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DontRaisePlz
08-20-2005, 09:15 PM
What are the chances that at least one deuce shows up?
masse75
08-20-2005, 09:33 PM
should be 1 minus (probability NO deuces hit):

1 - ([48....36]/[52....40])= ~66%

Not a stats guy...anyone check the math?
BruceZ
08-20-2005, 09:42 PM
[ QUOTE ]
should be 1 minus (probability NO deuces hit):

1 - ([48....36]/[52....40])= ~66%

Not a stats guy...anyone check the math?

[/ QUOTE ]

Right equation, but it's =~ 69.6%.
AaronBrown
08-21-2005, 10:51 AM
Aheravi's method is correct, and BruceZ has supplied the answer. Another way of doing this that may give more insight, and is faster in Excel, is to say you can pick 13 cards from 52 in C(52,13) ways. If you leave out the twos, there are only C(48,13) ways. So the probability of zero twos is C(48,13)/C(52,13) = 0.3038. One minus that is the probability of at least one two, 0.6962.