A fair coin is flipped repeatedly until it turns up tails five times. What is the expected number of heads before that happens? Based on the link given in the comment, I have found it can be solved using the recursion $E(n)=\frac{1}{2}(E(n)+1)+\frac{1}{2}(E(n-1))$ which is equivalent to $E(n)=E(n-1)+1$. Is it correct?
64740
A fair coin is tossed repeatedly until 5 consecutive heads occurs.   What is the expected number of coin tosses?
116460
A fair coin will be tossed until a heads results. You will then be paid $2^{n-1}$ dollars where $n$ equals the number of flips. Now why is the expected pay out infinite?  $$ \sum_{n \geq 1} (\frac{1}{2^n})2^{n-1} = \sum_{n \geq 1} \frac{2^{n-1}}{2^n}  $$ Why does this give the payout? The payout shouldn't be infinite, it should be a potential infinity.   For example, if I flip the coin once and it is a tails, then I will receive two dollars for certain. On the second flip there is a $25$ percent chance that I will receive a minimum of $4$ dollars. but a $75$ percent chance that I will receive only two dollars. And so on as the chances of receiving more money decreases and eventually approaches $0$. Why then is it suggested that an investor pay any amount of money for such an opportunity?           Source: A Brief History of the Paradox, Philosophy and Labyrinths of the Mind     Obviously this appears false by intuition, but I'd like to know the argument for the paradox; why one would pay a trillion dollars to enter such a game. The argument just seems far too fallacious to me.   How many times would one have to play? After almost twenty thousand plays, we are still making less than 8 dollars. IF every twenty thousand plays we increased our winnings by 8 dollars, we would have to play hundreds of billions of times just to start to have an average earnings equal to one trillion... we'd run out of time in the universe it seems...    Edit I am curious,  however,  if the overall slope of the simulation data would remain consistent,  if so why? If not, why? What accounts for the vertical jumps we see occurring less and less often? Why do they occur at all?
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We have a random walk which starts in state $0$. At each step, a coin is tossed with probability of heads: $P(H)=p$. If we get a heads, we go to the next higher integer state and on tails, we go to the next lower integer state (so state $n$ would go to $n+1$ on heads and $n-1$ on tails).   Now, I want to know the probability that we will reach state $k$ for the first time after exactly $n$ tosses of the coin. Turning out to be surprisingly hard for me.    Here is my attempt:  I define $a_n^{k}$ as the probability described above and $c_n^k$ as the probability that the random walk will be in state $k$ at toss $n$ (regardless of if it was there in a previous toss).  It's clear that we need $\left(\frac{n-k}{2}\right)$ tails and $\left(\frac{n+k}{2}\right)$ heads. So if $n-k$ is not even, the sequences for those $n$'s become $0$.   To get $a_n^k$, we need to identify all sequences where the cumulative number of heads stays less than $k$ + the cumulative number of tails for all tosses leading upto $n$. This is not so easy to solve.  On the other hand, I got an expression for $c_n^k$ and hoped I could use it to get $a_n^k$. I reasoned that the probability the walk reaches $k$ for the first time on toss $n$ is the probability it is in state $k$ on toss $n$ subtracted by the probabilities that it was in state $k$ in any previous toss. So,  $$a_n^k = c_n^k - \sum_{i=1}^{n-1} c_i^k$$  But this can't be right since this expression will become negative for many values of $n$.
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It is possible to choose between three equally desirable outcomes by tossing a fair coin as follows:  Choose option 1 if the first head appears on an even toss  Choose option 2 if the first tail appears on an even toss  Choose option 3 if the first head and the first tail both appear on an odd toss  Note that option 1 and option 2 are mutually exclusive, because we must have a head or a tail on the first toss.  This works because the binary expansion $$\frac 13 =0.010101010101 \dots$$ has a $1$ in all the even places. Or equivalently because $$\frac 13=\frac 14+\frac1{16}+ \dots +\frac 1{4^n}+\dots$$  This is quite a simple procedure (though perhaps not as well known as it might be). I was pondering in the bath what might be the most convenient set of options if there were five or seven possibilities.  For example I can pick off two fifths using the binary expansion of $\frac 15$ with heads and tails, and use my procedure for thirds to do the remaining three fifths - so it is possible. But can one do fifths or sevenths without resort to this kind of recursive procedure?
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When you flip a coin four times, what is the probability that it will come up heads    exactly twice?      My calculation:    we have $2$ results for one flip : up or down  so flip $4$ times, we have $4\cdot2 = 8$ results total    Thus the probability is: $2/ 8 = 0.25 $ but the correct answer is $0.375$. Can anyone explain why I'm wrong?
