The probability of having a tradeable set of cards when holding three cards is 33.3% (9/27), 74% holding four cards (60/81), and 100% holding 5 cards (243/243).
j1mathman wrote:Nietzsche is right on the first three,
Wiki gives 66 and 100 percent for 4 and 5 though
The probability of having a tradeable set of cards when holding three cards is 33.3% (9/27), 74% holding four cards (60/81), and 100% holding 5 cards (243/243).
While the numbers stated above are correct you could look at it a bit differently. The probability of having a set in 4 cards includes the odds that you already made it in 3. If, however, you have not made a set in 3 the probability of making a set on the next card is 66.7%. To simplify, as this is necessary for BBS, the only way to not match in 4 is to have 2 pair. That means there is only 1 chance in 3 for that to happen and 2 chances in 3 that it does not.
loutil wrote:While the numbers stated above are correct you could look at it a bit differently. The probability of having a set in 4 cards includes the odds that you already made it in 3. If, however, you have not made a set in 3 the probability of making a set on the next card is 66.7%. To simplify, as this is necessary for BBS, the only way to not match in 4 is to have 2 pair. That means there is only 1 chance in 3 for that to happen and 2 chances in 3 that it does not.
Nice, thanks. My feeble mind now needs a heavy dosage of kittens. Excuse me.
Great...someone says 60, someone says 61 and someone says 64. And nobody is correct.... It is not a difficult problem. And you do not need whatever program or strange method of writting down all odds you use. You have 1/3 chance after 3 cards In 4th card, if you do not have a set already (2/3) you have 2/3 chance to get one. That is 2/3 * 2/3 = 4/9 1/3+4/9=7/9. Which would have been 63/81 with that strange "I refuse to think, lets just writte them all down" method
Kaskavel wrote:Great...someone says 60, someone says 61 and someone says 64. And nobody is correct.... It is not a difficult problem. And you do not need whatever program or strange method of writting down all odds you use. You have 1/3 chance after 3 cards In 4th card, if you do not have a set already (2/3) you have 2/3 chance to get one. That is 2/3 * 2/3 = 4/9 1/3+4/9=7/9. Which would have been 63/81 with that strange "I refuse to think, lets just writte them all down" method
Kaskavel wrote:Great...someone says 60, someone says 61 and someone says 64. And nobody is correct.... It is not a difficult problem. And you do not need whatever program or strange method of writting down all odds you use. You have 1/3 chance after 3 cards In 4th card, if you do not have a set already (2/3) you have 2/3 chance to get one. That is 2/3 * 2/3 = 4/9 1/3+4/9=7/9. Which would have been 63/81 with that strange "I refuse to think, lets just writte them all down" method
64.
You insist on counting instead of thinking, even when this has already been done for you?
Kaskavel wrote:Great...someone says 60, someone says 61 and someone says 64. And nobody is correct.... It is not a difficult problem. And you do not need whatever program or strange method of writting down all odds you use. You have 1/3 chance after 3 cards In 4th card, if you do not have a set already (2/3) you have 2/3 chance to get one. That is 2/3 * 2/3 = 4/9 1/3+4/9=7/9. Which would have been 63/81 with that strange "I refuse to think, lets just writte them all down" method
64.
You insist on counting instead of thinking, even when this has already been done for you?
I tried to work it out from first principles, but it's been 30 years or more since last I studied probability, and I must admit it was fuzzy in my mind.
“Life is a shipwreck, but we must not forget to sing in the lifeboats.” ― Voltaire
Kaskavel wrote:Great...someone says 60, someone says 61 and someone says 64. And nobody is correct.... It is not a difficult problem. And you do not need whatever program or strange method of writting down all odds you use. You have 1/3 chance after 3 cards In 4th card, if you do not have a set already (2/3) you have 2/3 chance to get one. That is 2/3 * 2/3 = 4/9 1/3+4/9=7/9. Which would have been 63/81 with that strange "I refuse to think, lets just writte them all down" method
64.
You insist on counting instead of thinking, even when this has already been done for you?
Kaskavel wrote:Great...someone says 60, someone says 61 and someone says 64. And nobody is correct.... It is not a difficult problem. And you do not need whatever program or strange method of writting down all odds you use. You have 1/3 chance after 3 cards In 4th card, if you do not have a set already (2/3) you have 2/3 chance to get one. That is 2/3 * 2/3 = 4/9 1/3+4/9=7/9. Which would have been 63/81 with that strange "I refuse to think, lets just writte them all down" method
64.
You insist on counting instead of thinking, even when this has already been done for you?