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What is the probability to have a set in 3 cards...

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What is the probability to have a set in 3 cards...

Postby HardAttack on Tue Sep 02, 2008 2:05 am

Is it ?

The possibility of having 3 red cards in 3 random cards is as follows...

The 1st card you pick gonna be red, that s 1/3 probability.
The 2nd card you pick gonna be red too, that is again 1/3 probability.
But together with 1st card case, makes total 1/3 * 1/3 = 1/9 probability for the first 2 consequetives are red.
Same with the previous explanations,
The 3rd card you pick gonna be red, that s 1/3 probability as well.
Totally, 1/3 * 1/3 * 1/3 = 1/27 probability to have a red set in just 3 cards.

The cases are same for three greens, three blues.
Then this makes a sub total of 3 * 1/27 = 1/9 (three reds, greens or blues in just three cards) ==> Conclusion 1
The problem is going on, cos we are talking about a set in just three cards..
We need to calculate red green and blue situation as well....

The first card you pick is out of any care, whatever it is doesnt matter...The probability for it is 1. (May be green, blue or red...)
The following card you pick is going to be any 2 of 3, just will not be the same card with which you picked in the 1st turn....The probability is 2/3...
The third card is going to be 1 of the 3, that means can not be the first 2 colors....
Then the color set has the probability of 1 * 2/3 * 1/3 = 2/9 conclusion 2
Finally, we need to add conc 1 + conc 2 and get the final probability for getting a set in just 3 cards which is 1/9 + 2/9 = 3/9, in return % 33.3333....

But the calculations above are made based on the assumption that, the number of cards stay same in the pool whatever you or other players pick any color in their turns or not...Otherwise the calculation gets real complicated cos you need to include the probability of the other players, what might they have picked etc...But anyways, the above calculation will give some idea, but not show the real probability....I will calculate the probability of having set in 4 cards as well but of course under the assumption of cards stay same in number in the pool....The other case needs some computer program for real complex scenarios...
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Re: What is the probability to have a set in 3 cards...

Postby BENJIKAT IS DEAD on Tue Sep 02, 2008 2:08 am

There is always an equal possibility of the next card being any of the 3 colours, so your calculations above are correct (even if you could have saved yourself a lot of trouble with a little searching - it has been discussed a number of time before)
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Re: What is the probability to have a set in 3 cards...

Postby HardAttack on Tue Sep 02, 2008 2:14 am

i didnt know such an issue took place some time before... hmmm, i will check it..ty
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Re: What is the probability to have a set in 3 cards...

Postby HardAttack on Tue Sep 02, 2008 2:20 am

yep, now i got there and realised that same issue took place here around some before.....Then i appologise for repetition... :cry:

HardAttack wrote:i didnt know such an issue took place some time before... hmmm, i will check it..ty
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Re: What is the probability to have a set in 3 cards...

Postby lancehoch on Tue Sep 02, 2008 11:33 am

A more in-depth look at these probabilities (at least for the first 5 cards) can be found: here.
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Re: What is the probability to have a set in 3 cards...

Postby HardAttack on Tue Sep 02, 2008 11:58 am

ty lancehoch mate, you did good there, i checked it out..thanks

lancehoch wrote:A more in-depth look at these probabilities (at least for the first 5 cards) can be found: here.
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Re: What is the probability to have a set in 3 cards...

Postby FabledIntegral on Thu Sep 04, 2008 11:20 pm

Much easier to do the math as...

After having 2 cards, no matter what, you can always manage to get a 3-set the next turn. In any scenario of 2 cards you have, you have to get a specific third card in order to form a set. Thus since you need a specific card, and only one of them will work, there's a 33% chance of forming a set...

IE.

R-R ... you need an R not G/B
B-B ... you need a B not R/G
G-G ... you need a G not R/G
R-B ... you need a G not R/B

etc...
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