Probability question re: Ace Prediction

I appreciate your explanation and example, which is reminiscent of the infamous Monty Hall Problem. The implications of that problem, and your example, are not surprising to me, and I certainly agree with the math behind both.

However, I still disagree with their applicability to the situation described in Blackjack Ace Prediction. Like the advice columnist who discussed the Monty Hall problem, I'm sure I will be in the minority in this opinion.

In the BJAP situation, the player HAS keyed a particular ace. Whether or not the suit of this ace is remembered does not affect the composition of the unknown cards. Your derivation,

"For the case where the Ace of Clubs is known to be in the among the four cards, the remaining 51 cards are divided into a group of 3 aces and 48 non-aces. Following are the number of combination for having the AC + 0, 1, 2, or 3 aces. (C is combination function, COMBIN(n,r) in Excel).

AC + 0 A = C(3,0)*C(48,3) = 17,296
AC + 1 A = C(3,1)*C(48,2) = 3,384
AC + 2 A = C(3,2)*C(48,1) = 144
AC + 3 A = C(3,3)*C(48,0) = 1

Altogether, the number of combinations is 20,825 of which 3,529 have at least one more ace: 16.95%."

though irrefutable, would be identical whether the suit of the ace is clubs, spades, hearts, or diamonds. In all cases, the final probability is 16.95%. In the BJAP problem, since a PARTICULAR ace HAS been keyed (whether its suit is remembered or not), exactly one of these four identical derivations will apply. The probability of at least one additional random ace will thus be 16.95%.

I can hear the thoughts echoing in your brain, "This guy just doesn't get it," the very thoughts I think when one of my students refuses to grasp a math concept I know to be true. Rest assured, however, that I have given this problem a great deal of thought. Though I wouldn't think to rest on my laurels, I have a degree in mathematics with a focus on combinatorics and have dealt with similar problems in the past. My guess is that you're still sticking to your guns, however, just as I am.

Please don't misconstrue my stubbornness as contentiousness...I really do want to feel settled about this problem. Maybe a Monte Carlo solution would be the best way to determine an answer?

In any event, I truly appreciate the careful, thoughtful responses to my original question. Now if only MathProf can forgive my misguided John Patrick analogy, maybe he can lend his able mind...

David Spence

I'm aware of the effect of conditional probabilities. That is, I understand that an answer of "yes" to, "Is there an ace of clubs in these four cards?" results in a greater number of expected aces than an answer of "yes" to, "Is there any ace in these four cards?"

First of all, your example is not right. The questions should be "Are there more aces knowing that the ace of clubs is already in the group?" and "Are there more aces knowing at least one ace is in the group?"

Sorry for not being clear--my example was not meant to parallel precisely the question at hand. I just wanted to establish sort of a starting point of understanding. I think parallel questions would be, "Are there more aces knowing that at least one ace is in the group, and I remembered the ace's suit until a few seconds ago?" and "Are there more aces knowing that at least one ace is in the group, and I still remember the ace's suit?"

Anyway, I'd feel just as satisfied being convinced that I'm wrong as I would if I got confirmation that my analysis is correct (well, almost :-) )

David

I think parallel questions would be, "Are there more aces knowing that at least one ace is in the group, and I remembered the ace's suit until a few seconds ago?" and "Are there more aces knowing that at least one ace is in the group, and I still remember the ace's suit?"

Probability works on the basis of known information. If you introduce new information into a problem then probabilities change. If you look at the cards in front of you, then you know what they are, and prior calculations of probability are moot. But I think the assumption is meant to be you don't know what they are.

To clarify, here's a restatement of the problem:

A player notices a particular ace (say, the ace of clubs), and uses key cards to track this ace. In the next shoe, the player sees the key cards for this ace, and knows that this ace will appear in the next four cards.

The Question:
What is the probability that there will be one or more additional aces in this group of four cards?

McDowell claims that there are two distinct situations:
Situation 1: The player remembers the suit of the tracked ace.
Situation 2: The player does not remember the suit of the tracked ace.

McDowell claims that the two situations yield different answers to the question. Though I think the answers will be the same, other people whose math skills are clearly advanced disagree with me.

In deference to these opinions, I considered running a sim to find out the answer. However, consider what setting up this sim entails. In Situation 1, the sim program waits until the key cards appear, then notes the four cards that follow. In Situation 2, the sim program waits until the key cards appear, then notes the four cards that follow. Whether the suit of the tracked ace, which DOES have a particular, known suit when it's first associated with the key cards, is remembered or not, does not affect the results. I humbly ask that you consider what the differences between the two sims would really be.

