In a football​ game, the sides in the soccer field are determined on the basis of a single coin flip by the referee and the team which correctly guesses the flip result gets its desired side. What is the probability that a team will have to wait until its 8th game to get its desired side?  I am stuck with this question, any help would be much appreciated! I thought, $\frac{1}{2}^8$ but it is wrong.
122065
I am stuck with this question, I thought $(\frac{1}{2})^7$ but it is wrong. Any help would be much appreciated. In a football​ game, the sides in the soccer field are determined on the basis of a single coin flip by the referee and the team which correctly guesses the flip result gets its desired side.  a) What is the probability that a team will have to wait until its 7th game to get its desired side?
67722
An unfair three-sided die. Side $1$ will be rolled with probability $p_1$ , Side $2$ with probability $p_ 2$ and Side $3$ with probability $p_ 3$ . Find a formula for the probability that the $8^{th}$ $3$ will be rolled on the $10^{th}$ roll.  Attempt:  First 9 rolls there are 7 three's.  $$[\binom{9}{7}(p_3)^7(p_1+p_2)^2]\cdot p_3$$ $$=[84\cdot(p_3)^7(p_1+p_2)^2]\cdot p_3$$  any thoughts of this solution?
52378
I am trying to solve interesting probability problem which I found in the book "50 Mathematical Ideas You Really Need to Know" by Tony Crilly:  A group of people go to lunch and afterwards pick up their hats at random. What is the probability that no one gets their own hat?"  So he claimed that It can be shown that this probability is $1/e$ (about $37\%$) so that the probability of at least one person getting their own hat is $1 – 1/e$ $(63\%$)  I was thinking about asymptotic cases where $n \to \infty$. The probabilities of "no one gets their own hat" and "at least one person getting their own hat" should be $0$ and $1$ respectively, right? So why it is $1/e$ and $1-1/e$?  Thank you in advance.
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?
8873
I want to find the expected number of coin tosses to get $N$ heads in a row, where $p$ is the probability of getting a head in a single toss.    Let $F(N)$ be the expected number of tosses to get $N$ heads consecutively, so    $$F(N) = 1 + p F(N-1) + (1-p) F(N)$$   which gives   $$F(N) = 1/p + F(N-1)$$  With base condition : $F(1) = 1 + 1/p$   My logic is as follows:     if we get a head, in current toss we need to get N-1 more heads consecutively, but if we get a tail, we have to start over   This is what I thought, but is not correct. Can you please help me?
67366
A coin with heads probability $p$ is flipped $n$ times. A "run" is a maximal sequence of consecutive flips that are all the same. For example, the sequence HTHHHTTH with $n=8$ has five runs, namely H, T, HHH, TT,H. Show that the expected number of runs is  $$1+2(n-1)p(1-p).$$  I have tried to use some generating function on this but calculus got pretty messy and didn't work.
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A teacher wants to randomly form two teams of 5 students from a group   of 5 girls and 5 boys for a sports activity. Two of the girls, Ann and   Alice, are selected as team leaders. Find the probability that one   team has exactly 3 girls.   If i do this way $$\dfrac{{3 \choose 2}{5\choose 2}{1\choose 1}{3\choose 3} \times 2!}{{8 \choose 4}{4 \choose 4}} = \dfrac{6}{7}$$  Which turns out to tally with the answer. But i got one big question, is my way correct. Do i really need to times $2$ here? Please see my link below to my previous question cause the previous question make me doubt if i should times $2$.
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I am confused about how to proceed with solving 3rd and 4th question.  The question is :  Alice has two coins. The probability of Heads for the first coin is  ${1\over4}$, and the probability of Heads for the second is ${3\over4}$. Other than this difference, the coins are indistinguishable. Alice chooses one of the coins at random and sends it to Bob. The random selection used by Alice to pick the coin to send to Bob is such that the first coin has a probability  $p$  of being selected. Assume that  $0&lt;p&lt;1$. Bob tries to guess which of the two coins he received by tossing it  $3$  times in a row and observing the outcome. Assume that for any particular coin, all tosses of that coin are independent.  1. Given that Bob observed  $k$  Heads out of the  $3$  tosses $(where                     k=0,1,2,3 )$, what is the conditional probability that he received the first coin?  My answer = ${3^{3-k}⋅p \over3^{3-k}⋅p+3^k⋅(1−p)}$  2. We define an error to have occurred if Bob decides that he received one coin from Alice, but he actually received the other coin. He decides that he received the first coin when the number of Heads,  $k$ , that he observes on the  $3$  tosses satisfies a certain condition. When one of the following conditions is used, Bob will minimize the probability of error. Choose the correct threshold condition.  My answer = $k ≤ {3\over2}+{1\over2}log_3 {p\over1−p}$  3. For this part, assume that  $p={3\over4}$  3 a. What is the probability that Bob will guess the coin correctly using the decision rule from part  2?  3 b. Suppose instead that Bob tries to guess which coin he received without tossing it. He still guesses the coin in order to minimize the probability of error. What is the probability that Bob will guess the coin correctly under this scenario?  4. Bob uses the decision rule of Part 2. If  $p$  is small, then Bob will always decide in favor of the second coin, ignoring the results of the three tosses. The range of such  $p's$ is  $[0,t)$. Find $t$.
