Robinette
Without wild cards, wouldn't it be a 1 in 3 chance of making a set with 3 cards?
And 2 in 3 chance of making a set with 4 cards?
Not exactly, this would only be correct if there were infinite territories.
For example, if you have 2 green cards, the probability of getting another green card is smaller than 1/3, since there would be N/3 of blue cards, N/3 of red cards but only (N-2)/3 green cards to pick (N is the total number of territories).
In a similar way, if your first two cards are of a different colour (which is the most probable case), the probability of getting a turn in would be a bit bigger than 1/3.
The 81.70% probability of having a set holding 4 cards is probably correct. However, it includes the case of the player having a set with three cards. In case that the player didn't have
a set with 3 cards (two cards of one colour and one of a different colour is the only case), the probability of having a set with the 4th card would be 2/3 as you say (asuming infinite territories and no wild cards).
Note that knowing that the probability of having a set with 3 cards is 0.4228, we can estimate the probability of having a set with 4 cards by adding the probability of having a set with 3 cards and the probability of not having a set multiplied by 2/3:
0.4228 + (1-0.4228)*2/3=0.808
Which is very close to 0.8170.
In a similar way, if we assume that both the 0.8170 and 0.4228 probabilities are correct, we can calculate the real probability P of having a set with 4 cards in the case that you didn't have a set with the first three cards (without the assumption of infinite territories and no wild cards).
0.4228 + (1-0.4228)*P=0.8180
Which gives a probability of 68.4%, which is very close to 2/3.
[quote=Robinette]
Without wild cards, wouldn't it be a 1 in 3 chance of making a set with 3 cards?
And 2 in 3 chance of making a set with 4 cards?
[/quote]
Not exactly, this would only be correct if there were infinite territories.
For example, if you have 2 green cards, the probability of getting another green card is smaller than 1/3, since there would be N/3 of blue cards, N/3 of red cards but only (N-2)/3 green cards to pick (N is the total number of territories).
In a similar way, if your first two cards are of a different colour (which is the most probable case), the probability of getting a turn in would be a bit bigger than 1/3.
The 81.70% probability of having a set holding 4 cards is probably correct. However, it includes the case of the player having a set with three cards. In case that the player didn't have
a set with 3 cards (two cards of one colour and one of a different colour is the only case), the probability of having a set with the 4th card would be 2/3 as you say (asuming infinite territories and no wild cards).
Note that knowing that the probability of having a set with 3 cards is 0.4228, we can estimate the probability of having a set with 4 cards by adding the probability of having a set with 3 cards and the probability of not having a set multiplied by 2/3:
0.4228 + (1-0.4228)*2/3=0.808
Which is very close to 0.8170.
In a similar way, if we assume that both the 0.8170 and 0.4228 probabilities are correct, we can calculate the real probability P of having a set with 4 cards in the case that you didn't have a set with the first three cards (without the assumption of infinite territories and no wild cards).
0.4228 + (1-0.4228)*P=0.8180
Which gives a probability of 68.4%, which is very close to 2/3.