Probability (cards)

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July 18th 2012, 03:24 AM
Rikimaru

Probability (cards)

Just need confirmation for my answers (Nod)! Thank you!

1. The chance that the top card in a well shuffed deck is a two or a nine is:
8/52 = 15.4% (4 twos and 4 nines in deck)

2. The chance that the eighth card from the top of the deck is a heart or a diamond is:
26/52 = 50% (13 hearts & 13 diamonds in deck); assuming we don't know the first 7 cards; the position doesn't matter here.

3. The chance that the top card in the deck is a two or a nine given that the eighth card from the top is a heart or a diamond is:
Wouldn't it still be 15.4%? Unless they mention that the eighth card is a two or a nine, it wouldn't matter right?

4. Are the events that the top card is a two or a nine and that the eighth card is a heart or a diamond independent?
Yes (I'm not sure entirely on this one. An explanation would be great! Other options: No or Cannot Determined Without Knowing The Other Cards)

Thanks a bunch!
July 18th 2012, 10:55 AM
pratique21

Re: Probability (cards)

1 and 2 are fine
for 3, i think the chances will be affected. Consider 2 cases:
1:- the card we know of is neither 2 or 9
2:- the card we know of is either 2 or 9 (of H or D)
so
P= (top card 2 or 9)*(case 1) + (top card 2 or 9)*(case 2)
or (8/52)*(22/26)+(7/52)*(1/4)

where 22/26 is chances that ~(2,9) was selected from D and H
and 1/4 is chance that either of 2H,9H,2D or 9D was selected.

about 4, no they are not independent because if the H or D is a 2 or 9, the chances of top card being 2 or 9 are affected.

Hope it helps!
July 18th 2012, 12:36 PM
Rikimaru

Re: Probability (cards)

For 3, there could only be one answer... so I'm not sure... I understand that there could be 2 scenarios: (1) neither 2 or 9 is in the 8th position (2) either 2 or 9 is in the 8th position. That being said, the information is limited? Could we assume that it's not 2 or 9 then? I'm confused :(! Help please!

Thanks you for the confirmation on 1, 2, 4 :)!!!
July 18th 2012, 12:36 PM
Soroban

Re: Probability (cards)

Hello, Rikimaru!

Quote:

3. The probability that the top card in the deck is a 2 or a 9, given that the 8th card from the top is a $\heartsuit$ or a $\diamondsuit.$

Bayes' Theorem: . $P(\text{top }2 \vee 9\,|\,\text{8th }\heartsuit \vee \diamondsuit) \;=\; \frac{P\big([\text{top }2 \vee 9]\:\wedge\:[\text{8th }\heartsuit \vee \diamondsuit]\big)}{P(\text{8th }\heartsuit\text{ or }\diamondsuit)}$ .[1]

The numerator requires some work.
The probability that the top card is a 2 or 9 depends whether the 8th card is the $2\heartsuit,\,9\heartsuit,\,2\diamondsuit,\text{ or }9\diamondsuit.$

[1] If the 8th card is not one of those 4 cards: . $P = \frac{48}{52} \,=\,\frac{12}{13}$
. . .Then the top card is a 2 or 9 with probability $\frac{8}{51}$
. . . $P(\text{Case 1}) \:=\:\frac{12}{13}\cdot\frac{8}{51} \:=\:\frac{96}{663}$

[2] If the 8th card is one of those 4 cards: . $P = \frac{4}{52} \,=\,\frac{1}{13}$
. . .Then the top card is a 2 or 9 with probability $\frac{7}{51}$
. . . $P(\text{Case 2}) \:=\:\frac{1}{13}\cdot\frac{7}{51} \:=\:\frac{7}{663}$

The numerator is: . $\frac{96}{663} + \frac{7}{663} \:=\:\frac{103}{663}$

The denominator is: . ${(\text{8th is }\heartsuit\text{ or }\diamondsuit) \:=\:\frac{26}{52}\:=\:\frac{1}{2}$

Therefore, [1] becomes: . $\dfrac{\frac{103}{663}}{\frac{1}{2}} \;=\;\frac{206}{663}$

Quote:

4. Are the events that the top card is a 2 or a 9 and that the 8th card is a $\heartsuit$ or a $\diamondsuit$ independent?

