What are the odds...

[Crazyirishman]

of getting 3 double cashes with 6 cards in the same round for a 7 player escalating game?
Game 7107884

my guess is that it would be quite rare

[ljex]

I would think somewhere around 1 in 27. This is because you are guaranteed to get one set so the odds of you having a 3 card set are 1/3. Now you need this to happen 3 times so 1/3^3. Someone please correct me if im wrong. I dont remember how to do the odds of having 2 sets with 6 cards and dont feel like writing down all the combination so if someone would explain that i would appreciate it.

[jammyjames]

Explained.

[squishyg]

i could give you inaccurate odds, but that would make me a cheap tactician in addition to be an incorrect statistician

[Army of GOD]

I know for a fact that there is a .0000000000000000000000000000003% chance of having a set with 0 cards.

[Caymanmew]

I know for a fact that there is a .0000000000000000000000000000003% chance of having a set with 0 cards.

take away the 3 and your right

[Army of GOD]

No, it's .0000000000000000000000000000003%.

[Caymanmew]

I would think somewhere around 1 in 27. This is because you are guaranteed to get one set so the odds of you having a 3 card set are 1/3. Now you need this to happen 3 times so 1/3^3. Someone please correct me if im wrong. I dont remember how to do the odds of having 2 sets with 6 cards and dont feel like writing down all the combination so if someone would explain that i would appreciate it.

incorrect i think

i think the chance is 1/64

the chance of getting a set with 3 cards is 1/4 so 1/4 times 1/4 times 1/4 would be 1/64 that should be right (have not done math i a couple of weeks so i may be off)

[ljex]

I would think somewhere around 1 in 27. This is because you are guaranteed to get one set so the odds of you having a 3 card set are 1/3. Now you need this to happen 3 times so 1/3^3. Someone please correct me if im wrong. I dont remember how to do the odds of having 2 sets with 6 cards and dont feel like writing down all the combination so if someone would explain that i would appreciate it.

incorrect i think

i think the chance is 1/64

the chance of getting a set with 3 cards is 1/4 so 1/4 times 1/4 times 1/4 would be 1/64 that should be right (have not done math i a couple of weeks so i may be off)

this would work if the chance in getting a 3 card set is 1/4 however it is 1 in 3. Think about it like this, no matter what your first 2 cards are there is one card that can give you a 3 card set and 2 that wont, so therefore your chance of a 3 card set is 1/3

thus 1/27 because (1/3)(1/3)(1/3)=1/27

[Caymanmew]

I would think somewhere around 1 in 27. This is because you are guaranteed to get one set so the odds of you having a 3 card set are 1/3. Now you need this to happen 3 times so 1/3^3. Someone please correct me if im wrong. I dont remember how to do the odds of having 2 sets with 6 cards and dont feel like writing down all the combination so if someone would explain that i would appreciate it.

incorrect i think

i think the chance is 1/64

the chance of getting a set with 3 cards is 1/4 so 1/4 times 1/4 times 1/4 would be 1/64 that should be right (have not done math i a couple of weeks so i may be off)

this would work if the chance in getting a 3 card set is 1/4 however it is 1 in 3. Think about it like this, no matter what your first 2 cards are there is one card that can give you a 3 card set and 2 that wont, so therefore your chance of a 3 card set is 1/3

look at total possiblitys of having cards

red is the first then green then blue

1-1-1
1-2-0
1-0-2
2-1-0
2-0-1
3-0-0
2-1-0
0-1-2
1-2-0
0-2-1
0-3-0
2-0-1
0-2-1
1-0-2
0-1-2
0-0-3

that 16 possibility and 4 possible set. that 1/4

[ljex]

I would think somewhere around 1 in 27. This is because you are guaranteed to get one set so the odds of you having a 3 card set are 1/3. Now you need this to happen 3 times so 1/3^3. Someone please correct me if im wrong. I dont remember how to do the odds of having 2 sets with 6 cards and dont feel like writing down all the combination so if someone would explain that i would appreciate it.

