If I flip a fair coin 10 times, and all of them land on heads, according to the Law of Large Numbers I am more likely to get tails on the next flip. However, this is clearly the Gambler's Fallacy.   How am I misinterpreting the Law of Large Numbers?
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I assume the following is true: assuming a fair coin, getting 10 heads in a row whilst tossing a coin does not increase the chance of the next coin toss being a tail, no matter what amount of probability and/or statistical jargon is tossed around (excuse the puns).   Assuming that is the case, my question is this: how the hell do I convince someone that is the case?   They are smart and educated but seem determined not to consider that I might be in the right on this (argument).
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I have little knowledge of probability and have recently been thinking of a math problem I am sure you guys could answer  If for example I tossed a coin ten times and got all heads, would the odds of the next coin toss beeing heads be 50 50 or s very low as the odds of getting 11 in a row is tiny?   Thanks
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First some context, I'm not a mathematician, not even close (as you will soon see) I do grasp some things about it but in a need to know basis, so plain english answers are appreciated (too).  I can't figure out this one even though some have tried very hard to explain/convince me.  So let's say I'm on a roulette (let's leave out the zero, or make it a coin for that matter).  The roulette hits 5 (or any amount of) times red on a row.  My question is why should't I bet to black (or better said why it is the same the one I pick).  I DO UNDERSTAND THE BASIC PRINCIPLE, that there is a material reality which makes the ball randomly fall on any of the two colors (which gives the options a 50-50 chance).  But, I also understand that if you "say": I bet red will come out six times on a row, you do have a very low probability of that happening so:  How is it that if you have seen the ball fall 5 times on a row on red, and you bet on red, you are NOT betting on 6 times on a row on red (that does have lower probability).  Thanks in advance!    I hope I was clear enough with my question, if not, please ask for any clarifications needed.  Sorry for the VERY plain English, feel free to modify or suggest a change to anything that may be misleading.
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A coin is tossed $n$ times. What is the probability of getting odd number of heads?   I started this chapter sometimes ago and faced in front of a tough problem. At first I started considering cases. Case-I : The probability of getting 1 head.Case-II : The probability of getting 3 head and so on. But there are many cases. So how can I solved this . Please help me. Thank you!
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You throw a fair $k$-sided dice repeatedly and keep track of all the results. You stop if you have seen every face exactly the same number of times. What is the expected number of times you have to throw the dice? Is that number even finite?  Start with the simplest example of a $2$-sided dice, ie a fair coin. There is a 50% chance that after 2 flips you will have one head and one tail so you stop. There is a 12.5% chance that after 2 flips you will be at 2:0 and continue but after 4 flips you will be at 2:2, so you stop. One can compute a few more terms but I don't see enough of a pattern to get a summable series.  It is not too hard to compute the probability that after $k\cdot m$ throws of a $k$-sided dice you get every face exactly $m$ times but these probabilities are not independent for different $m$ so I'm not sure whether that is helpful.  The relation to random walks looks useful. One can map the results of throwing a coin repeatedly to a random work on the integers. Seeing the same number of heads and tails is equivalent to coming back to the starting point (at zero). This is a well studied problem and it is known that you come back to zero with probability one in a finite time. The number I'm looking for would be the expected time when you get back to zero for the first time but I haven't found any results on that.  For $k$ bigger than $2$ (and even) the mapping to random walks is not bijective anymore. If you map the outcome of a $6$-sided dice throw to a random walk on the $3$-dimensional integer lattice with each face corresponding to a step in one of the 6 directions that a return to the origin is necessary but not sufficient for having seen each face the same number of times. However, it is known that for a random walk on the integer lattice in dimension of at least $3$, the chance to return to the origin in finite time is strictly smaller than $1$. So this should prove that for a dice with at least $6$ faces, the expected number of throws is not finite.  Is this a finite number for $k=2$ and $k$ between $3$ and $5$? Are there estimates for it?
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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?
