My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Reply
 
LinkBack Thread Tools Display Modes
March 16th, 2017, 05:58 AM   #1
Newbie
 
Joined: Mar 2017
From: Usa

Posts: 5
Thanks: 0

Partially known cards

I posted this yesterday but it hasn't shown up, apologies if it ends up being double (don't know if there's a waiting period here.)

Randomly shuffled deck of 52 cards. Someone removes the top 2 and tells you truthfully that the 2 cards are not both aces. (They could be one ace or zero aces in those 2 cards). What is the probability that the next card off the deck is an ace?

It seems to me that while you don't know what either of the first 2 cards are, you know one of them is not an ace. Therefore, you can treat the problem as if there is one known card where an ace is not, and therefore the 4 aces must be in the remaining 51 unknown cards. So the chance the next card is an ace is 4/51? Is there any hole in this thinking? Is there another more "concrete" way of calculating this?
Jeffnc is offline  
 
March 16th, 2017, 09:11 PM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 934
Thanks: 504

so you know that the top two cards contain 0 or 1 ace.

$P[\text{3rd card is an ace}] = P[\text{3rd card is an ace}] | \text{ 0 aces in first two cards}]P[\text{0 aces in first two cards}] +

P[\text{3rd card is an ace}] | \text{ 1 ace in first two cards}]P[\text{1 ace in first two cards}]$

$P[\text{3rd card is an ace}] | \text{ 0 aces in first two cards}] = \dfrac{4}{50}$

$P[\text{0 aces in first two cards}] = \dfrac{48}{52}\dfrac{47}{51}$

$P[\text{3rd card is an ace}] | \text{ 1 ace in first two cards}] = \dfrac{3}{50}$

$P[\text{1 ace in first two cards}] = \dfrac{4}{52}\dfrac{48}{51}$

So combining all these we get

$P[\text{3rd card is an ace}] =\dfrac{4}{50}\dfrac{48}{52}\dfrac{47}{51} + \dfrac{3}{50}\dfrac{4}{52}\dfrac{48}{51} =\dfrac {16}{221}$
romsek is offline  
March 17th, 2017, 08:42 AM   #3
Newbie
 
Joined: Mar 2017
From: Usa

Posts: 5
Thanks: 0

romsek, thanks for your answer. I'm going over it, but I think it's incorrect. This is logical if we look at the final result.

If we have no knowledge at all of the first 2 cards off the deck, then if we peel the 3rd card off the deck, the chance it's an ace is 4/52.

4/52 = 7.69%

If all we know is that it is not the case that the first 2 cards are both aces, then according to your calculation, P[3rd card is ace] is 16/221.

16/221 = 7.24%

It is not possible that P[3rd card is an ace] decreases when all we know is that the first 2 cards were not both aces. Logically, this value must be higher than 7.69%, even if only slightly.

(My answer of 4/51 = 7.84%.)

Specifically, it appears as though your calculation for P[1 ace in first 2 cards] is not correct. It must rather be (4/52)(48/51) + (48/52)(4/51) (ace first then non-ace, or non-ace first then ace), or equivalently 2*(4/52)(48/51).

Further, you're not accounting for the fact that we know that the first 2 cards are not aces. You are calculating P[0 aces in first 2 cards] and P[1 ace in first 2 cards]. If you add up your values (including the correction just noted), you get a chance of 99.55%. This would be true if we had no additional knowledge. However in this case, these 2 probabilities have to add up to 100% because we know these are the only possible cases. P[both cards are aces] = .45% ( (4/52)(3/51) ), and this needs to be accounted for.

Last edited by Jeffnc; March 17th, 2017 at 09:05 AM.
Jeffnc is offline  
March 17th, 2017, 10:24 AM   #4
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 934
Thanks: 504

Quote:
Originally Posted by Jeffnc View Post
romsek, thanks for your answer. I'm going over it, but I think it's incorrect. This is logical if we look at the final result.
I'll take a look at it later but it sounds like you are on top of this problem enough to solve it yourself.

I think maybe what I didn't do was scale the final answer by

$P[\text{0 or 1 aces in first two cards} = 1-\dfrac{4}{52}\dfrac{3}{51}$
romsek is offline  
March 17th, 2017, 01:34 PM   #5
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 934
Thanks: 504

I think I made two errors. One in not scaling by the probability of getting 0 or 1 aces and one in not including both ways of getting a single ace.