incorrect i think

i think the chance is 1/64

the chance of getting a set with 3 cards is 1/4 so 1/4 times 1/4 times 1/4 would be 1/64 that should be right (have not done math i a couple of weeks so i may be off)

this would work if the chance in getting a 3 card set is 1/4 however it is 1 in 3. Think about it like this, no matter what your first 2 cards are there is one card that can give you a 3 card set and 2 that wont, so therefore your chance of a 3 card set is 1/3

look at total possiblitys of having cards

red is the first then green then blue

1-1-1
1-2-0
1-0-2
2-1-0
2-0-1
3-0-0
2-1-0
0-1-2
1-2-0
0-2-1
0-3-0
2-0-1
0-2-1
1-0-2
0-1-2
0-0-3

that 16 possibility and 4 possible set. that 1/4

This is wrong but i dont know why, think about it this way you have a 3/3 chance of your first card contributing to a set, a 3/3 chance of your second card contributing to a set and then a 1/3 chance of your 3rd card contributing to a set, thus 9/27 (1/3) chance of getting a set with 3 cards.

[Caymanmew]

it should be about possibility have what colour cards you can get.

there 16 possibility and 4 set (all red all blue all green 1 of each) that would be 1/4 your way is not looking at all the possibility's just the last card


edited

we need a math guy here

[Commander62890]

The way I see it, it's gotta be 1/3 to get a set with 3 cards.

No matter what combination of cards you get with your first two, that third card will give you a set 1/3 times, because there are 3 different colors.



For example... your first 2 cards are Red + Blue. A Green card will get you a set, while Red or Blue will not. 1/3.

[Blinkadyblink]

I would think somewhere around 1 in 27. This is because you are guaranteed to get one set so the odds of you having a 3 card set are 1/3. Now you need this to happen 3 times so 1/3^3. Someone please correct me if im wrong. I dont remember how to do the odds of having 2 sets with 6 cards and dont feel like writing down all the combination so if someone would explain that i would appreciate it.

incorrect i think

i think the chance is 1/64

the chance of getting a set with 3 cards is 1/4 so 1/4 times 1/4 times 1/4 would be 1/64 that should be right (have not done math i a couple of weeks so i may be off)

this would work if the chance in getting a 3 card set is 1/4 however it is 1 in 3. Think about it like this, no matter what your first 2 cards are there is one card that can give you a 3 card set and 2 that wont, so therefore your chance of a 3 card set is 1/3

look at total possiblitys of having cards

red is the first then green then blue

1-1-1
1-2-0
1-0-2
2-1-0
2-0-1
3-0-0
2-1-0
0-1-2
1-2-0
0-2-1
0-3-0
2-0-1
0-2-1
1-0-2
0-1-2
0-0-3

that 16 possibility and 4 possible set. that 1/4

Okay, here's the problem. Not all of the 16 possibilities have the same probability of occurring. You can get 1-1-1 six different ways (red, then green, then blue; red, then blue, then green; blue, then red, then green, etc.) You can get 1-2-0 (or 1-0-2, or 2-0-1 etc.) three different ways (red, green, green; green, red, green; green, green, red,) and you can only get 3-0-0 (or 0-3-0, or 0-0-3) one way (red, red, red.) So you are twice as likely to get 1-1-1 as 1-2-0, and six times as likely to get 1-1-1 as 3-0-0. There really are 27 possible permutations: 6 ways to get 1-1-1, 3 ways each to get 1-2-0, 1-0-2, 2-1-0, 2-0-1, 0-1-2, and 0-2-1, and one way each to get 3-0-0, 0-3-0, and 0-0-3. Of those 27, the six 1-1-1s, 3-0-0, 0-3-0, and 0-0-3 will all get you a set. So the odds of a three card set are 6+1+1+1 divided by 27, or 1/3

