Let $X_1, X_2, X_3$ be the numbers that appear when three balanced dice are rolled. Find the expected value of the largest number that you will see. My attempt: $P[X_i=x] = 1/6$ for $i=1,2,3$. $E[X] = \sum{xp(x)} = 1×1/6 + 2×1/6 + 3×1/6 + 4×1/6 + 5×1/6 + 6×1/6$ Is my attempt correct?
121526
You toss a fair die three times. What is the expected value of the largest of the three outcomes?   My approach is the following: calculate the probability of outcome when $\max=6$, which is $$P(\text{at least one $6$ of the three rolls}) = 1 - P(\text{no }6) = 1 - (5/6)^3$$ and then calculate the probability of outcome when $\max=5$, which is $$P(\text{at least one $5$ of the three rolls &amp; $5$ is max}) = 1 - P(\text{no $5$ &amp; $5$ is max}) = 1 - (4/6)^3.$$  I wonder if this approach is right.
70994
A fair 6-sided die is rolled repeatedly in till a 6 is obtained. Find the expected number of rolls conditioned on the event that none of the rolls yielded an odd number  I had tried to figure out what will be the conditional distribution of $\frac{X}{Y}$ but I can't solved it yet Where $X $ is the face numbered 6 is obtained and $Y$ is the event only even number is occured
18104
I've just played a game with my kids that basically boils down to: whoever rolls every number at least once on a 6-sided die wins.  I won, eventually, and the others finished 1-2 turns later. Now I'm wondering: what is the expectation of the length of the game?  I know that the expectation of the number of rolls till you hit a specific number is  $\sum_{n=1}^\infty n\frac{1}{6}(\frac{5}{6})^{n-1}=6$.  However, I have two questions:   How many times to you have to roll a six-sided die until you get every number at least once?  Among four independent trials (i.e. with four players), what is the expectation of the maximum number of rolls needed? [note: it's maximum, not minimum, because at their age, it's more about finishing than about getting there first for my kids]   I can simulate the result, but I wonder how I would go about calculating it analytically.    Here's a Monte Carlo simulation in Matlab  mx=zeros(1000000,1); for i=1:1000000,    %# assume it's never going to take us &gt;100 rolls    r=randi(6,100,1);    %# since R2013a, unique returns the first occurrence    %# for earlier versions, take the minimum of x    %# and subtract it from the total array length    [~,x]=unique(r);     mx(i,1)=max(x); end  %# make sure we haven't violated an assumption assert(numel(x)==6)  %# find the expected value for the coupon collector problem expectationForOneRun = mean(mx)  %# find the expected number of rolls as a maximum of four independent players maxExpectationForFourRuns = mean( max( reshape( mx, 4, []), [], 1) )  expectationForOneRun =    14.7014 (SEM 0.006)  maxExpectationForFourRuns =    21.4815 (SEM 0.01)
45975
The problem:  X is a binomial random variable, find $E[\frac{1}{X+1}]$ n and p are not given  PDF for a binomial distribution is $\binom{n}{k}p^k(1-p)^{n-k}$  Expected value is $\sum{x_ip(x_i)}$  But this is where I get stuck, I'm really rusty on my statistics and I'm not sure exactly how to structure it in the next step? I think I want to get the form of the following out of the summation  $\sum _{k=0}^{n} \binom{n}{k}p^k(1-p)^{n-k} = (p + 1 - p)^n = 1$  But I'm not sure if it should look like  $\sum \frac{1}{xp(x)+1} $  and if it should where to go from here?
116172
Can someone please guide me to a way by which I can solve the following problem. There is a die and 2 players. Rolling stops as soon as some exceeds 100(not including 100 itself). Hence you have the following choices: 101, 102, 103, 104, 105, 106.  Which should I choose given first choice.  I'm thinking Markov chains, but is there a simpler way?   Thanks.  EDIT: I wrote dice instead of die. There is just one die being rolled
92114
Given a 7 sided dice, 2 players roll it independantly simultaneously and when a player rolls a '6' he gets a prize.  What is the probability of the first player getting a prize before the second?  So my solution is to sum over all the turns j in which player 1 could win, and make the second player lose all these turns:  $$P(A) = \sum_{i=1}^{\infty}(6/7)^{j-1}*(1/7)*(6/7)^j = 0.461$$ My question is if my solution is correct, and if so, why is the probability so low? It seems to me that either player 1 wins before the 2nd or the other way around, and the probability of them playing forever is 0, so it should be 0.5?
