[quote name=\'chad1m\' post=\'228479\' date=\'Oct 14 2009, 01:29 AM\']I am not a bitch.
/And if I was, I'm not
your bitch.[/quote]
[quote name=\'clemon79\' post=\'228480\' date=\'Oct 14 2009, 01:39 AM\'][quote name=\'chad1m\' post=\'228479\' date=\'Oct 13 2009, 10:29 PM\']I am not a bitch.[/quote]
Approves/was very happy with the winner this season
[/quote]
I'm guessing this is an inside joke concerning Hell's Kitchen or something, a show I do not watch. When I'm excited, I tend to say "bitches" a lot. :-)
[quote name=\'Steve McClellan\' post=\'228483\' date=\'Oct 14 2009, 02:13 AM\']I'm a sucker for a math problem. Here's what I did.
To
lose, you must lack a CAR card
and you must lack at least one of C, A, and R. Ergo, the final probability of losing equation will be:
P(lose) = P(no CAR card) * P(no C, A, or R, given no CAR)
To not have a CAR card, you must miss them five times. At the start, there are 28 non-CARs out of 30, then (if you miss the first) 27 out of 29, etc. To find the chance of missing all five times, multiply:
P(no CAR card) = (28/30) * (27/29) * (26/28) * (25/27) * (24/26) = 11 793 600/17 100 720 = 20/29 (about .690)
Finding the probability of missing a C, A, or R is a bit more involved, but uses the same form of equation. Since we already took into account the chance of finding a CAR card, we deal with only the 28 remaining cards from here on.
P(no C) = (17/28) * ... * (13/24) = 17/270 (about .063)
P(no A) = P(no C) = 17/270
P(no R) = 209/780 (about .268)
Now comes the not-immediately-obvious part. Anytime you add probabilities, you have to check whether any situation would meet two or more of the conditions, and compensate for the fact that you would be counting it twice. In this case, five Cs would be counted under no A and no R; five As and five Rs give you similar issues. Since they're counted twice, subtract them once to correct.
P(no C, A, or R, given no CAR) = 17/270 + 17/270 + 209/780 - 11/2340 - 11/2340 - 1/16380 = 1889/4914 (about .384)
So the final equation looks like this:
P(lose) = P(no CAR card) * P(no C, A, or R, given no CAR)
P(lose) = 20/29 * 1889/4914
P(lose) = 18890/71253 (about .2651)
P(win) = 1 - P(lose) = 52363/71253 (about .7349)
Ergo, unless I have royally screwed something up, there is a roughly 73.49% chance that a player with five cards will win, ignoring the possibility of quitting early and taking cash. By the same math, the chances of winning are:
5 cards: 52363/71253 = 73.49%
4 cards: 1067/1827 = 58.40%
3 cards: 151/406 = 37.19%
2 cards: 19/145 = 13.10%[/quote]
THANK YOU STEVE! Much appreciated. I knew the odds were higher than they "look," so to speak, but it is nice to see it written out. And I can actually follow it!
Regards,
Brent