Two piece of gold are contained in two same-looking black boxes respectively. It is known that one piece weights twice as the other, but do not know which is which.    Two persons, say A and B, randomly choose a box. One person, say A, opened his box, but he does not known whether it is lighter or weightier.   Question: Is A willing to exchange his box with B?  This might be a well-known problem, but I do not know the proper name to search it online.   Intuitive, it make no difference to exchange the boxes.  On the other hand, if we compute the expectation for A, it seems that A should change the box (the expectation is 1.25 of the current holding).  I would like to hear your answer to the question and preferably fuller story about this paradox.
31814
Suppose you're given two envelopes. Both envelopes have money in them, and you're told that one envelope has twice as much money as the other. Suppose you pick one of the envelopes. Should you switch to the other one?  Intuitively, you don't know anything about either envelope, so it'd be ridiculous to say that you should switch to the other envelope to maximize your expected money.   However, consider this argument. Let $x$ be the amount of money in the envelope you picked. If $y$ is the amount of money in the other envelope, then the expected value equals  $$E(y) = \frac{1}{2}\left(\frac{1}{2}x\right) + \frac{1}{2}\left(2x\right) = \frac{5}{4} x$$  But $5x/4 &gt; x$, so you should switch!    The Wikipedia article says that $x$ stands for two different things, so this reasoning doesn't work. I say this is not a valid resolution.   Consider opening up the envelope that you pick, and finding $\$10$ inside. Then you can run the expected value calculation to get $$E(y) = \frac{1}{2} \cdot \$5+\frac{1}{2} \cdot \$20 = \$12.50$$  This means that if you open one of the envelopes and find $\$10$, you should switch to the other envelope. The $\$10$ doesn't stand for two different things, it literally just means $\$10$.   But you don't have to open up the envelope to run this calculation, you can just imagine what's inside, and run the calculation based on that. This is what "Let $x$ be the amount in the envelope" means. The problem with the argument is not that $x$ stands for two different things.   So what is the problem?    Previous questions on stack exchange have given the resolution that I just said I wasn't satisfied by, so please don't mark this as a duplicate. I want a different resolution, or a more satisfying explanation of why $x$ does stand for two different things.  Apparently there is still research being published about this problem - maybe it isn't so obvious?    I think there's something subtle wrong with the premise. Because there's no uniform probability distribution on $\mathbb{R}$, statements like "random real number" are not well-defined. Likewise, I think "one envelope has twice as much money as the other" assumes some probability distribution on $\mathbb{R}$, and perhaps our expected value calculation assumes that this distribution is uniform, which it cannot be ...
121345
My question comes from different modes of probability law, which I met. I see two options in equality: set belonged to $B(\mathbb R^n)$ and cartesian product of B belonged to $B(\mathbb R)$.
55865
Are both the Venn diagram's (i and ii) showing dependent properties (showing that A and B are dependent on one another), I am told that the quartered one is independent (Venn diagram i), however I do not see any difference between the 2 diagram's. Surely P(A|B) (probability of B given A), will come out to be the same in both diagrams. This is because we are constrained to A and AB (highlighted red in Venn diagrams) as we know the outcome will be either AA or AB. Further, the ratio of N(AB) to N(A) will give us our probability P(A|B), which can be seen to lie in both diagrams to be in the region A and AB.  I know that if P(A|B) = P(B) , we have independent events, however how can we deduce if the Venn diagrams are showing independent events or not. Thanks
8548
From Wikipedia :     Suppose you're on a game show, and   you're given the choice of three   doors: Behind one door is a car;   behind the others, goats. You pick a   door, say No. 1, and the host, who   knows what's behind the doors, opens   another door, say No. 3, which has a   goat. He then says to you, "Do you   want to pick door No. 2?" Is it to   your advantage to switch your choice?   The answer is, of course, yes - but it's incredibly un-inituitive. What misunderstanding do most people have about probability that leads to us scratching our heads -- or better put; what general rule can we take away from this puzzle to better train our intuition in the future?
119678
Suppose we had a normal 52 count deck of cards. If two cards are randomly picked, what is the probability that both are Hearts given at least one Heart was chosen.  I think the answer would be $ \frac{12}{51}$ because there are twelve Hearts left after one is chosen with 51 total cards, but I might be approaching this problem the wrong way. Any idea on how I should think about it, or what I did wrong?
5564
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 :)
72323
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.