[MichelSableheart]

The mistake you are making, caymanmew, is that the chance of getting 2 reds 1 blue isn't the same as getting 1 red 1 blue 1 green, because there are more different orders in which you can get them. For the first card, there are 3 possible colours, for the second card, there are 3 possible colours, and for the third card, there are 3 possible colours. In total, there are 3x3x3=27 possibilities:

1. RRR
2. RRG
3. RRB
4. RGR
5. RGG
6. RGB
7. RBR
8. RBG
9. RBB
10. GRR
11. GRG
12. GRB
13. GGR
14. GGG
15. GGB
16. GBR
17. GBG
18. GBB
19. BRR
20. BRG
21. BRB
22. BGR
23. BGG
24. BGB
25. BBR
26. BBG
27. BBB

out of those combinations, the following combinations form a set:

1. RRR
2. RGB
3. RBG
4. GRB
5. GGG
6. GBR
7. BRG
8. BGR
9. BBB

That's 9/27 combinations, so 1/3.

Not entirely sure if the reasoning of 100% chance of getting one set, so 1/3 chance of getting 2 is correct though.

[Caymanmew]

ok i was wrong
ljex you were right


.... i do need to touch up on my math .... i used to be better

[MichelSableheart]

Not going to write this one fully down, but:

6x red: 1 combination (two sets)
5x red, 1x green: 6 combinations
5x red, 1x blue: 6 combinations
4x red, 2x green: 15 combinations
4x red, 1x green, 1x blue: 30 combinations (2 sets)
4x red, 2x blue: 15 combinations
3x red, 3x green: 20 combinations (2 sets)
3x red, 2x green, 1x blue: 60 combinations
3x red, 1x green, 2x blue: 60 combinations
3x red, 3x blue: 20 combinations (2 sets)
2x red, 4x green: 15 combinations
2x red, 3x green, 1x blue: 60 combinations
2x red, 2x green, 2x blue: 90 combinations (2 sets)
2x red, 1x green, 3x blue: 60 combinations
2x red, 4x blue: 15 combinations
1x red, 5x green: 6 combinations
1x red, 4x green, 1x blue: 30 combinations (2 sets)
1x red, 3x green, 2x blue: 60 combinations
1x red, 2x green, 3x blue: 60 combinations
1x red, 1x green, 4x blue: 30 combinations (2 sets)
1x red, 5x blue: 6 combinations
6x green: 1 combination (2 sets)
5x green, 1x blue: 6 combinations
4x green, 2x blue: 15 combinations
3x green, 3x blue: 20 combinations (2 sets)
2x green, 4x blue: 15 combinations
1x green, 5x blue: 6 combinations
6x blue: 1 combination (2 sets)

Total number of combinations: 729
Total number of combinations with 2 sets: 243
Chance: 243/729 = 1/3

Conclusion: the reasoning of chance of double set = chance of 1 set in 3 cards is correct.

[jrh_cardinal]

as the two guys before me have said, it's definitely 1/3 with 3 cards. That's pretty crazy that he would get it 3 times in one round. What's crazier is that he'd get to exactly 6 cards 3 times in one round, card count obviously worked out perfectly

[ljex]

as the two guys before me have said, it's definitely 1/3 with 3 cards. That's pretty crazy that he would get it 3 times in one round. What's crazier is that he'd get to exactly 6 cards 3 times in one round, card count obviously worked out perfectly

And he eliminated people in the right order...

[thebest712]

have a question, if you have a deck of cards, with all a territ on it, do you have on equal number of red, green and blue cards.

if so you don't have 1/3 cause for example: if you have 2 blue cards, then change of getting a third is smaller( realy on a small map with not a lot of cards in play) so it depends on the map and the number of cards in play :p

[MichelSableheart]

On CC, the colour of the card is determined randomly when the card is given to a player. There are no fixed colour-territory combinations.