Re: 2+2 Poker Probability Question

From: Stephen J. Herschkorn (sjherschko_at_netscape.net)
Date: 03/10/05 

Date: Wed, 09 Mar 2005 19:58:09 -0500

[cc'd to posters whom I quote]

irchans wrote:

> At twoplustwo.com, ajizzle posted the following probability
>question: "You are playing with a 52 card deck. Assuming a random
>shuffle, you deal cards face up from the top until the first Ace is
>dealt. Which has a higher probability of being the next card off the
>deck, the Ace of Spades or the Deuce of Clubs? "
>
> Several responders (pzhon, BruceZ, and icepoker) proved that the
>probability of the next card being the Ace of Spades is 1/52 and the
>probability of the next card being the Deuce of Clubs is 1/52.
>

The answer of 1/52 in each case is correct.

Let a = P{card after first ace is AS};
let c = P{card after first ace is 2C}.

Note that, by symmetry, a is the probability for any particular ace
and c is the probability for any particular non-ace. Thus,

4a + 48c = 1. (*)

A priori, it is not clear that a = c.

Let A be the event that the card after the first ace is AS; let X =
the position (1 through 49) of the first ace, and let B be the event
that the first ace is the AS. [Aside: Determine EX without much
computation. That has always been a favorite exercise for me.]

PA = EP(A | X) = E[(3/4) P(A | X, not B)] = 3/4 E[1/(52-X)]

P{X=n} = P(nth card is ace | first n-1 cards are not aces) P{first n-1
cards are not aces}
= 4/(53-n) * C(48, n-1) / C(52, n-1)
= (52-n) (51-n) (50-n) / 1624350

Thus, PA = sum(n=1..49, n (n+1)) / 2165800 = 1/52. (I did a change of
variable in the summation.)

(*) then implies that c = 1/52 as well.

It is surprising that a = c, and off-hand, I have no intuition as to
why this should be so. (J. Moriss's calculation is nice, but I still
get no intuition from it.)

Some comments on others' responses:

Singletoned wrote:

>If you deal until the first Ace and then look at the next card, there
>is a 1/68 chance that it is the Ace of Spades, but only a 1/51 chance
>that it is the 2 of clubs.
>
>
>There is a 1/4 chance that the first ace dealt will be the Ace of
>Spades, after which, the next card can't be the Ace of Spades. After
>that there is a 1/51 chance that the next card is an Ace of Spades,
>which I work out as being a 1/68 chance.
>
>

False. To get your figure of 1/51, you are ignoring the fact that all
the cards before the first ace are not aces.

>When you deal the first ace the next card can't be that first ace, so
>there are 51 unknown cards left, and there is an equal chance that the
>next one is the 2 of clubs, so 1/51.
>
>
>The odds vary wildly depending on what your stopping criteria is. If
>you say "deal until you have dealt the Ace of Spades, what is the
>chance that the next card is the Ace of Spades" you have a 0 chance.
>If you say "deal until you have dealt an ace, what is the chance that
>the next card is an ace" then you have a 3/51 chance.
>
>
Ditto.

>I'd love to see what proof they came up with to say that it was 1/52.
>

And now you have seen several.

The Qurqirish Dragon wrote:

>Let us assume that the first ace is at position x, so the card we are
>checking is at position x+1.
>The probability of it being the ace of spades is:
>(52-x)/52 * 1/(52-x) = 1/52
>
>The first term is the probability that the ace of spades is not among
>the first x cards, the second is the probability that it is the next
>card up (a specific card out of the 52-x remaining)
>
>

Huh?? Again, you ignore the fact that there are no aces amongst the x-1
first cards.

>Technically, you also need a term for "what if the ace of spades IS in
>the first x cards," but then the probability that the next card is the
>AS is 0, so this term can be ignored.
>
>It is obvious that your 1/68 is incorrect, and here is why:
>Assuming your reasoning is correct:
>Probability of any specific ace= 1/68
>Probability of any other specifc card = 1/51
>Probability of ANY card = 3*1/68 + 48/51 < 3/51 + 48/51=1 (note that
>there are only 3 aces left, so it is NOT 4*1/68 for the first term,
>since one of the aces has probability zero.)
>
>

Not quite. Every card has the possibility of being the card after the
first ace, so you want to look at
4/68 + 48/51, which equals unity. Thus, this check does not help us here.

MS wrote:

> This sounds like an application of the negative-binomial distribution.

Nope. Negative binomial applies to cases of independent trials. This
problem is related *somewhat* to the hypergeometric distribution.
(Compare the number of cards until the first ace with the number of
coin-tosses until the first head.)

Poker Scott wrote:

> Unless there are two Ace of spades in the deck there is an equal
> probability that the next card is and Ace of Spades or 2 of clubs
> regardless on whether you stopped after the first ace came up or any
> other card.

Although it turns out the two probabilities are equal, it is not clear a
priori. You ignore the difference that AS could be the first ace, while
the 2C certainly cannot.

> If you deal and the 10th card is an ace, then it's 1:41 that the next
> card is an ace of spades and 1:41 that the next card is a two of clubs.

Huh? Given that the 10th card is an ace, the conditional probability
that the 11th card is 2C is 1/51. Given than the 10th card is an ace,
the conditional probability that the 11th card is AS is 3/4 * 1/51 =
1/68. Given that the 10th card is the *first* ace, the conditional
probability that 11th card is 2C is C(47,9) / C(48,9) * 1/42 = 13/672.
Given that the 10th card is the first ace, the conditional probability
that the 11th card is AS is 3/4 * 1/42 = 1/56.

> If a coin is flipped 100 times and it lands on heads 100 straight
> times, although the odds of flipping a coin and having it land on
> heads 101 straight times is astronomical, the odds of the 101st flip
> being a head or tail is 50/50. Thus if 3 aces came up in a row,
> althought the overall odds of dealing 4 aces very slim, the actual
> odds of the next card being an ace is 1 out of how many ever cards are
> remaining.

While your last sentence is correct, it is misleading to compare this
example to the case of coin-tosses, which are independent trials.

> My bankroll is full of stories about folks who think the deck has a
> "memory".

While it is true that many gamblers see memory where there is none
(e.g., "f**king the deck" in blackjack and "hot tables" in craps), a
deck of cards, qua sampling without replacement, *does* have memory.
For example, given that four aces have been dealt and cards remain in
the single deck, what is the probability that one of the remaining cards
is an ace? 

-- 
Stephen J. Herschkorn                        sjherschko@netscape.net
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