Re: [Phys-l] probability problem

[Stefan Jeglinski <jeglin@4pi.com>]

> 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:

If ones want to do simple counting (paper and pencil version of JM's 
spreadsheet), imagine a deck that has 3 suits (spade/heart/club), and 
only the Ace and 2. You can then actually write out all the 
combinations (30 deals). I did this and then generalized to 52 cards 
and 4 suits. Solution far below.

> IOW, what odds would you take to bet on it?

I do have a quibble here: I understand the intent of the rephrased 
question, but in practice it depends on what odds are given as to 
what bet I would take. A perfectly fair game pays out in inverse 
proportion to the probabilities (1 in 10 odds pays 10:1). But like in 
Vegas, the dealer might give out odds that match the gambler's 
intuition but not the cold math. I would have to ask the dealer what 
odds he was paying on each scenario, and judge whether each scenario 
was in my favor or his. And then, just like in Vegas, I'd probably 
place my money on the wrong one, because of a stupid hunch :-)


Also, I'd like to see the JDenkerian approach, wherein one measures 
entropy by asking (for example, I think) what is the fewest number of 
binary questions required to discover the extent of the dealer's 
knowledge (am I even in the ballpark here?). In bits, this should be 
somehow related (inversely?) to each one of the probabilities 
calculated. For example, in scenario 1, I don't know how many 
questions need to be asked, but I'm pretty sure it's the highest 
number of all the scenarios. I'm sitting here now trying to figure 
that all out, but it could take me weeks of confusion.


Stefan Jeglinski


=================================

> 1. He tells you nothing.

short deck:  (3/6) *  (2/5) =  6/30 [1]
full  deck: (4/52) * (3/51) = .00452

[1] only 6 hands out of 30 have 2 aces

> 3. He says, "Wow, you've got the ace of spades."

It's either the 1st card or the 2nd card. To that we can assign a 
probability of 1 because you have been told so. The dealer may know 
that you have a second ace, but he hasn't given any information about 
it, so you may in a sense assume that he doesn't know. Thus:

short deck: 1 *  (2/5) =  (2/5) * 1 =  4/10 [2]
full  deck: 1 * (3/51) = (3/51) * 1 = .05882

[2] only 10 hands have Ace of spades, of which only 4 have 2 Aces


> 2. He tells you, "You've got at least one ace."

This one's tricky. You don't know if you have 2, but you do know you 
have 1, so from your point of view you might think the answer is the 
same as for 3. But the dealer is telling you he may actually further 
know that you have 2 aces. There is more information but it is not 
straightforward to measure (so I just counted):

short deck: 6/24 [3]
full deck : 12/396 = 1/33 [4]


[3] 2*3*1*3+6 (24) hands out of 30 have at least 1 ace,
	of which only 6 have both

[4] 2*4*12*4+12 (396) hands out of 51*52 have at least 1 ace,
	of which only 12 have both