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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? 
March 16th, 2017, 09:11 PM  #2 
Senior Member 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 $P[\text{3rd card is an ace}] =\dfrac{4}{50}\dfrac{48}{52}\dfrac{47}{51} + \dfrac{3}{50}\dfrac{4}{52}\dfrac{48}{51} =\dfrac {16}{221}$ 
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 nonace, or nonace 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. 
March 17th, 2017, 10:24 AM  #4  
Senior Member Joined: Sep 2015 From: CA Posts: 934 Thanks: 504  Quote:
I think maybe what I didn't do was scale the final answer by $P[\text{0 or 1 aces in first two cards} = 1\dfrac{4}{52}\dfrac{3}{51}$  
March 17th, 2017, 01:34 PM  #5 
Senior Member 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 $\large \frac{\frac{48\ 47\ 4}{52\ 51\ 50}+\frac{48\ 4\ 3}{52\ 51\ 50}+\frac{4\ 48\ 3}{52\ 51\ 50}}{\frac{48\ 47}{52\ 51}+\frac{48\ 4}{52\ 51}+\frac{4\ 48}{52\ 51}} = \dfrac{106}{1375} \approx 7.709\%$ 
March 18th, 2017, 04:45 AM  #6  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 112 Thanks: 20  Quote:
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:
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 twochild problem to demonstrate:
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.
Last edited by JeffJo; March 18th, 2017 at 04:54 AM. Reason: Add in an answer to the card problem  
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. 

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