If I am rolling I die until I roll every number at least once, what is the expected value of times that I will need to roll the die?  After a brief computer simulation, I got 15. But why is this the answer?
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?
18025
Ello, my teacher gave me a simple exercise but I'm stuck.     Rolling a die 100 times, denote the outcome of roll $i$ by $X_i$. Estimate the probability $$ \Pr\{\prod_{i=1}^{100}X_i\leq a^{100}\}$$ for real $1&lt;a&lt;6$   I think that $$ \Pr\{\prod_{i=1}^{100}X_i\leq a^{100}\}=(\Pr\{X_1\leq a\})^{100}$$ by trying to find a solution numerically. Is it correct and why ?  Thanks, Herosix
51315
Is there a way to calculate on average, the maximum amount of times we can expect a coin to land heads during 1,000 flips?  So the answer (and formula if one exists) I am looking for would be something like: during 1,000 flips we can expect a maximum run of 12 heads in a row.
92945
If I roll an n-sided die m times, what is the probability of getting only one face, two faces, .... or n faces?  Let us assume that I have a 6-sided die and I roll it 10 times. I want to calculate different probabilities. The probability of getting only one face from all rolls. The probability of getting only 2 faces in all rolls and so on until calculating the probability of the appearance of the 6 faces (at least one time for each face) in all rolls.
39036
"An event less frequent in the long run is likely to happen before a more frequent event!"  How can I show that THTH is more likely to turn up before HTHH with a probability of  9/14, even though waiting time of THTH is 20 and HTHH, 18!  I would be very thankful if you could show me the way of  calculating the probability of turning up earlier,   and the waiting time. Thank you!
6251
Choose a random number between $0$ and $1$ and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds $1$. What's the expected value of the number of random numbers needed to accomplish this?
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
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?
42879
A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $\{2,3,5\}$. What is the probability the element set has exactly $2$ elements? how about $3$ elements?   I know the sample space is $6^6$. The counting for $2$ elements is $6C2 \cdot 6!$? I would really appreciate the help! :)
99549
A red die, a blue die, and a yellow die (all six-sided) are rolled. We are interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die which is less than that appearing on the red die $P(B &lt; Y &lt; R)$.  I know one way to calculate this is:  E = event that no two dice land on the same number:  $$P(E) = \frac{6 \cdot 5 \cdot 4}{6 \cdot 6 \cdot 6} = \frac{5}{9}$$  F = event that B &lt; Y &lt; R and no two dice land on the same number  $$P(F) = P(E) \cdot P(B &lt; Y &lt; R|E) = \frac{5}{9} \cdot \frac{1}{6} = \frac{5}{54}$$  Is there another way to find the number of outcomes where $B &lt; Y &lt; R$ using the formula $N(E)/N(S)$, where $N(S) = 6 \cdot 6 \cdot 6 = 216$?
72323
Ok, I know this is wrong, but I want someone to tell me why.   Let's take a normal heads tails example of a fair coin. The probability of getting head = 1/2. And I write this is because, either it will be heads, or not. Hence two cases that makes it 1/2.  Now, I know I can't do this to all the cases. For example, the probability of getting a 2 when I roll a dice. I know the answer is 1/6 but why can't I do, either the outcome will be 2 or not. And in that case, my probability is 1/2.
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.
106836
If I roll the dice 50 times, how do I calculate the chance that I will roll 6 at least 5 times in a row? Why this problem is hard  With 5 tries this would be easy: take $(1/6)$ to fifth power. With 6 tries this is manageable; take the probability of rolling 6 the first five times, add the probability of rolling 6 the last five times, then subtract the overlap (all six results are 6). Given two overlapping sets of 5 rolls, the probability that one is all 6's is not independent from the probability that the other is all 6's. In principle this could be continued, but inclusion-exclusion gets out of hand. There has to be a better way; what is it?
20329
Suppose we flip a coin until we see a head. What is the expected value of the number of flips we will take?  I am pretty new to expected value, so I tried to evaluate it by multiplying the probability of each scenario with the number of flips it took to get there (like taking the arithmetic mean). This didn't work though, because of the infinite possibilities. I'm really confused, and if someone were to provide an answer, I would really appreciate it if they could go into detail.
115465
I was wondering about finding expected value of dice throw. I know that in case of single dice throw,the expected value is 3.5 = ((1+2+3+4+5+6)/6). So for purpose of betting the base price could be 3.5 to ensure no profit and no loss scenario. But how can I extend this logic for double dice throw or N dice throw. I though that for case of double dice out of the 36 cases, I could take the maximum value for each pair which would give 6 eleven times, 5 nine times, 4 seven times, 3 five times , 2 three times and 1 once. But this gave expected value to be 4.47, whereas the correct answer in book is 4.25. How can I determine the fair value in this case?
73766
I was recently commenting on game mechanics, when I realized that I actually underestimated this particular part.  Given an unbalanced die (in this case, some of the sides are 0), what is the expect number of rolls to reach a total sum of N.  As an example. What is the expected number of dice rolls with a die having sides [0,0,0,1,2,6] to reach the sum 24.  I can do this for a balanced die, because that is uniform distribution, so the expected number of rolls is ceil(N/((1+2+3+4+5+6)/6)) so in this example it would be ceil(24/3.5)=7 rolls.
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.