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unJon
unJon
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December 24th, 2021 at 2:06:35 PM permalink
Quote: ThatDonGuy

Quote: unJon

Can someone explain to me what Iím missing? I assume we can drop spoilers at this point.

Wizís table above says the probability of 0 points is 0%. But how is that possible? If the place bets are working from the come out, then thereís a chance of a 7 killing them all before a point number is rolled. Likewise if they are off until a point is established, there is still likewise a probability of a 7 out killing them before another point number rolled.


No, it says the expected value of rolling 0 point numbers is 0, and the probability of doing so is 1/5.
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Yup. That makes sense.
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Ace2
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December 27th, 2021 at 6:06:52 PM permalink
If you randomly pick two numbers (integers) between 1 and 1000, what is the probability they are coprime (largest common divisor is 1) ?
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teliot
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December 27th, 2021 at 6:30:46 PM permalink
Quote: Ace2

If you randomly pick two numbers (integers) between 1 and 1000, what is the probability they are coprime (largest common divisor is 1) ?
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A great classical problem and very deep.
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Wizard
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Wizard
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December 27th, 2021 at 6:59:17 PM permalink
Can the two integers be the same?
ďExtraordinary claims require extraordinary evidence.Ē -- Carl Sagan
ChesterDog
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December 27th, 2021 at 7:44:15 PM permalink
Quote: Ace2

If you randomly pick two numbers (integers) between 1 and 1000, what is the probability they are coprime (largest common divisor is 1) ?
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Brute force in Excel using GCD function yields 608,383 / 1,000,000.
Ace2
Ace2
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December 28th, 2021 at 7:26:14 AM permalink
Quote: ChesterDog

Quote: Ace2

If you randomly pick two numbers (integers) between 1 and 1000, what is the probability they are coprime (largest common divisor is 1) ?
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Brute force in Excel using GCD function yields 608,383 / 1,000,000.

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Correct. If you randomly pick two numbers between 1 and n, the probability they are coprime is 6 / π^2 =~ 60.79% as n approaches infinity. For n=1000, the probability is within 5 basis points of that at 60.84%.

I donít know how to prove this, I only read about it. Interestingly, the reciprocal of 6 / π^2 equals the infinite series 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2ÖI posted a math puzzle using that series a while back

This is another example of how all math seems linked together at some level, often involving e and/or π. Iíve never really thought about primes except when factoring down numbers to simplify a formula. Iíd have never guessed that π (or any formula) would come up in this coprime scenario.
Last edited by: Ace2 on Dec 28, 2021
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Wizard
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Wizard
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December 28th, 2021 at 10:28:36 AM permalink
Quote: Ace2

Correct. If you randomly pick two numbers between 1 and n, the probability they are coprime is 6 / π^2 =~ 60.79% as n approaches infinity. For n=1000, the probability is within 5 basis points of that at 60.84%.
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I find things like this fascinating. I almost look at it as evidence of some kind of higher power. Is there a term for this particular limit?
Last edited by: Wizard on Dec 28, 2021
ďExtraordinary claims require extraordinary evidence.Ē -- Carl Sagan
Gialmere
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December 29th, 2021 at 10:37:13 AM permalink


Take a standard deck of cards, and pull out the numbered cards from one suit (the cards 2 through 10). Shuffle them, and then lay them face down in a row. Flip over the first card. Now guess whether the next card in the row is bigger or smaller. If youíre right, keep going.

If you play this game optimally, whatís the probability that you can get to the end without making any mistakes?


Extra credit: What if there were more cards ó 2 through 20, or 2 through 100? How do your chances of getting to the end change?
Have you tried 22 tonight? I said 22.
aceside
aceside
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December 29th, 2021 at 11:27:37 AM permalink
Interesting! This puzzle makes me think about wizardí glass ceiling problem in Squid game. Let me make a guess.
The probability for 9 cards is 0.5^4=6.25%.
The probability for 21 cards is 0.5^10=0.098%
Last edited by: aceside on Dec 29, 2021
ksdjdj
ksdjdj
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Thanks for this post from:
aceside
December 29th, 2021 at 3:24:57 PM permalink
When played optimally, I got ~77.78% (7/9) chance of getting to the 2nd card
and a ~66.52% chance of getting to the 3rd card (if you made it to the 2nd card)
for a combined total of ~51.74% of making it to the 3rd card.

Is this correct so far?

Note: At my current "math level " I am not going to go past this even if I am correct so far, because it would take me a real long time to get to the "4th", "5th" and "nth" card average values.
Last edited by: ksdjdj on Dec 29, 2021

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