SOLUTION: its a word problem .. if out of a deck i use 20 cards ... 5 suits and 4 of each card.. what is the propability that i will chose 4 cards in the exact order as 4 random generated ca

Question 460207: its a word problem .. if out of a deck i use 20 cards ... 5 suits and 4 of each card.. what is the propability that i will chose 4 cards in the exact order as 4 random generated cards

and is the propability greater than if i use 5 cards of 4 suits ? is it less than or is it equal to
also this is to neither homework or test . i am trying to make a board game
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!
The probability would be less if the randomly generated sequence
of ranks coutains duplications such as 2,1,2,3. So I will assume
the randomly generated sequence are 4 different ranks.
`
Suppose the randomly generated sequence is 3,4,1,2 and you use 4 ranks
and 5 suits:
`
There are 5 3s and 20 cards, so you will pick a 3 first 5/20ths or 1/4 of the
time.
`
Now of those 1/4ths of the times when we successfully pick the 3 first:
`
There are 5 4s and 19 cards, so you will pick a 4 second 5/19ths
of the time.
`
So the probability of successfully picking a 3 first and a 4 second is
5/19ths of 1/4th of the time or 5/76ths of the time
`
Now of those 5/76ths of the time when we successfully pick the 3 first
and the 4 second:
`
There are 5 1's and 18 cards, so you will pick a 1 third 5/18ths of
the time.
`
So the probability of successfully picking a 3 first, a 4 second, and a
1 third is 5/18ths of 5/76ths of the time or 25/1368ths of the time
`
Now of those 25/1368ths of the time when we successfully pick the 3 first,
the 4 second, and the 1 third:
`
There are 5 2's and 17 cards, so you will pick a 2 fourth 5/17ths of
the time.
`
So the probability of successfully picking a 3 first, a 4 second, a
1 third and a 2 fourth is 5/17ths of 25/1368ths of the time or 125/23256ths of the time.
`
That works out to a probability of about .005 or only 1 chance in 200.
`
---------------------------------------------------------------
`
Now suppose you use 5 ranks of 4 suits.
`
Suppose the randomly generated sequence is 3,4,1,2
`
There are 4 3s and 20 cards, so you will pick a 3 first 4/20ths or 1/5 of the
time.
`
Now of those 1/5ths of the time when we successfully pick the 3 first:
`
There are 4 4s and 19 cards, so you will pick a 4 second 4/19ths
of the time.
`
So the probability of successfully picking a 3 first and a 4 second is
4/19ths of 1/5th of the time or 4/95ths of the time
`
Now of those 4/95ths of the time when we successfully pick the 3 first
and the 4 second:
`
There are 4 1's and 18 cards, so you will pick a 1 third 4/18ths or 2/9ths
of the time.
`
So the probability of successfully picking a 3 first, a 4 second, and a
1 third is 2/9ths of 4/95ths of the time or 8/855ths of the time
`
Now of those 8/855ths of the time when we successfully pick the 3 first,
the 4 second, and the 1 third:
`
There are 4 2's and 17 cards, so you will pick a 2 fourth 4/17ths of
the time.
`
So the probability of successfully picking a 3 first, a 4 second, a
1 third and a 2 fourth is 4/17ths of 8/855ths of the time or 32/14535ths of the time.
`
That works out to a probability of about .002 or only 1 chance in 500.
`
----------------------------------------------------------
`
So the probability is higher if you use four ranks and 5 suits, but
either way, the probability of succeding is very low, and even lower
if you allow repetitions in the randomly selected sequence.
-----------------------------------------------------------
Edwin

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