A person randomly chooses a coin with probability 0.5: Find which coin he chose according to the outcome of 3 tosses: Tossing coin 1 yields a head with a probability P(X1 = H) = .3 (and tail with P(X1 = T) = .7). Tossing coin 2 yields a head with a probability P(X2 = H) = .6 (and tail with P(X2 = T) = .4). You earn $1 if you correctly guessed the coin and $0 otherwise. Design the optimum decision rule and estimate your average earning.  Hello folks, I have this problem in hand and I want to understand the following :   Are P(X1)=P(X2)=0.5?  P(H|X1)=0.3 , P(H|X2)=0.7? How does the Decision Table look like in this problem?  How to find the average earning?
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A person randomly chooses a coin with probability 0.5: Tossing coin 1 yields a head with a probability P(X1=H)=.3 (and tail with P(X1=T)=.7). Tossing coin 2 yields a head with a probability P(X2=H)=.6 (and tail with P(X2=T)=.4). You earn 1 dollar if you correctly guessed the coin and 0 dollars otherwise. Design the optimum decision rule and estimate your average earning.  Can anybody explain me ,how should i go about solving this problem?
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When flipping a coin to make important decisions in life you can flip once to choose between 2 possible outcomes. (Heads I eat cake, Tails I eat chocolate!)  You can also flip twice to choose between 4 outcomes. (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads)  Can you use a coin to choose evenly between three possible choices? If so how?  (Ignoring slight abnormalities in weight)
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You have a six-sided die. You keep a cumulative total of your dice rolls. (E.g. if you roll a 3, then a 5, then a 2, your cumulative total is 10.) If your cumulative total is ever equal to a perfect square, then you lose, and you go home with nothing. Otherwise, you can choose to go home with a payout of your cumulative total, or to roll the die again.  My question is about the optimal strategy for this game. In particular, this means that I am looking for an answer to this question: if my cumulative total is $n$, do I choose to roll or not to roll in order to maximize my cumulative total? Is there some integer $N$ after which the answer to this question is always to roll?  I think that there is such an integer, and I conjecture that this integer is $4$. My reasoning is that the square numbers become sufficiently sparse for the expected value to always be in increased by rolling the die again.  As an example, suppose your cumulative total is $35$. Rolling a $1$ and hitting 36 means we go home with nothing, so the expected value of rolling once is:  $$E(Roll|35) = \frac 0 6 + \frac {37} 6 + \frac {38} 6 + \frac{39} 6 + \frac {40} {6} + \frac{41}{6} = 32.5$$  i.e.  $$E(Roll|35) = \frac 1 6 \cdot (37 + 38 + 39 + 40 + 41) = 32.5$$  But the next square after $35$ is $49$. So in the event that we don't roll a $36$, we get to keep rolling the die at no risk as long as the cumulative total is less than $42$. For the sake of simplification, let's say that if we roll and don't hit $36$, then we will roll once more. That die-roll has an expected value of $3.5$. This means the expected value of rolling on $35$ is:  $$E(Roll|35) = \frac 1 6 \cdot (40.5 + 41.5 + 42.5 + 43.5 + 44.5) = 35.42$$  And since $35.42 &gt; 35$, the profit-maximizing choice is to roll again. And this strategy can be applied for every total. I don't see when this would cease to be the reasonable move, though I haven't attempted to verify it computationally. I intuitively think about this in terms of diverging sequences.  I recently had this question in a job interview, and thought it was quite interesting. (And counter-intuitive, since this profit-maximizing strategy invariably results in going home with nothing.)
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Here is the problem setup:  Two people play a game of throwing balls at a pin. Sally hits the pin 70% of the times she throws, and Bill hits the pin 40% of the times he throws. The game is won by the person who hits the target first. If Bill throws first each time, find the probability he will win.  With simulation, I get Bill's probability of winning to be around 0.52 (rounding to two decimal places). How to solve this problem analytically?
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An interview problem is like the following:     Given a coin you don’t know it’s fair or unfair. Throw it 6 times and    get 1 tail and 5 head. Determine whether it’s fair or not. What’s your   confidence value?   I came out the following solution:     $H_0:$ the coin is fair   $H_a:$ the coin is unfair      $X$: is the number of heads      Rejection region: $|X - 3| &gt; 2$, i.e., $X = 0,1,5,6$      Significance level alpha:      $\alpha = P(\text{reject }H_0 \mid H_0 \text{ is true})$   $=P(X=0,1,5,6 \mid H_0 \text{ is true})$   $= (\binom{6}{0}+\binom{6}{1}+\binom{6}{5}+\binom{6}{6})*(1/2)^6$    $= (1+6+6+1)*(0.5^6) = 0.21875$      because $\alpha &gt; 0.05$, we do not have enough evidence to reject $H_0$, and   we  accept $H_0$, so the coin is fair   Is the above test a good one? And I did not know how to calculate the confidence value?
