My Math Forum Basic probability - what am I misunderstanding?

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 April 1st, 2013, 09:50 AM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Basic probability - what am I misunderstanding? Hey all, long time no see Here's the question - friends are playing a game of poker. What's the chance that player A will have a hand of 5 cards of different values (not colors, ace of hearts and ace of spades for example, are the same value)? Here's my solution: there are 13 overall values to choose from (ace,2,3,4,5,6,7,8,9,10,jack,queen,king), but once player A draws a card, he reduces the number of overall values to 12, and then 11, and so on, so I think the solution is the number of different combinations I can get when all the cards have different values, divided by all the combinations available whatsoever. $\frac{{13 \choose 1}*{12 \choose 1}*{11 \choose 1}*{10 \choose 1}*{9 \choose 1}}{{52 \choose 5}}$ which is the same as: $\frac{13*12*11*10*9}{{52 \choose 5}}= \frac{154440}{2598960} \approx 0.0594$ Is this correct? Doesn't seem right.
 April 1st, 2013, 11:12 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Basic probability - what am I misunderstanding? Is player A getting 5 cards from the full deck; nobody else gets any while this goes on?
 April 1st, 2013, 12:40 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Basic probability - what am I misunderstanding? Actually, you can show that as long as you do not have knowledge of what cards the other players get, that does not affect the probability. There are 52 cards, thirteen values with four of each value. The first card can be anything. There are then 51 cards left, with 51- 3= 48 not of the same value as the first. Your chance of getting anything other than the value of the first card is $\frac{48}{51}$. There are now 50 cards left, 50- 6= 44 not of the same value as either of the first two cards. Your chance of getting a new value card is $\frac{44}{50}$. There are now 49 cards left, 49- 9= 40 not of the same value as any of the first three cards. Your chance of getting a new value card is now $\frac{40}{49}$. There now 48 cards left, 48- 12= 36 not of the same value as any of the first four cards. Your chance of getting a fifth card not the same value as any of the first four is $\frac{36}{48}$. The probability that the five cards are of 5 different values is $\left(\frac{48}{51}\right)\left(\frac{44}{50}\righ t)\left(\frac{40}{49}\right)\left(\frac{36}{48}\ri ght)$.

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