Author Topic: Fun with math (Read 6223 times)

The Guardian · « **on:** December 08, 2016, 03:31:22 PM »

Suppose you have a deck of 4 cards: Son of God, New Jerusalem, Christian Martyr and Burial.

You draw 2 of the cards, but don't look at them.

I look at the two remaining cards. Suppose I told you that one of the cards you drew was a good dominant.

Knowing that, what are the odds you drew both good dominants?

Now suppose I told you that you for sure drew Son of God.

Knowing that, what are the odds you drew both Son of God and New Jerusalem? Are the odds the same or different?

Ironisaac · « **Reply #1 on:** December 08, 2016, 04:35:57 PM »

1/3, and the odds stay the same?

The Guardian · « **Reply #2 on:** December 08, 2016, 04:52:36 PM »

Perhaps start with the second question...

If I tell you for certain that you drew Son of God, then there are three possible choices for what the other card could be (New Jerusalem, Christian Martyr or Burial). Since there's three possibilities and only one satisfies the desired result (two good dominants) then your odds are clearly 1 in 3.

Getting back to the first question--I only tell you for certain that you drew one good dominant (but not which one), are the odds still 1 in 3 that you drew both good dominants?

Red Dragon Thorn · « **Reply #3 on:** December 08, 2016, 05:34:29 PM »

Yes?

I feel like you're trying to set up a Monty Hall, but I'm fairly certain it doesn't work

EmJayBee83 · « **Reply #4 on:** December 08, 2016, 06:00:48 PM »

@Students--There is really no need to guess. Simply write down the combinations and count.

@(YM)Teachers -- Fun with math?!?!? Unpossible.

YourMathTeacher · « **Reply #5 on:** December 08, 2016, 07:54:44 PM »

I always tell my students to avoid "common errors." The common error here is to try to treat each selection separately, rather than just skip to the output of two drawn cards. Since this is a combination not a permutation (as MJB stated), the number of possibilities is very limited.

In the end, the real question is whether knowing you have one of two different cards yields more successful outcomes than knowing you only have one of those cards in particular.

The Guardian · « **Reply #6 on:** December 09, 2016, 04:42:31 PM »

Anyone going to give it a shot before MJB or YMT reveal the answer??

Red Dragon Thorn · « **Reply #7 on:** December 09, 2016, 08:40:06 PM »

I still maintain that as the draw is a static event the odds are always 1 in 6. You can evaluate the second card on the basis of the first. It's not a Monty Hall because there isn't a choice involved.

YourMathTeacher · « **Reply #8 on:** December 09, 2016, 10:10:57 PM »

I would disagree based on his wording "knowing that." The decision about probability computation is not made until after the disclosure of known information. This would eliminate some of the original six possibilities, much like the Hat Puzzle.

Red Dragon Thorn · « **Reply #9 on:** December 09, 2016, 10:28:08 PM »

Feels like semantics

YourMathTeacher · « **Reply #10 on:** December 10, 2016, 07:22:57 AM »

It's actually a fun version of "Conditional Probability."

kariusvega · « **Reply #11 on:** December 10, 2016, 10:43:41 PM »

So which door doesn't have a goat?

YourMathTeacher · « **Reply #12 on:** December 11, 2016, 08:50:21 AM »

Math would be fun if not for those that make it not so fun.

kariusvega · « **Reply #13 on:** December 11, 2016, 03:46:44 PM »

I'm going with the odds are 1 in 2 and yes they are the same

YourMathTeacher · « **Reply #14 on:** December 11, 2016, 08:43:46 PM »

Quote from: Master Q on December 11, 2016, 07:49:01 PM

Calculus killed any interest I had in mathematics long ago. Disliked and Unsubscribed.

And yet your inner calling for mathematics brought you into this thread. Suppressing that innate love for math will only leave you unfulfilled and incomplete.

ChristianSoldier · « **Reply #15 on:** December 12, 2016, 03:33:16 AM »

Quote from: The Guardian on December 08, 2016, 03:31:22 PM

Suppose you have a deck of 4 cards: Son of God, New Jerusalem, Christian Martyr and Burial.

You draw 2 of the cards, but don't look at them.

I look at the two remaining cards. Suppose I told you that one of the cards you drew was a good dominant.

Knowing that, what are the odds you drew both good dominants?

Now suppose I told you that you for sure drew Son of God.

Knowing that, what are the odds you drew both Son of God and New Jerusalem? Are the odds the same or different?

The first thing I'd have to do is figure out the probability that you are lying
If I decided to assume that you weren't lying, I could get on with the problem.
There are 24 ways the deck can be arranged, in 4 of them do you draw both Good Dominants. Of the 24 options 4 are removed if you know you drew at least 1 good dominant, leaving you with 4 options out of 20, or 1/5. From there an additional 8 options are removed if you know one of the dominants is Son of God, leaving you with 4 options out of 12, or 1/3.

The Guardian · « **Reply #16 on:** December 12, 2016, 08:45:09 AM »

Quote

The first thing I'd have to do is figure out the probability that you are lying
If I decided to assume that you weren't lying, I could get on with the problem.
There are 24 ways the deck can be arranged, in 4 of them do you draw both Good Dominants. Of the 24 options 4 are removed if you know you drew at least 1 good dominant, leaving you with 4 options out of 20, or 1/5. From there an additional 8 options are removed if you know one of the dominants is Son of God, leaving you with 4 options out of 12, or 1/3.

Winner!!!

