Conquer Club

What's the chance of having a set (escalating) if you have 3 cards? 4 cards?


(I've searched the forum and couldn't find an answer).


(Yes, I'm awful at setting up probability questions).

9/1129812391283

1 card 0%

2 cards 0%

3 cards 33%

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4 cards 59% (81^-1 x 48 I believe, to lazy to do it, but it seems low, I was always assuming with 4 cards I had around 75% of chances of having a set)

Edit: my ego was hurt, so fixed. 60 out of 81 indeed. 74%

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Nietzsche is right on the first three,

Wiki gives 74 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).

http://en.wikipedia.org/wiki/Risk_%28game%29

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).

http://en.wikipedia.org/wiki/Risk_%28game%29

yeah, that's what i thought.

Thanks, gentlemen. This thread is now about kittens as well.

I fucking love kittens

I'd feel bad about drowning a 50-gallon bag full of kittens.

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.

nietzsche wrote:
1 card 0%

2 cards 0%

3 cards 33%

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4 cards 59% (81^-1 x 48 I believe, to lazy to do it, but it seems low, I was always assuming with 4 cards I had around 75% of chances of having a set)

Edit: my ego was hurt, so fixed. 60 out of 81 indeed. 74%

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You're wrong. for 4 cards is 75%. 61 out of 81.

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i find your new avatar offensive. please take it down.

wtf.. now i'm lost, i'm counting 64 out of 81.. 79%

Damn you for making me think BBS was back for a mere second.

Serbia wrote:Damn you for making me think BBS was back for a mere second.

He was back a month ago. I have a BBS trap.

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.

nietzsche wrote:

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:

nietzsche wrote:

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?

My bad, i missed one. it's 63, 78%.

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but I don't appreciate the way you bullied me. you should be ashamed of yourself.

(i tried to work this out with a formula for combinations and didn't work, I must say your way was clever.)

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.

nietzsche wrote:

Kaskavel wrote:

nietzsche wrote:

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?

My bad, i missed one. it's 63, 78%.

show

but I don't appreciate the way you bullied me. you should be ashamed of yourself.

(i tried to work this out with a formula for combinations and didn't work, I must say your way was clever.)

You are obviously right. I appologize

Kaskavel wrote:

nietzsche wrote:

Kaskavel wrote:

nietzsche wrote:

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?

My bad, i missed one. it's 63, 78%.

show

but I don't appreciate the way you bullied me. you should be ashamed of yourself.

(i tried to work this out with a formula for combinations and didn't work, I must say your way was clever.)

You are obviously right. I appologize

lol no need man, i was joking. i'm not easily offended.

(i was joking too when i replied 64 without really reading your post)

bet i can offend you, easily.

Your face offends me.