I'm a bit lost how to handle this. The situation is the following:
Card game with 36 cards (9 of each suit), 6 cards are known from the start. 5 more cards will be revealed.
If I start with 3 of one suit, what are the odds of getting at least 2 more of the same suit, in the final five cards?
So if I have 3 clubs, there are 6 clubs left out of 30 unknown cards.
I found this on another forum
Quote:
There are several ways to calculate this.
The probability of winning the game at least twice is equal to the sum of the probabilities of winning the game 2, 3, 4 and 5 times out of 5 trials.
P ( W twice) =(0.20)^2(0.80)^3 * 5C2=0.2048
P (W thrice) =(0.20)^3(0.80)^2 *5C3=0.0512
P (W 4 times) =(0.20)4(0.80)5C4=0.0064
P (W 5 times) =(0.20)5=0.0003
Therefore, P (at least twice)=0.2048+0.0512+0.0064+0.0003=0.2627
But since the trials are dependent on each other, I cant figure out which formula to use?
