Task:     In box are three cards. First card is red on both sides. Second card is black on both sides. Third card is red on one side and black on other side. I random choose one card and I see that color on one side is red. What is the probability that second side is red?   My solution is $P=\frac{1}{2}$ because on other side should be black or red color. Is this correct?
103136
I have $3$ coins, $1$ coin has $2$ heads (HH), 1 coin has $2$ tails (TT), $1$ coin has $1$ head and $1$ tail (HT). I toss the coin, it fells on my hand, and the side i see is a tail. What's the chance that the other side is also a tail?  I got this as a teaser from a friend, possible from , as you can see he is insisting on 1/2 as not being the correct answer, I got 1/3 as my answer, am I right?
52717
Ok, game is as follow, with spanish cards (you can do it with poker cards using the As as a 1)  You shuffle, put the deck face bottom, and start turning the cards one by one, saying a number each time you turn a card around ---> 1, 2, 3; 1, 2, 3; etc. If when you say 1 a 1 comes out, you lose, same with 2 and 3. If you finish the deck without losing, you win.  I know some basics of probabilities, but is there a way to calculate the probability of winning the game, given a random shuffled deck?
23898
Edit: Sorry for not providing more on this question the first time. I tried to solve it. When choosing 13 cards from a deck of 52 cards, what is the probability of choosing precisely a card with a number 6,7,8 of the same suit? For example a 6,7,8 of heart  or a 6,7,8 of diamonds, or a 6,7,8 of spades, or a 6,7,8 of clubs. All three cards must be the numbers of 6,7,8 and only of a single suit. These must be among the hand of  13 cards. My first thoughts on this question that I only have 4 possibilities of this happening for the entire deck of 52 cards. So I thought $(_4 C_1)(_{13}C _{3})$ will account for the possible number of decks that will include cards of 6,7,8 of all the same suit out of 13 cards in the hand. And $(_{52} C_{13})$ gives me all the possible combinations of 13 cards from 52 cards.  So this probability that I got would be: $\frac{(_4 C_1)(_{13}C _{3})}{(_{52} C_{13})}\approx 5*10^{-7}$. However I feel something is off.
19374
Have been working on this problem for a while. No clue how to approach! Player A and Player B are playing a game with 3 possible results: win, tie and lose. Player A’s probabilities for each outcome are:  Win: 0.3  Tie: 0.2  Lose: 0.5   A win gives 2 points, tie gives each 1 point, and loss gives 0 points. The first of the two players to win 3 or more points than the other will win the match. What is the probability that Player A will eventually win? Thanks!
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.
8434
I have a very simple problem.  Lets assume that I have a well shuffled deck of 52 cards. I start drawing the top card always and when the card matches its rank I lose. J=11 Q=12 K=13.  If there were only 13 cards I could easily use the $\ \frac{!n}{n!}$ for derangements in order to solve this. The problem is that there are 52 cards so when I pass 13 I start from 1 again so I don't know what is the probability to win. Example of the game     1st card: 4 - Continue   2nd Card: A - continue   3rd Card: K - Continue   4th Card: K - Continue   5th Card: 6 - Continue   6th Card: 9 - Continue   7th Card: 10 - Continue   8th Card: A - Continue   9th Card: J - Continue   10th Card: 3 - Continue   11th Card: 2 - Continue   12th Card: 8 - Continue   13th Card: A - Continue   1st card: 5 - Continue   2nd Card 2 LOSE   So actually I have to count from 1 to 13 4 times and if I draw all 52 cards then I win. What's the probability?
73347
So for a bit of recreational spreadsheet creation I've been pondering probabilities.  Let's say I have two types of trial, A and B, with a 40% and 25% chance of success respectively. How can I calculate the probability, for instance, that I get at least two successes if I run trial A three times and B twice?  As I understand it, I can use a binomial distribution to calculate each one individually: 35.2% for at least two successes from A, or 6.25% from B. In addition, I would reckon the probability of getting at least one success in the combined set to be 87.85%, as I calculate a 12.15% chance of getting exactly zero successes on all five trials.  I'd greatly appreciate a little help working this out, because it's been bothering me for a couple of days.
79981
Bob rolls $3$ six-sided die and sums the numbers facing up. Bill rolls a single $20$ sided dice and records the number. What is the probability that Bob's number is greater than Bill's number.  I started the problem trying to come up with an equation, but that didn't work, so I resorted to creating $6$ six by six tables with all of the possible sums for Bob's die. Then, I counted the number of each number and created a chart and calculated the probability of each of those numbers occurring. Then, I multiplied the probability of each number occurring by the probability that Bill's number is greater than that number. Finally, I added them all up.  Obviously, this was very tedious and time-consuming. Is there a more elegant/less tedious way to do this problem.   PS: Is there away to do a $n$ die vs $m$ dice problem without listing them all out? Is there a general formula? Up until now, that's what I've been doing.
