Winning at least two out of 5 trials, trials dependent on each other.
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
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?
2 Replies
This can be solved using combinations:
For exactly 2 clubs out of the 6 remaining when drawing from a 30 card deck, the probability is C(6,2)*C(24,3)/C(30,5) = 21.3%
On calculating 2 or more clubs , I get 25.4%
The "trials" (individual cards) being dependent is essentially a red herring.
As statmanhal posted, in problems like this the "dependence" vanishes if you consider drawing the final 5 cards all at once (as a set).
Then standard combinatorics apply such as combinations (getting a certain number of clubs out of the 6 remaining clubs when you choose a set of 5 cards of the remaining 30 cards).
