Let's say we bet on 10 different soccer matches. The probability of our team winning, for each game, is 0.17, 0.78, 0.41, 0.79, 0.47, 0.08, 0.29, 0.30, 0.35, and 0.54. What is the probability of winning at least 3 of these bets?  I know how to deal with this problem when the probability of each event is the same (coin tosses etc), but get confused when they are different.
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I am interested in how uncertainty can be accounted for when considering the risk of extinction of a species. Forgive me for extending a rather tired thought experiment, but at least it's familiar territory and I hope it illustrates what I am trying to understand.  Let's say that Schrödinger was not satisfied with killing and not killing only one cat, so he went out and collected the last 15 remaining Himalayan Snow-Cats. He put each one in a box with a vial of poison, hammer, and a trigger device for releasing the hammer. Each trigger device has a known probability of releasing the hammer within any given hour, and the poison will take 5 hours to kill a cat (things got a bit trippy when he used radioactive decay, so this time he steers clear of quantum mechanics). After one hour, Schrödinger receives a notice from the ethics board telling him he's nuts and ordering him to release the cats immediately. He pushes a button which opens a cat-flap at the back of each box, thus releasing them all back into the wild. By the time the humane society check the boxes, all the cats have bolted. Before anyone can restrain him, Schrödinger detonates the remaining triggers, so nobody knows how many of the cats were poisoned.  The probability that each of the cats is poisoned is as follows:   0.17 0.46 0.62 0.08 0.40 0.76 0.03 0.47 0.53 0.32 0.21 0.85 0.31 0.38 0.69   After 5 hours have elapsed and all poisoned cats have died, what is the probability that there are:   (A) More than 10 Himalayan Snow-Cats still alive  (B) 10 or less still alive (C) 5 or less still alive (D) 2 or less still alive
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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.
107751
Two players $A$ and $B$ are playing the final of chess championship which contains a series of matches.   The probability that $A$ wins is ${2\over 3}$.  The probability that $B$ wins is ${1\over 3}$.  The winner will be the one who is  ahead by 2 games as compared to the other and wins at least 6 games.  If the player $B$ wins the first 4 matches,  find the probability that $A$ wins the championship.  My attempt:  We should find the following probabilities:  $A$ wins 6 matches given $B$ has won the first 4. $A$ wins 7 matches given $B$ has won the  first 4 and wins one out of the next 8.  But I am not getting any patterns.
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I'm struggling to find where I made a mistake on the way to solving the following problem.  Problem description: A grocery sells chocolates bars. There are $5$ kinds of stickers. Each chocolate is sold with $1$ sticker of one of these $5$ kinds. The probability to find any kind of stickers in any chocolate bar is the same.   What's the probability of collecting all $5$ types of stickers if you buy $7$ chocolates at once?  My solution: since the order of chocolate bars that I've bought is irrelevant, the total number of 7-chocolate bar sets that I can buy equals   $$\left(\binom{5}{7}\right)=\binom{7+5-1}{7}=\binom{11}{7}=330$$  Now I'm solving the reverse problem: let's calculate the probability of failing to collect all $5$ kinds of stickers. To do that I need to calculate the number of 7-chocolate bar sets that have less than $5$ kinds of stickers, which means that I need sets with only $1$ kind of stickers, $2$, $3$ and $4$. Since, again, the order of chocolate bars doesn't matter, the number of such sets is   $$\left(\binom{4}{7}\right)=\binom{7+4-1}{7}=\binom{10}{7}=120$$  Finally, my answer to the initial problem should be  $P = 1 - \frac{120}{330}= 0.6363636364$  The answer key says it's $\approx 0.215$  Where's the flaw in my solution?  I appreciate any help.
