Why are mixed sets worth the most?

[AK_iceman]

The colors are chosen randomly, there isn't a real deck. The odds of getting a green/red/blue card are the same regardless of how many are already out.
So therefore, the odds of getting a mixed set are the same as the regular sets.

I will accept the first part of your statement as correct: namely, the odds of getting a green/red/blue cared are the same. That is, the cards are selected independently (there being no real deck).

But, the second part of your statement is not correct, as several previous posters have demonstrated. Assuming you have an equal chance of getting a green/red/blue card, the result when you consider sets of three cards is that you are much more likely to get a mixed set as you are to get a pure (monochromatic) set. This has to do purely with combinatorics and the assumption that the first part (about the independent nature of the card selection) is true.

I think if you go back and read the earlier posts, that they will make this argument quite convincingly.

Ok, I didn't read the thread before I posted that and it makes sense now, thanks.
I think the reason that the mixed set is worth more is because thats how it is played in the board game. I know this isn't RISK, but certain similarities are there and this is one of them.

[yowzer14]

The colors are chosen randomly, there isn't a real deck. The odds of getting a green/red/blue card are the same regardless of how many are already out.
So therefore, the odds of getting a mixed set are the same as the regular sets.

Is this the same random number generator that spawns complaints about the dice as well....

[AAFitz]

The colors are chosen randomly, there isn't a real deck. The odds of getting a green/red/blue card are the same regardless of how many are already out.
So therefore, the odds of getting a mixed set are the same as the regular sets.

Is this the same random number generator that spawns complaints about the dice as well....

just stop

[tahitiwahini]

Arrgggghhhhh!!!!!!

No, this is not National Talk-Like-A-Pirate Day.

That's frustration.

Forget about CC. Forget about the deck, forget about cards.

You have three things, lets call them A, B, and C. I will put 9 of each letter in a special machine.

OK so far?

So when I go to this special machine, the likelihood that I will get an A is expected to be the same as the liklihood that I will get a B, which is the same as the liklihood that I will get a C.

No one disagrees with this I hope?

Now, you have a bunch of special boxes that can hold exactly three balls. You open a box and put the next three things produced by the special machine into the box.

You do this process 27 times in total. That leaves you 27 special boxes, each of which holds 3 things produced by the special machine in a random fashion.

You can expect that some boxes will have three A's, you can expect that some boxes will have three B's, you can expect that some boxes will have three C's, and you can expect that some boxes will have some combination of A's, B's, and C's.

[Edit]

Mathematically, you can expect (based on probability and the fact that the special machine is truly random) that one of those 27 boxes will have AAA, one of those boxes will have BBB, one will have CCC, and six will have ABC. The other 18 boxes will have some combination of A's B's and C's other than the ones previously described.

[/Edit]

That's just the way it is mathematically. It's really not subject to debate. It's not conjecture, it's not opinion, it's a fact.

So I hope you can see that you can expect to get twice as many boxes containing ABC (6 in this case) as you can boxes where the items contained therein are all the same (AAA, BBB, and CCC).

Still not convinced? You want to see what's in the boxes? Here are the 27 boxes:

AAA --> all A
AAB
AAC
ABA
ABB
ABC --> one of each
ACA
ACB --> one of each
ACC
BAA
BAB
BAC --> one of each
BBA
BBB --> all B
BBC
BCA --> one of each
BCB
BCC
CAA
CAB --> one of each
CAC
CBA --> one of each
CBB
CBC
CCA
CCB
CCC --> all C

Any special box filled with three letter from the special machine will look exactly like one of those 27 boxes listed above.

It works the same way with green/red/blue as it does with A/B/C.

OK?

Please don't make me talk like a pirate anymore, it hurts my throat.

[AK_iceman]

The above example is NOT random, that is all the possible combinations.

And who said that there was 9 of each color in the machine to begin with?
Since the cards are randomly assigned, it could have been 20 green, 3 red, and 4 blue balls in your machine. Then what kind of results would you get?

[AAFitz]

Arrgggghhhhh!!!!!!

No, this is not National Talk-Like-A-Pirate Day.

