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Copy pasted from wikipedia:

The probability of having a tradable set of cards when holding three cards is 42.28%, 81.70% holding four cards, and 100% holding 5 cards.

The exact probabilities can differ a little bit per map, because of wildcards.

Note: a double turn in with 7 cards has a probability of about 93%. I don't know the probability of a double turn in with 6 cards, but it's low. Too low to gamble upon.

Edit: According to this topic, the odds for a double turn in with 6 cards is 33%.

To summarize:
- A turn in with 3 cards: 42.28%
- A turn in with 4 cards: 81.70%
- A turn in with 5 cards: 100%
- Two turn ins with 6 cards: 33%
- Two turn ins with 7 cards: 93%
"Strength doesn't lie in numbers, strength doesn't lie in wealth. Strength lies in nights of peaceful slumbers." ~Maria
Matty
Copy pasted from wikipedia:

The probability of having a tradable set of cards when holding three cards is 42.28%, 81.70% holding four cards, and 100% holding 5 cards.

The exact probabilities can differ a little bit per map, because of wildcards.

Note: a double turn in with 7 cards has a probability of about 93%. I don't know the probability of a double turn in with 6 cards, but it's low. Too low to gamble upon.

I don't think those wikipedia numbers are correct... unless it includes wild cards..

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?

Well one thing is for sure, the odds are 100% for 2 sets with 6 cards when 1 is wild The odds of me having a set with 3 are about 20% and a set with 4 when the turn in is high 35-40% lol
A Harmonic Generator Intermodulator
Σ
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.
Well, now we know why you call yourself Einstein lolol I would be curious to know how it works in general. Take the classic map. Is there 1 green, 1 red, 1 blue for each territory? Or is it territories divided by 3 and only one color per territory? What about wilds?

When they are assigned is it a random pull from the deck? When they are turned in the go back in the deck, or at the bottom, etc?
dough_boy is online.
We use the number of cards of each kind (including wildcards) from the original Risk game. All the other maps have cards divided among the colours on a pro rata basis from this "golden standard". If we don't have enough cards, we duplicate the deck until we do.

When cards are turned in, they are shuffled into the remaining deck.
“This is how humans are: We question all our beliefs, except for the ones that we really believe in, and those we never think to question.”
- Speaker for the Dead, O.S. Card
From the source code:
\$numWilds = round(\$territories->count() / 21);
"Strength doesn't lie in numbers, strength doesn't lie in wealth. Strength lies in nights of peaceful slumbers." ~Maria
Edit: Sorry, indeed looks like since there's 42 territories, I should've written 1:7 ratio to each color in the set (there's 14 of each "color" [or symbol],, and 2 wild cards)

So that does match the formula used here. Great to know, mighty thanks for the insight Cireon/Matty :-D
Couple other questions I guess, while I'm at it:

I'm guessing if the tert count is not modulo 3, then it's random which color shows up extra?

If we don't have enough cards, we duplicate the deck until we do.
Have seen mention of the duplicated decks before. So does this mean that the same territory may show up multiple spots at the same time? If so, does anything prevent the same player from getting both in his hand at the same time? And they're guaranteed to be the same color?
I'm guessing if the tert count is not modulo 3, then it's random which color shows up extra?
Yes.

So does this mean that the same territory may show up multiple spots at the same time?
Yes. We mark them with 1, 2, etc.

If so, does anything prevent the same player from getting both in his hand at the same time?
No, but they can be distinguished by the number annotations.

And they're guaranteed to be the same color?
Yes, since the colours are generated once.

EDIT: I should implement that BB-code XD
“This is how humans are: We question all our beliefs, except for the ones that we really believe in, and those we never think to question.”
- Speaker for the Dead, O.S. Card