EmJayBee83 · « **Reply #17 on:** December 12, 2016, 08:54:31 AM »

Quote from: ChristianSoldier on December 12, 2016, 03:33:16 AM

There are 24 ways the deck can be arranged...

You can save yourself some time if--instead of starting with the full deck--you only worry about the two drawn cards (12 ways to draw the two cards). This can also be simplified by realizing that the order of the draw is *not* important in this case. So there are actually only six combinations to worry about: SN, SC, SB, NC, NB, and CB.

The Guardian · « **Reply #18 on:** December 12, 2016, 09:43:23 AM »

Indeed.

ChristianSoldier · « **Reply #19 on:** December 12, 2016, 10:09:10 PM »

Quote from: EmJayBee83 on December 12, 2016, 08:54:31 AM

Quote from: ChristianSoldier on December 12, 2016, 03:33:16 AM
There are 24 ways the deck can be arranged...
You can save yourself some time if--instead of starting with the full deck--you only worry about the two drawn cards (12 ways to draw the two cards). This can also be simplified by realizing that the order of the draw is *not* important in this case. So there are actually only six combinations to worry about: SN, SC, SB, NC, NB, and CB.

I knew instinctively that a bunch of them were redundant (3/4 to be exact) because the order in the deck and of the draw didn't matter (and there are 4 ways to order the set while still keeping the same drawn combinations) but because I was only dealing with 24 combinations, it wasn't really that much longer to write down all the combination than to prove to myself the exact combinations.

I actually debated posting my proof (which was essentially the list of the options, highlighted in different colors depending on whether they were what I was looking for, dropped from knowing there was a good dominant or dropped because I knew I had Son of God) but I decided it wasn't important enough.

But this sort of thing is easy math, I one time figured out an equation to figure out how to maximize your chance of drawing a specific number of copies of a card (or theoretically set of cards) in an opening hand (by inputting the number of cards you want, the number of cards you draw, the number of the desired cards in the deck and the total number of cards in the deck). It turned out to be mostly useless, except to prove what Magic: the Gathering players already knew about how many lands to include in a normal deck.

The Guardian · « **Reply #20 on:** December 13, 2016, 09:29:55 AM »

I used an equation like that for calculating odds of having one of a certain set of cards in my opening draw, which I actually found pretty useful. It was interesting to analyze how much the probability went up by adding just one more card to the set, even if I didn't take another card out. For example, if I used a 51 card deck with 8 Heroes instead of a 50 card deck with 7 Heroes, my odds of having a Hero in the opening draw of 8 improved by almost 4%.

Ivek · « **Reply #21 on:** December 13, 2016, 10:00:43 AM »

Quote from: The Guardian on December 12, 2016, 08:45:09 AM

Quote
The first thing I'd have to do is figure out the probability that you are lying
If I decided to assume that you weren't lying, I could get on with the problem.
There are 24 ways the deck can be arranged, in 4 of them do you draw both Good Dominants. Of the 24 options 4 are removed if you know you drew at least 1 good dominant, leaving you with 4 options out of 20, or 1/5. From there an additional 8 options are removed if you know one of the dominants is Son of God, leaving you with 4 options out of 12, or 1/3.

Winner!!!

Not winner, but looser. The chances are in both cases 1/3. You made a typical mistake in complicating the simple things and overcounting yourself.

The Guardian · « **Reply #22 on:** December 13, 2016, 11:13:11 AM »

It seems intuitively that the answer should be 1 in 3 for both. If I tell you one card is SoG, the odds are 1 in 3 you also drew NJ. Likewise, if I told you one card is NJ, the odds are 1 in 3 you also drew SoG. It's therefore reasonable to think that if I tell you one card is a good dominant, that the odds would remain 1 in 3 that you drew the other one. However, because we don't know which good dominant was drawn, there a couple more "losing" combinations that are possible (which are not possible when we know exactly which good dominant we drew).

Here are the 12 two card combinations you can draw: (S=SoG N=NJ C=CM B=Burial)
S-N
S-C
S-B
N-S
N-C
N-B
C-S
C-N
C-B
B-S
B-N
B-C

If we know we drew SoG, we can see there are 6 possible combos that involve SoG. Two of those combos also include NJ so when you know one of the cards is SoG, there is a 2 in 6 (1 in 3) chance that you also hold NJ.

If we only knew we drew a good dominant (but not which one), we see there are 10 combinations that involve either SoG or NJ. Two of those include both SoG and NJ therefore we have 2 "winning" combinations out of a possible 10 which gives us odds of 2 in 10 (1 in 5).

The Guardian · « **Reply #23 on:** December 13, 2016, 11:30:10 AM »

When we know that one of the cards is a good dominant, we can only eliminate the combos of:
C-B
B-C

When we know specifically that one of the cards is SoG, we can also eliminate the combos of:
N-B
N-C
C-N
B-N

Therefore our odds improve when we know we have SoG. Also, remember that the question being asked is not "What are the odds of drawing both dominants?" The question is "What are the odds that we drew both dominants?"

YourMathTeacher · « **Reply #24 on:** December 13, 2016, 04:28:21 PM »

And, as MJB pointed out, whether you drew SoG first and NJ second, or NJ first and SoG second, does not affect the result of what you have in your hand when Guardian makes the initial claim. Thus S-N and N-S are not counted as distinct draws for the purpose of this probability computation.

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Author Topic: Fun with math (Read 6223 times)

kariusvega

kariusvega