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Let X1,X2,X3... be an endless sequence of random variables,
uniformly distributed on the set {1,2,3....10}.
Index i will be called "King" if X_i is greater than all of the numbers before it in the sequence.
Calculate the expected value of the number of indices that are to be crowned Kings.

please help.

A term whose index is what you call a "king" is ordinarily called a record, so you are asking for the expected number of records in an iid sequence of Uniform${1,...,m}$ random variables, e.g. with $m=10$. (This assumes the terms are mutually independent.)

Shorter derivation

The probability that the $k$th distinct term is a record is ${1over k},$
and the number of records in the infinite sequence can be written as the sum of indicators
$$begin{align}sum_{k=1}^m mathbb{1}_{text{$k$th distinct term is a record}}
end{align}$$

so the expected number of records in the infinite sequence is

$$begin{align}
Eleft(sum_{k=1}^m mathbb{1}_{text{$k$th distinct term is a record}} right)&=sum_{i=1}^m E(mathbb{1}_{text{$k$th distinct term is a record}})\[2ex]
&=sum_{k=1}^m{1over k}\[2ex]
&=H_m
end{align}$$

where $H_m=sum_{k=1}^m{1over k}$ is the $m$th harmonic number. (E.g., $H_{10}={7381over 2520}$.)

Longer derivation

The probability that the $i$th term is a record is
$$begin{align}P_i&=sum_{k=1}^mP((X_i=k)cap(X_{j}< k text{ for all }j<i)) \[2ex]
&=sum_{k=1}^m{1over m}left({k-1over m}right)^{i-1}
end{align}$$
and the number of records in the infinite sequence can be written as the sum of indicators
$$begin{align}sum_{i=1}^infty mathbb{1}_{text{$X_i$ is a record}}
end{align}$$

so the expected number of records in the infinite sequence is

$$begin{align}
Eleft(sum_{i=1}^infty 1_{text{$X_i$ is a record}}right)&=sum_{i=1}^infty E(1_{text{$X_i$ is a record}})\[2ex]
&=sum_{i=1}^infty P_i\[2ex]
&=sum_{i=1}^inftysum_{k=1}^m{1over m}left({k-1over m}right)^{i-1}\[2ex]
&= {1over m}sum_{i=1}^inftysum_{k=1}^mleft({k-1over m}right)^{i-1}\[2ex]
&={1over m}sum_{k=1}^msum_{i=1}^inftyleft({k-1over m}right)^{i-1}\[2ex]
&={1over m}sum_{k=1}^m{mover m-k+1}\[2ex]
&={1over m}(mcdot sum_{j=1}^m{1over j})\[2ex]
&=H_m.
end{align}$$

This problem can be modeled using an absorbing Markov chain. Let ${Y_n:ngeqslant 1}$ be defined by $Y_1$ uniformly distributed over ${1,ldots,10}$ and for $ngeqslant 1$:
$$
mathbb P(Y_{n+1}=jmid Y_n=i)=begin{cases}
frac1{10-i},& i<jleqslant 10\
1,& i=j=10.
end{cases}
$$
Let $P$ be the transition matrix of this Markov chain, then $$P=pmatrix{Q&R\0&I} $$ where $Q$ is the substochastic matrix corresponding to transitions between transient states, $R$ that corresponding to transitions from a transient state to an absorbing state, and $I$ that corresponding to transitions between absorbing states. (Here we have only one absorbing state.) For pair of transient states $i,j$ the probability of transitioning from $i$ to $j$ in exactly $k$ steps is the $(i,j)$ entry of $Q^k$. Summing for all $k$ yields the fundamental matrix
$$
N = sum_{k=0}Q^k.
$$
Since the rows of $Q$ sum to strictly less than one, the Neumann series converges and we have $N=(I-Q)^{-1}$, with $I$ the identity matrix. The expected number of transitions until being absorbed starting in transient state $i$ is the $i^{mathsf{th}}$ entry of the vector $t=Ncdotmathbf 1$, where $mathbf 1$ is a column vector whose entries are all one. Here
$$
t=left(
begin{array}{c}
frac{9649}{2520} \
frac{1041}{280} \
frac{503}{140} \
frac{69}{20} \
frac{197}{60} \
frac{37}{12} \
frac{17}{6} \
frac{5}{2} \
2 \
end{array}
right),
$$
and since the initial distribution was uniform over ${1,ldots,10}$, we weight each of these entries, along with $1$ (for the case when $X_1=10$), by $frac1{10}$ and sum to obtain $$frac{7381}{2520}. $$

Given an infinite sequence of uniform draws from {1,...,10}, since we are only counting records, there is nothing changed by deleting any draw that is equal to a previous draw -- the number of records will be the same.

The edited sequence is finite, and with probability 1, is a linear ordering of {1,...,10} (since the probability is $0$ that any value is omitted in the original infinite sequence). By iid uniform, all $10!$ such orderings are equally likely .

Looking at a random ordering of {1,...10}, the expected number of (left-to-right) records in location $j$ is $1/j$, since among the first $j$ entries, each one of them is equally likely to be the maximum among them.

Using linearity of expectation the expected number of records in the original sequence will be $1 + 1/2 + 1/3 + .... + 1/10$.