Probability addicts

Suppose you are playing a game of Hold'em Poker alone . You are holding 2 Spades in your hand and on the table are other 2 Spades and 1 Diamond. What is the probability of drawing another Spade in 2 tries from your deck?

Details and Assumptions

  • A deck of cards contains 52 cards (total)
  • There are 13 spades/diamonds/clubs/hearts in a deck of cards
  • Answer is up to 4 decimal places (0.abcd)


The answer is 0.3496.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Radinoiu Damian
Jul 11, 2015

This is supposed to be a hypergeometric distribution problem (probability without replacement). Let:

  • N = 52 5 = 47 N = 52-5 = 47
  • k = 13 4 = 9 k = 13-4=9
  • n = 2 n = 2
  • x 1 x \geqslant 1

All we have to do is calculate h ( x , N , n , k ) h(x, N, n, k) which translates to h ( x 1 , 47 , 2 , 9 ) = h ( 1 , 47 , 2 , 9 ) + h ( 2 , 47 , 2 , 9 ) = ( 9 1 ) ( 38 1 ) ( 47 2 ) + ( 9 2 ) ( 38 0 ) ( 47 2 ) = 0.3496 h(x\geqslant1, 47, 2, 9) = h(1, 47, 2, 9) + h(2, 47, 2, 9) = \frac{\left(\!\begin{array}{c}9\\1\end{array}\!\right) *\left(\!\begin{array}{c}38\\1\end{array}\!\right)} {\left(\!\begin{array}{c}47\\2\end{array}\!\right)} +\frac{\left(\!\begin{array}{c}9\\2\end{array}\!\right) *\left(\!\begin{array}{c}38\\0\end{array}\!\right)} {\left(\!\begin{array}{c}47\\2\end{array}\!\right)} = 0.3496

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...