Cactus Game Design Message Boards
Open Forum => Off-Topic => Topic started by: The Guardian on December 08, 2016, 03:31:22 PM
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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?
8)
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1/3, and the odds stay the same? :dunno:
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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? :o
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Yes?
I feel like you're trying to set up a Monty Hall, but I'm fairly certain it doesn't work
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@Students--There is really no need to guess. Simply write down the combinations and count. :police:
@(YM)Teachers -- Fun with math?!?!? Unpossible. :P
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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. ;)
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Anyone going to give it a shot before MJB or YMT reveal the answer?? ;D
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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.
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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.
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Feels like semantics
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It's actually a fun version of "Conditional Probability." ;D
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So which door doesn't have a goat?
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Math would be fun if not for those that make it not so fun.
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I'm going with the odds are 1 in 2 and yes they are the same
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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. :'(
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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?
8)
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.
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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!!! 8)
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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.
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Indeed. :)
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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.
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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%. 8)
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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!!! 8)
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.
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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).
8)
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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?"
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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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For the odds of drawing both good dominants, there is a 1 in 2 chance that you drew a good dominant the first time since there are 2 good dominants, and 2 evil dominants. The odds of drawing a good dominant the first time is 2 out of 4 which is also 1 in 2, then the odds of pulling the 2nd good dominant assuming that you have successfully already drawn the first good dominant there is a 1 in 3 chance.
So all in all, the chance of pulling both good dominants is 1 in 6.
That's the math behind the problem, and yes I do find mathematics fun, and that's half the battle in learning.
Steven
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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).
8)
Yes you are right, my apology, I got it after a little bit of considering the problem. It's just "frustrating math" and not "fun with math".
The whole thing is about that the odds make sense when it comes to the statistics, i.e. repeating the process. So if we would constantly repeat the process of drawing two cards and have a computer/friend to tell us wheter we drew the two evil cards or at least one of the good dominants, then we could consider the cases in which computer told us we drew at least one good dominant and see that statistically in one fifth of that cases we drew both good dominants.
Really interesting fact that it statistically makes difference if the source of information names the specific card from the set of similar cards...
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Calculus killed any interest I had in mathematics long ago. Disliked and Unsubscribed. ;)
And yet your inner calling for mathematics Jesus brought you into this thread. Suppressing that innate love for math Jesus will only leave you unfulfilled and incomplete. :'(
I fixed it for you. You are welcome ;D
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I wrote this post up after only reading the first post. I have no idea who got it right, or if my answer is right (but I think it is).
There are 6 combinations that Justin could have drawn:
- SoG/NJ
- SoG/CM
- SoG/Burial
- NJ/CM
- NJ/Burial
- CM/Burial
If Justin tells me he knows I drew a good Dominant, that means Justin drew at most 1 good Dominant. That means he didn’t draw SoG/NJ – but he does have one of the other 5 combinations. Of those 5 possibilities for Justin, only 1 of them leaves me with SoG/NJ – that’s if Justin drew CM/Burial. Therefore, the odds I drew SoG/NJ is 20% (1 in 5).
If Justin tells me he knows I drew SoG, that means Justin has 2 cards that are not SoG. There are 3 possible combinations for Justin. Of those 3 possibilities for Justin, once again 1 of them leaves me with SoG/NJ. Now my odds of having SoG/NJ is 33% (1 in 3).
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For the odds of drawing both good dominants, there is a 1 in 2 chance that you drew a good dominant the first time since there are 2 good dominants, and 2 evil dominants. The odds of drawing a good dominant the first time is 2 out of 4 which is also 1 in 2, then the odds of pulling the 2nd good dominant assuming that you have successfully already drawn the first good dominant there is a 1 in 3 chance.
So all in all, the chance of pulling both good dominants is 1 in 6.
This is correct, but this is the unconditional probability of drawing SoG/NJ. Once Justin gives you information, it changes the odds of what actually happened, and you are dealing with conditional probability.
Here's how you arrive at the unconditional 1/6 odds, using one of Justin's examples for conditional probabilities:
There's a 5/6 chance that Justin can say (truthfully) "You have drawn at least 1 good Dom". When this happens, you have a 1/5 chance of getting SoG/NJ. 5/6 * 1/5 = 1/6
There's also a 1/6 chance that Justin can NOT say (truthfully) "You have drawn at least 1 good Dom". When this happens, you have a 0% chance of getting SoG/NJ, because Justin drew them. 1/6 * 0 = 0
1/6 + 0 = 1/6
QED
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I think what everyone is missing here is that Justin would never willingly give you accurate information in a game of Redemption unless it was to his advantage. Not to say he would lie, as he wouldn't do that either, but really the question you should be asking is not "what are the odds that I drew both dominants?", but rather, "why is he telling me this?"
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I've always thought these message boards could benefit from a "Fun with English" thread. ;)
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We have a whole board for that called "Rulings Questions." Mayhem delenda est.