That's frustration.

the conjecture i was reffering to was only whether there was a deck or whether they were pulled randomly...each card is pulled randomly is all i was saying....if it were a deck the odds would change on each pull and after each player pulled...

i have fully accepted that the chances of a mixed are more likely...

the question was why are mixed sets worth more....the answer has nothing to do with the odds....my answer was...thats how it is because it makes sense conceptually...for one because of the original game....and just in general it just makes sense...even if the odds are more likely you get a mixed set......

[Aimless]

tahitiwahini's result is correct, but his methodology is flawed.

If we have three classes of object, and we choose from among those classes three times, there are 27 possible results, as listed above. Three of those results represent the "pure" sets - one for each set. Six of those results represent "mixed" sets.

Now, here is the tricky part. If the selection of an object from among the three classes is random, if the probability for selection from any one class is identical to the probability for selection from each of the others, and if each selection is independent of the next, then if we perform this selection three times, each of the 27 possible outcomes has equal probability. This is to say that your chance of getting a specific pure set is 1/27, and your chance of getting a mixed set is 6/27.

Beyond that, I am relatively certain that the math I posted on page three of this thread is an accurate representation of the probabilities and expectation values of each hand size. I would appreciate it if someone would check me - the math for a four card hand was somewhat complicated, especially in the case dealing only with "surviving" hands - but barring any odd Algebraic mistakes my methodology should be correct.

And, forgive me for tooting my own horn, but I do know what I'm talking about here.

[tahitiwahini]

The above example is NOT random, that is all the possible combinations.

And who said that there was 9 of each color in the machine to begin with?
Since the cards are randomly assigned, it could have been 20 green, 3 red, and 4 blue balls in your machine. Then what kind of results would you get?

Because I put 9 of each letter in my special machine, that how I know.

You're quite right that the example above is not random. It is, in fact, all the possible combinations that you can get with three letters in a box.

Since the three letters are independent, over time one would expect the distribution produced by the special machine to approximate all possible combinations. If it did not, why would it favor one combination over another combination? Remember we agreed the letters are independent.

When you throw a dice a million times you would expect to get roughly equal numbers of 1's as 2's as 3's, etc. Would you not? What other expectation could you reasonably have? If you get lots of ones and never get a six, I think you would throw the dice away, right?

It's the same here.

I'm trying to make a mathematical argument in an informal intuitive way, since the more rigorous mathematical explanation (see Aimless) didn't seem to do the trick.

Of course, this may not either.

I hope I've convinced you. If not, I'm content to let you continue to belive whatever it is that you believe. We'll just have to agree to disagree.

[detlef]

I hope I've convinced you. If not, I'm content to let you continue to belive whatever it is that you believe. We'll just have to agree to disagree.

As long as we can also agree that whomever doesn't think that the one-color sets are less frequent than mixed sets given 3 cards is simply wrong.

[detlef]

the question was why are mixed sets worth more....the answer has nothing to do with the odds....my answer was...thats how it is because it makes sense conceptually...for one because of the original game....and just in general it just makes sense...even if the odds are more likely you get a mixed set......

Guys, I don't think the "it's how it is in the original game" argument is valid. the original game, according to the rules I found on the Hasbro Risk site say that cards pay-off in an escalating manner. There is nothing about flat rates.

Honestly, I think the only valid answer given is that it makes the game more fun because people are more inclined to make big sets, which puts more armies in place, and creates more havoc.

That may be all there is to it. [/b]

[AAFitz]

the question was why are mixed sets worth more....the answer has nothing to do with the odds....my answer was...thats how it is because it makes sense conceptually...for one because of the original game....and just in general it just makes sense...even if the odds are more likely you get a mixed set......

Guys, I don't think the "it's how it is in the original game" argument is valid. the original game, according to the rules I found on the Hasbro Risk site say that cards pay-off in an escalating manner. There is nothing about flat rates.

Honestly, I think the only valid answer given is that it makes the game more fun because people are more inclined to make big sets, which puts more armies in place, and creates more havoc.

