So the problem is this: A flea hops about at random on the vertices of a triangle. Find the probability that after $n$ hops the flea is back where it started.  So let $X_n$ be the position of the flea at the $n$th hop. I want to find $P(X_n = 1 | X_0 = 1)$.   Now \begin{align} p_{1,1}(n) = P(X_n = 1 | X_0 = 1) &amp;= \displaystyle \sum_{k=1}^3 P(X_n = 1|X_1 = k, X_0 = 1)P(X_1 =k|X_0 = 1)  \\&amp;=\sum_{k=1}^3 P(X_{n} = 1|X_1=k) P(X_1=k|X_0=1) \\&amp;= \sum_{k=1}^3 P(X_{n-1}=1|X_0=k)P(X_1=k|X_0=1) \\&amp;= \sum_{k=1}^3 p_{1,k}(n-1)p_{k,1}(1) \\&amp; = \sum_{k_1,\cdots,k_{n-1}} p_{1,k_1}(1)p_{k_1,k_2}(1)\ldots p_{k_{n-1},1}(1) \end{align}  So I just need to find the $(1,1)$ entry of $\begin{pmatrix} 0 &amp;1/2&amp;1/2\\ 1/2 &amp; 0 &amp; 1/2\\ 1/2 &amp; 1/2 &amp; 0\\ \end{pmatrix}^n$ which after diagonalising and a lot of work turns out to be $\dfrac 1 3 \left(1 + \dfrac{(-1)^n}{2^{n-1}}\right)$.  Obviously there is a nicer combinatorial way of deriving this expression, but I'm rubbish at this kind of thing. Please help.  Please read the FULL question before marking as duplicate. Is this question going to get reopened then or shall I ask it again?
120435
"A flea hops randomly on the vertices of a triangle with vertices labeled 1,2 and 3, hopping to each of the other vertices with equal probability. If the flea starts at vertex 1, find the probability that after n hops the flea is back to vertex 1."  Could someone provide a hint to help me start this?
119561
I was trying to solve a problem in which I was given a matrix containing probabilities of person s[0] to go s[n] as s[0][n].   Input:  0   1   0   0   0   1     4   0   0   3   2   0     0   0   0   0   0   0     0   0   0   0   0   0     0   0   0   0   0   0     0   0   0   0   0   0   Tracing the probabilities: [0,1,0,0,0,1]: #s0 the initial state, goes to s1 and s5 with equal probability [4,0,0,3,2,0]: #s1 can become s0,s3 or s4, but with different probabilities [0,0,0,0,0,0]: #s2 is terminal, and unreachable [0,0,0,0,0,0]: #s3 is terminal [0,0,0,0,0,0]: #s4 is terminal [0,0,0,0,0,0]: #s5 is terminal   So we can consider different paths to terminal states such as: s0->s1->s3 s0->s1->s0->s1->s0->s1->s4 s0->s1->s0->s5  Tracing probabilities of each, we find that: s2 has probability 0 s3 has probability 3/14 s4 has probability 1/7 s5 has probability 9/14    Output: [s2.numerator,s3.numerator,s4.numerator,s5.denominator]  which is: [0,3,2,9,14]   I'm trying to apply Regular Markov chain rule but couldn't succeed. Can anyone give me insights about how to solve this problem. Or at least any method or directions which I can use to understand this problem.  Any help would be really appreciated
86055
Say you have a stick which breaks randomly into three pieces (we can choose the points randomly). What is the probability that the area is greater than or equal to $0.4$?   I can see it has something to do with Heron's formula but I just can't put t together.
4142
Given a random square matrix of size $n\times n$ in the field $\mathbb F_2$, what is the probability that its determinant is $1$? (This is also the probability that the matrix is non-singular, since $\mathbb F_2$ only has the elements $0$ and $1$.)
56137
There is a probabilistic method to solve it. But I am not familiar with probability. I am trying to compute it by analytic method, such as using L Hospital's rule or Stolz formula, but they are not working.
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Problem: This is problem 33 of section 2.6 (conditioning) in Bertsekas, Tsitsiklis intro to probability text.  We are given that a coin has probability of heads equal to p and tails equal to q and it is tossed successively and independently until a head comes twice in a row or a tail comes twice in a row. What isthe expected value of the number of tosses?  Solution:  Let $X$ be a random value that tracks the number of tosses. In addition, let $H_k, T_k$ be the event that a head or tail comes up on the kth flip respectively. The solution I am looking at involves using total expectation theorem.  I outline the following steps in the solution.  We can partition the sample space into $H_1$ and $T_1$, allowing us to invoke the total expectation theorem. Hence,  $E[X] = pE[X|H_1] + qE[X|T_1]$  Again by total expectation theorem, we can utilize another partition to obtain  $E[X|H_1] = pE[X|H_1,H_2] + qE[X|H_1,T_2] = 2p + q(1 + E[X|T_1])$  $E[X|T_1] = qE[X|T_1,T_2] + pE[X|T_1,H_2] = 2q + p(1+E[X|H_1])$  Stuck: The solution follows up by saying that we can combine the above two relations and use the fact that p+q = 1 to obtain  $E[X|T_1] = \frac{2+p^2}{1-pq} , E[X|H_1] = \frac{2+q^2}{1-pq}$  I have been staring at this for a while now, but I cannot seem to see the steps of the computation. I'd be very grateful if someone could fill in the details regarding the computation. Thank you.
