Conquer Club

The odds of getting a spoil set with 3 cards is exactly 1/3. The odds of getting a set with 5 cards is 100% of course. Has anyone computed the odds of getting a spoil set with 4 cards? I would guess somewhere around 60%? I'm too lazy to get my stat class notes out from the attic for that (involves binomials)...

7/9 i think

To not get a set

1)any card
2)any card
3)2/3 of a card
4) 1/3 of a card

2/3*1/3=2/9

Nip, that looks correct. I've computed it in another way and got the same result, 7/9 it is. Guess we didn't need binomials for this after all.

7/9 is correct

idk why i have trouble getting them though

Yep, 7/9 is correct.

Here's some analysis:

If you know that the other person has neither joker, the odds he has a set with 3 cards are 33.41%; the odds he has a set if he has 4 cards are 77.80%.

macbone wrote:Yep, 7/9 is correct.

Here's some analysis:

If you know that the other person has neither joker, the odds he has a set with 3 cards are 33.41%; the odds he has a set if he has 4 cards are 77.80%.

33,41?? why is that? i would v'e said 33.34%.

also about the 4 cards odds, i disregard the 3 first ones, and i just think of the odds of getting a set before receiving my 4th, which would be 66.66%. But if you look at it as a whole, you are right, 77.78%

Whoops, forgot the link: http://www.kent.ac.uk/smsas/personal/od ... aq.htm#3.5

He's evaluating the board game Risk, a world domination game that uses cards instead of spoils, but the principles are the same. =)

kylegraves1 wrote:7/9 is correct

idk why i have trouble getting them though

I second that. It often feels like the chance is 10%, especially when it could make the difference between sweeping the map or not getting another turn.

Lx

pr(set, given 3 cards) = 1/3, so
pr(no set, given 3 cards) = 2/3

If you have no set with your 3 cards, then you have 2 cards of one color and 1 card of a second color. Without loss of generality, say you have 2 blue cards and 1 green card. Thus from that point the only way you don't have a set on your 4th card is if you draw another green card, and pr(green card) = 1/3.

So we have the path to no set on 4 cards = pr(no set, given 3 cards) * pr(not making set on 4th card, given started with 3 cards that don't form a set) = 2/3 * 1/3 = 2/9.

Therefore pr(set with 4 cards) = 1 - pr(no set with 4 cards) = 1 - 2/9 = 7/9.

If you consider the generic odds then it is indeed 7/9, if on the other hand the first 3 don't give you a set, the odds for the 4th making a set are 6/9.

macbone wrote:
He's evaluating the board game Risk, a world domination game that uses cards instead of spoils, but the principles are the same. =)

i have to check out this "risk" you talk about..

/

anonymus wrote:

macbone wrote:
He's evaluating the board game Risk, a world domination game that uses cards instead of spoils, but the principles are the same. =)

i have to check out this "risk" you talk about..

/

If someone had told me earlier about this "Risk" i would not have wasted my time here

Interesting, it certainly doesn't SEEM like 7/9...

And since when do you get to play Risk (the board game) with people from all over the world? It's much more fun this way. Although now that I think about it, I'm much better at beating friends and family members than all-comers.

TDK wrote:Interesting, it certainly doesn't SEEM like 7/9...

And since when do you get to play Risk (the board game) with people from all over the world? It's much more fun this way. Although now that I think about it, I'm much better at beating friends and family members than all-comers.

+1

betiko wrote:

macbone wrote:Yep, 7/9 is correct.

Here's some analysis:

If you know that the other person has neither joker, the odds he has a set with 3 cards are 33.41%; the odds he has a set if he has 4 cards are 77.80%.

33,41?? why is that? i would v'e said 33.34%.

also about the 4 cards odds, i disregard the 3 first ones, and i just think of the odds of getting a set before receiving my 4th, which would be 66.66%. But if you look at it as a whole, you are right, 77.78%

He is getting his statistics from board risk analysis, where "colours" are fixed for each country. In CC, it is 1/3 exactly.

Note that if you already have 3 cards and you do not have a set, chances that you get it on next spoil are 2/3.

betiko wrote:

macbone wrote:Yep, 7/9 is correct.

Here's some analysis:

If you know that the other person has neither joker, the odds he has a set with 3 cards are 33.41%; the odds he has a set if he has 4 cards are 77.80%.