That may be all there is to it. [/b]

it could be to keep the game taking forever....and every other site ive ever played does it this way.....so to me it just makes sense....having them be worth four just seems too wierd...

but to the poor statiticians...great work...i believe it fully...when i originally posted the odds were even it was just quoting someone else...my main point was that there wasnt a deck....i never took into account anything after the third card....

im sorry if it seemed i didnt believe the odds of mixed were more...i just didnt mention it after i looked at your data...never put any thought into it before....

im always on the dice arent rigged side of the argument, so i know how frustrating it can be to know something and not be able to convey it...well done....i only scanned what you wrote...and knew it was true immidiately...i should have posted that...

[reverend_kyle]

They are, after all, far more common.

For instance, after 3 cards your chances of drawing the following sets:
Red: 1 in 27
Green: 1 in 27
Blue: 1 in 27
Mixed: 2 in 9

I can't think of any other chance-oriented game where the most common result pays the most.

My Answer: Because your penis is so big.

See urban dictionary's detlef schrempf definition.

[flashleg8]

They are, after all, far more common.

For instance, after 3 cards your chances of drawing the following sets:
Red: 1 in 27
Green: 1 in 27
Blue: 1 in 27
Mixed: 2 in 9

I can't think of any other chance-oriented game where the most common result pays the most.

Perhaps you are looking at this the wrong way. You are not being rewarded for getting a mixed (polychromatic) set, as this is the most common result (as you statistically proved), it is better to consider this the defacto bonus rate and a failure to get a mixed set will in effect penalise you with respect to the other players.

The effect of making one colour (or troop type) worth less than another is merely to increase the penalisation rate and introduce a further bad luck factor.

The mathematics to work out the increased probability of achieving a mixed set with the introduction of the two wild cards is beyond me, but I am almost certain that this increases the rate at which "the norm" (of mixed sets) is achieved greater than a matching (monochromatic) set.

On a side note, is it correct to state that if a mixed set is not achieved in the first try (i.e. three turns) the likelihood of achieving it reduces each turn until at the end of the fourth turn it has an equal probability you will achieve a lower rated set as the mixed set? Or have I got that wrong?

[tahitiwahini]

On a side note, is it correct to state that if a mixed set is not achieved in the first try (i.e. three turns) the likelihood of achieving it reduces each turn until at the end of the fourth turn it has an equal probability you will achieve a lower rated set as the mixed set? Or have I got that wrong?

I should probably wait for Aimless on this as his reasoning tends to be sounder and more elegantly expressed then mine, but in the meantime I'll take a stab at it.

You have not acheived a mixed set in the first three cards, so one of two things are ture.

1) You have 3 of a kind. In which case (if I understood Aimless's argument about opportunity costs fully) you are probably justified in turning in. So let's ignore this branch.
2) You have a double and a single. Let's take the case of RRG.

So you have RRG and you get a fourth card which has an equal probability of being a R, G, or B. There are three equally likely possibilities:

1) RRGR, so you turn in the RRR.
2) RRGG, not exactly what you were hoping for, more about this branch later.
3) RRGB, good! You turn in the RGB set.

Further pursuing what happens when you have RRGG (i.e., two doubles) and you get a fifth card. Again there are 3 possibilities, each equally likely:

1) RRGGR, you turn in the RRR.
2) RRGGG, you turn in the GGG.
3) RRGGB, you turn in the RGB.

So, I think what you've stated is correct:

[If] a mixed set is not achieved in the first try (i.e. three turns) the likelihood of achieving it [relative to the liklihood of achieving a pure set] reduces [with each card received] until at the end of the fourth turn it has an equal probability you will achieve a lower rated set as the mixed set.

Maybe Aimless can provide an elegant proof now if he's done with his statistical mechanics mid-term.

[detlef]

The fact that the odds of each merge as you go to 4 and 5 cards seems logical to me. But, since the odds of making any single pure set ever exceeds the chance of making a mixed one, they never overcome the overwhelming odds of making the mixed set after 3.

If I were king, I would set it up as follows: Mixed sets are worth 5 and any pure set, regardless of color is worth 10.

[yeti_c]

They are, after all, far more common.

For instance, after 3 cards your chances of drawing the following sets:
Red: 1 in 27
Green: 1 in 27
Blue: 1 in 27
Mixed: 2 in 9

I can't think of any other chance-oriented game where the most common result pays the most.

