Comment author:Flumber
08 September 2013 06:16:56PM
1 point
[-]
The first reply eliminates the no-ace case from six equally likely cases, leaving two aces as one of five equally likely cases. So the probability is one fifth. (The second question is irrelevant, by symmetry.)
= 27d567bc2d48d9f40e9ccc20ea370b15)
= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
= f037147d6e6c911a85753b9abdedda8d)
The first reply eliminates the no-ace case from six equally likely cases, leaving two aces as one of five equally likely cases. So the probability is one fifth. (The second question is irrelevant, by symmetry.)