What is the probability to have a set in 3 cards...
Posted: Tue Sep 02, 2008 2:05 amIs 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...
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...
