Probability question

p(z) = y/x

1 - (1/x)^y

1 - ((x-1)/x)^y

it's 50:50, either you roll a 4 or you don't.

I haven't done a problem like this in a long time

but wouldn't be something similiar to Chaz answer

Say it is a six headed dice and you wanted to know the odds was rolling a 4

1 - (5/6) for one roll

1 - 5/6 - (5/6)^2 for two rolls

1- 5/6 - (5/6)^2 - (5/6)^3 for 3 rolls ect.

I don't know how to type the sum of sign on my keyboard but you get the idea.

chaz22 wrote:

1 - ((x-1)/x)^y

bingo

probability help please wrote:

Lets say I have a dice with x sides that I decide to roll y times. How do I calculate the odds of rolling at least 1 time where it lands on side z?

For example lets say I had a ten sided dice that I rolled 6 times. How would I figure out my odds of rolling at least one 4? Thanks!

It sounds like you need a full explanation since you can't get this easy problem.

Think about what the complement of the event {rolling at least 1 time where it lands on side z} is. It's {never rolling a z}. Hopefully you learned at some point that the probability of a complement of an event is 1 minus the probability of the event.

The probability of not rolling a z on a single roll is 1 minus the probability of rolling a z. The probability of rolling a z is 1/x, so the probability of not rolling a z is (x-1)/x. We assume that each roll is independent, so the probability of not rolling any z's in y rolls is:

P(not rolling any z's in y rolls) = [(x-1)/x]^y

Then the probability of rolling at least one z is 1 minus that probability:

P(rolling at least one z) = 1 - [(x-1)/x]^y

If you copied this question verbatim, then you want the odds of this event, which is something different than the probability but not hard to calculate. If you have an event that has probability p, then the odds of that event are:

odds = p/(1-p)

probability help please wrote:

Lets say I have a dice...

You have A dice?

Good explanation. I see where I went wrong now. Thanks. This was fun.

chaz22 wrote:

1 - ((x-1)/x)^y

This is correct, but a more general form to find the probability of rolling z any number of times (where a=number of times z occurs) is:

P(z occurring a times)= (y choose a)*[(1/x)^a][(x-1)/x]^(y-a)

and (y choose a) = y!/[a!(y-a)!]

If you plug in a=0 it simplifies to [(x-1)/x]^y as stated above. Thought this might be helpful.

I think it's y^z or something like that.

spelling help please wrote:

probability help please wrote:

Lets say I have a dice...

You have A dice?

I have a square triangle.

The 3:30 Thonster wrote:

spelling help please wrote:

You have A dice?

I have a square triangle.

Will someone please explain this post? What is its relevance to the previous one?

spelling help please wrote:

The 3:30 Thonster wrote:

I have a square triangle.

Will someone please explain this post? What is its relevance to the previous one?

You really need an explanation?

Saying "I have a square triangle" is a lot like saying "I have A dice." A square triangle is impossible. So is A dice. Dice is a plural word, therefore you cannot have A dice, you can have a die. A triangle is three-sided, therefore you cannot have a square trianlge, you can only have a triangular triangle.

Perhaps you "spelling help please," are confused because you saw thought that the OP misspelled "die" as "dice," while "The 3:30 Thornster" thought that the OP intentionally wrote "dice," not knowing that it was a plural form of the word "die."

remedial poster wrote:

the probability of a complement of an event is 1 minus the probability of the event.

The probability of not rolling a z on a single roll is 1 minus the probability of rolling a z. The probability of rolling a z is 1/x, so the probability of not rolling a z is (x-1)/x.

Some of us need even more help.

If the probability of rolling a z is 1/x, why is the probability of its complement not 1-1/x, rather than (x-1)/x?

Some of us need even more help.

If the probability of rolling a z is 1/x, why is the probability of its complement not 1-1/x, rather than (x-1)/x?[/quote]

1-(1/x)=(x-1)/x

if you multiply the solo 1 by x/x to get a common denominator, then you can combine the two fractions. That is how you get to (x-1)/x.

Stat Matt wrote:

Some of us need even more help.

If the probability of rolling a z is 1/x, why is the probability of its complement not 1-1/x, rather than (x-1)/x?

1-(1/x)=(x-1)/x

if you multiply the solo 1 by x/x to get a common denominator, then you can combine the two fractions. That is how you get to (x-1)/x.[/quote]

Damn it, I knew it would be something that simple. I'm going to go punch myself in the face. Thank you.

Dr. Frankenstein wrote:

You really need an explanation?

Saying "I have a square triangle" is a lot like saying "I have A dice." A square triangle is impossible. So is A dice. Dice is a plural word, therefore you cannot have A dice, you can have a die. A triangle is three-sided, therefore you cannot have a square trianlge, you can only have a triangular triangle.

Perhaps you "spelling help please," are confused because you saw thought that the OP misspelled "die" as "dice," while "The 3:30 Thornster" thought that the OP intentionally wrote "dice," not knowing that it was a plural form of the word "die."

Thanks. That straightens that out. My "You have a dice?" was not a statement; it was a question. A display of bafflement. Like, "What's a dice?"

I was also baffled as to why the other guy aimed his sarcasm at me. Maybe he didn't see the question mark and understand its significance (at the 99% level).

Nice monster, by the way.

your still wrong and a dumbass.......