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I don't know what I am doing wrong. So intuitively I am calculating the probability of getting a full-house like this :

  1. The first card can be any card out of the $52$ cards so probability = $1$
  2. The second card has to be the same number as the first card so probability = $3/51$
  3. The third card also has to be the same number as the previous two cards so probability = $2/50$
  4. Now the fourth card can be any card except for the number drawn previously so probability = $48/49$
  5. The fifth card has to be the same number as the fourth card so probability = $3/48$ So finally I get a probability of = $0.000144$ but the answer is $0.00144$. Can anyone suggest what I am doing wrong?
Bob
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Rupam Baruah
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1 Answers1

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Because the order in which the cards appear doesn't matter, it is safer to use combinations throughout

$\Large\frac{\binom{13}1\binom43\cdot\binom{12}1\binom42}{\binom{52}5}$

If you start with one particular order as you did by writing

$\Large\frac {52}{52}\frac3{51}\frac2{50}\cdot \frac{48}{49}\frac3{48}$, you are saddled with the problem of what you should multiply it with to consider other possible orders, which is $\Large\left(\frac{5!}{3!2!}\right)$ where you may quite possibly slip-up.

true blue anil
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