The sample space consists of seven possible outcomes: "1" through "5" on the die, "6" and "tails", and "6" and "heads." Let's abbreviate these as $\Omega=\{1,2,3,4,5,6T,6H\}$.
The events will be generated by the atoms $\{1\}, \{2\}, \ldots, \{6H\}$ and therefore all subsets of $\Omega$ are measurable.
The probability measure $\mathbb{P}$ is determined by its values on these atoms. The information in the question, together with the (reasonable) assumption that the coin toss is independent of the die throw, tells us those probabilities are as given in this table:
$$\begin{array}{lc}
\text{Outcome} & \text{Probability} \\
1 & \frac{1}{6} \\
2 & \frac{1}{6} \\
3 & \frac{1}{6} \\
4 & \frac{1}{6} \\
5 & \frac{1}{6} \\
\text{6T} & \frac{1-p}{6} \\
\text{6H} & \frac{p}{6}
\end{array}$$
A sequence of independent realizations of $X$ is a sequence $(\omega_1, \omega_2, \ldots, \omega_n, \ldots)$ all of whose elements are in $\Omega$. Let's call the set of all such sequences $\Omega^\infty$. The basic problem here lies in dealing with infinite sequences. The motivating idea behind the following solution is to keep simplifying the probability calculation until it can be reduced to computing the probability of a finite event. This is done in stages.
First, in order to discuss probabilities at all, we need to define a measure on $\Omega^\infty$ that makes events like "$6H$ occurs infinitely often" into measurable sets. This can be done in terms of "basic" sets that don't involve an infinite specification of values. Since we know how to define probabilities $\mathbb{P}_n$ on the set of finite sequences of length $n$, $\Omega^n$, let's define the "extension" of any measurable $E \subset \Omega^n$ to consist of all infinite sequences $\omega\in\Omega^\infty$ that have some element of $E$ as their prefix:
$$E^\infty = \{(\omega_i)\in\Omega^\infty\,|\, (\omega_1,\ldots,\omega_n)\in E\}.$$
The smallest sigma-algebra on $\Omega^\infty$ that contains all such sets is the one we will work with.
The probability measure $\mathbb{P}_\infty$ on $\Omega^\infty$ is determined by the finite probabilities $\mathbb{P}_n$. That is, for all $n$ and all $E\subset \Omega^n$,
$$\mathbb{P}_\infty(E^\infty) = \mathbb{P}_n(E).$$
(The preceding statements about the sigma-algebra on $\Omega^\infty$ and the measure $\mathbb{P}_\infty$ are elegant ways to carry out what will amount to limiting arguments.)
Having managed these formalities, we can do the calculations. To get started, we need to establish that it even makes sense to discuss the "probability" of $6H$ occurring infinitely often. This event can be constructed as the intersection of events of the type "$6H$ occurs at least $n$ times", for $n=1, 2, \ldots$. Because it is a countable intersection of measurable sets, it is measurable, so its probability exists.
Second, we need to compute this probability of $6H$ occurring infinitely often. One way is to compute the probability of the complementary event: what is the chance that $6H$ occurs only finitely many times? This event $E$ will be measurable, because it's the complement of a measurable set, as we have already established. $E$ can be partitioned into events $E_n$ of the form "$6H$ occurs exactly $n$ times", for $n=0, 1, 2, \ldots$. Because there are only countably many of these, the probability of $E$ will be the (countable) sum of the probabilities of the $E_n$. What are these probabilities?
Once more we can do a partition: $E_n$ breaks into events $E_{n,N}$ of the form "$6H$ occurs exactly $n$ times at roll $N$ and never occurs again." These events are disjoint and countable in number, so all we have to do (again!) is to compute their chances and add them up. But finally we have reduced the problem to a finite calculation: $\mathbb{P}_\infty(E_{n,N})$ is no greater than the chance of any finite event of the form "$6H$ occurs for the $n^\text{th}$ time at roll $N$ and does not occur between rolls $N$ and $M \gt N$." The calculation is easy because we don't really need to know the details: each time $M$ increases by $1$, the chance--whatever it may be--is further multiplied by the chance that $6H$ is not rolled, which is $1-p/6$. We thereby obtain a geometric sequence with common ratio $r = 1-p/6 \lt 1$. Regardless of the starting value, it grows arbitrarily small as $M$ gets large.
(Notice that we did not need to take a limit of probabilities: we only needed to show that the probability of $E_{n,N}$ is bounded above by numbers that converge to zero.)
Consequently $\mathbb{P}_\infty(E_{n,N})$ cannot have any value greater than $0$, whence it must equal $0$. Accordingly,
$$\mathbb{P}_\infty(E_n) = \sum_{N=0}^\infty \mathbb{P}_\infty(E_{n,N}) = 0.$$
Where are we? We have just established that for any $n \ge 0$, the chance of observing exactly $n$ outcomes of $6H$ is nil. By adding up all these zeros, we conclude that $$\mathbb{P}_\infty(E) = \sum_{n=0}^\infty \mathbb{P}_\infty(E_n) = 0.$$ This is the chance that $6H$ occurs only finitely many times. Consequently, the chance that $6H$ occurs infinitely many times is $1-0 = 1$, QED.
Every statement in the preceding paragraph is so obvious as to be intuitively trivial. The exercise of demonstrating its conclusions with some rigor, using the definitions of sigma algebras and probability measures, helps show that these definitions are the right ones for working with probabilities, even when infinite sequences are involved.
probability that i get 0 times even number is equal to (0.5)^4∗200
No, that would be the expected count of the same sequence of outcomes, after 200 throws. But there are more than 1 sequence, leading to an even outcome. This is where it goes wrong.
To obtain your frequencies, you need to evaulate first how many different sequences you can throw at all, irrespective of odd or even outcomes.
Then look for the sequences which give odd outcomes, which you already found: sequences with exactly 1 odd or sequences with exactly 3 odd (or 1 even).
The probability of finding 1 odd outcome is the number of sequences to throw 1 odd:
oeee, eoee, eeoe, eeeo
divided by the total number of sequences.
The total number of sequences is $2^4=16$.
Since the are 4 sequences with 1 odd, the probability of finding 1 odd is $4/16=1/4$. So the expected frequency will be $200\cdot\,1/4=50$.
Best Answer
This is a straightforward binomial distribution. B(n,pi) with n= number of trials and pi=probability of an event. n=5 and pi=P(even number)=.5 n and pi are parameters of the distribution. they are independent of one another, so n=5 (or more or less) will not affect the value of pi.