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Context: I need to solve one variation of the . Problem: Two players ($A$ and $B$) throw a coin until one of the victory sequences appears. For player $A$ the victory sequence is $HTT$, for player $B$ the victory sequence is $TTH$. What is the probability that $A$ wins and what is the expected number of throws given that $A$ won? My attempt: Intuitively, it seems that player $B$ is more likely to win. Basically, I reliase that we can use the Markov-Chains, but my idea is only draw a binary tree and consider the sequences on the n-th step. My another idea was to calculate the expected number of throws to get, say, $HH$ that would be equal to $r_{HH}$: $$r_{HH} =  \frac{1}{4}E(number|HH) + \frac{1}{4}E(number|HH)+ \frac{1}{4}E(number|TT)+ \frac{1}{4}E(number|TH) = \frac{1}{4}(9 + \frac{5}{2} r_{HH})$$  And get $r_{HH}$. My issues: I don't really understand how to join these ideas to get a proper solution.
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I want to find the expected number of coin tosses to get $N$ heads in a row, where $p$ is the probability of getting a head in a single toss.    Let $F(N)$ be the expected number of tosses to get $N$ heads consecutively, so    $$F(N) = 1 + p F(N-1) + (1-p) F(N)$$   which gives   $$F(N) = 1/p + F(N-1)$$  With base condition : $F(1) = 1 + 1/p$   My logic is as follows:     if we get a head, in current toss we need to get N-1 more heads consecutively, but if we get a tail, we have to start over   This is what I thought, but is not correct. Can you please help me?
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A two-sided coin has just been minted with two different sides (heads and tails). It has never been flipped before. Basic understanding of probability suggests that the probability of flipping heads is .5 and tails is .5. Unexpectedly, you flip the coin a very large number of times and it always lands on heads. Is the probability of flipping heads/tails still .5 each? Or has it changed in favor of tails because the probability should tend to .5 heads and .5 tails as you approach an infinite number of trials?  I understand that flipping coins is generally a stochastic process, but does that change at all if you see a large number of trials bias to one side?
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Assume a coin has a probability p to get a head H. Suppose a coin is tossed until the partern T,T appear in the last 2 tosses. Once he got T,T then the game is finished. What is the expected number of flips is expected for the game?  I tried to find out the distribution of $N$ where $N$ stands for the expected value of the no. of tosses before getting T,T, but i don't know what kind of distribution is that, seems need another apporaches
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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.
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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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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?
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this is the scenario.  roll one dice 6 times and note the result of each roll.  lets say the results are the following. 3, 3, 6, 2, 1, 4.   this will become the standard sequence.  begin to roll the dice again.  As soon as the next 3 is rolled, then what is the probability that the next 5 rolls will be exactly has follows.  3, 6, 2,  1, 4 ? The remainder of the standard sequence.
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I was working on problems on expectation and found this one as a question from a well-known exam  Assume that you are flipping a fair coin, i.e. probability of heads or tails is equal. Then the expected number of coin flips required to obtain two consecutive heads for the first time is.  (A)4  (B)3  (C)6  (D)10  (E)5  I worked up like Let N be the number of tosses required untill 2 consecutive heads are obtained for the first time.  Let the probability of obtaining heads be p and tails be q and for a coin p=q=1-p=1-q.  For N=2, the probability will be $p^2$  For N=3, the probability will be $p^2.q$  For N=4, the probability will be $q^2.p^2$  For N=5, the probability will be $q^3.p^2$  and so on...  Now, $E[N]=2(p^2)+3(p^2.q)+4(q^2.p^2)+5(q^3.p^2)+......$  and this comes out to be $\frac{3}{2}$ but it matches with none of the options.  Please tell me where I am wrong and what should I do to reach the correct solution?
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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?
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A fair coin is tossed $10$ times. What is the probability that no two   consecutive tosses are heads?   Possibilities are (dont mind the number of terms):  $H TTTTTTH$, $HTHTHTHTHTHTHT$.  But except for those,  let $y(n)$ be the number of sequences that start with $T$  $T _$, there are two options, $T$ and $H$ so,  $y(n) = y(n - 1) + x(n-1) = y(n - 1) + y(n - 2)$  Let $x(n)$ be the number of sequences that start with $H$,   $H _$ the next option is only $T$ hence,  $x(n) = y(n - 1)$   The total number of sequences $F(n)$ is:  $$F(n) = y(n) + x(n) = 2y(n - 1) + y(n - 2)$$  We are after $F(10)$,  $$F(10) = 2y(9) + y(8)$$  $$F(3) = 2y(2) + y(1) = 1 + 4 = 5$$  $$F(4) = 2y(3) + y(2) = 6 + 2 = 8$$  But I'm not quite sure where this will take me though.
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If you toss a coin $n$ times, there are $2^n$ possible sequences of heads and tails. Let $E_n$ be the set of sequences which do NOT contain two consecutive heads and $e_n$, the number of sequences in $E_n$.  Thus $E_3$ $=$ $\{$ $TTT$, $TTH$, $THT$, $HTT$, $HTH$ $\}$ and $e_3$ $=$ $3$.  How many elements of $E_n$ have $T$ as the first toss?  Please can anyone help me out here?