
Probability puzzle  Patti

 Maybe I'm dumb (don't answer that), but it would seem to me that as long as the Ace that appeared first was not the ace of spades, they are equally likely show up as the next card. If it was the ace of spades, there is a much stronger likelyhood that the two of clubs is the next card. :)
R  (Reply) (Thread) 
 Likelier to be the ace of spades. The two of clubs might already have been turned, but the ace of spades has not.  (Reply) (Thread) 
 Haven't seen other answers, but the one that seems obvious to me is that there's a 1/4 chance you just burned the ace of spades, and you know nothing about the 2c, so the 2c is more likely.
Thinking for a few more minutes, I guess I need to actually calculate. 1) p(first ace is spade and 2c has passed) = 1/4 * 1/2 2) p(spade and 2c not passed) = 1/4 * 1/2 3) p(nonspade) 3/4 sum = 1. good.
For case 1, there is a 0% chance of As or 2c. For case 2, there is a 0% chance of As, and some chance of 2c. I _think_ that chance is 1/25.5 on average, but haven't thought it through yet. For case 3, there is an equal chance of As and 2c.
This would sum to 1/8 * 1/25.5 = about halfpercent more likely to be 2c.
I'm not sure what your last sentence means. The probability before the first card is turned is conditional on seeing the first ace, isn't it?
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 I just pulled the puzzle verbatim off of RGP, so your assessment of the last sentence is as valid as mine.  (Reply) (Parent) (Thread) 
 Hm, no.
If the As is first, then chance of it being next is 0. Check
But if another A was first, then the As is guaranteed not to have come yet, but the 2c could already have gone by.
I like your 51! decks, but I think they're actually 52*50! decks. Even so... you're doublecounting. Some of your As2c decks are also (first) AxAs decks.
I think your implied answer is correct but your reasoning isn't quite complete. ;)  (Reply) (Parent) (Thread) 
 Consider the limiting cases.
You pull all but 4 cards off the deck without seeing an A. Now you must turn over an Ace. Now with P=3/4 the next card will be the As, since 1/4 of the time you already pulled it, and P=0 that the next card will be the 2c, since only A's are left.
You pull the first card off the deck and it is an Ace. There are two cases: It is the As, it is not the As. If it is the As, P=0.0 that the next card is the As, P=1/51 it is the 2c. If it is not the As, then P=1/51 that it is the As, and P=1/51 that it is the 2c.
Weighting the two possibilities in the latter case, P= 1/51 * 3/4 + 0.0 * 1/4 that it is the As, and P= 1/51 that it is the 2c. So it is more probable in the latter case that it is the 2c, while in the former case it is less probable.
Therefore the answer to the question is a function of the number of cards you have pulled. At some depth there is a crossover point.  (Reply) (Thread) 
 And the question (sort of) specifies that you have no idea how many cards you'll pull before you get there.
I have every reason to believe that the problem can be reduced to a fivecard deck.  (Reply) (Expand) (Parent) (Thread) 
 The 2c is more likely, 16/31 vs 15/31 for the As.
The probability it will be the 2c:  There is a 20% chance the 2c will arrive before any ace, in which case there is 0% probability the 2c will be the first card after the first ace.  An ace arrives before the 2c (80% probability), say with N cards remaining in the deck. There is a 1/N probability the next card is the 2c. So the probability the first card after the first ace will be the 2c is 0% + 80%/N = 80%/N.
The probability it will be the As: There is a 25% chance the first ace will be As, it which case the probability is zero that the next card is As. There is a 75% chance the first ace is not the As. If so, and there are N cards remaining, the probability the next card is As is 1/N. So the probability the first card after the first ace is As is 75%/N.
S0 80%/N divided by 75% over N makes the 2c a 16/15 favorite.  (Reply) (Thread) 
 I'm pretty sure this is wrong, and the 2c is a more slender favorite, and I think it's because your Ns aren't the same. For the 2c reasoning, N is the cards remaining in the deck, guaranteed to include the 2c and the other 3 aces. For the As reasoning, N is the cards remaining in the deck, only possibly including the 2c, but including the other 3 aces.
