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#1
12-10-2002, 01:44 PM
 Huh Senior Member Join Date: Dec 2002 Posts: 385
Hold\'em Probability Puzzler

How many two-card hands, on average, will it take for you to see every card in a standard deck? I'll post my answer in a couple of days in a separate post. I'm curious to see how people get there.

Huh?

#2
12-10-2002, 08:19 PM
 Carl_William Member Join Date: Dec 2002 Location: CA & Ohio USA Posts: 70
Re: Hold\'em Probability Puzzler

Dear Huh,
You wrote:

"How many two-card hands, on average, will it take for you to see every card in a standard deck? I'll post my answer in a couple of days in a separate post. I'm curious to see how people get there.

Huh? "

I appreciate the challenge -- maybe I can solve it. Anyway -- I enjoy thinking about things like this -- maybe an excellent clue would be to solve it for a very small deck -- say 2 cards (too simple), 3 cards, 4 cards (etc.) and then proceed to 52 cards. Thanks again for the excellent post.

Most warm regards,
Carl William

PS: one could do a Bernoulli trial (binomial random variable) and determine the probability that 2 specific cards will or will not be dealt. But I think this is a counting technique problem not unlike the birthday classroom problem where the probability is determined (on average) that two or more people in a class were born on the same birthdate (i.e., month and day of month).
#3
12-10-2002, 08:33 PM
 PseudoPserious Senior Member Join Date: Oct 2002 Posts: 151
Possible Solution

I'll give it a shot...

Let's call N the number of cards you've already seen. Basically, we're looking for the average number of hands it takes to go from N=0 to N=52.

Call A(N) the average number of hands it takes to see a new card, given that you've already seen N cards.

Call P(N) the probability of having seen exactly N cards at some point during the game.

A(N)*P(N) is the expected number of hands you'll be dealt for each value of N.

If you sum A(N)*P(N) over N=0 to 51, you'll get the average number of hands it takes to see the entire deck.

Tackling A(N) first:

Given that you've seen N cards, the probability of being dealt two cards you've already seen is (N/52)*((N-1)/51). Thus, the probability on any one deal of seeing a new card is:

M(N) = 1 - (N/52)*((N-1)/51)).

We're looking for the average number of independent trials (hands) it takes for us to see our first success (our first hand with a new card). This is the geometric distribution. The mean of the geometric distribution with probability of success p is 1/p.

So, A(N) = 1/M(N)

Now for P(N):

You will NOT see exaclty N cards at some time during the game if and only if (a) you have seen exactly N-1 cards and (b) the next time you are dealt an unseen card, you are dealt two unseen cards instead of just one.

We've already defined the probability of (a) as P(N-1).

The probability of (b) is the fraction of times that you receive two unseen cards when you are dealt at least one unseen card, given that you've seen N-1 cards. We'll call this T(N-1).

Given that you've seen (x) cards, on any one deal the probability of being dealt two unseen cards is P2(x) = ((52-x)*(51-x))/(52*51). The probability of being dealt exactly one unseen card is P1(x) = 2x(52-x)/(52*51).

So, T(N-1) = P2(N-1) / (P1(N-1) + P2(N-1))

Thus, we can find the chance of skipping a value of N by P(N-1)*T(N-1). This gives us an iterative formula for the chance of seeing exactly N cards:

P(N) = 1 - P(N-1)*T(N-1)

Along with P(0) = 1 (since we have seen no cards before the first deal) and P(1) = 0 (since the first deal will always give us two unseen cards), this allows us to calculate P(N) for all values of N.

Using the above definitions, if you sum A(N)*P(N) from N=0 to 51, you get 117.09.

So, I'd have to say just over 117 hands or so...

Cya,
PP

P.S. A quickie computer sim of 20,000 games gave an average of 117.65 hands.
#4
12-10-2002, 09:02 PM
 Carl_William Member Join Date: Dec 2002 Location: CA & Ohio USA Posts: 70
Re: Possible Solution

Dear Cya,
PP

So you solved the problem for a one card hand not a two card (holdem) hand....
You indicated about 117 hands and wrote,

"P.S. A quickie computer sim of 20,000 games gave an average of 117.65 hands."

Cya, of course if your solution is valid than you can use the same approach for holdem hands.
Unless I don't understand Huh's problem statement -- I thought Huh was talking about 2 card holdem hands. There are 1326 distinct holdem hands in a 52 card deck. Therefore it has to be 1326 or more "usually way more" deals to get dealt every holdem hand combination.

