Difference between revisions of "Talk:2328: Space Basketball"

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(The meteorite probability)
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On the other hand,  suppose that any 30 consecutive success counts. In that case the waiting time is shorter, but not much shorter. [This](https://math.stackexchange.com/questions/893941/distribution-of-maximum-run-length-of-independent-multinomial-trials) suggests the average time for any 30 consecutive is the same  as the average time for batches of 29 when you need to get all 29 in a batch. So the difference is smaller than uncertainties/approximations we're already ignoring
 
On the other hand,  suppose that any 30 consecutive success counts. In that case the waiting time is shorter, but not much shorter. [This](https://math.stackexchange.com/questions/893941/distribution-of-maximum-run-length-of-independent-multinomial-trials) suggests the average time for any 30 consecutive is the same  as the average time for batches of 29 when you need to get all 29 in a batch. So the difference is smaller than uncertainties/approximations we're already ignoring
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Since almost all meteors are incinerated and reduced to dust upon contact with the earth's atmosphere, it stands to reason that there may already be a (teeny-weeny bit of a) meteor already passing through the hoop. [[User:RAGBRAIvet|RAGBRAIvet]] ([[User talk:RAGBRAIvet|talk]]) 02:40, 4 July 2020 (UTC)

Revision as of 02:40, 4 July 2020


I'd just like to point out that this assumes cueball's odds of sinking a basket remain at 30% after hundreds/thousands of shots. One would think he would improve with practice. 162.158.62.75 23:53, 3 July 2020 (UTC)Duban

Randall expresses as much in the title text. --NotaBene (talk) 00:00, 4 July 2020 (UTC)

Cueball's odds of 30 consecutive baskets are 0.3^30 = 2.06*10^-16. Earth is hit by about 6100 meteors per year, and a basketball hoop has a radius of 9 inches. Using that it will be hit about once every 5.09*10^11 years. In order for it to be even, Cueball would have to do approximately one trial every 55 minutes. Since he'll start over each time he misses, it works out to once attempt every 38.6 minutes. DanielLC (talk) 00:36, 4 July 2020 (UTC)

(?Almost) no-one in recorded history has been killed by a meteor, so the estimate of 1 in 250,000 is based on a very small chance of a very large number of people dying from something like a "Dinosaur Killer" object, which would not fit through the hoop.

Ok. Area of hoop: 0.166 square meters. Area of earth: 510 million square kilometers, or about 3x10^15 hoops. The Planetary Science Institute thinks 500 meteorites per year; Cosmos magazine think 6100 per year (which will essentially all be small enough to go through the hoop). So we get 5x10^11 or 6x10^12 years for space to score. If Cueball had to do multiple sets of 30 throws and wait until one of those sets was all successes he'd take 5x10^15 attempts, so 1000 or 10,000 attempts per year for a fair game. Which seems ok.

On the other hand, suppose that any 30 consecutive success counts. In that case the waiting time is shorter, but not much shorter. [This](https://math.stackexchange.com/questions/893941/distribution-of-maximum-run-length-of-independent-multinomial-trials) suggests the average time for any 30 consecutive is the same as the average time for batches of 29 when you need to get all 29 in a batch. So the difference is smaller than uncertainties/approximations we're already ignoring

Since almost all meteors are incinerated and reduced to dust upon contact with the earth's atmosphere, it stands to reason that there may already be a (teeny-weeny bit of a) meteor already passing through the hoop. RAGBRAIvet (talk) 02:40, 4 July 2020 (UTC)