The 5 card deck makes the difference clear. There, your percentages are correct, but the first N is definitely 4 and the second N is 4 or 3, 80% chance of 4.
IOW, your second number is not 75%/N, but 75%/M, which turns out to equal 75%/N * N/(N+1) + 75%/(N1) * 1/(N+1) ... I think. ;)  (Reply) (Parent) (Thread) 
 So even though I have done 0 math, apparently I have the answer 100% correct as evidenced by all the math people below? Yay me.
Where is your Tom Bayes now bitches?
R  (Reply) (Thread) 
 I don't see that you gave an actual answer; you simply made two statements which don't remotely constitute a quantitative answer :)  (Reply) (Expand) (Parent) (Thread) 
 Well, I missed it on first glance too. Thought 2c was more likely, because you've eliminated the As by it being the first ace sometimes. But the arguments others made convinced me, and then (as usual, much later) I saw (an) easy way to look at it.
If you concentrate on what the first of the five interesting cards is, you can see that everything is the same if it's any of AcAdAh. So we're left with the cases where the first of the five is
a) the As (in which case 2c wins if it's the first of the remaining four interesting cards, and there are no uninteresting cards between the As and the 2c)
b) the 2c (in which case As wins if it's the second of the remaining four interesting cards, and there are no uninteresting cards between the first of the remaining four interesting cards and the As.)
Ok, that doesn't spell out as cleanly as I'd like, but it's "clear" that the two cases have equal probability.
I bet there's an even easier way to see it. I'll think about it...  (Reply) (Thread) 
 I think this is still wrong.
We're talking about the As or the 2c being the *next* card, not the next *interesting* card.
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 When you have reached an Ace, the chance of drawing any of the cards that remain is equally likely. So the problem is to compare the chance that you've already passed the Ace of Spades to the chance that you've already passed the 2 of clubs.
What's the chance that you've already passed the Ace of Spades? This happens exactly when the first of the four aces to appear in the deck is the Ace of Spades, so 1/4.
What's the chance that you've already passed the Deuce of Clubs? This happens exactly when the first of the five "interesting cards" to appear in the deck is the Decue of clubs, so 1/5.
Since the chance that you've already passed the Ace of Spades is greater than the chance that you've already passed the Deuce of Clubs, the chance that the next card is the Deuce of Clubs is greater than the chance that the next card is the Ace of Spades.  (Reply) (Thread) 
 Once we get to our first Ace, it of course doesn't matter which of the remaining cards we turn over next, since the deck has been shuffled. So instead of turning over the next card, let's turn over the bottom card.
If we turn over cards until we reach an Ace, then turn over the bottom card, what's the chance that it's the Deuce of Clubs? 1/52.
If we turn over cards until we reach an Ace, then turn over the bottom card, what's the chance that it's the Ace of Spades? 1/52
So the two probabilities are equal.  (Reply) (Thread) 
 I've just posted persuasive arguments for two different answers, so at least one of them must be wrong. If you really understand the problem, you should not only be able to determine what the right answer is: you should be able to clearly explain the fallacy in the incorrect argument. Otherwise, when solving similar problems in the future, you can be misled by arguments similar to the incorrect argument I've posted, when you're not fortunate enough to stumble on a more persuasive argument that leads to the correct answer.
I think that clearly explaining the fallacy in the incorrect argument is considerably more difficult than figuring out the correct answer to the original problem.
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 Absolutely!
The flaw in the first argument is that you're double counting some situations, where the card order is 2c...As...the rest of the aces. That is, the two situations you give are not mutually exclusive. To make them exclusive, you need to rephrase the first one as "what's the chance that you've already passed *only* the As (and not the 2c yet.)"
At which point it's clear that the probability for the first one is 1/5, just like the second one.  (Reply) (Expand) (Parent) (Thread) 

Probability puzzle  Patti

