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No questions are required!
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(a) I claim that the expected typing length are the same for both monkeys. I guess something in my argument will be incorrect, as jafe's answer has 9 approvals, but finding that incorrectness would be helpful to me.
I prove that the probabilty that a monkey ends his typing after exactly $n$ letters is the same for both monkeys, which then implies that the ...
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(a)
Edit: This is incorrect, see comments
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OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 ...
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Monkey problem
To settle down which monkey is faster on average, I'll use Markov chains and Mathematica. Define a state $i$, for $i = 0..6$, as that the monkey 1 has currently written $i$ correct subsequent characters of COCONUT, but has never written all of COCONUT. Define a state 7 as the monkey 1 having written COCONUT at least once. Since state 7 cannot ...
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The answer is
Proof
Alternative proof
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Answer:
Explanation:
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A difference between the Monty Hall scenario and the train platform scenario is that
After you remember that the train will definitely not be leaving from platform 2,
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I'd
Then
Then
So
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(a) has been adequately answered by
In (b), though:
Consider:
In this case:
Now consider:
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This gambling problem is the famous St. Petersburg paradox. It is a paradox because
The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the ...
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@Ingix provides a good proof, one certainly worthy of the tick, but one that could perhaps use some reinforcement. The answer is that the N number of letters would have the same expected value. A few points that might help @jafe realise the flaw in his proof.
The order of independent events occurs has no directional bias. For example, drawing a King Of ...
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I have a solution with a success rate of 93.5%, according to my simulations.
The reason this solution works so well is
Here's my code that I used to verify my solution:
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For clarify:
No matter who you ask, he will answer the correct door.
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There is a
chance that the other jar is empty. This puzzle is asking the equivalent of
which can be calculated by
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Suppose $t(n)$ is the average number of spins you get if you start with $\$10n$. We want $t(2)$. If you start with $n$ ten-dollar bills, put $n-1$ in your pocket and play until you are broke, then take the next \$10 out and play until you are broke, and so on: this is exactly equivalent to just starting with $\$10n$ and playing until you're broke, so $t(n)=...
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I would pay
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Originally there are;
10 S, 6 V and 5C
For the first case where he wants to have 7 S, 5 V and , 4 C. (so we dont want to have 3S, 1V and 1C)
In the worst case scenario;
so
I do not want to continue for the rest since the same methodology works for them too:
This will give you the number of cookies to guarantee to have some specific number of cookies....
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Here's how the original Monty Hall worked:
You pick a door. (2/3 chance of it being a goat, 1/3 chance of it being a car)
Monty Hall opens a door that he knows has a goat in it, and then asks you if you'd like to switch to the other door.
2/3 of the time, the door Monty Hall opened is the only other door with a goat - because 2/3 of the time, you picked a ...
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To do this,
Finally,
I can give you the Python code if you want. Not sure it can be spoilered (I've never learned how to have multi-line spoilers).
Python Code
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Proceeding along the same lines as hexomino, I get a different result. The expectation
The place where I (ha!) diverge from hexomino's answer is
We can try to deal with this inconvenient situation in three ways. First,
Second,
Third,
The upshot of all this is
Credit where due: hexomino did a lot of this before I did (with, unfortunately, an error, but ...
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Ask the following question of all three guards:
Now the number of Yeses (Y) will be between 0 and 3 inclusive.
If Y=1, go through that door. The position may either be
1.
in which case you go to heaven, or it may be one of
in which case you go to hell.
If Y=2, namely
3.
then pick one of the Yeses at random and ask the utterer the same question again. ...
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You should
because of
In numbers:
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First of all, let's do the obvious calculation.
Now note that
Hence
But of course
So the answer to the question as stated is
But it's important to notice
One common way to deal with the second of these is
Let's consider both of those together
In fact
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One strategy:
Probability of success:
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There are several nuances to this question. First of all, it asks how much you are willing to pay, not what price is fair. Second, you have to understand, that even if a game is fair, that does not mean that It is reasonable to play it.
For example, if someone offers me a one in a million chance to win a million dollars for 1, I will take it. It seems ...
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I would not pay anything. I would not play. I would encourage you to not play. Are you doing okay? I'm willing to help you out of you need help. I would offer you a hug.
You are my best friend, and you live right next to me. Any outcome of this game that would be monetarily meaningful to either of us would also most likely be highly damaging to our ...
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There is a clever trick to finding the kind of expected times that appear in part (a). It turns out that both for COCONUT and TUNOCOC, the expected time is just $26^7$ (so the monkeys have equal expected times to win), but for this to be convincing, I will demonstrate the method on a case where the expected value is unexpected: the word ABRACADABRA.
Suppose ...
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N=1,677
Even with all of the complex mathmatics being done, probability only gives us a way to calculate the answer devoid of other information.
Since you gave an example, but didn't specifically say it WASN'T your example, the most likely hypothesis is that the site you listed as an example was the site you visited.
Random.cat actually lists directly on ...
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