I believe the final probability is

$\large \frac{\frac{48\ 47\ 4}{52\ 51\ 50}+\frac{48\ 4\ 3}{52\ 51\ 50}+\frac{4\ 48\ 3}{52\ 51\ 50}}{\frac{48\ 47}{52\ 51}+\frac{48\ 4}{52\ 51}+\frac{4\ 48}{52\ 51}} = \dfrac{106}{1375} \approx 7.709\%$
romsek is offline  
March 18th, 2017, 04:45 AM   #6
Senior Member
 
Joined: Apr 2015
From: Planet Earth

Posts: 112
Thanks: 20

Quote:
Originally Posted by Jeffnc View Post
Randomly shuffled deck of 52 cards. Someone removes the top 2 and tells you truthfully that the 2 cards are not both aces. (They could be one ace or zero aces in those 2 cards). What is the probability that the next card off the deck is an ace?
This is just a more complicated version of the two-child problem. I'll address a version of it, to show you where you are wrong in your "4/51" answer.
I have exactly two children, and they are not both girls. What is the probability that both are boys?
Your logic, applied to this problem would go something like this:
You know that one of my two children is not a girl. Therefore, you can treat the problem as if there is one known child that is not a girl, and so is a boy. The other child has a 50% chance to be a boy, and if it is I have two boys. So the answer is 50%.
There are four possible combinations for my family, {BB, BG, GB, GG}. The statement "you can treat the problem as if there is one known child that ... is a boy" is saying that you know I have BX of XB, where X is the unknown. So BX could be BB or BG, and XB could be BB or GB.

Did you notice something wrong there? I counted BB twice. In your solution to the card problem, you essentially counted two cases:
  1. C1 was not an ace, and C2 could be anything including an ace.
  2. C2 was not an ace, and C1 could be anything including an ace.
Each possible set of 2 cards that has 1 ace is counted in one case, but each set that has none is counted in both. You have to count then only once each.

Finally, you have not given enough information to answer the question. We need to know why this person told you that the first two cards were not a pair of aces. Again, I'll retreat to the two-child problem to demonstrate:
  1. Me: "I have two children." You: "Are any of them boys?" Me: "Yes."
    You know that I could have any of three equally-likely combinations, {BB,BG,GB}. So the answer is 1/3.
  2. Me: "I have two children." You: "At least one of the statements 'at least one is a boy' or 'at least one is a girl' must be true. Make one such true statement." Me: "At least one is a boy."
    Since you don't know how I would choose if both statements were true, you must assume I'd say "boy" with probability 1/2. So BG and GB count as only half a case each, and the answer is 1/(1+1/2+1/2)=1/2.
  3. Me: "I have two children, and at least one is a boy."
    You decide which of the previous cases is a better model for this.

Now, if you asked the person "are both aces?" which is the version you addressed, then there are 1326 (52*51/2) possible combinations for the first two cards. But the 6 ways to get a pair of aces are eliminated, leaving 1320: 192 with one ace, and 1128 with none.
  • In the 192 with one ace, there is a 3/50 chance that the third card is an ace. Total probability for this case, (192/1320)*(3/50)=12/1375.
  • In the 1128 with no aces, there is a 4/50 chance that the third card is an ace. Total probability for this case, (1128/1320)*(4/50)=94/1375.
  • The chances that the third carD is an ace are (12+94)/1375=106/1375~=7.7%.

Last edited by JeffJo; March 18th, 2017 at 04:54 AM. Reason: Add in an answer to the card problem
JeffJo is offline  
March 18th, 2017, 06:12 AM   #7
Newbie
 
Joined: Mar 2017
From: Usa

Posts: 5
Thanks: 0

romsek, Jeff, while waiting and hoping to hear back here, I went about the problem this way. The "problem" with this approach is that it's not clean because it doesn't involve the actual fractions.

Without knowing anything about the first 2 cards, the probabilities are as follows:
- P(AA) = (48/52)(47/51) = .85068
- P(Ax) = (4/52)(48/51) = .07240
- P(xA) = (48/52)(4/51) = .07240
- P(xx) = (4/52)(3/51) = .00452

Since we know P(AA) is actually 0, we need to redistribute that probability among the other 3 answers, and that needs to be done proportionally, so that they now add up to 1. (I believe this is all correct, but comments welcome.) P(xx) represents 85.454% of the total amount of these 3, and P(Ax) and P(xA) 7.240% each (significant digit error). Therefore P(xx) gets .00386 added to it, and the other two .00033 each. So now
- P(xx) = .85454
- P(Ax) = .07273
- P(xA) = .07272

These do add up to 1. We now use these to weight the probabilities of the next card coming off the deck.

.85454(4/50) + (.07273 + .07273)(3/50) = 7.709%

Which is precisely romsek's new answer, and Jeff's as well. Which does satisfy my "condition" that the answer must intuitively be greater than 4/52 = 7.692%!

Obviously I don't like the method of this solution, nor did I like the fact that it didn't = (4/51)!

Thanks both for your explanations!! Well done.

Last edited by Jeffnc; March 18th, 2017 at 06:25 AM.
Jeffnc is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
cards, partially



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
partially inelastic collisions rath Physics 0 August 19th, 2016 01:33 PM
lattices and partially ordered sets dksellou Applied Math 0 October 19th, 2013 09:43 AM
Partially ordered sets and relations g.s.t Applied Math 1 November 1st, 2012 12:20 AM
Subset of a partially ordered set is partially ordered? xYlem Applied Math 1 January 27th, 2012 05:24 PM
Horizontal vessel partially filled problem abc420 Algebra 1 April 8th, 2010 08:09 AM





Copyright © 2017 My Math Forum. All rights reserved.