dice

[aroup]

On the average how many times a die must be thrown until one gets a 6?

[Moon]

If you meant an unbiased dice then the answer should be $6$ on average. However, this is completely probabilistic estimation, so this does not mean that in real world you'll get a $6$, once in every 6 times. However, if you roll the dice for many times (for example 1200 times) the average should be around 6.

[aroup]

in the solution they used a term named,"mean number of trials".what does it indicate?

trails are indicated by 1,2,3,...... and probability of success in trial are $p,pq,pq^2,\cdots$
(where p is the probability of 6 on a given trial and p+q=1)

mean number of trial(say m)
then equals to $m=p+2pq+3pq^2+4pq^3.........$
then the solver played the trick of geometric progression and got

$m*p=1$,and $m=(1/p)$

$p=(1/6)$ and so $m=6$


whats the meaning of mean number of trails and how they solver got the value of m?


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Mod Edit: Please write the whole equation within two dollar signs.

[tanvirab]

The mean here is expectation value. That means what is the expected outcome of an experiment given that you know the possible results and their probabilities. This is a weighted mean of the possible results, where each possible result is given an weight equal to it's probability.

In this case, possible results are possible numbers of trials to get a $6$, i.e. any positive integer. The probability of getting the $6$ on the $n$-th trial i.e. the probability of result $= n$ is $pq^{n-1}$.

The expected value is then $\sum_{n=1}^{\infty} = npq^{n-1}$

[tanvirab]

To make the weighted mean a bit clear.

Say you have a coin and your friend says that if you flip the coin it always shows tail. You do not agree, you say that it always shows head. So, you decide to do a experiment. You flip the coin many times and calculate the following,

$S =$ Number of heads - Number of tails
If $S$ is positive, you win, and your friend gives you $S$ taka. If $S$ is negative, you lose and you give your friend $-S$ taka.

So, you can think that you get one point for every head and your friend gets one point for every tail. What is the expected value of $S$? If the coin is unbiased it should be $0$, right? Let's see how we get that.

At each stage you can get $+1$ (if head) or $-1$ (if tail). The probability of each happening is $\frac{1}{2}$. So expected value is

$(+1)*$Probability of head $+$ $(-1)*$Probability of tail $=$ $0$

So, the basic idea is, if you flip the coin $100$ times then expected is, half of the time you get head $(+1)$, and half of the time you get tail$(-1)$. So, in total, you expect to get $0$.

But, if the coins were biased, say, Probability of head $= \frac{3}{4}$ and probability of tail $= \frac{1}{4}$ then you expect that $\frac{3}{4}$-th time i.e. $75$ times you will get head and the rest will be tail. So, expected value after $100$ trials is,

$\frac{3}{4}*100*(+1) + \frac{1}{4}*100*(-1) = 50$
This means when you play the game, if you know that the coin is biased as stated, then you will be confident that you will win (i.e. you expect to win). So, if you were gambling, then you would want to gamble very high amount of money, because the expectation value is very high.

The expectation value for $1$ trial is then $\frac{50}{100} = \frac{1}{2}$. This is the wighted (by probability) mean, commonly known as expected value.


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[aroup]

thanks,its a great help

[aroup]

thanks.its clear....