Easy math puzzles
Poll
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| | 6 votes (13.33%) | ||
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| | 5 votes (11.11%) | ||
| | 12 votes (26.66%) | ||
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45 members have voted
Quote: ThatDonGuyQuote: unJonCan someone explain to me what I’m missing? I assume we can drop spoilers at this point.
Wiz’s 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 there’s 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.
A great classical problem and very deep.Quote: Ace2If 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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Quote: Ace2If 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.
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%.Quote: ChesterDogQuote: Ace2If 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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I don’t 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 + 1/4^2…I 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. I’ve never really thought about primes except when factoring down numbers to simplify a formula. I’d have never guessed that π (or any formula) would come up in this coprime scenario.
Quote: Ace2Correct. 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?
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 you’re right, keep going.
If you play this game optimally, what’s the probability that you can get to the end without making any mistakes?
Extra credit: What if there were more cards — 2 through 20, or 2 through 100? How do your chances of getting to the end change?
The probability for 9 cards is 0.5^4=6.25%.
The probability for 21 cards is 0.5^10=0.098%
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.