But what are the chances after 4 and 5 cards? That might give you your answer?!

PS...

RRR ALL RED
RRG
RRB
RGR
RBR
RGG
RBG MIXED
RBB
RBG MIXED
GGG ALL GREEN
GGR
GGB
GRG
GBG
GRR
GBR MIXED
GBB
GRB MIXED
BBB ALL BLUE
BBR
BBG
BRB
BGB
BRR
BRG MIXED
BGR MIXED


C.

[detlef]

They are, after all, far more common.

For instance, after 3 cards your chances of drawing the following sets:
Red: 1 in 27
Green: 1 in 27
Blue: 1 in 27
Mixed: 2 in 9

I can't think of any other chance-oriented game where the most common result pays the most.

But what are the chances after 4 and 5 cards? That might give you your answer?!

PS...

RRR ALL RED
RRG
RRB
RGR
RBR
RGG
RBG MIXED
RBB
RBG MIXED
GGG ALL GREEN
GGR
GGB
GRG
GBG
GRR
GBR MIXED
GBB
GRB MIXED
BBB ALL BLUE
BBR
BBG
BRB
BGB
BRR
BRG MIXED
BGR MIXED


C.

What was the point of posting those chances yet again? They simply illustrated what I said. FYI 2 in 9 was just a tidy way of saying 6 in 27 which is what you showed.

Once again, it appears that it becomes as likely that you'll make any single pure set as a mixed one after 4 and 5 cards but never exceeds it. So, overall the odds never even equalize, let alone favor making pure sets. Yes, assuming that you have 2 each of any color after 4, you are twice as likely to make any pure set, but not either one individually.

[Aimless]

Maybe Aimless can provide an elegant proof now if he's done with his statistical mechanics mid-term.

What you've posted looks correct. Considering only the case of surviving hands (that is, turning in any time a set is made), if no set is complete after 3 cards then there is a 1/3 chance of making a mixed set and a 1/3 chance of making a pure set. On the fifth card, assuming no set was made with the fourth, there is a 1/3 chance of making a mixed set, and a 2/3rds chance of a pure set.

So, counting from the beginning, and again considering only surviving hands, there is a 2/9 chance of a mixed set with the first three cards, a 2/3*1/3 = 2/9 chance of a mixed set on the fourth card, and a 2/3*1/3*1/3 = 2/27 chance of a mixed set with the fifth card. (Well, that's the probability of making a mixed set on each card counting from the beginning. The probability of making a mixed set on the fifth card is low only because it is likely that a set will be made sooner, and thus few hands survive to the fifth card.)

BTW, I should note that the "surviving hands" rule alters the total probability of making a mixed set; it's now 2/9+2/9+2/27 = 14/27 = 0.518. Still the most probable set, but less so.

[MOBAJOBG]

If pure sets are cashed with 10 armies and mixed sets with lesser armies, since there is only a 2 out of 9 occurences... this game would be on a large part determined by card colors and could last for far longer periods/rounds.

I do believe the current compensation scheme encourages practical playability. In a 6-player game, this ensures that 4.67 players may play for the win assuming that skills, initial positions & dice rolls remain balance. In other words, the game is not over yet by any means.

[tahitiwahini]

If pure sets are cashed with 10 armies and mixed sets with lesser armies, since there is only a 2 out of 9 occurences... this game would be on a large part determined by card colors and could last for far longer periods/rounds.

I do believe the current compensation scheme encourages practical playability.

I think I agree with you here. We can assume that the game designers knew the relative odds of receiving the various card set combinations. I think it's also probably true that the influx of armies when turning in card sets leads to shorter duration games. I think the average length of a flate rate card game compared to that of a no-card game bears this out. So, if it's true that the game designers want to pump in relatively large blocks of armies into the game in fairly regular intervals it does make sense that they would assign to the most common card set combination, the mixed set, the highest number of armies, and to the less frequently occurring card set combinations lesser numbers of armies. The range of allotments: 4, 6, 8, 10; seems to provide another element of uncertainty into the game. There's a pretty big difference between the low end and the high end. These decisions on the part of the game designers help make the game as interesting as it is. Whether they hit upon them by accident or through trial and error they do seem to work very well: keeping the game reasonably balanced, encouraging attacks, etc.

