probability question

[HardAttack]

in 2v2 escalating game,
what is the probability of having at least 1 set when me and my partner holding 3 cards each ?

[timogl]

well, since the two are independent, aren't the odds the same for either of you as when you play alone?

[HardAttack]

well, since the two are independent, aren't the odds the same for either of you as when you play alone?

i dont think so...
single player to catch a set in 3 cards is 1/3...
do you mean it is same probability for 2 players with 3 cards each, any of them or both of them, probability still 1/3 ?
i dont think so...
now playing/taking couple of turns,
i ll study on this, lol.

[DoomYoshi]

The odds for either of you to have a set is 2/3. The odds for you both to have a set is 1/9.

[HardAttack]

The odds for either of you to have a set is 2/3. The odds for you both to have a set is 1/9.

odds for both of us is ok, it is 1/9

but the first part is wrong i guess,
take this, what if the game was not 2v2 but 3v3,
then probability, is it 3/3 ?

[Swifte]

since they are independent events, they multiply, they do not add. the math goes like this:
in 2v2, if odds of getting a set in 3 cards is 1/3:
Neither has a set: (2/3)*(2/3) = 4/9
Exactly 1 has a set: (2/3)*(1/3)*2 = 4/9
Both have a set: (1/3)*(1/3) = 1/9
Total = 9/9 (100%)

in 3v3:
none have a set: (2/3)*(2/3)*(2/3) = 8/27
exactly 1 has a set: 3*(1/3)*(2/3)*(2/3) = 12/27
exactly 2 have a set: 3*(1/3)*(1/3)*(2/3) = 6/27
all 3 have a set: (1/3)*(1/3)*(1/3) = 1/27
Total = 27/27

[HardAttack]

cool answer swifte,
answer is 1-(4/9) = 5/9 it is.

[timogl]

there are three cards colors. you have three cards.

when you have three cards, the possible combinations (4) that match are

red red red
blue blue blue
green green green
red blue green

that don't match (6) are

red red blue
red red green
green green red
green green blue
blue blue red
blue blue green

so, in the first place, that is not one in three, it is 4 in 10.

and your partner odds to match are precisely the exact same as yours are, you are drawing from the same exact set of possibles aren't you? and why do they multiply? explain that. i understand you applying some law of probability, but i don't believe it applies here. this is three random draws that can fall in any one of 10 combinations. two vs two, three vs three, 100 vs 100, everybody still has the same chance. four in 10. it doesn't change.

[timogl]

(i won't delete. but i do aplogize if the odds actually are one in three because of how the blue, green and red relate to each other, i don't keep up with that, but, i assume red is the highest number, follwed by green and blue (i suppose they are distributed the same way green, red and blue sets relate in flat rate, that is, 8-6-4?). so if that works out to one in three, thats fine. but i don't understand why changing the number of players changes the chances for any one player, your partner or any other player matching up. any player's chance of matching has to stay the same, doesn't it?)

[Swifte]

yes. the odds of any player individually having a set with 3 cards is constant. the question was how to determine if at least 1 of a set of players has a set - surely you'd agree that the odds of at least 1 player having a set would vary by the number of players involved. more players, greater chance at least one has a set, as the math i posted above shows. if i told you to flip a coin twice, and then flip it 4 times, which time do you think you have a greater chance of flipping heads at least once? each flip may be 50/50, but give me 4 flips for that bet, please.

[Swifte]

As far as an individual player, the odds of having a 3 card set are 1/3, which is most easy to think about in the following way:

if you get red and red, only red makes a set (1 color out of 3)
if you get blue and blue, only blue makes a set (1 color out of 3)
if you get green, green, only green makes a set (1 color out of 3)
if you get 2 mixed cards, only the remaining color will make a set. (1 color out of 3)
so, regardless of what 2 cards you get to start, only 1 color out of the 3 will get you a 3 card set. Hence the use of 1/3 for an individual player.

Why does your estimate come out high? Because you have not considered all permutations. In fact there are 27 (3*3*3) ways of drawing cards.

show: Possible combinations

Of the 27 combinations, only the 9 highlighted in green give you a 3 card set. 9 of 27 is 1/3.

[timogl]

if there are 27 combinations then list them.

oh ok, i looked at your combinations. i am sorry, but red red blue is the same thing as blue red red (and so on). there are not 27 unique combinations.

[timogl]

(the 3! thing, 3*3*3, that is true for when you have three numbers, you say, how many 3 number combinations can you make. when you have, say, 1, 2 and 3, yes, 113 is different from 311. but these are colors. when you change their position in the order (red red blue and blue red red for intance, it doesn't change their value. they either match or they don't. there are 10 unique cominations, i listed them. four of them are matches. the one in three part i won't disupte, it makes perfect sense. as far as three people drawing cards, if you say the probability that one of them will draw a set is higher because there are higher because there are more of them, ok, if you say so.)

[SirSebstar]

(the 3! thing, 3*3*3, that is true for when you have three numbers, you say, how many 3 number combinations can you make. when you have, say, 1, 2 and 3, yes, 113 is different from 311. but these are colors. when you change their position in the order (red red blue and blue red red for intance, it doesn't change their value. they either match or they don't. there are 10 unique cominations, i listed them. four of them are matches. the one in three part i won't disupte, it makes perfect sense. as far as three people drawing cards, if you say the probability that one of them will draw a set is higher because there are higher because there are more of them, ok, if you say so.)

huh?!?
you start of with red red red, then constantly chance one card till you have also had green green green and blue blue blue.

if you like, you can use three sides dice.
1,1,2 and 2,1,1, are then different combinations. both dont give you a set, but its about the possibility, so you need to list all possibilities. as in, you are more likely to see 1,1,2 (=also 2,1,1)then 1,1,1 in a series of 3 spoils.

show

27 posibilities, 9 set 27/9= one in 3
so each person has a 1 in 3 chance to get a set.

now, back to your comment. swifte said that if you are playing with multiple people(more then 1),the amount of chances to get a set (per individual are 1/3) are equal per person, but as a group, if 3 peoples have 3 spoils, then its very likely that one of them has a set (and the other 2 have not)

play 27 games on paper for 3 players. get to 3 spoils, write down all the spoil results (not only the "unique" variants)and again its 1/3 chance. the chances everybody(all 3) have a spoil happens 1/9 times,one player getting aspoils but the other 2 not, 1/3.

does this help?

[greenoaks]

(the 3! thing, 3*3*3, that is true for when you have three numbers, you say, how many 3 number combinations can you make. when you have, say, 1, 2 and 3, yes, 113 is different from 311. but these are colors. when you change their position in the order (red red blue and blue red red for intance, it doesn't change their value. they either match or they don't. there are 10 unique cominations, i listed them. four of them are matches. the one in three part i won't disupte, it makes perfect sense. as far as three people drawing cards, if you say the probability that one of them will draw a set is higher because there are higher because there are more of them, ok, if you say so.)

when it comes to permutations the order matters however for the determination of the odds of 1 player having a set it does not.

it does not matter if you got red then blue or blue then red, there is only 1 card out of 3 that can give you a set.