33,41?? why is that? i would v'e said 33.34%.

also about the 4 cards odds, i disregard the 3 first ones, and i just think of the odds of getting a set before receiving my 4th, which would be 66.66%. But if you look at it as a whole, you are right, 77.78%

I suspect he's talking about the board game where you actually draw cards from a limited pool, which will have a slight effect on the outcome. However cards on cc are randomly generated so the math is easier because the pool is unlimited and each new card has an exactly 1/3 chance of being either color.

Ok; I follow this thread so far but can any one calculate the odds of getting a rainbow (one card of each color) in 4 cards? Now that I would like to see done.

Boys, break out your calculators?

Oh; And how come we don't have joker spoils? Does anyone know?

I think that is a good question.

Adding jokers would complicate the random card generator program. I'm sure it's doable, but why fix what isn't broken?

Viceroy63 wrote:Ok; I follow this thread so far but can any one calculate the odds of getting a rainbow (one card of each color) in 4 cards? Now that I would like to see done.

Boys, break out your calculators?

Oh; And how come we don't have joker spoils? Does anyone know?

I think that is a good question.

That is 1/3*2/3*1/3+2/3*(1/3+2/3*1/3)=2/27+10/27=12/27=44,44% unless I am too sleepy to calculate right.

Viceroy63 wrote:Ok; I follow this thread so far but can any one calculate the odds of getting a rainbow (one card of each color) in 4 cards? Now that I would like to see done.

Boys, break out your calculators?

Oh; And how come we don't have joker spoils? Does anyone know?

I think that is a good question.

Given that you don't already have a rainbow? Then it's easy cuz you either have 2 of 1 color and 1 of another color or all 3 of the same color. If it's the latter, which only occurs 1/9 of the time, you know you won't be getting a rainbow. But in the 8/9 times that's not the case, you need 1 of the missing color giving you a 1/3 chance of getting a 4-card rainbow, given that you don't have a rainbow in 3 cards. So, you'd have an 8/27 or 30%

If you are starting from scratch and just want to know the odds of getting a rainbow given any four cards, then odds are slightly more complicated.

Your first card is X-colored. The other two colors are now Y and Z.

You need at least 1 Y and at least 1 Z in the next 3 cards.

1/3 chance of getting another X, in which case you need a Y and a Z to come out: odds 2/3*1/3 = 2/9
2/3 chance of getting a Y (i.e. anything but X), in which case you need a Z to come out. The chances of not getting a Z in these next two cards is 2/3*2/3 = 4/9. That means you have a 5/9 chance of getting a Z.

So, 1/3*2/9 + 2/3*5/9 = 2/27 +10/27 = 12/27.

You have a 12/27 or 44% chance of getting a rainbow in 4 cards.

So, you have a 2/3*1/3=2/9=22% chance of finding a rainbow set in 3 cards.

You have a 37% chance of finding one in any 4 cards.

Carrying on with the 4 card math, we need to add some number to that probability to figure out what your chances of finding a rainbow in a 5 card hand.

If you have one of the 15/27 or 63% percent of 4 card hands that don't already contain a rainbow AND you are NOT holding one of the 1/27 or 4% of hands that are all the same color, then you have a 1/3 chance of making a rainbow with your next card. Thus, the chances of making a rainbow WITH your fifth card are 14/27*1/3 or 14/81. Add this to the chance that you already had a rainbow in your hand and you see that:

You have a 50/81 or 62% chance of finding a rainbow in any given 5 card hand.

Oddly enough, I think that I understand.

So If I already have two different colors in three cards then the odds are the same (one in three) of getting a rainbow as getting a set of the same colors.

So I would have a one in three chance of getting something either a rainbow or a flush on that fourth card. One in three is still pretty low.

Thank you for that enlightening comment. =)

Viceroy63 wrote:Oddly enough, I think that I understand.

So If I already have two different colors in three cards then the odds are the same (one in three) of getting a rainbow as getting a set of the same colors.

So I would have a one in three chance of getting something either a rainbow or a flush on that fourth card. One in three is still pretty low.

Thank you for that enlightening comment. =)

Yes, you are right about that. Some of the above analysis only applies to getting a rainbow set though. But if you have a 2 pair, you are equally likely to end up cashing 3 of color A, 3 of color B, or a rainbow.