If 2 players toss a fair die, one tosses it 999 times and the other tosses it 1000 times, what is the probability that one gets more even numbers than the other? It's clearly comparing binomial distributions but I'm not sure how to do it.
108922
First player rolls a 6-sided die 100 times and the second player 101 times. What are the odds of the second player getting more odd numbers than the first one?  I tried solving this via listing the options which led to a nasty sum that I wasn't able to solve neither by hand nor by using wolfram alpha. Clearly there has to be a more clever solution using the properties of a binomial distribution, but so far I haven't come across anything useful.
51599
I found on this website  proof that the probability of two people in a room having the same birthday equates to 50% when when there are 23 people in the room  The only thing is, the explanation is too complicated for the average Middle School student to comprehend. So, my question is, is there any simplifies explanation of how this works? I tried Wikipedia but the explanations were even more complicated.  Thanks!
63948
Suppose three players play the following game: Player 1 picks a number in $[0,1]$. Then player $2$ picks a number in the same range but different from the number player $1$ picked. Player $3$ also picks a number in the same range but different from the previous two. We then pick a random number in $[0,1]$ uniformly randomly. Whoever has a number closer to the random number we picked wins the game. Assume all players play optimally with the goal of maximizing their probability of winning. If one of them has several optimal choices, they pick one of them at random.  1)If Player 1 chooses zero, what is the best choice for player 2?  2)What is the best choice for player 1?  I have some trouble seeing how this problem is well-defined. For instance, if Player $1$ picks $0$ and Player $2$ picks 1, then I cannot see what the optimal choice would be for the last player since he has to pick different numbers. Can someone help?  EDIT: I now understand better how the problem works, but I still have no idea how to approach this. Can someone give me some hints?
22292
Yesterday night, while I was trying to sleep, I found myself stuck with a simple statistics problem. Let's imagine we have a "magical coin", which is completely identical to a normal coin but for a thing: every time you toss the coin, the probability to get a head halves. So at t = 1 we have 50:50, then 25:75, then 12.5:87.5 and so on. At t = ∞ we are going to have 0:100.  My question is: if I toss this magical coin infinite times, can I say I am sure I am going to get at least one head? On one hand, I thought, the law of large numbers states that if an event is repeated infinite times, every state that is possible is bound to happen. On the other side, however, at t = ∞ the probability to get a head is zero.  Surely the solution of the problem is fairly easy, so what am I doing wrong?
103001
Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?  Since you need exactly two to be the same, there are three possibilities: 1. First and second, not third 2. First and third, not second 3. Second and third, not first  For 1) The first die, you have $\frac{6}{6}$. The second die needs to be equal to the first, so you have probability of $\frac{1}{6}$. Then the third die can't be equal to the first and second dice, so it's $\frac{5}{6}$.  All together you get $1 \cdot \frac{1}{6} \cdot \frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3 \cdot \left(1 \cdot\frac{1}{6} \cdot \frac{5}{6}\right)=\frac{5}{12}$$  Did I do this correctly? Thank you.
89448
A fair die is thrown 3 times. Find the probability  that the largest outcome of the three throws is a 3.    I understand that the problem isn't as simple as $ \frac{3^3}{6^3}=1/8$, being the answer. That would mean getting any of the values from $1-3$ for all of the throws is favorable, which includes series of throws that do not even have $3$ as an outcome.  What's the solution to ensure that there is at least one throw with $3$ as the outcome, with the $2$ other throws being anywhere from $1-3$?
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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.
107193
The dice are fair.  You have a $1\over6$ chance of getting the first number. A $1\over6$ chance of the second and so on.  Is it just $({1\over6})^3$ (1/216) or is that not accounting for the second and third roll properly?
91646
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 $t$.
112326
A pair of dice is rolled repeatedly until each outcome (2 through 12) has occurred at least once. What is the expected number of rolls necessary for this to occur?  Notes: This is not very deep conceptually, but because of the unequal probabilities for the outcomes, it seems that the calculations involved are terribly messy.  It must have been done already (dice have been studied for centuries!) but I can't find a discussion in any book, or on line. Can anybody give a reference?
36835
I have a d20 that seems to be, well, remarkably lucky.*  How can I determine whether it's really just luck, or whether the die is in fact unfairly biased?  *) Well, I don't, really.  This is actually a spin-off from , which is specifically about determining whether a die is loaded.  This one is intended as a more general question about how to detect any kind of bias in dice, since we apparently don't have one yet.  I've posted my own answer below, but feel free to add more.
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My problem is: suppose I have an urn containing balls of $n = 10000000$ (i.e., $10^7$) different colors, with $1000$ balls of each color (so the total number of balls is $1000n = 10^{10}$). Suppose I draw $100000000$ (i.e. $10^8 = 10n$) balls. My question is:  how do I calculate the probability that I have drawn at least 90% (in this case $1000000$ or $10^6$) of the different colors?  Many thanks in advance
124011
So a run of $x$ heads has the probability $0.5^x$.  Thefore a run of $9$ would be $0.5^9 = 0.001953125$.  But what is the chance of that run of $9$ occuring in $1000$ coinflips? Can anyone please explain how to work this out?
9103
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?
44908
Let's say that I want to calculate the probability of achieving more than 8 heads in 10 coin tosses. The easiest and most accurate way will be to simply use the binomial distribution and sum the probability of getting 9 an 10 heads.  However, it is also possible to simulate numerous coin tosses and then see which proportion of those answers our question. In which general scenario (doesn't have to be restricted to coin tosses) would this method of simulation be better than using the binomial and why?
14953
Suppose the winning combination consists of $7$ digits, each digit randomly ranging from $0$ to $9$. So the probability of $1111111$, $3141592$ and $8174249$ are the same. But $1111111$ seems (to me) far less likely to be the lucky number than $8174249$. Is my intuition simply wrong or is it correct in some sense?
66299
Consider a six-sided die with numbers from $1$ to $6$. Imagine you have a jar with $99$ of such dice. You throw all dice on the floor randomly. You look at one of the dice on the floor at a time. For each die, you do the following:   If it landed at an even number $(2,4,6)$, you turn the die so that it lands on the number $1$. If the die landed on an odd number $(1,3,5)$, you throw the die up in the air, so it can land on any number.   After you finish doing the above for all dice on the floor, you come back to the first die and repeat the entire process again. You keep on doing this until eternity (for a billion years, let’s say). If I come into the room after a billion years, how many dice on the floor will have even numbers up?