83383
This is not homework. I am interested in understanding if my logic is correct with this simple stats problem.  Let's say I have a 2 sided coin where the probability of flipping a head is $P(H)$ and the probability of flipping a tail is $1-P(H)$. Let's assume all flips have independent probabilities. Now, let's say I want to maximize my chances of predicting whether the coin will be a head or tail on the next flip. If $P(H) = 0.5$, I can guess heads or tails at random and the probability of me being correct is $0.5$.  Now, suppose that $P(H) = 0.2$, if I want to maximize my chances of guessing correctly, should I always guess tails where the probability is $0.8$?  Taking this one step further, if I had a 3-sided die, and the probability of rolling a 1, 2, or, 3 was $P(1)=0.1$, $P(2)=0.5$, and $P(3)=0.4$, should I always guess 2 to maximize my chances of guessing correctly? Is there another approach that would allow me to guess more accurately?
66589
This is a dice problem  1) I want to calculate the probability to have more than X throwing 3 dice of 6 faces. X = addition of the result of the 3 dice.  2) This is the first step but if you can also provide me a solution to calculate the probability to have more than X with Y dices of Z faces it would be really great.
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?
48084
Given $X \sim N(1,2)$, what is $E[X^3]$?  I've been told I should use the moment generating function, but I'm not sure how to apply it here in this instance.  Using the given mean and variance, I subbed the values into the normal density function $f(x)$, then tried to compute $E[X^3]$ by integrating $\int_{-\infty}^{\infty} f(x) \, x^3 \mathrm{d}x$ . At this point, I have no idea how to compute the integral, which leaves me pondering if I've veered way off track. I'd appreciate any help!
580
What is the average number of times it would it take to roll a fair 6-sided die and get all numbers on the die?  The order in which the numbers appear does not matter.  I had this questions explained to me by a professor (not math professor), but it was not clear in the explanation.  We were given the answer $(1-(\frac56)^n)^6 = .5$ or $n = 12.152$  Can someone please explain this to me, possibly with a link to a general topic?
109592
Suppose that there is a man standing on the origin of the real line and plays heads or tails. Everytime he gets a head, he moves $1$ unit right and everytime he gets a tail, he moves $1$ unit left. What is the expected value of the number of his steps when he reached to $-7$ or $10$ for the first time? A friend of mine asked me this question today and my approach to solve it was quite complex. So any help is appreciated.
59477
Over the holidays I played a dice game where each player had 3 to 7 d6 to roll each turn.  The game gave certain advantages to doubles and triples.  I wanted to know the odds of rolling doubles or triples given N dice to understand the importance of "upgrading" to more dice.  My question is really about how to reason about such problems:  It's easy to see that the odds of rolling doubles with 2 d6 are 1/6 (first die can be anything, and 1/6 of the time the second will match).  It's also easy to see that with 7 dice the odds are 1.0 (6 dice could be 1..6, but the 7th must match one of them).  The middle cases are fuzzier to me.  I ended up writing a program to enumerate all of the possibilities (even 6^7 is only 279,936 cases to evaluate) because that was the only way I felt I could verify any closed formula I came up with.  This also enabled me to distinguish cases of multiple doubles (or triples).  I would like to know how to derive closed forms and how to validate such solutions without brute-force simulation.
101747
I know that if events $A$ and $B$ are independent, then $P(A\cap B) = P(A).P(B)$.  Lets say I roll a dice, and define the following events:  $x$ = Number is a multiple of $3 = \lbrace 3, 6\rbrace$  $y$ = Number is more than $3 = \lbrace 4, 5, 6\rbrace$  $z$ = Number is even  =  $\lbrace 2, 4, 6 \rbrace$  Therefore,  $P(x) * P(y)$: 0.166666666667  $P(x \cap y)$   : 0.166666666667  $P(x) * P(z)$ : 0.166666666667  $P(x \cap z)$    : 0.166666666667  $P(y) * P(z)$ : 0.25  $P(y \cap z)$    : 0.333333333333  My question is, why are $xy$ and $xz$ independent events but $yz$ is not? I had expected all combinations of events to be independent as neither events affect any other.
75983
I have no idea how to do this question. If you have a loaded dice such that the probability to obtain an outcome of 1 is 2p/3, the probability of obtaining 2, 3, 4 or 5 is p each, and the probability of obtaining 6 is 3p/2. Find p.   Any help would be much appreciated!
34894
So the question is really hard I think. I tried using a simple way by calculating the probability of each combination that makes a sum divisible by six, but it would take forever. Does anyone have any ideas?  Suppose that we roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by six?
104226
If I throw a fair dice  $12$ times, the expected number of $6$ is $2$  i.e $6$  is expected to happen  $2$ times when the dice is thrown $12$ times. But the probability of getting $6$ exactly $2$ times is ${12}\choose{2}$$(1/6)^{2} (5/6)^{10}$ which is less than $1$.    I am a first reader of probability. I am quite curious to know why this is happening?    you are expecting $6$ appearing two times . On the other hand you are telling that the possibility of appearing 6 ,  two times is very low. How these two fact go simultaneously? I can not understand this.  I feel that the probability of getting six, two times should have been close to $1$. Can anyone please explain me where I am going wrong?