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I have a bet with a co-worker that out of 50 ping pong games (first to win 21 points, win by 2), I will win all 50. So far we've played 15 games and on average I win 58% of the points, plus I've won all the games so far. So we're wondering if I have a 58% chance of winning a point and he has a 42% chance of winning a point, what's the percent chance that I would win the game? Is there a formula that we can plug in difference % chances?  We've googled all over and even asked the data scientists at our company but couldn't find a straight answer.  Edit: Wow, I am blown away by the thoroughness of responses. Thank you all so much!!! In case people are curious, I have an update to how my bet is going: I've now won 18 out of 50 games, so I need to win 32 more games. I've won 58.7% of all points and my opponent has therefore won 41.3% of points. The standard deviation for my opponent is 3.52, his average score is 14.83, and his median score is 15.50. Below is a screenshot of the score of each game so far. I can keep updating as the bet goes on, if people are interested.  Edit #2: Unfortunately we've only been able to play a few more games, below are the results. I'm just going to keep replacing the picture so I don't have a bunch of screenshots of the score.  Final Update: I finally lost to my co-worker on game #28. He beat me 21-13. Thanks for all of your help!
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I'd like to ask for help with a task that I've been given.  The task goes like this: In a poker game (52 cards), 5 players are given 5 cards each. What is the probability that both the first and the third player will get a full-house?  It's apparent that the probability of the first player getting a full house is  $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}}$  Then the second player gets 5 random cards, so $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5}$  Now the third player is supposed to get a full house, I'm assuming it should look like this: $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5} * \frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{42}{5}}$  The fourth and the fifth player both receive 5 random cards, so the final equation should be:$\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5} * \frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{42}{5}} * \binom{37}{5}* \binom{32}{5}$  This is how I would solve the task, but I'm not sure that my solution for the third player is correct. Could anyone more experienced than me have a look at my solution?  Also I'm sorry if there is a solution for this type of task on this site, I tried looking for it and I found a lot of poker probability questions, most of them were however focused on only one player and not multiple players, which is my case.
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Suppose that we have the situation as depicted in the figure: a random experiment which has 4 outcomes $x_1, ..., x_4$ and two events $A$ and $B$. Also assume that $P(x_i)=0.25$.  Now, since $P(A \cap B) = P(A)P(B) = 0.25$, by definition the events $A$ and $B$ are independent. This does not make sense to me. Why are they independent? Is there an intuitive explanation?  Another thing is, suppose that there is a 5th outcome, $x_5$, outside of $A \cup B$, in that case the events $A$ and $B$ are no longer independent since  $P(A \cap B) \ne P(A)P(B)$. This result also does not make sense to me.
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One can calculate the probability of a correct score of a football match by Poisson(Number of Team1 goals, Mean average Team1 goals) x Poisson(Number of Team2 goals, Mean average Team2 goals)  So for 0:0 scoreline, Poisson(0, 1.05) x Poisson(0, 2.5)  if mean Team1 is 1.05 and mean Team2 is 2.5.   But how does one calculate the probability/odds of "home team to win by exactly one goal"? Adding double Poissons for each 2:1, 1:0, 3:2, etc scorelines might be one of the methods but it will be a tedious task and I am trying to avoid Excel for the time being to find the simplest method to calculate this. How do you involve the other team in your calculations?   Was hoping an easier, simpler way was there (something like calculating P=0 and then 1-P(0) to find odds of P>=1).
