Re: [Phys-l] probability problem

["Ann Reagan" <areagan@csmd.edu>]

WARNING: Spoiler below for those still thinking about Carl's card 
problem. 

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Kudos to John for his analysis. I had a much-less elegant, low-tech approach to the same result:

Case 2: Consider two cards dealt from a deck of 52. Call the cards' positions "A" and "B". There are 51 possible combinations with the ace of Spades dealt into the "A" position, 51 combinations with the ace of Hearts in the "A" position, 51 with the ace of clubs, and 51 with the ace of diamonds in the "A" position. Likewise there are 51 combinations possible with the ace of spades in the "B" position, minus the three already counted (i.e., the three combinations already counted because there is an ace in the "A" spot), gives 48 unique combinations with the ace of spades in the B position. Likewise, there are 48 previously-uncounted combinations with the ace of hearts, 48 with the ace of clubs, and 48 with the ace of diamonds in the "B" position. Thus, there are (4 x 51) + (4 x 48) = 396 unique combinations with at least one ace in the A or the B spot. Of these possibilities, there are exactly 12 ways that both the "A" and the "B" spot would contain an ace. The odds that both cards are aces, given that at least one is an ace, is therefore 12/396 = .030303

Case 3: Given that one card is the Ace of Spades, there are 51 possible combinations of two cards, 3 of which result in two aces in the hand. Odds are therefore 3/51 = .0588

> I easily programmed a spreadsheet to deal me a million two
> card hands in 20 sets of 50,000. After throwing out the 19407
> of 'em that happened to consist of two identical cards, I found
> that 
>
> Case 2) 0.0304 +/- 0.0006 of the hands with at least one ace
> had two aces. (Expected value = 1/33 = 0.0303) 
> Case 3) 0.0592 +/- 0.0014 of the hands with the ace of spades
> had two aces. (Expected value = 3/51 = 1/17 = 0.588) 
>
> (Quoted uncertainties are standard deviations of the mean for the
> twenty trials.) 

> John Mallinckrodt 
> Cal Poly Pomona 

On Jun 28, 2010, at 6:35 AM, Carl Mungan wrote: 

> The following problem comes out of Boas. I have a solution on my 
> website. From time to time, someone will email me and say my solution 
> is nonsense. In any case, it's a cute problem: 
>
> --- 
>
> You're sitting across from a dealer. He shuffles a single deck of 
> cards and deals you two cards face down. He then looks at them 
> without showing them to you. Consider the following three distinct 
> scenarios: 
>
> 1. He tells you nothing. 
> 2. He tells you, "You've got at least one ace." 
> 3. He says, "Wow, you've got the ace of spades." 
>
> For each of these three scenarios, what is the probability that if 
> you now turn over the two cards you'll find that you've got two aces? 
> IOW, what odds would you take to bet on it? 

Dr. Ann M. Reagan
Adjunct Faculty
Department of Math/Physics/Engineering
College of Southern Maryland, Leonardtown Campus