parts, just think how many losses you need before you need a win. e.g. say you want to find the probability of winning on the 5th roll. This has probability (5/6)^4 x (1/6) = 625/7976, because we need exactly 4 losses and exactly 1 win, in that order. (order is important, we win MUST be the last roll, as the game ends then)

In case you didn't know, as I can't remember if this is on the S1 course, (if it is and you are familiar with it I apologise if this sounds patronising) this is an example of a geometric distribution, which models the probability of it taking a certain number of attempts to achieve a success once, and then stopping. It is called geometric because every time we increase the number of trials before a success, the probability decreases by the same factor, just like geometric sequences.

For example, for Chris to win on his first roll, we need the first two people to NOT win, and for the third person TO win. The probability of any roll resulting in a six is 1/6, so the probability of a win is 1/6 and the probability of a loss is 5/6. so the probability that the first roll is a loss is 5/6, the second roll is a loss is also 5/6, and the third roll is a win is 1/6. Multiply these to get 25/216, which is your answer for the first part.

Similarly for the next three parts, just think how many losses you need before you need a win. e.g. say you want to find the probability of winning on the 5th roll. This has probability (5/6)^4 x (1/6) = 625/7976, because we need exactly 4 losses and exactly 1 win, in that order. (order is important, we win MUST be the last roll, as the game ends then)

In case you didn't know, as I can't remember if this is on the S1 course, (if it is and you are familiar with it I apologise if this sounds patronising) this is an example of a geometric distribution, which models the probability of it taking a certain number of attempts to achieve a success once, and then stopping. It is called geometric because every time we increase the number of trials before a success, the probability decreases by the same factor, just like geometric sequences.

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#12

(Original post by

It can be simplified greatly by considering what events must occur in each of the 4 cases.

For example, for Chris to win on his first roll, we need the first two people to NOT win, and for the third person TO win. The probability of any roll resulting in a six is 1/6, so the probability of a win is 1/6 and the probability of a loss is 5/6. so the probability that the first roll is a loss is 5/6, the second roll is a loss is also 5/6, and the third roll is a win is 1/6. Multiply these to get 25/216, which is your answer for the first part.

Similarly for the next three parts, just think how many losses you need before you need a win. e.g. say you want to find the probability of winning on the 5th roll. This has probability (5/6)^4 x (1/6) = 625/7976, because we need exactly 4 losses and exactly 1 win, in that order. (order is important, we win MUST be the last roll, as the game ends then)

In case you didn't know, as I can't remember if this is on the S1 course, (if it is and you are familiar with it I apologise if this sounds patronising) this is an example of a geometric distribution, which models the probability of it taking a certain number of attempts to achieve a success once, and then stopping. It is called geometric because every time we increase the number of trials before a success, the probability decreases by the same factor, just like geometric sequences.

**Johnw77**)It can be simplified greatly by considering what events must occur in each of the 4 cases.

For example, for Chris to win on his first roll, we need the first two people to NOT win, and for the third person TO win. The probability of any roll resulting in a six is 1/6, so the probability of a win is 1/6 and the probability of a loss is 5/6. so the probability that the first roll is a loss is 5/6, the second roll is a loss is also 5/6, and the third roll is a win is 1/6. Multiply these to get 25/216, which is your answer for the first part.

Similarly for the next three parts, just think how many losses you need before you need a win. e.g. say you want to find the probability of winning on the 5th roll. This has probability (5/6)^4 x (1/6) = 625/7976, because we need exactly 4 losses and exactly 1 win, in that order. (order is important, we win MUST be the last roll, as the game ends then)

In case you didn't know, as I can't remember if this is on the S1 course, (if it is and you are familiar with it I apologise if this sounds patronising) this is an example of a geometric distribution, which models the probability of it taking a certain number of attempts to achieve a success once, and then stopping. It is called geometric because every time we increase the number of trials before a success, the probability decreases by the same factor, just like geometric sequences.

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