Question 1198936
A hand of 13 cards is dealt from a standard deck of 52 cards. What is the
probability that it contains more aces than tens? 
<pre>
There are 4 aces, 4 tens, and 44 cards which are neither aces nor tens.
I've done the calculation for two of the cases.  Do the others, then add them:

ACES  TENS  NEITHER  PROBABILITY
  1     0     12      (4C1)(4C0)(44C12)/(52C13) = <u>0.1328518567</u>
  2     0     11      (4C2)(4C0)(44C11)/(52C13) = ____________
  2     1     10      (4C2)(4C1)(44C10)/(52C13) = ____________
  3     0     10      (4C3)(4C0)(44C10)/(52C13) = ____________
  3     1      9       (4C3)(4C1)(44C9)/(52C13) = ____________
  3     2      8       (4C3)(4C2)(44C8)/(52C13) = <u>0.0066984129</u>
  4     0      9       (4C4)(4C0)(44C9)/(52C13) = ____________
  4     1      8       (4C4)(4C1)(44C8)/(52C13) = ____________
  4     2      7       (4C4)(4C2)(44C7)/(52C13) = ____________
  4     3      6       (4C4)(4C3)(44C6)/(52C13) = ____________
---------------------------------------------------------------
                              Total probability = ____________
</pre>
How does this probability change when you have the information that the hand
contains at least one ace?<pre>It doesn't change at all, because that's automatically given. If there are
more aces than tens, there has to be at least one ace.

Edwin</pre>