Attention mathematical geniuses...part II

[the liquidator]

What are the odds of having a set in 3 cards? How about 4?

[ConquererKing]

A set in three is a 33.33333...% chance and a set in 4 is a 66.66666...% chance. And of course a set in 5 is 100%

[Georgerx7di]

Yes, he's absolutely right. 1/3 change with 3 cards. 2/3 with 4, and 3/3 or 100% with 5. Also, if two players have 3 cards, the chances that at least one of them has a set is 5/9 or 55.555%. This is a little harder to calculate, but if you think of the nine possibilities it makes sense. I hope I don't have to post again to show the math. Also, the odds of both having a set would obviously be 1/9.

[chessplaya]

Yes, he's absolutely right. 1/3 change with 3 cards. 2/3 with 4, and 3/3 or 100% with 5. Also, if two players have 3 cards, the chances that at least one of them has a set is 5/9 or 55.555%. This is a little harder to calculate, but if you think of the nine possibilities it makes sense. I hope I don't have to post again to show the math. Also, the odds of both having a set would obviously be 1/9.

[ParadiceCity9]

Yes, he's absolutely right. 1/3 change with 3 cards. 2/3 with 4, and 3/3 or 100% with 5. Also, if two players have 3 cards, the chances that at least one of them has a set is 5/9 or 55.555%. This is a little harder to calculate, but if you think of the nine possibilities it makes sense. I hope I don't have to post again to show the math. Also, the odds of both having a set would obviously be 1/9.

thanks for makin me feel bad about myself
now im gonna go cut my self

[Ishiro]

Yes, he's absolutely right. 1/3 change with 3 cards. 2/3 with 4, and 3/3 or 100% with 5. Also, if two players have 3 cards, the chances that at least one of them has a set is 5/9 or 55.555%. This is a little harder to calculate, but if you think of the nine possibilities it makes sense. I hope I don't have to post again to show the math. Also, the odds of both having a set would obviously be 1/9.

Unless I'm wrong, color of cards is random, there is no fixed deck like in the board game, so each player's chance to have a set is independent.

[Velvecarrots]

And the chance of having a top set in a flat rate game after 3 cards is 2/9.

[maniacmath17]

Actually, the chances of having a set with 4 cards (assuming players don't automatically cash with a set at 3) is 7 out of 9 or 77.7%

You just take the chances of having a set with 3 plus the chances of having a set with 4. The 2/3 chance of making a set with 4 cards is under the assumption that the player does not already have a set when at 3 cards.

[tahitiwahini]

What are the odds of having a set in 3 cards? How about 4?

This subject is discussed at length here, you may be interested.

The discussion eventually led to the Estimated Armies from Cards calculation that's currently available in stocksr's greasemonkey script which I again highly recommend (get it from the Utilities and Plug-ins for Conquer Club thread at the top of this forum).

[AAFitz]

What are the odds of having a set in 3 cards? How about 4?

This subject is discussed at length here, you may be interested.

The discussion eventually led to the Estimated Armies from Cards calculation that's currently available in stocksr's greasemonkey script which I again highly recommend (get it from the Utilities and Plug-ins for Conquer Club thread at the top of this forum).

I love the new plug-in...its great, i dont use some of the features, but when I need them they are great...but in regards to the expected number of armies from cards...I ignore it completely...the only way to really play....is to see 3 cards and assume there is a mixed set, and be prepared for it....the odds of the set are irrelevant, because you have to be ready for it either way...and i mean this mostly for flat rate doubles....

and in escalating when the cards are worth high figures, the expected armies is listed as a percentage of the total...but in reality, you either have to gamble there isnt a set, that the player wont use it, or that he has a set....if the cash in is worth 70....you can expect either 70 or 0 with 3 or 4 cards....planning for 57 armies is not going to get you much

edit....and this isnt knocking the idea, Im more commenting on my particular strategy, and how it doesnt fit....Im sure if used correctly its very useful

[mibi]

yeah the 'expected armies' feature is pretty useless.

[Georgerx7di]

After reviewing the play, maniacmath is absolutely right.

For a more in depth explanation:

You have 4 cards, each has 3 possibilities. So the total number of possible combinations is 81.
The only way you don't have a set is if you have 2 pair.
There are 6 ways that you can have 2 red and 2 green, 6 ways to have 2 blue and 2 green, and 6 ways that you can have 2 blue and 2 red, for a total of 18 possible ways you won't have a set. Leaving 63 ways that you can.

So, the probability is 63/81, which reduces to 7/9.

I apologize for my earlier mistake.

Note: Ishiro's comment is rediculous and shows that he knows nothing about probability.