Say that Amy tosses a coin 6 times, and Bob tosses a coin 5 times. What's the probability that Amy gets more heads than Bob does?
103426
Say Bob tosses his $n+1$ fair coins and Alice tosses her $n$ fair coins. Lets assume independent coin tosses. Now after all the $2n+1$ coin tosses one wants to know the probability that Bob has gotten more heads than Alice.    The way I thought of it is this : if Bob gets $0$ heads then there is no way he can get more heads than Alice. Otherwise the number of heads Bob can get which allows him to win is anything in the set $\{1,2,\dots,n+1\}$. And if Bob gets $x$ heads then the number of heads that Alice can get is anything in the set $\{0,1,2,..,x-1\}$. So\begin{align}P(\text{Bob gets more heads than Alice})&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} P( \text{Bob gets x heads }\cap \text{Alice gets y heads }) \\[0.2cm]&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \left(C^{n+1}_x \frac{1}{2}^{x} \frac{1}{2}^{n+1-x}\right)\left( C^n_y \frac{1}{2}^y \frac {1}{2}^{n-y}\right)\\[0.2cm]&amp; = \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \frac{C^{n+1}_x C^n_y}{2^{2n+1}}\end{align}   How does one simplify this?      Apparently the answer is $\frac{1}{2}$ by an argument which looks like this, Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2.
53633
Question: whats the probability that at least 3 out of a group of n people have the same birthday.  I am confused as to where to start. I know that it would be complementary probability, but even then I dont know exactly.  Would it be;  P(at least 3 have the same bday) = 1 - (|all 3 have different| + (2 have the same|)? Any directions and suggestions?
84531
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?
9639
What is the probability of obtaining THT before HHH when a toss is tossed sequentially? Could someone solve using state diagram and law of total probability?
69471
The answer to this is supposedly close to $0.63$. However, I get approximately $0.9765625$ for the following reason:  The probability of a fair coin flipped $N$ times resulting in all heads is $1/2^N$. In this case, $1/2^{10}=1/1024$. If I flip $M$ coins, $N$ times each, there are $M$ ways for some coin to result in all heads (from the binomial coefficient "$M$ choose $1$"), so the probability of some coin resulting in all heads is $M(1/2^N)$. In this case, $1000\times 1/1024 \approx 0.9765625$.  Can someone please explain the flaw in my reasoning?
64616
I was working on problems on expectation and found this one as a question from a well-known exam  Assume that you are flipping a fair coin, i.e. probability of heads or tails is equal. Then the expected number of coin flips required to obtain two consecutive heads for the first time is.  (A)4  (B)3  (C)6  (D)10  (E)5  I worked up like Let N be the number of tosses required untill 2 consecutive heads are obtained for the first time.  Let the probability of obtaining heads be p and tails be q and for a coin p=q=1-p=1-q.  For N=2, the probability will be $p^2$  For N=3, the probability will be $p^2.q$  For N=4, the probability will be $q^2.p^2$  For N=5, the probability will be $q^3.p^2$  and so on...  Now, $E[N]=2(p^2)+3(p^2.q)+4(q^2.p^2)+5(q^3.p^2)+......$  and this comes out to be $\frac{3}{2}$ but it matches with none of the options.  Please tell me where I am wrong and what should I do to reach the correct solution?
74073
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?
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 :)
54971
A coin is tossed three times: The probability of zero heads is 1/8 and the probability of zero tails is 1/8.  And my question is: What is the probability that all three tosses result in the same outcome?   So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  If so, we could use the Addition Rule which is $P(A \cup B) = P(A) + P(B)$  where A and B are disjoint events, i.e. $A \cap B = \emptyset$, A is the event of tossing all heads and B is the event of tossing all tails.  I'm not sure how to continue after that... would the complement be used?
55435
If coin $X$ and coin $Y$ are biased, and have the probability of turning up heads at $p$ and $q$ respectively, then given one of these coins at random, how many times must coin A be flipped in order to identify whether we're dealing with coin $X$ or $Y$? We assume a 0.5 chance that we can get either coin.
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.
82881
Before reading the Wikipedia article, my idea to calculate the probability was as follows:  $1-\left(\frac{364}{365}\right)^{_nC_{2}}$  Basically, I thought to compare all combinations of pairs ($_nC_{2}$) and require them to have different birthday ($\frac{364}{365}$). It gave a very similar result to the wiki method, but is slightly different and I can't tell how  my way of thinking is wrong.  Can anyone tell me what I'm missing?
92474
Two players P and Q take turns, in which they each roll two fair and independent dice. P rolls the dice first.  The first player who gets a sum of seven wins the game. What is the probability that player P wins the game?
20556
My professor assigned a completely random question on our problem set. Basically, it goes:  Sophie is 30 years old and majored in philosophy. As a student, she participated in anti-nuclear demonstrations and was deeply concerned with social justice. Which is more probable?  1. Sophie is a bank teller. 2. Sophie is a bank teller and is active in the feminist movement.  I'm guessing the point the prof was trying to get across is that assuming you have two different events $A$ and $B$,  $A\cap B$ would always have a smaller probability than either. Is this correct?
84249
A coin is tossed three times. The probability of zero heads is 1/8 and the probability of zero tails is 1/8. What is the probability that there is at least one head and at least one tail?  So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  We would use the complement rule and Multiplication Rule?   P(at least one head) = 1 - P(no heads) = 1 - 1/8= 7/8 P(at least one tail) = 1 - P(no tails) = 1 - 1/8= 7/8  would I multiply the two values?
73966
Here is the problem setup:  Two people play a game of throwing balls at a pin. Sally hits the pin 70% of the times she throws, and Bill hits the pin 40% of the times he throws. The game is won by the person who hits the target first. If Bill throws first each time, find the probability he will win.  With simulation, I get Bill's probability of winning to be around 0.52 (rounding to two decimal places). How to solve this problem analytically?
107990
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.