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Probability with Martingales        This was answered  using bounds and  using PGF. I would like to try inclusion-exclusion.  Let $H_1, H_2, ...$ be independent events with $P(H_k) = p$ where $H_k$ means the kth toss is heads.  It seems that ($\cap$ omitted)  $$A_0 = H_1$$  $$A_1 = H_2 \cup H_3$$  $$A_2 = (H_4H_5) \cup (H_5H_6) \cup (H_6H_7) := \bigcup_{i=1}^{3} E_{2,i}$$  $$A_3 = (H_8H_9H_{10}) \cup ... \cup (H_{13}H_{14}H_{15}) := \bigcup_{i=1}^{6} E_{3,i}$$  $$\vdots$$  $$A_k = (H_{2^k}...H_{2^k + k-1}) \cup ... \cup (H_{2^{k+1} - k}...H_{2^{k+1} - 1}) := \bigcup_{i=1}^{2^k - k + 1} E_{k,i}$$  So by inclusion-exclusion, we have  $$P(A_0) = p$$  $$P(A_1) = p + p - p^2$$  $$P(A_2) = 3p^2 - 2p^3$$  $$P(A_3) = \ldots$$  It gets tricky starting $A_3$  $$P(A_3) = P\left(\bigcup_{i=1}^{6} E_{3,i}\right)$$  Attacking that directly is going to lead to problems.  What about this?  $$P\left(\bigcup_{i=1}^{6} E_{3,i}\right)$$  $$ = P\left(\bigcup_{i=1,4} E_{3,i} \cup \bigcup_{i=2,5} E_{3,i} \cup \bigcup_{i=3,6} E_{3,i} \right)$$  $$ := P(F_{3,1} \cup F_{3,2} \cup F_{3,3})$$  Each $F$ is a union of disjoint sets.   Also, I notice independence isn't being used. I guess the $H_k$'s and the $A_k$'s are independent. What about some subset of the the $E$'s? The $F$'s?
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Given that I have a fair coin which I toss three times, how do I:  a) Find three events $A$ , $B$, and $C$ such that no two are independent but $P( A  \cap B \cap C) = P( A ) P( B) P( C)$ ?  b) Find three events $A$ , $B$, and $C$ such that every two are independent but all three are not independent: $P( A  \cap B \cap C) \not= P( A ) P( B) P( C)$ ?  How would I approach a problem like this ? I started by drawing a binary tree that ended up with 8 leaves but I have no idea how to use it to get the answers. Any help would be appreciated.
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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?
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If A and B play a game with a fair coin such that A wins by getting consecutive TT for the first time, while B wins by getting a consecutive TH.  Who has higher probability of winning the game?   I tried the following method but not sure:  For A to win,     TT, HTT, HHTT ... that is, 2-k for each game length k   For B to win,     TH, HTH, HHTH ... that is, 2-k for each game length k   So ultimately I am getting same probability of each to win.
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Problem: Players A,B and C toss a fair coin in order, the first to throw a head wins what are their respective chances of winning?  Attempt: Let X = event that A throws a head on the first toss, and Y = event that B throws a head on first toss, similarly let Z = event that C throws a head on first toss.  Then there are eight different triple for tossing a coin.  S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}   where H is a head, and T is a tail.  Can anyone please help me? I don't know how to continue. Any feedback/help would be really appreciated. Thank you.
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I need some clarification for why the probability of the union of three events is equal to the right side in the following:  $$P(E\cup F\cup G)=P(E)+P(F)+P(G)-P(E\cap F)-P(E\cap G)-P(F\cap G)+P(E\cap F\cap G)$$  What I don't understand is, why is the last term(intersection of all) added back just once, when it was subtracted three times as it appears from a Venn Diagram?  Here on page 3, this is explained but not in enough details that I can understand it:
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I'm working on some problems from Pitman's Probability.  Let events $A$ and $B$ be independent, with $P(A)=0.1$ and $P(B)=0.3$. Find: $(a)$ Neither of the events occur $(b)$ At least one of the events occur $(c)$ Exactly one of the events occur  Attempt: $(a)$ If neither of the events occur, this is the complement of both events occurring. Hence $1-P(A\cap B)=1-P(A)P(B)=1-\frac{1}{30}=\frac{29}{30}$ $(b)$ At least one of the events can be represented as the union of the two events. Hence $P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)=\frac{4}{10}-\frac{1}{30}=\frac{11}{30}.$ $(c)$ Not sure what this would be. Maybe $P(A^c\;\cap\;B)+P(A\cap B)$?
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...2 to 5 questions answered correctly, out of 20 of them? Each question has 5 choices. Probability of getting one right is 1/5. Probability of getting exactly 1 right is ${20 \choose 1} p^1  q^{19}$, with $p=P(\mathrm{right})$ and $q=P(\mathrm{wrong})$  (which I managed to understand and calculate). However how do I calculate for the problem above?
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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.