Timeline for Alternative solution to finding P(ace of spades is immediately drawn after first ace).
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 23, 2021 at 9:50 | comment | added | threelinewhip | Ok thanks - I was just checking for where I went wrong with brute force. As ugly as it is, brute force usually can work (if done correctly) and in a time-pressured situation, I might not be able to think of a slick sol. | |
| Mar 22, 2021 at 19:39 | comment | added | true blue anil | As a last comment, have a look at the solutions to this link. So many approaches have been given, you have to use whatever you are comfortable with. math.stackexchange.com/questions/1345413/… | |
| Mar 22, 2021 at 19:28 | comment | added | true blue anil | Yes, but from the dictum, it follows that all ordered pairs are equally likely, and we directly get the answer of 1/52. But yes, sometimes you can get misled without some counting | |
| Mar 22, 2021 at 18:17 | comment | added | threelinewhip | Doesn't the author's slick solution make use of orderings? (51!) | |
| Mar 22, 2021 at 18:14 | comment | added | true blue anil | Well, if you are counting, fine, I just meant that you should keep that dictum in mind for a possibly simpler approach for many problems. | |
| Mar 22, 2021 at 17:55 | comment | added | threelinewhip | And since I'm working out number of orderings that verify the event and divide this by total orderings (although I guess you can work this out with an unordered approach - nevertheless, an ordered approach can work as well), then cards would have a preference for position? | |
| Mar 22, 2021 at 17:55 | comment | added | threelinewhip | Otherwise the AoS wouldn't be directly after the first ace? My idea was to write an exhaustive list of all possible orderings that verify the event 'AoS after first ace' just to test my understanding of combinatorics (ik the author's sol is v slick, but I tried this before reading the solution). | |
| Mar 22, 2021 at 17:09 | history | edited | RobPratt |
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| Mar 22, 2021 at 15:32 | comment | added | true blue anil | Why other than an ace in the second line? You could even have all the aces together. And why go for a more cumbersome proof where there are more chances of error when a simple proof satisfies you ? Anyway, always remember the dictum "Cards don't have preference for positions." | |
| Mar 22, 2021 at 14:36 | history | asked | threelinewhip | CC BY-SA 4.0 |