In a 6-player game, this ensures that 4.67 players may play for the win assuming that skills, initial positions & dice rolls remain balance. In other words, the game is not over yet by any means.

Here you've lost me. I'm not sure where the 4.67 figure comes from or what its significance is.

[tahitiwahini]

BTW, I should note that the "surviving hands" rule alters the total probability of making a mixed set; it's now 2/9+2/9+2/27 = 14/27 = 0.518. Still the most probable set, but less so.

Thanks for the update.

[MOBAJOBG]

If pure sets are cashed with 10 armies and mixed sets with lesser armies, since there is only a 2 out of 9 occurences... this game would be on a large part determined by card colors and could last for far longer periods/rounds.

I do believe the current compensation scheme encourages practical playability.

I think I agree with you here. We can assume that the game designers knew the relative odds of receiving the various card set combinations. I think it's also probably true that the influx of armies when turning in card sets leads to shorter duration games. I think the average length of a flate rate card game compared to that of a no-card game bears this out. So, if it's true that the game designers want to pump in relatively large blocks of armies into the game in fairly regular intervals it does make sense that they would assign to the most common card set combination, the mixed set, the highest number of armies, and to the less frequently occurring card set combinations lesser numbers of armies. The range of allotments: 4, 6, 8, 10; seems to provide another element of uncertainty into the game. There's a pretty big difference between the low end and the high end. These decisions on the part of the game designers help make the game as interesting as it is. Whether they hit upon them by accident or through trial and error they do seem to work very well: keeping the game reasonably balanced, encouraging attacks, etc.

In a 6-player game, this ensures that 4.67 players may play for the win assuming that skills, initial positions & dice rolls remain balance. In other words, the game is not over yet by any means.

Here you've lost me. I'm not sure where the 4.67 figure comes from or what its significance is.

Formula -> 6players-(2/9*6players)=4.67 (feel free to correct it)

[tahitiwahini]

Formula -> 6players-(2/9*6players)=4.67 (feel free to correct it)

Thanks, now I see where it comes from.

I certainly agree with the point behind it. Namely, that at each turn the possiblity that each player could receive such a relatively large influx of armies means that on any given turn a certain number of players are in a position to win the game or short of that to dethrone the currently dominant player.

Not sure how that notion could be quantified in a formula though...

[moo_lol]

I haven't read all of the replies, so somebody may have mentioned this already, but here goes:

It's arbitrary. It doesn't make sense to you that the mixed set is worth the most, but it makes sense that three of one color is worth more than 3 of another? Why should 3 blue cards be worth twice as much as 3 red cards when you have the same chance to get both sets? It's just arbitrary and the only thing that makes it fair is the fact that everyone is in the same boat.

[kev82]

BTW, I should note that the "surviving hands" rule alters the total probability of making a mixed set; it's now 2/9+2/9+2/27 = 14/27 = 0.518. Still the most probable set, but less so.

The one thing I learnt about combinatorics, back when I was studying for my masters degree is that you should never use it unless you absolutely have to. The reason being because it is so easy to get wrong.

I just want to check that we are talking about the same thing first, are you claiming that upto a maximum of 5 cards that the probability of geting a mixed set is 14/27, under the assumptions that any set you get is cashed in straight away, at that point we stop, and that the probability of all cards is equally likely?

Well, it is quite simple to write a computer program to loop through all the combinations, I am happy to supply it if you wish. Anyway, the output, listing all the combinations can be found at:

http://www.lancs.ac.uk/postgrad/martink ... /cards.txt

As I stated quite far up in this thread, under the assumptions above the probability of a mixed set is 42/99 (14/33) and the probability of any particular single colour set is 19/99, making the probability of any single colour set 57/99(19/33) so the probability of getting any single colour set is higher than the probability of geting a mixed set.

This of course is assuming I haven't done anything silly or have completely misunderstood something. But I think I'm OK.

I will attempt to verify your expected values later, once I understand what it is you're actually trying to work out.