smile101

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Does anyone know any good websites to practice some hard permutation and combination problems...especially when probability is involved. I am horrible at these types of problems and want to practice some hard ones. I did some from kaplan online QR and they were really good questions, except for I got most of them wrong. So, please help!!!
 

Streetwolf

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If you can't find a site, try using playing cards or dice.

I'll start you off with an easy one.
How many unique 5 card poker hands could you be dealt with a standard 52 card deck?

Answer is (52 choose 5) = 52!/47!5! = 52*51*50*49*48 / 120 = 2,598,960.

Here's a harder one.
How many ways can you be dealt a full house (3 cards of one rank, 2 cards of another rank... example QQQ99 where Q = queen)?

Answer: You first choose the two ranks that you want to use. Keep in mind that they are NOT equal since one is used for 3 cards and the other is used for 2 cards. So you need a permutation (order matters). This is (13 P 2). Now you choose the suits for the rank with 3 cards. There are 4 suits of which you need 3. This is a combination: (4 C 3). Now choose the 2 suits of the other rank. This is (4 C 2).

Multiply them all together: (13 P 2) * (4 C 3) * (4 C 2) = 156*4*6 = 3,744.

I did it that way to demonstrate the difference between a permutation and a combination. You could have also first picked the 3 card rank by doing (13 C 1) and then later picked the 2 card rank by doing (12 C 1). You get the same answer.

To find the probability of making a full house, just do 3,744 / 2,598,960.

===

I suggest doing it in this order:

Royal flush, straight flush, straight, flush, 4 of a kind, 3 of a kind, two pair, one pair, high card (none of the others).

Remember that straight flush = all straight flushes EXCEPT for the royal flushes.
Remember that a straight = all straights EXCEPT for the straight flushes and the royal flushes.
Remember that a flush = all flushes EXCEPT for the straight flushes and the royal flushes.
Remember that high card = all hands EXCEPT for those that have one pair or better.

===

As for the dice, try doing things like:
What are the odds that when you roll 5 dice, that 3 of them show a '4'?

Answer: You first select the 3 dice that will have a '4' on them. Do this with (5 C 3). The probability of 3 of them having a '4' is (1/6)^3. The probability of the other two NOT having a '4' is (5/6)^2. Multiply them all together and your answer is 10*(1^3)*(5^2) / 6^5 = 250/7,776 = 3.22%.

Just come up with some examples. If you can't solve something you come up with, or if you want to verify the answer, just post it here and I'll keep an eye out for it.
 
OP
smile101

smile101

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Thanks streetwolf...you are awesome. I am going to try some and will bother you with some more.
 

mashinka

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Feb 17, 2010
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just wanted to bump this thread cuz it totally helped me with probability, it anyone else is having issues.