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Say Bob tosses his $n+1$ fair coins and Alice tosses her $n$ fair coins. Lets assume independent coin tosses. Now after all the $2n+1$ coin tosses one wants to know the probability that Bob has gotten more heads than Alice.    The way I thought of it is this : if Bob gets $0$ heads then there is no way he can get more heads than Alice. Otherwise the number of heads Bob can get which allows him to win is anything in the set $\{1,2,\dots,n+1\}$. And if Bob gets $x$ heads then the number of heads that Alice can get is anything in the set $\{0,1,2,..,x-1\}$. So\begin{align}P(\text{Bob gets more heads than Alice})&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} P( \text{Bob gets x heads }\cap \text{Alice gets y heads }) \\[0.2cm]&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \left(C^{n+1}_x \frac{1}{2}^{x} \frac{1}{2}^{n+1-x}\right)\left( C^n_y \frac{1}{2}^y \frac {1}{2}^{n-y}\right)\\[0.2cm]&amp; = \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \frac{C^{n+1}_x C^n_y}{2^{2n+1}}\end{align}   How does one simplify this?      Apparently the answer is $\frac{1}{2}$ by an argument which looks like this, Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2.
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Part of a series of trying to understand Bayesian vs frequentist:         I think I get the difference of how Bayesians and frequentists approach , but I'm not quite sure if or how that is supposed to explain to me how they view probability.  From what I understand, according to , a frequentist "defines" probability as follows:  Given probability space $(\Omega, \mathscr{F}, \mathbb{P})$, $\forall A \in \mathscr{F}$, $\mathbb{P}(A) \approx \frac{n_A}{n_t}$, where $n_t$ is the number of trials conducted and $n_A$ is the number of times A has occurred in those trials.  Furthermore, $\mathbb{P}(A) = \lim_{n_t \to \infty} \frac{n_A}{n_t}$.  Okay, so how do Bayesians define probability? The above seems to be one approach to computing probability of an event in addition to defining a probability.  Bayesians seems to assume a prior probability, conduct some trials and then update their probability, but that doesn't really seem to explain how they define what probability is.   says 'Bayesian probability is a quantity that we assign for the purpose of representing a state of knowledge, or a state of belief.'  What exactly does that mean? Is state synonymous to degree? For example, Walter's state of belief that a particular coin is fair is represented with the number 0.1 while Jesse's state of belief that the same coin is fair is represented with the number 0.2. Given new information, Walter's state of belief could become 0.96 while Jesse's state of belief could become 0.03. So, initially, Walter was less inclined to believe the coin is fair, but later on Jesse was more inclined to believe the coin is fair?  I'm hoping for something in terms of  like the frequentist one above.  Same Wiki page says 'The Bayesian interpretation of probability can be seen as an extension of propositional logic that enables reasoning with hypotheses, i.e., the propositions whose truth or falsity is uncertain.', it seems like Bayesian and frequentist probability are analogous to fuzzy and Boolean logic, respectively.
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For each positive even integer $n$, set   $$P_n = \displaystyle {{n} \choose {\frac{n}{2}}}\frac{1}{2^n}.$$  Show that $\displaystyle \lim_{n \to \infty} P_n$ exists and determine its value.   Here's what I have so far: each $P_n$ can be thought of as the probability of tossing a fair coin $n$ times, and obtaining exactly $\frac{n}{2}$ heads. My expectation is that $P_n \to 1$, since in any large trial, one would expect to record just about as many heads as tails. But I'm not sure how to mathematically justify this hunch.  Hints or solutions are greatly appreciated.
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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?
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There is an unfair coin. It tends to land on one side more than on the other. It is unknown which side is it.  There is Mr. A and Mr. B. They argue about something and they want to use that coin to decide who is right.  Is there any technique of using that unfair coin to get fair result?  My solution: I can think of only that there is going to be 2 rounds. First time Mr. A chooses HEAD, than Mr. B chooses HEAD if any of them wins twice they win.