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I flip a coin until I get $m$ more heads than tails, or $n$ more tails than heads. Let the expected number of flips of the coin before stopping be $f(m,n)$.  I obtained $f(m,n)=mn$ from the recursion $f(m,n)=1+\frac{f(m-1,n+1)+f(m+1,n-1)}2$ with $f(k,0)=f(0,k)=0$ for all $k$.   Other than going through this recursion (and either solving by inspection or by writing as linear recurrence in single variable and solving brute force), is there an intuitive reason you should expect this process to take $mn$ flips? I was thinking about the more general problem with probability $p$ of getting heads and was struck by how simple the formula became when handling, what turned out to be a special case (general formula broke down) of $p=\frac12$.
105191
Context: I need to solve one variation of the . Problem: Two players ($A$ and $B$) throw a coin until one of the victory sequences appears. For player $A$ the victory sequence is $HTT$, for player $B$ the victory sequence is $TTH$. What is the probability that $A$ wins and what is the expected number of throws given that $A$ won? My attempt: Intuitively, it seems that player $B$ is more likely to win. Basically, I reliase that we can use the Markov-Chains, but my idea is only draw a binary tree and consider the sequences on the n-th step. My another idea was to calculate the expected number of throws to get, say, $HH$ that would be equal to $r_{HH}$: $$r_{HH} =  \frac{1}{4}E(number|HH) + \frac{1}{4}E(number|HH)+ \frac{1}{4}E(number|TT)+ \frac{1}{4}E(number|TH) = \frac{1}{4}(9 + \frac{5}{2} r_{HH})$$  And get $r_{HH}$. My issues: I don't really understand how to join these ideas to get a proper solution.
67947
I'm in a computer algorithms course and have a question about basic probability. My math background includes no more than discrete math and a little calculus, so this probability question left me completely stumped. I checked for similar questions but didn't see any. If this question has been answered already I apologize in advance but I don't even know the correct terms to search to find an answer.  The question goes like this:   Suppose you're playing a set of games and the first to win 10 games  wins the set. Each player has an even chance (1/2) of winning a  game. What is the probability that the first player will win if she  has already won i games and the second player has already won j  games? How do you calculate this for different values of i and j?    Please word the explanation for a non-maths student. I need to provide a recursive formulation for a function that computes this probability then solve it using dynamic programming. No worries on recursion or dynamic programming. I can solve that once I understand how to solve for probability.
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Suppose, we roll a fair die $n$ times. The expected value of how many times each side of the die will appear is $\frac{n}{6}$. The question is, what is the probability that any side of the die will come up more than $k$ times, where $n&gt;k&gt;\frac{n}{6}$? Or you could put it this way: what is the probability that no side will come up more than $k$ times? Clearly these two probabilities add up to 1.
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Assume that we have a fair coin. We flip it 100 times. The outcome is all heads. Why is it surprising? Doesn't all outcomes have the same probability? Any particular outcome, including irregular ones, would have the same very small probability. Why is it that observing an irregular outcome is less surprising to us than a regular one?
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A two-sided coin has just been minted with two different sides (heads and tails). It has never been flipped before. Basic understanding of probability suggests that the probability of flipping heads is .5 and tails is .5. Unexpectedly, you flip the coin a very large number of times and it always lands on heads. Is the probability of flipping heads/tails still .5 each? Or has it changed in favor of tails because the probability should tend to .5 heads and .5 tails as you approach an infinite number of trials?  I understand that flipping coins is generally a stochastic process, but does that change at all if you see a large number of trials bias to one side?
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I am told that a fair coin is flipped $2n$ times and I have to find the probability that it comes up heads more often that it comes up tails.     Please, how do I find the required probability?
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Question: Players A and B play a sequence of independent games. Player A starts and throws a fair die and wins on a "six". If she fails then player B throws the same die and wins on a "six" or fails otherwise. If she fails then player A gets the die and throws again, and so on. The first player to throw a "six" wins. Find the probability that player A wins.  Here is what I have done:  A wins right away: $\dfrac{1}{6}$  A fails B wins: $\dfrac{5}{6}\times \dfrac{1}{6}$  A fails B fails A wins: $\dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{1}{6}$  Unlike another similar question which the player B can win on "five" and "six" after player A fails on the first throw, in this case, the games do not seem to have an endpoint. How should I calculate the probability that player A wins then?
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Suppose you ﬂip a fair coin repeatedly until you see a Heads followed by a Tails. What is the expected number of coin ﬂips you have to ﬂip?  By manipulating an equation based on the result of the first flip, shown at this link:    the answer is 6. This also makes sense intuitively since the expected value of the number flips until HH or TT is 3. But is there a way to tackle this problem by summing a series of probabilities multiplied by the values?  Thank you!