04-19-2024 , 03:35 PM
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?
04-19-2024 , 04:00 PM
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%
04-19-2024 , 04:33 PM
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).

m