No, as expained in part 3, the two events are dependent.
July 18th 2012, 12:46 PM
Rikimaru

Re: Probability (cards)

So the final answer for 3 would be 206/663? What I'm confused here is that there are multiple scenarios... I think I might need to reread Bayes' Theorem since I recall seeing it in the book. Thanks for the clear up...!

Um... I just resubmit the assignment with 206/663... and it seems like my previous answer: 15.38461% was right O_O!? Now I'm even more confused :(!
July 18th 2012, 01:47 PM
tom@ballooncalculus

Re: Probability (cards)

Quote:

Originally Posted by Rikimaru

it seems like my previous answer: 15.38461% was right O_O!? Now I'm even more confused :(!

Yes, your intuition was right. How could it matter??

Quote:

Originally Posted by Soroban

[1] If the 8th card is not one of those 4 cards: . $P = \frac{48}{52} \,=\,\frac{12}{13}$

[2] If the 8th card is one of those 4 cards: . $P = \frac{4}{52} \,=\,\frac{1}{13}$

But don't forget that it's given that the 8th card is ${\heartsuit\text{ or }\diamondsuit}$ , so we want

[1] If the 8th card is not one of those 4 cards: . $P = \frac{22}{26} \,=\,\frac{11}{13}$

[2] If the 8th card is one of those 4 cards: . $P = \frac{4}{26} \,=\,\frac{2}{13}$

... as pratique21 had it... although that poster seems to have somehow cancelled the second fraction down to a quarter, and also neglected to reduce the denominators under the 8 and 7 to 51 instead of 52.

But for those slips, their calculation would have produced a fraction equal to 8/52.

And, consequently, the two events are independent.
July 18th 2012, 09:15 PM
pratique21

Re: Probability (cards)

tom@ballooncalculus , thanks for pointing that out.
However, could you explain how you came to conclude that the events are independent ?
July 19th 2012, 02:40 AM
tom@ballooncalculus

Re: Probability (cards)

Quote:

Originally Posted by pratique21

tom@ballooncalculus , thanks for pointing that out.
However, could you explain how you came to conclude that the events are independent ?

Quote:

Originally Posted by tom@ballooncalculus

But for those slips, their calculation would have produced a fraction equal to 8/52.

And, I might have added, they were assuming...

Quote:

Originally Posted by pratique21

2 cases:

(of H or D)

... i.e. that the 8th card is Hearts or Diamonds, which will happen in precisely half of the outcomes. So we have...

$P\big([\text{top }2 \vee 9]\:\wedge\:[\text{8th }\heartsuit \vee \diamondsuit]\big)\ =\ \frac{8}{52} \cdot \frac{1}{2}\ =\ P(\text{top }2 \vee 9)\cdot P(\text{8th }\heartsuit \vee \diamondsuit)$

... satisfying...

$P(A \cap B) = P(A) \cdot P(B)$

Independence (probability theory) - Wikipedia, the free encyclopedia

Just in case a picture helps...

http://www.ballooncalculus.org/draw/misc/prob3.png
July 19th 2012, 03:10 AM
pratique21

Re: Probability (cards)

tom , kindly refer to the post by soroban, you will find that p(A and B) not equal to p(A)*p(B), so the events are dependent

Logically, whether the eighth card that is drawn is a 2H,2D,9H or 9D is affecting the chances of top card being a 2 or a 9.
So these events ca not possibly be dependent !
July 19th 2012, 05:15 AM
tom@ballooncalculus

Re: Probability (cards)

Quote:

Originally Posted by pratique21

Logically, whether the eighth card that is drawn is a 2H,2D,9H or 9D is affecting the chances of top card being a 2 or a 9.

True. But that isn't the question. The question is...

Quote:

Originally Posted by Rikimaru

3. The chance that the top card in the deck is a two or a nine given that the eighth card from the top is a heart or a diamond:

(Happy)
July 25th 2012, 05:51 AM
markdavis938

Re: Probability (cards)

For 3, there could only be one answer... so I'm not sure... I understand that there could be 2 scenarios:

Loans and Grants