Conquer Club

I just have one question about card set values, and this may have been asked before, I'm not entirely sure...anyway:

In a flat rate game, a red set is 4, green 6, blue 8, and mixed is 10.

My question is, why is a mixed set worth more? Technically if you look at the probability of getting a mixed set versus a colored set, you're more likely to obtain a mixed set:

Colored Set: [Any 3 Colors], [1 Color], [1 Color]
Mixed Set: [Any 3 Colors], [2 Colors], [1 Color]

Therefore, the probability of getting a colored set is (1) * (1/3) * (1/3) or (1/9), and the probability of getting a mixed set is (1) * (2/3) * (1/3) or (2/9).

So if you're twice as likely to obtain a mixed set, why are they worth more than a colored set?

NW.Traveler wrote:I just have one question about card set values, and this may have been asked before, I'm not entirely sure...anyway:

In a flat rate game, a red set is 4, green 6, blue 8, and mixed is 10.

My question is, why is a mixed set worth more? Technically if you look at the probability of getting a mixed set versus a colored set, you're more likely to obtain a mixed set:

Colored Set: [Any 3 Colors], [1 Color], [1 Color]
Mixed Set: [Any 3 Colors], [2 Colors], [1 Color]

Therefore, the probability of getting a colored set is (1) * (1/3) * (1/3) or (1/9), and the probability of getting a mixed set is (1) * (2/3) * (1/3) or (2/9).

So if you're twice as likely to obtain a mixed set, why are they worth more than a colored set?

assuming you always cash in when you get your first set (no holding out for a better set), you have a 14/27 probability of your first set being a mixed set and a 13/81 probability of red and equivalently blue or green being your first set. 13/81 + 13/81 + 13/81 + 14/27 =1.

You are only 1/13th more likely to get a mixed set ((14/27)/(13/27)) over any colored set BUT you are 1/3 more likely to get a mixed set over any SPECIFIC color AKA you are almost 3x as likely to get a mixed set, than a set of reds.((14/27)/(5/27))=2.8

I think it is intentional that mixed is more. I mean why even have different values for reds, blues, and greens if they are all as equally likely. I think it is done to keep the game moving quicker.

Edit: This accounts for possibility of not getting a set on your 3rd or fourth draw instead of only focusing on the 3rd draw.

Well I figured the different values for the colored sets would to be to add variety

assuming you always cash in when you get your first set (no holding out for a better set), you have a 14/27 probability of your first set being a mixed set and a 13/81 probability of red and equivalently blue or green being your first set. 13/81 + 13/81 + 13/81 + 14/27 =1.

How did you come up with these numbers?

Let me go ahead and ask this, just in case: When you're about to receive a card, that card does have a 1/3 chance to be red, 1/3 chance to be green, and 1/3 chance to be blue right? Or are the cards "drawn out of a random deck", thus changing the probability of the color every time a card is taken out?

NW.Traveler wrote:Well I figured the different values for the colored sets would to be to add variety

assuming you always cash in when you get your first set (no holding out for a better set), you have a 14/27 probability of your first set being a mixed set and a 13/81 probability of red and equivalently blue or green being your first set. 13/81 + 13/81 + 13/81 + 14/27 =1.

How did you come up with these numbers?

Let me go ahead and ask this, just in case: When you're about to receive a card, that card does have a 1/3 chance to be red, 1/3 chance to be green, and 1/3 chance to be blue right? Or are the cards "drawn out of a random deck", thus changing the probability of the color every time a card is taken out?

no, it is always 1/3

as you pointed out, when you have 3 cards there is a 2/9 chance of a mixed set and a 1/9 chance of a colored set. That means there is a 6/9 chance that you will not have a set on your 3rd card. If you have 3 cards and no set, you must have 2 of one color and one of another. The chance of you getting the color you need on the fourth card is 1 in 3 for a mixed set and 1 in 3 to add to the color you have 2 of. There is also a 1 in 3 chance of you getting another of the color you have only one of thereby giving you 2 pair. The probability that you need a 5th card to make a set GIVEN that you didn't have one when you were getting your 4th card is 1 in 3 (the same as the probability that you get 2 pair on the 4th card). The 5th card has a 1 in 3 chance of giving you a mixed set and a 2/3 chance of completing one of your pairs.

Probability of first set being a mixed set = (probability of mixed set on 3rd card) + (probability of mixed set on 4th card)*(probability of no set on 3rd card) + (probability of mixed set on 5th card)*(probability of no set on 4th card)*(probability of no set on 3rd card) = 2/9 + (1/3)(6/9) + (1/3)(1/3)(6/9) = 14/27

probability of colored set = 1 -14/27 = 13/27
probability of red = probability of green = probability of blue =(13/27)/3 = 13/81

It's simple Bayesian probabilities.

ummmm ya what he said

simple

Oh ok, I see. I thought you were calculating the probability of something else.