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Gialmere
Gialmere
Joined: Nov 26, 2018
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June 14th, 2021 at 8:10:43 AM permalink
It's easy Monday...



What is the missing number in this sequence?

20, 22, 24, 26, 30, 33, 40, 44, 120, ___, 11000
Have you tried 22 tonight? I said 22.
charliepatrick
charliepatrick
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Gialmere
June 14th, 2021 at 8:23:02 AM permalink
It's all about the base!
Ace2
Ace2
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June 14th, 2021 at 8:48:10 AM permalink
Quote: ThatDonGuy

Okay, I'll ask: how do you simplify this?

What do you mean by simplify?
Itís all about making that GTA
Ace2
Ace2
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June 14th, 2021 at 8:48:13 AM permalink
Quote: ThatDonGuy

Okay, I'll ask: how do you simplify this?

What do you mean by simplify?
Itís all about making that GTA
Ace2
Ace2
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June 14th, 2021 at 8:48:15 AM permalink
Quote: ThatDonGuy

Okay, I'll ask: how do you simplify this?

What do you mean by simplify?
Itís all about making that GTA
gordonm888
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gordonm888
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June 14th, 2021 at 9:40:55 AM permalink
Quote: Ace2


You can also get the answer by integrating the following formula over all t:

(1-(1-1/e^(t/52))^13)^4



Quote: ThatDonGuy


Okay, I'll ask: how do you simplify this?



Quote: Ace2

What do you mean by simplify?



I think he means that the analytical expression to be integrated has 2^13^4 = 5E15 terms if you straightforwardly expand it. So, can you analytically simplify the integration or must you do it numerically?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
ThatDonGuy
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June 14th, 2021 at 9:56:20 AM permalink
Quote: Ace2

What do you mean by simplify?


I mean, is there a way to get from the integral to the rational number result without using approximation?

I assumed that, since you expressed the answer as a rational number, that there was a way to do it.
Ace2
Ace2
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June 14th, 2021 at 12:10:09 PM permalink
Quote: ThatDonGuy

I mean, is there a way to get from the integral to the rational number result without using approximation?

I assumed that, since you expressed the answer as a rational number, that there was a way to do it.

I use https://www.integral-calculator.com/ for the integration of something like this. If you put that formula in, it shows all the steps (several pages) and it does come out to the rational answer. To check the answer for reasonableness, I either do a simulation or a numerical integration in excel. In this case, if you do a numerical integration with spacing of 0.1 the first four digits of the answer have already converged.
Itís all about making that GTA
Wizard
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Wizard
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June 14th, 2021 at 12:43:07 PM permalink
Quote: Ace2

You can also get the answer by integrating the following formula over all t:

(1-(1-1/e^(t/52))^13)^4 dt

Which gives us the answer of 712830140335392780521 / 6621889966337599800 =~ 108 draws

Starting from the middle, 1/e^(t/52) is the probability at any time t that any single card has not been drawn. The complement of that to the 13th power (1-1/e^(t/52))^13 is the probability that all 13 cards of any one suit have been drawn at least once. The complement of that to the 4th power (1-(1-1/e^(t/52))^13)^4 is the probability that none of the 4 suits meet that condition (of all 13 cards being drawn at least once).

The formula works due to one of the most useful properties in probability: the expected time for an event to happen is equal to the sum of the probabilities over all time that it has not happened yet. This is basically the geometric series 1 + r + r^2 + r^3...= 1 / (1-r) carrying over from the binomial distribution to the exponential one.



I'm finally catching up with another classic Ace2 calculus problem. I plan to make this a future Ask the Wizard question. Here are some notes for now on my solution. I welcome all comments and corrections.

Question asked: https://wizardofvegas.com/forum/questions-and-answers/math/34502-easy-math-puzzles/181/#post809012

Solution: https://wizardofvegas.com/forum/questions-and-answers/math/34502-easy-math-puzzles/183/#post809294

Integral calculator: https://www.integral-calculator.com/

Q: A card is draw with replacement from a deck of 52 cards. What is the expected number of draws needed until every card of any one rank has been drawn? Please use calculus for your solution.

Hint: The expected time for an event to happen equals the sum over all time that it hasn't happned yet. This is true for both discreet and continuous variables.

Solution: Instead of a card being drawn exactly once per unit of time, the answer will be the same if a card is drawn with a random period of time between draws if that average time follows an exponential distribution with mean of 1.

The time between any given card being drawn will have a mean of 52. Given the properties of the exponential distribution, the probability the card will not have been drawn after t units of time is exp(-t/52).

After t units of time, the probability any specific card will have been drawn at least once is 1-exp(-t/52).

After t units of time, the probability 13 specific cards will have been drawn at least once is (1-exp(-t/52))^13.

After t units of time, at least one of 13 specific cards will cards will NOT have been drawn is 1-(1-exp(-t/52))^13.

After t units of time, four groups of 13 cards each, having no overlapping cards, all will have at least one missing card is (1-(1-exp(-t/52))^13)^4.

Putting this equation into an integral calculator, being careful to set to set the bounds of integration from 0 to infinity, yields 712830140335392780521 / 6621889966337599800 =~ 107.6475362712258
It's not whether you win or lose; it's whether or not you had a good bet.
DogHand
DogHand
Joined: Sep 24, 2011
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Gialmere
June 14th, 2021 at 12:52:44 PM permalink
Quote: Gialmere

It's easy Monday...



What is the missing number in this sequence?

20, 22, 24, 26, 30, 33, 40, 44, 120, ___, 11000



With thanks to Charlie for the hint, I get...

220 = 24 in base 3. Each number is 24 in based decreasing from 12 to 2.


Dog Hand

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