65290
There are $2 \cdot n$ people in the queue to the theater office; n people on only banknotes worth $20$ zlotys, and the remaining n people only have banknotes worth $10$ zlotys . At the beginning of the sale at the box office there is no money. Each person buys one ticket worth 10 zlotys.   If one with only $20$-zlotys banknotes is in the first of the queue, then he/she needs to wait for another guy with only 10-zlotys banknote to complete his/her transaction, because the ticket office does not have any change to offer at that time.  What is the probability that no one will wait for the change?  $A$ = no one will wait for the rest. $P (A) = 1-P (A ')$, that is, it subtracts the waiting persons from the whole and will leave me without waiting, but I do not know how to calculate it.
28517
Consider two different shaped containers having same area: one is cylinder, the other is like an inverted pyramid (roughly). Both have the same level of water, the weight of the inverted pyramidal container will therefore be greater than that of the cylinder. But I know from Pascal's law that the pressure should be same in both containers. If the base areas are the same then the force (weight as measured by a scale) should also be the same. Why are the weights different?
72516
Having read a little about exchangeability, I went back to thinking about the iid condition required for the central limit theorem. It struck me that if two random variables are drawn from an identical distribution, the occurrence of one event does not make it more or less probable for the occurrence of another event. I have no doubts that I am wrong and there is a reason why both independence and identical distribution are required. I just don't know why!
99777
Two players, $A$ and $B$, alternately and independently flip a coin and the first player to get a head wins. Assume player $A$ flips first.  If the coin is fair, what is the probability that $A$ wins?   So $A$ only flips on odd tosses. So the probability of winning would be $$ P =\frac{1}{2}+\left(\frac{1}{2} \right)^{2} \frac{1}{2} + \cdots+ \left(\frac{1}{2} \right)^{2n} \frac{1}{2}$$  Is that right? It seems that if $A$ only flips on odd tosses, this shouldn't matter. Either $A$ can win on his first toss, his second toss, ...., or his $n^{th}$ toss. So the third flip of the coin is actually $A$'s second toss. So shouldn't it be $$P = \frac{1}{2} + \left(\frac{1}{2} \right)^{2} + \left(\frac{1}{2} \right)^{3} + \cdots$$
30513
I was recently asked this question in an interview, but was completely stumped as to how to even begin answering it - it's been bugging me ever since, and I thought it was quite a nice question, so hopefully someone on here can help me out. Any help would be appreciated! Here goes:  You start off with £100 and you toss a coin 100 times. Before each toss you choose a stake $S$ which cannot be more than your current balance $x$ (so your maximum stake for the first toss is £100). If the coin comes up heads, you win $2S$ and your new balance is $x+2S$. If it comes up tails, you lose your stake and have $x-S$. How do you choose your stake so as to maximise your expected winnings from the game, not including the initial balance?  Cheers,  Boris
20556
My professor assigned a completely random question on our problem set. Basically, it goes:  Sophie is 30 years old and majored in philosophy. As a student, she participated in anti-nuclear demonstrations and was deeply concerned with social justice. Which is more probable?  1. Sophie is a bank teller. 2. Sophie is a bank teller and is active in the feminist movement.  I'm guessing the point the prof was trying to get across is that assuming you have two different events $A$ and $B$,  $A\cap B$ would always have a smaller probability than either. Is this correct?
119731
I am mainly confused about my reasoning/solution  Problem: I have a bag of numbers ranging from $1,2... n$. I randomly choose a number and then I throw out all numbers less than that. Then I keep choosing numbers until I've found the largest one. What's the expected number of things I have chosen?  My Argument: I expect on average to have chosen a number of $n/2$ on each pick. Hence, everytime I choose a number, I expect to discard half of them. Hence, the $E(X) = \log_2(n)$  Note that I devised this problem as an equivalent one while trying to solve the "cluster of cars on one long road" question. If this is not equivalent, please let me know but I am confident that it is. As a result of this equivalency, I know my solution is incorrect. However, I cannot pinpoint where the issue arises.
112751
That is to say, the value of a second roll, how much of a bonus value is it worth.  For example: when you make a saving throw, attack or check roll, when does +bonus to the roll and rolling twice equal the same worth.  I believe the value of a second roll would be different depending on if it was 'Choose the Highest', 'Choose the Lowest', 'Choose any Result' and 'Reroll on Failed'.  Though in this case I'm thinking for 'Choose the Highest'.  Edit in response to comments; For the intentions of this question the dice is a d20.
39036
"An event less frequent in the long run is likely to happen before a more frequent event!"  How can I show that THTH is more likely to turn up before HTHH with a probability of  9/14, even though waiting time of THTH is 20 and HTHH, 18!  I would be very thankful if you could show me the way of  calculating the probability of turning up earlier,   and the waiting time. Thank you!
107193
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?