- #1

Cosmossos

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## Homework Statement

in a poker game we want to calculate the Probability to get differnet combinations. in a card deck there are 52 cards from 4 different series. each series has 13 cards. we assume that each card get 5 random cards.

What is the number of combinations to get the cards? (The order isn't important)

my answer: 52*51*50*49*48

what is the number of combinations if we don't distinguish?

My answer: 52C5-13*12*4 (the number of combinations to choose 5 cards - the number of combinations which repeat themselves)

What is the probability to get K,J,Q,A,Flush and all the cards are from the same series?

My answer: (52/52)(12/52)(11/52)(10/52)(9/52)

what is the probability to get a one pair (two card of the same kind)?

My answer: 3!*(52/52)(3/52)(48/52)(47/52)(46/52)

We play N times. What's the probabilty that until the Nth game we won't get one pair and in the last game we will get one pair?

My answer:[(1-3!*(52/52)(3/52)(48/52)(47/52)(46/52))^(n-1)]*(3!*(52/52)(3/52)(48/52)(47/52)(46/52))

Is it correct?

thanks