10845
Suppose there are $m$ red balls and $n$ blue balls in an urn. We randomly choose $p:m&lt;p&lt;n$ balls uniformly from the urn. What is the probability that exactly $q$ red balls are chosen?  Note:- Normally the answer would be $\frac{{m}\choose{q}}{{m+n}\choose{q}}$. However, since the number of balls that are chosen are provided, it is confusing me out. Any hints for the answer will also be appreciated.
36415
Question I know this is a duplicate of , i am still posting it because i want to know flaw in my approach.  Candidates were asked to come to an interview with $3$ pens each. Black, blue, green and red were the permitted pen colours that the candidate could bring. The probability that a candidate comes with all $3$ pens having the same colour is  $?$  My Approach Let $E=\text{chossing all three pen of same colour}=\binom{4}{3} \times 1 \times 1 \times 1=4$ i.e select which colour among red,blue,green and black and then there will be only $1$ choice Let $F=\text{possible way of selecting 3 pen}=4 \times 4 \times 4$ As there are $4$ ways to select colour of the pen and we have to choose for $3$ pens. Hence reqd probability$$=\frac{E}{F}$$ $$=\frac{4}{4 \times 4 \times 4}=\frac{1}{16}=0.0625$$ Where am i wrong? Please help!
72003
You have 6 red balls, 6 blue, and 6 white. You randomly select a sample of 3 balls without replacement. The balls, save for the color, are indistinguishable from each other. What are the odds that your sample contains at least a ball from each color?  I thought about finding the chance that the sample will not contain at least one color and subtracting that from one:  $$P(\text{sample is missing at least one color})=P(\text{sample is missing one color})+P(\text{sample is missing two colors}) $$  Essentially the second probability statement on the right hand side is the probability that all the balls in the sample are the same color. The case where the sample does not contain three colors doesn't make sense, so I didn't include it. $$=\binom{3}{1} \left[\left(\frac{12}{18}\right) \cdot \left(\frac{11}{17}\right) \cdot\left(\frac{10}{16}\right)\right]+\binom{3}{2}\left[\left(\frac{6}{18}\right) \cdot \left(\frac{5}{17}\right) \cdot\left(\frac{4}{16}\right)\right] \approx 0.882353$$  When all of this is done, we simply subtract the probability from one to get $0.117647$.  What am I doing wrong?
116033
Needs to be solved using conditional probability. I was thinking using the formula for total probability and having the hypothesis $H_i$ - they get to the $i$-th round and ($A$-they meet) $$P(A)=\sum P(A|H_i)P(H_i)$$ I reach a dilemma when trying to find these values, unsure whether to take into account which part of the tournament tree do they meet..  each player has an equal chance of winning.
26788
There are 50 numbered objects from 1 to 50. We extract 3 of these objects one after the other without reinsertion. Calculate the probability P(a&lt;b&lt;c)  Given 3 objects, they can be ordered in $3!$ different ways. Of all these ways only 1 satisfies a&lt;b&lt;c so the probability should be:  $P(a&lt;b&lt;c) = \frac{1}{3!} = \frac{1}{6}$  Is it solved in the right way?
114497
Calculate $$A={1\over {n\choose 0}}+ {1\over {n\choose 1}}+ ...+{1\over  {n\choose n}}$$  and  $$B={1\over {n\choose 0}}- {1\over {n\choose 1}}+ ...+{(-1)^n\over  {n\choose n}}$$     My idea for $A$ is some probabilistic reasoning. Color the sets $$\{\}, \{1\}, \{1,2\}, \{1,2,3\},...\{1,2,...,n\}$$ and ask our self what is the probability that I choose colored set among all sets. Clearly this is exactly ${n+1\over 2^n}$ and on the other side it is $A$:  probability that I take empty set is ${1\over {n\choose 0}}$,   probability that I take colored set with 1 element is ${1\over {n\choose 1}}$  probability that I take colored set with 2 elements is ${1\over {n\choose 2}}$  and so on ...   So $A ={n+1\over 2^n}$. But I have no idea how to attack $B$.
95398
I am looking for the probability of an outcome (x in Win, y in Opponent Actions and z in Opponent Cards) in a poker scenario.  I need a mathematical proof to simplify  $\Pr (x \mid y, z)$ to $\Pr (x \mid z)$
7355
I found this math "problem" on the internet, and I'm wondering if it has an answer:     Question: If you choose an answer to this question at random, what is the probability that you will be correct?      a. $25\%$      b. $50\%$      c. $0\%$      d. $25\%$   Does this question have a correct answer?
6239
Need clarification on the Draw 4 rule. - Player 1 lays down a Draw 4 Card and says the color is Red. - Player 2 does not have a Red card but has a Draw 4 Card. Can Player 2 lay down his Draw 4 Card and have Player 3 draw 8 cards?