Chances of getting one particular number one or more times in 6 dice rolls.

[Pereza0]

Its something ive been asking myself. Im studying physics, but I havent studied anything related to probability since before high school

I mean, for example, the chance of getting tails after one throw is 50%, after two throws 75%, after three throws 87,5% and so on..... (right?)

Say, you roll a dice 6 times. You trying to get a number, say, 2.

What is the chance of getting a single 2 after the six rolls?
What is the chance of getting two 2s after the six rolls?

I just want to know how probability stacks up, and if there's a formula to determine it.

Now for something a little more complicated.

Imagine theres some woman giving away candy to children.

now say each kid gets five chances to pick from a bag with one sweet and a changing number pieces of coal. If you pick up the sweet, she'll put another one for you to pick up the next time. (So theres a chance youll end up with five sweets)

1st time
50 sweet
2nd time
40 sweets
3rd time
35 sweets
4 time
30 sweets
5th time
20 sweets

I made up the problems, so feel free to change the data, or not even comment about them.
I just want to get a decent grasp on it

[Monachus]

You need to revise probability trees, they will explain things much better and give a better visual of how probabilities in these situations work.

With your coin flip scenario you are going the wrong way, you have a 50% chance the first time, but to calculate the probability the second time you have to multiply the first probability of a tails by the second probabilty, i.e. 1/2 X 1/2 (a coin flip is independant therefore the probabilities stay the same), which would give you 25%. The third time you would have 1/2 X 1/2 X 1/2 giving a probability of 12.5% for getting tails three times in a row.
This can also be applied to your dice question in the same way just with different probabilites (1/6). So rolling 2 twice in a row would have a probability of 1/6 X 1/6 = 1/36.

With the bag thing you said there is just 1 sweet the sentence but in the list there is a varying number of sweets... but I hope this simple scenario will help:
You have a biased coin, 3/4 heads, 1/4 tails say. You flip (to keep things simple) twice. To find the probabilty of getting a head and a tails you have to consider there are two routes in getting this result; heads then tails and vice versa. The probabilty of each route is calculates by multiplying the probabilities both giving an answer of 3/16. But as both routes give you a head and a tails you have to add them together to get 6/16.

Sorry if this wasnt very clear!