more difficult question

[ryan1986]

Okay, on a statistics test today a question came up along these lines, IF you draw 4 cards from a 52 card deck and do not replace them what is the probability of drawing AT LEAST one heart.
My Reasoning:
4 hearts (13/52)(12/51)(11/50)(10/49)=.00264
3 hearts (13/52)(12/51)(11/50)(39/49)=.01030
2 hearts (13/52)(12/51)(39/50)(38/49)=.03558
1 heart (13/52)(39/51)(38/50)(37/49)=.11564

Then add all the probabilities of making hands with at least one heart. Since you can hit anyone of those and have a heart.
Giving you: roughly .16 or 16%

The books way did it using a complement.
(1-The odds of making an all Non-heart hand)
(1- 39/52*38/51*37/50*36/49)= .696 or roughly 70%

I am not contesting the books answer, that is a correct way to do it but id like to know why my method didnt come out correct, if anyone can help it will be greatly appreciated. Thanks

[RocketManJames]

I think it is because you are missing the permutations within each of your categories. So, you need to multiply your percentages by (4 choose 4), (4 choose 3), (4 choose 2), and (4 choose 1). Then, I think it should work.

In Edit: Also, your last term for 1 heart, you multiplied wrong. It should be: 0.1097.

-RMJ

[irchans]

You almost got it right! If drawing four cards,


no hearts = 39/52*38/51*37/50*36/49 = 0.303818
1 heart = (13/52)(39/51)(38/50)(37/49)* 4 = 0.438848
2 hearts = (13/52)(12/51)(39/50)(38/49)* 6 = 0.213493
3 hearts = (13/52)(12/51)(11/50)(39/49)* 4 = 0.041201
4 hearts = (13/52)(12/51)(11/50)(10/49) = 0.002641.

[Crooked Paul]

Yeah, the problem is your calculations give the odds of a much more specific case, which if you put it into words would be something like:

What is the probability that ANY ONE of the following will happen when drawing four cards (without replacement)?
a. The first is a heart and the following three are not
b. The first and second are hearts and the follwing two are not
c. The first, second, and third are hearts and the last is not
d. All four are hearts.

Your figures don't cover the chance that, say, you could draw a club, a club, a heart, then a diamond.

All in all, figuring the complementary probability (no hearts) and subtracting from 1 is way easier. =)


Crooked

[gamble4pro]

Your reasoning is correct, but your probability calculus is wrong.
Using combinations in such calculations spares you of a lot of errors. So:
The favorable 4-card combinations for that event are:
(Hxxx) - in number of 13*C(39,3) = 118807
(HHxx) - in number of C(13,2)*C(39,2) = 57798
(HHHx) - in number of C(13,3)*39 = 11154
(HHHH) - in number of C(13,4) = 715.
(I denoted heart by H, x different from H - as symbol)
Totally, you have 188474 favorable combinations from C(52,4)=270725 possible. This means a 69.618% probability, so the book is right.