Probability Question

We don't do homework questions here.

Please can u explain ur answer in details

He just did!

Please do your homework and at least think through what has been said.

I didn't understand it, but I didn't understand your original question either. n seems to mean different things in different places, unless I'm missing something.

the probability of anyone here doing your homework is a solid ZERO

Besides the pribibility is undefined in the limits of your equation.

This should help....

https://www.youtube.com/watch?v=fIPVM9jbu18

THanks for given the link for basic understsndibg if probability

8 Top Websites that Students Use to Cheat - CBS News

DO YOUR WON HOMEWORK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

READ THE RULES for posting!!!Do your own homework. CR4 is not a homework cheat site; however, if you have questions about understanding concepts or how a portion of a problem is derived, these types of questions will be accepted.

Did CR4 make it into the eight?

No. I think that finding CR4 requires more organizational and search skills than is possessed by the average loser who is just trying to cheat their way to an "education".

I think it's somewhat of a random happenstance that they find their way to this forum.

Once here, they seem to keep on returning to the well instead of doing their own research.

But hey, What do I know?

The OP, Mr. Haris Aslam, graduated in 2010, as indicated in his earlier posts dating back then.

He could be working on a Masters, or PHD, but may just be curious to this probability outcome.

Sure, I'll buy that if you'd like.

It's Haris Siddiqui.

More then one name?

Haris

Must be a text book question about an unfair coin toss.

If he graduated, why can't he either

a) do what he learned when he got his (flaunted) degree? or

b) check the books that are too available to see what he forgot?

I'm not impressed with "he graduated" if he didn't learn how to do what he studied. What's so great about a degree that does nothing better than stick letters after your name?

We don't do homework, unless it's our own. And we don't do THAT here!!

Show your working and show you're working.

I probably won't answer you, and if I did, the answer would probably be wrong.

It is quite amusing how people react to a challenge !

I introduced a flaw in my explanation but nobody made the effort to analyse the problem, the only effort was to discard any intellectual effort and say "go away with your questions we are not here to solve your problems".

In fact the reaction was exactly as I expected it is the one I see during last times.

It was different before but times change.

I expect a lot of aggressive comments but I had to say what I think as usual.

How do you think this goes down giving the wrong answer here?

If there is one?

50% for the first go is clear.

Multiple throws that are not head lead to n. If there is one head in the middle then it is not possible to obtain n.

So what's the catch? It is most probable that with the second throw you are off.

I toss this. OP has not come back so why bother.

I am out.

The answer is not wrong but incomplete. It was a test and it worked perfectly.

for one thing, n was not specified as even or odd. if n is odd, then no amount of wrong throws will provide the correct answer. In other words, OP should stay single, or if he picks an even number of wives, he will need life insurance.

I really hope this fellow is not an acuary.

since you define the present probability of getting n points from the previous two throws of the coin, what happened in the case it takes more than three tosses? Is this probability density general for all integer n, or only for even numbered n?

General case examination:

n=0, then P0=1/2(P-1-P-2), does not exist, rather Po=0. (previous throws do not exist), and also previous values do not exist???? also 2>n

n=1, then P1 = 1/2(Po-P-1), does not exist, P1=0, previous values and throws either do not exist, or have zero value. 2>n

n=2 P2=1 since no matter the outcome, he will get two points on the first throw.

n=3 P3=1/2 using the equation, since P2=1, and P1=0

etc etc ad nauseum.

"if Pn is the probability of getting exactly n points.

Pn=1/2 (Pn-1 - Pn-2)"

Pn is the probability of getting exactly n points. A probability is a number x such that 0≤x≤1.

Now, as someone who works with probability, this part confuses me: 1/2 (Pn-1 - Pn-2)

As written, I want to reduce what is in the parentheses (Pn-1 - Pn-2) = (-3). Then I am not sure if the equation has reduced to Pn = 1/(2*-3) or Pn = 0.5 * -3. In either case, Pn is a negative number, so it is not a probability.

Please restate the problem using standard notation. It may help to provide a link to the book it came from.

The problem is not one of algebra but it is one of notation. I don't particularly like the formulism, as the OP's formula is not inclusive as a recursion probability that starts with the 0th or the first iteration, but starts at (n-2)^th iteration. I thing the whole thing is rubbish.

Maybe I'm being dim, but I still don't understand the OP's question, and tend to agree with your last sentence. Perhaps you or somebody could rephrase it?

I think the equation is unlikely to fit the situation he is referring to. It cannot work if n is odd since thrower gets 2 points for every tails, and n points for every heads, and the target is n. It appears that points accumulate until n is reached. If n = 0 , there is 50% chance on first throw, and 0% chance on succeeding throws. If n=1, the situation is exactly the same. If n=2, there is 100% probability of success with the first throw, since 2 pts awarded either outcome. if n=3, then 50% on first throw, 0% on all succeeding throws. Once again, and so on ad nauseum.

The whole thing is just silly rubbish.

OK I'm still puzzled, that's my problem, but thanks for trying!

I agree that notation is the problem. As written, my mind will not go in Probability directions. (all those classes in probability and statistics petrified my mind?) I suspect the desired equation is more like P_n = ½(P_n-1 - P_n-2). Or, in English, when is the probability of n equal to one half the difference of the probabilities of n-1 and n-2?

Is that what the OP's question described? I'll have to go back to the beginning and look.

That can't be the correct interpretation, either. P_n < P_n-1 < P_n-2. That is, it is less likely to get n=3 heads than n-1=2 heads, and less likely to get n-1=2heads than n-2=1 head. So, since P_n-1 < P_n-2, the difference (P_n-1 - P_n-2) is always negative, so it can not define a probability.

I believe that math can be applied to this equation. Although there are a few variables that I can think of that can affect the results: How high does the coin go before it loses momentum and drops to the ground? How high is the coin flippers thumb from the ground when the coin is first flipped ? How many flip revolutions does the coin make ? Does the coin flipper flip the coin easier or harder when the coin flipper start flipping the coin? Is there a learning curve where the coin flipper learns how to make the coin land on heads or tails more than one side or less ?

Here's my suggestion:

It should be a simple computer program to write in your favorite language. Loop it, say 100 times, and see what comes up. Run it multiple times to see if you get the same answer. If you get a consistent result, at least you will know what the right answer is and then you can work out why.

GA from me. But, if he can even DO this, it was the proverbial "stupid question" in the first place.

Reminds me of the time I came home from my Calculus course, and wrote a Basic (yeah, THAT language!) program to accept all the variables in my homework by text prompt, then figure out what I needed to know from what I didn't provide, and provide me the answer. My wife, who isn't a computer user/geek/nerd/dork asked me how I was going to learn Calculus if I made the computer do all the work. I assure her that if I could write a program to do it for me, I first had to know how to do it myself.

Near perfect score on the finals, so I must have been right. At least that time.

100 times could be not sufficient and by the way how do you simulate the "fate" ? With a number generator of excel for instance ? It is a pseudo stochastic generator so that results are only partly valid.