16735
For each positive even integer $n$, set   $$P_n = \displaystyle {{n} \choose {\frac{n}{2}}}\frac{1}{2^n}.$$  Show that $\displaystyle \lim_{n \to \infty} P_n$ exists and determine its value.   Here's what I have so far: each $P_n$ can be thought of as the probability of tossing a fair coin $n$ times, and obtaining exactly $\frac{n}{2}$ heads. My expectation is that $P_n \to 1$, since in any large trial, one would expect to record just about as many heads as tails. But I'm not sure how to mathematically justify this hunch.  Hints or solutions are greatly appreciated.
5564
Given a box that contains 90% fair coins and 10% loaded coins, (a loaded coin gives heads 90% of the time), what is the probability for a randomly drawn coin to give 5 heads in a row?  P(HHHHH | random coin from box of mixed coins) = ?   I used Bayes Theorem to calculate the probability I got .219 or 21.9% probability for the random coin to give 5 heads in a row!    But I am still doubting if my answer is correct !!!  Any help appreciated :)
92296
Most of the books I read, they only state some examples of tail events. One of them is  $[\sum_n X_n \ converges]$. My main problem is to show that this is indeed a tail event.
19374
Have been working on this problem for a while. No clue how to approach! Player A and Player B are playing a game with 3 possible results: win, tie and lose. Player A’s probabilities for each outcome are:  Win: 0.3  Tie: 0.2  Lose: 0.5   A win gives 2 points, tie gives each 1 point, and loss gives 0 points. The first of the two players to win 3 or more points than the other will win the match. What is the probability that Player A will eventually win? Thanks!
31887
A biased coin has probability p of showing heads. We toss the coin repeatedly until three heads or three tails occur in row. What is the probability that three heads occur before three tails?  I know how to solve this for two heads and two tails.  $$ P(A) =  p^2 + pq \cdot P(A|HT) + pq \cdot P(A|TH)  $$  Then, it is easy to relate $$ P(A∣HT)=p \cdot P(A∣TH) $$ and $$P(A∣TH)=p+qP(A∣HT) $$  And find solution. However for three we will have equation like this:  $$ P(A) = p^3 + p^2q \cdot P(A|HHT) + p^2q \cdot P(A|HTH) + p^2q \cdot P(A|THH) + q^2p \cdot P(A|HTT)  + q^2p \cdot P(A|THT) + q^2p \cdot P(A|TTH)  $$ How can we relate these conditional probabilities? The link for case with 2 tosses:
69762
If $P(A_n) \rightarrow 0$ and $\sum_{n=1}^{\infty}{P(A_n^c\cap A_{n+1}})&lt;\infty$ then $P(A_n \text{ i.o.})=0$.  How to prove this? Thanks.
95159
Given a random permutation $\sigma \in S_n$ from $[n] \to [n]$ in a uniform probability space, what is the probability that $\sigma $ has exactly $k$ fixed points for a given $k$ between $1$ and $n$?  In other words: what is the probability that $\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i  $ for $\ i\in \{1,...,k\}$ and for every $y \notin \{x_1 , ... , x_k\}$ we get $\sigma(y) \neq y$.  I saw that $\lim_{n \to \infty } prob(A_0) = e^{-1}$ using Inclusion–exclusion principle and i belive that for a given k : $\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$ but I am not sure how to show it.  *$A_k$ stands for the event "k".
116612
I have the following sentence:  If I toss a biased coin with 2/3 chance of landing on heads, given that there was at least one head in 3 flips, what is the probability that there is only 1 head?  If we let $X$ represent the number of heads, then I can use $Baye's$ formula to find out the probability.     Which of the following would be the formula?  $$P(X=1 | X \geq 1)$$ $$P(X \geq 1 | X=1)$$  I think its the first one, but my sheet is saying the second one. I think it might have made a mistake. Which is the right probability?
25462
Let $\pi_1,...,\pi_n$ be a random permutation of numbers $1,...,n$. If you are told that $\pi_k &gt; \pi_1,...,\pi_k &gt; \pi_{k-1}$, what is the probability that $\pi_k = n$?  What I've tried:  Let $A_j$ be the event that $\pi_k = j$. Let $H$ = be event that $\pi_k &gt; \pi_1,...,\pi_k &gt; \pi_{k-1}$.  Then $P(A_n|H) = \frac{P(H|A_n)P(A_n)}{\sum_{j = 1}^n P(H|A_j)P(A_j)} = \frac{\frac{1}{n}}{\sum_{j=1}^n P(H|A_j)\frac{1}{n}} = \frac{1}{\sum_{j=1}^n P(H|A_j)} = \frac{1}{\sum_{j=1}^n P(j &gt; \pi_1,...,j &gt; \pi_{k-1})}$.  How to calculate that last sum?
112318
what is wrong with this??? please,can anyone help?  \begin{equation} \label{probabilityy} P^{t}({E_{0})=P_{0}+(1-P_{0})P_{1}+ (1-P_{0})(1-P_{1})P_{2}+.......+P_{n}\Pi_{k_{0}}^{n-1}(1-P_{k}) \end{equation}
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we flip a fair coin N times. the probability getting a head or a tail is of course 0.5. after N tossings, we record a sequence of heads and tails. let A be the total number of times that we get a head right after we get a tail. let B be the total number of times that we get a tail right after we get a head. for example, if we flip the coin 7 times and record a result as HHTTHTH, then A=2 and B=2. find E(A) and E(B).  i relate this problem to the famous coin changeover problem. i got stuck right from the beginning. thanks for any help.