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I was watching the movie 21 yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the wrong answer) which revolves around theoretical probability.  The question goes a little something like this (I'm paraphrasing, but the numbers are all exact):  You're on a game show, and you're given three doors. Behind one of the doors is a brand new car, behind the other two are donkeys. With each door you have a $1/3$ chance of winning. Which door would you pick?  The character picks A, as the odds are all equally in his favor.  The teacher then opens door C, revealing a donkey to be behind there, and asks him if he would like to change his choice. At this point he also explains that most people change their choices out of fear; paranoia; emotion and such.  The character does change his answer to B, but because (according to the movie), the odds are now in favor of door B with a $1/3$ chance of winning if door A is picked and $2/3$ if door B is picked.  What I don't understand is how removing the final door increases the odds of winning if door B is picked only. Surely the split should be 50/50 now, as removal of the final door tells you nothing about the first two?  I assume that I'm wrong; as I'd really like to think that they wouldn't make a movie that's so mathematically incorrect, but I just can't seem to understand why this is the case.  So, if anyone could tell me whether I'm right; or if not explain why, I would be extremely grateful.
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If you were to flip a coin 150 times, what is the probability that it would land tails 7 times in a row?  How about 6 times in a row?  Is there some forumula that can calculate this probability?
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I was thinking of this problem.  I believe the solution and I think I understand it, but if I take the following approach I'm completely confused. Problem 1:  I will offer you the following game. You pay me \$10 and I will flip a fair coin. Heads I give you \$5 and Tails I give you \$20. The expectation is \$12.5 so you will always play the game.  Problem 2:  I will give you an envelope with \$10, the envelope is open and you can check. I then show you another envelope, closed this time and tell you: This envelope either has \$5 or $20 in it with equal probability. Do you want to swap? I feel this is exactly the same as problem 1, you forgo \$10 for a \$5 or a \$20, so again you will always switch.  Problem 3:  I do the same as above but close the envelopes. So you don't know there are $10 but some amount X. I tell you the other envelope has double or half. Now if you follow the same logic you want  to switch. This is the envelope paradox.  What changed when I closed the envelope?? EDIT: Some have argued that problem 3 is not the envelope problem and I'm going to try and provide below why I think it is by analysing how each views the game. Also, it gives a better set up for the game. Providing some clarification for Problem 3: From the perspective of the person organising the game:  I hold 2 envelopes. In one I put \$10 close it and give it to the player. I then tell him, I have one more envelope that has either double or half the amount of the envelope I just gave you. Do you want to switch? I then proceed to flip a fair coin and Heads I put \$5 in and Tails I put \$20. And hand him the envelope. I then ask him. The envelope you just gave me has twice or half the amount of the envelope you are holding. Do you want to switch?  From the perspective of the player:  I am given an envelope and told there is another envelope that has double or half the amount with equal probability. Do I want to switch. I think sure I have $X$, hence $\frac{1}{2}(\frac{1}{2}X + 2X) &gt; X$ so I want to switch. I get the envelope and all of a sudden I am facing the exact same situation. I want to switch again as the other envelope has either double or half the amount.
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I am told that a fair coin is flipped $2n$ times and I have to find the probability that it comes up heads more often that it comes up tails.     Please, how do I find the required probability?
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I'm trying to figure out what is the probability with 3 D10's if I were to roll them one after another."001" is one and "000" is a thousand. Would every number have the same probability or doubles and triples be less common?  And please explain.
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I have a pretty basic probability question, but I'm just having difficulties remembering what distribution this is.  A coin is tossed until a head appears two times in a row. Given that we are using a fair coin, what is the probability that we toss the coin exactly 4 times such that the two consecutive heads are the 3rd and 4th trials?  I know that this would be fairly easy to solve by brute force and listing out the different probabilities, but I'm a little confused as to how to use a "shortcut" such as identifying this as a binomial distribution. I also don't know how to account for the probability that we are using 4 tosses.  Any help is appreciated. Thanks!