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Could someone help me with how the following is calculated:  What is the probability of rolling a die seven times and getting at least one six?  My instinct told me it would be $1/6+\cdots + 1/6$ but this ends up being $7/6$.
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Given a box that contains 90% fair coins and 10% loaded coins, (a loaded coin gives heads 90% of the time), what is the probability for a randomly drawn coin to give 5 heads in a row?  P(HHHHH | random coin from box of mixed coins) = ?   I used Bayes Theorem to calculate the probability I got .219 or 21.9% probability for the random coin to give 5 heads in a row!    But I am still doubting if my answer is correct !!!  Any help appreciated :)
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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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Problem: This is problem 33 of section 2.6 (conditioning) in Bertsekas, Tsitsiklis intro to probability text.  We are given that a coin has probability of heads equal to p and tails equal to q and it is tossed successively and independently until a head comes twice in a row or a tail comes twice in a row. What isthe expected value of the number of tosses?  Solution:  Let $X$ be a random value that tracks the number of tosses. In addition, let $H_k, T_k$ be the event that a head or tail comes up on the kth flip respectively. The solution I am looking at involves using total expectation theorem.  I outline the following steps in the solution.  We can partition the sample space into $H_1$ and $T_1$, allowing us to invoke the total expectation theorem. Hence,  $E[X] = pE[X|H_1] + qE[X|T_1]$  Again by total expectation theorem, we can utilize another partition to obtain  $E[X|H_1] = pE[X|H_1,H_2] + qE[X|H_1,T_2] = 2p + q(1 + E[X|T_1])$  $E[X|T_1] = qE[X|T_1,T_2] + pE[X|T_1,H_2] = 2q + p(1+E[X|H_1])$  Stuck: The solution follows up by saying that we can combine the above two relations and use the fact that p+q = 1 to obtain  $E[X|T_1] = \frac{2+p^2}{1-pq} , E[X|H_1] = \frac{2+q^2}{1-pq}$  I have been staring at this for a while now, but I cannot seem to see the steps of the computation. I'd be very grateful if someone could fill in the details regarding the computation. Thank you.
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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?
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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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Suppose you roll one fair six-sided die and then flip as many coins as the number showing on the die. (For example, if the die shows 4, then you flip four coins.) Let Y be the number of heads obtained. Compute $E(Y)$.  So, $E(Y)=\Sigma_{y=1}^{y=6}yP_{y}(y)$  We will define $X$ as the number on the die. Therefore:  $$E(Y)=\sum_{y=1}^{y=6}yP_{y}(y)=\sum_{y=0}^{y=6}\sum_{x=1}^{x=6}(y)P(X=x,Y=y)=\sum_{y=0}^{y=6}y\sum_{x=1}^{x=6}(1/6)\binom{x}{y}(1/2)^x$$  This inner equation is valid: $(1/6)\binom{x}{y}(1/2)^x$  Basically, it says that if I roll a 3, and I get 1 head only, what is the probability of that occurring = $(1/6)\text{[for the roll]}\binom{3}{1}\text{[from 3 rolls, we choose one head]}(1/2)^3=1/16$  Now, I need to simplify the above double sum, but I am not sure how to do it. The answer in my textbook is $7/4$, but it does not show steps. I think I am right until now, I just need help doing this summation.
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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!
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A fair coin is tossed until a head comes up for the first time. The probability of this happening on an odd number toss is?  How do i approach this problem ?
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A coin is tossed three times. The probability of zero heads is 1/8 and the probability of zero tails is 1/8. What is the probability that there is at least one head and at least one tail?  So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  We would use the complement rule and Multiplication Rule?   P(at least one head) = 1 - P(no heads) = 1 - 1/8= 7/8 P(at least one tail) = 1 - P(no tails) = 1 - 1/8= 7/8  would I multiply the two values?
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Say you have an unfair 3-sided die.  Side 1 will be rolled with probability $P_1.$ Side 2 will be rolled with probability $P_2.$ Side 3 will be rolled with probability $P_3.$  Find a formula for the probability that the 8th 3 will be rolled on the 10th roll.  Here is how I started:  $$ \sum_{i=8}^{10}P(T_8 = i) $$  Im just having trouble with that expansion?
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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?