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 ...
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 ...
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)=...
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;
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....
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 ...
A set of coins is fair in the relevant sense if and only if
Proof (slightly highbrow, sorry):
Alternative kinda-equivalent proof (simpler ideas but needs you to know a theorem):
(Neil W suggested, in comments, taking that second approach. I'd avoided it because the other way seemed quicker and more first-principles-y, but the second way may well be easier ...
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,
The upshot of all this is
Credit where due: hexomino did a lot of this before I did (with, unfortunately, an error, but ...
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
in which case you go to heaven, or it may be one of
in which case you go to hell.
If Y=2, namely
then pick one of the Yeses at random and ask the utterer the same question again. ...
First of all, let's do the obvious calculation.
Now note that
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
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 ...
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 ...
The Sleeping Beauty problem itself is a famous problem in the philosophy of probability, and obviously we aren't going to resolve it here. Fortunately, the question here is more concrete, so let's just do it.
Of course, you need not say the same thing every time.
You always have a simple 50% strategy:
Whatever the opponent chooses, there are exactly 3 good spin results and 3 bad ones for you.
Your opponent always has a simple 50+% strategy:
Doing otherwise is always worse, or equal at best; there are no possible bullet configurations where there are more than 3 "good" spin results for the player going first.
The answer is quite simple when you know the concept of a derangement, since the only time this doesn't happen is if the first deck is a derangement of the second.
In other words, we are looking for
There are online calculators if you want the exact value, which is approximately
Edit: As pointed out by Gareth McCaughan, there is a mistake in this answer: $(N-8)$ in the numerator of the normalisation constant should be $(N-10)$ in which case the sum diverges, so this approach will not work.
I think the expectation value for the number of pictures in the database is
Point of Information
For both of the puzzles you link to, unless I am mistaken, the following is true:
Define the "N-restricted" version of the problem to be what you get by replacing "a random positive integer" with "a random positive integer <= N" and assuming that different integers are chosen independently. Then the limit, as N tends to infinity, of the answer to the N-...
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 ...