Again, I appreciate the well-reasoned responses, and I don't question the validity of any of the derivations or formulas used. The only area of difference is whether the formulas used are actually appropriate for the given situation.

David Spence

though irrefutable, would be identical whether the suit of the ace is clubs, spades, hearts, or diamonds. In all cases, the final probability is 16.95%. In the BJAP problem, since a PARTICULAR ace HAS been keyed (whether its suit is remembered or not), exactly one of these four identical derivations will apply. The probability of at least one additional random ace will thus be 16.95%.

I think it is a question about how the original problem is worded. I don't have access to the book, so I don't know for sure. I'm also not very familiar with the concepts regarding ace prediction.

For the quote you gave: "if a small group of cards is known to contain an ace, simply knowing its suit increases the probability of one or more additional random aces in the group." my calculations are (I think) correct. However, the point made in this quote might be misappropriate for the problem of ace prediction, which is why you are having a problem with it.

Please don't misconstrue my stubbornness as contentiousness...I really do want to feel settled about this problem. Maybe a Monte Carlo solution would be the best way to determine an answer?

No, Monte Carlo simulations are approximations. They are appropriate only when a combinatorial analysis is intractable. That is not the case here. The problem is specifying what we are looking for.

You're absolutely right--there is some ambiguity in the statement of the problem. I think "re: Probability works on the basis of known information" clears this up, and, obviously, the reality of actual play is the primary concern.

I never doubted your calculations--they seem sound, and your previous posts make it clear that you know what you're talking about :-)

Also, as you mentioned, an a priori solution is preferable to Monte Carlo if possible, but sometimes Monte Carlo is a good idea when the calculation method is in dispute.

Anyway, let me know if my restatement of the problem in the above-referenced post is still ambiguous. And thanks again for the thoughtful reasoning. As I mentioned in my original post on Don's Domain, this has been keeping me up at night :-)

David Spence

For the quote you gave: "if a small group of cards is known to contain an ace, simply knowing its suit increases the probability of one or more additional random aces in the group." my calculations are (I think) correct.

And there's no doubt about it...for the two example quotes (which, as I mentioned, do NOT precisely parallel the problem in BJAP), your calculations are correct.

McDowell claims that there are two distinct situations:
Situation 1: The player remembers the suit of the tracked ace.
Situation 2: The player does not remember the suit of the tracked ace.

McDowell claims that the two situations yield different answers to the question. Though I think the answers will be the same, other people whose math skills are clearly advanced disagree with me.

If I understand this correctly, then in case 1 you have more information than case 2. So, yes you would get a different answer to the question. Certainly the introduction of new information always forces a reassessment of the conditional probabilities.

In Situation 1, the sim program waits until the key cards appear, then notes the four cards that follow. In Situation 2, the sim program waits until the key cards appear, then notes the four cards that follow.

Did you mean to write this? The sentences are exactly the same apart from the situation number. So, yes, obviously the sims would give the same results.

In Situation 1, the sim program waits until the key cards appear, then notes the four cards that follow. In Situation 2, the sim program waits until the key cards appear, then notes the four cards that follow.

Did you mean to write this? The sentences are exactly the same apart from the situation number. So, yes, obviously the sims would give the same results.

Yes, I meant to write this. The point of doing so is to illustrate that despite the fact that in one situation the player remembers the ace's suit and in the other he doesn't, the situations lead to identical sims, and consequently, identical results. New information doesn't ALWAYS lead to different results. Clearly, some information (such as the color of the player's pants) is irrelevant to the calculation. Though not as ridiculous as the example just given, the player's memory of the keyed ace's suit, in this case, is equally immaterial.

Take three random cards from a deck. Don't look at them. Find an ace somewhere else in the deck. Memorize its suit. Put it in amongst the three you had. What is probability one or more of the other three is an ace?

Repeat the experiment. This time don't memorize the suit. Have you changed the probability one of the original three was an ace? Of course not.

ETF

Thanks, ET Fan. Your thought experiment is an elegant way to make the point.

I'm afraid the link got lost beneath the flashing ad, so I'll repeat it one more time.

http://www.blackjackforumonline.com/content/convexingcalculations.htm

Sorry to say, the math in BJAP is faulty, and ace prediction is no get rich quick scheme.

ETF