Image via WikipediaA long, metal slide resembling a sliding board has been constructed with 3 holes spaced out along the length of the slide. Coins are placed at the top of the slide and released one after another. For each coin that approaches the first hole the chances are 50 percent that it will fall through the hole. If it makes it past the first hole the chances are 50 percent that it will fall through the second hole. The third hole has the same chances.
How many coins need to be released so that chances are that one coin will make it all the way down the slide?
Chances for a coin to get past the first hole are 1:2, chances to pass the second hole are then 1:2 x 1:2 = 1:4, and finally chances to pass the third hole are 1:4 x 1:2 = 1:8.ReplyDelete
We should release 8 coins and theoretically one of them should make it all the way down.
Why do I get a feeling it can't be that simple?
Hmmm.... I think it depends how you interpret "...the chances are...".ReplyDelete
For example, with seven roles there is a 39% chance (i.e (7/8)^7) that no coins would make and, therefore, a 61% (1-0.39) that at least one coin would make it. So the chances are that at least one coin would make it.
Similarly, for 6 coins there is a 45% that no coins will make it and a 55% chance that at least one will. So, agian, the chances are that at least one coin would make it.