Most warm regards,

Carl William

#5
12-10-2002, 10:00 PM
 PseudoPserious Senior Member Join Date: Oct 2002 Posts: 151
Re: Possible Solution

Hey Carl,

1) I'm PP, not Cya...I was saying 'see you' [img]/forums/images/icons/wink.gif[/img]

2) I think he was asking how many hold'em hands it would take on average to see every card in the deck, not how many to see every possible hold'em hand. To quote:

"How many two-card hands, on average, will it take for you to see every card in a standard deck?"

But anyways, if that's not what he was asking, that's the problem I solved [img]/forums/images/icons/smile.gif[/img]

Take care,
PP
#6
12-12-2002, 12:46 AM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: Possible Solution

Note in my solution below I essentially only compute A(N) and set P(N) = 1. All the work you did of accounting for 2 card hands instead of 1 card hands only makes a difference of less than 1 hand in the final answer. Just thought I'd make you feel good [img]/forums/images/icons/grin.gif[/img]
#7
12-14-2002, 03:42 PM
 Carl_William Member Join Date: Dec 2002 Location: CA & Ohio USA Posts: 70
Re: Possible Solution

Hi, (BruceZ, Huh, PseudoPserious)

Maybe some "FOOD for Thought"

I again want to thank you people for the posts. I am way behind in comprehending some or most of these posts (I will take my time trying to grasp them). Anyway – I wanted your opinions (I wonder) if any of these solutions are exact or are they all excellent approximations. Myself (initially), without any feedback from Huh, Bruce, or PP: I would have used brute force (Monte Carlo Techniques) to get an answer -- using hopefully an accurate algorithm for pseudo random numbers.

Food for thought:
So far I get the impression from the posts that the original Huh problem statement as an expanding tree type problem where the probabilities at each level are to be calculated and summed. An individual deal is a 2-tuple sample without replacement, and each deal is a sample with replacement. This generates a tree type sequence with an infinite series of terms. I don’t know how to handle this on an exact basis. Maybe I can eventually understand it better. I wanted to mention something to you guys and see if you considered it….

From the posts, the consensus is that on average 117 deals are required to complete the process. Also since the different (or slightly different) techniques used to come up with the answer all predict about 117, I feel that any errors introduced by the various techniques are relatively small. That said: I wanted to see if any of you guys considered working the problem (getting an answer) by starting at the end and working backwards (reverse the tree). I’m just curious if you have considered any merit to this thought – maybe it is trivial and doesn’t deserve considering. Anyway, as you know near the end of the process….

At the end of the process when only one card is missing, on average, 26 deals are required to complete the process. If two cards are missing, about 13.26 or so deals are required to get it down to one card or zero cards. Probably the sequence is about 9.02 deals req’d to get down to one, two, or remain at three cards; and about 6.9 deals req’s to get four cards down to two, three, or remain at four cards (and so on and on). The numbers of deals required at each level reduces eventually (as you know) to about 1.92307 to generate the tree at the initial level. My interest in this was that most of the deals are req’d near the end of the process which may give some additional incite to the nature of the problem.

Also I had a question. Can some this problem be considered a Markoff chain (process), that is does it have Markovian properties.

Most warm regards,

Carl
#8
12-10-2002, 09:51 PM
 Carl_William Member Join Date: Dec 2002 Location: CA & Ohio USA Posts: 70
Re: Hold\'em Probability Puzzler

Huh,

I don't know how to solve your problem definition yet -- at least in a closed analytical form. I do know how to solve the problem using brute force (Monte Carlo technique). Just randomly deal out integer numbers from 1 to 1326 until all 1326 numbers are in this sequence sample. Then count how many (deals) it took to get this sample. Repeat this sequence sample process over and over, each time recording how many deals were required. Keep a running average of the deals required for each sequence -- when the running average converges within say plus or minus one deal you will have a pretty accurate answer to your problem.

Regards
Carl William
#9
12-10-2002, 10:04 PM
 PseudoPserious Senior Member Join Date: Oct 2002 Posts: 151
Re: Hold\'em Probability Puzzler

Heya Carl,

I'm in a bit of a rush, so I haven't thought about it very much, but why can't you can solve your interpretation of Huh's problem using the same method I did above?

PP
#10
12-11-2002, 01:20 AM
 Carl_William Member Join Date: Dec 2002 Location: CA & Ohio USA Posts: 70
Re: Hold\'em Probability Puzzler

PP, you are correct -- I misread the problen definition. One of my faults. You wrote:

"I'm in a bit of a rush, so I haven't thought about it very much, but why can't you can solve your interpretation of Huh's problem using the same method I did above?"

I will think about the correct definition. You are probably right all the way. Thank you.

Carl

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