In short, the probability of a 7-card straight when drawing 7 random cards from a standard deck of 52 is $0.000979$.
To calculate this value, we note that all 7-card hands are equally likely, of which there are ${52 \choose 7} = 133,784,560$ possibilities.
Next, we compute the number of 7-card straights. Ignoring suit, we note that there are $8$ possible straights (starting with {A, 2, 3, 4, 5, 6, 7} through {8, 9, 10, J, Q, K, A}). For each card in the straight, there are 4 possibilities for the suit, such that there are $4^7 = 16384$ ways to assign the suits to the 7 cards. However, $4$ of these suit assignments yield straight flushes (all clubs, all diamonds, etc.), so the actual number of suit assignments that can yield a straight (but not a straight flush) is $16384 - 4 = 16380$.
Putting all this together, there are $8 \times 16380 = 131,040$ possible 7-card straights out of $133,784,560$ possible 7-card hands, yielding a probability of $\approx 0.000979$.
I think that one problem in creating an AI solution to what should be bid in pinochle, it that humans have a hard time answering the same question.
Two thoughts:
- I like the way you isolated it to 3 player single deck. This avoids the more complex calculation of "what does my partner have?" or "should I let my partner have the bid? (does she have a better hand than mine). it also eliminates various strategies of partner bidding that are used to signal. (For example, some players use 51 as an opening bid to let the partner know that they have aces)
- The basic question you are trying to answer is: what I have in my hand + what can I expect from the 3 random cards? + what I can expect to win during the trick phase.
What I have in my hand is easy (you have already given the list)
What can I expect from the 3 random cards? the low end of the range is zero and the top end of the range depends on what is in your hand. I think this would be easier to model than the trick phase. In a single deck, their are no doubles, so the max would be 43, in the case where you had an ace, king and queen in a single suit, named that suit trump and the ace, king and queen also completed sets (one of each suit).
For the trick phase, my mom taught me that safe was 1/2 the points available (since you named trump) with upward adjustments if you have aces, a majority of trump or a "short suit" with a strong backup suit. (one suit with zero or very few cards with a non trump suit where you have ace, ten, king for example). There are 24 points in a singe deck, so 12 is a good starting point.
For humans, bidding is a combination of addition and instinct. I imagine that if I were programming a machine that I would put ranges in. Probabilities are no match for reality - what actually happens.
Best Answer
You have 6 cards (out of 52) and you want to know if another set of 12 (from the same 52) have at least one of three particular cards which you do not have. It is easier to work out the probability they do not have any of the three, which is
$$\frac{43}{46}\times\frac{42}{45}\times\frac{41}{44}\times\frac{40}{43}\times\frac{39}{42}\times\frac{38}{41}\times\frac{37}{40}\times\frac{36}{39}\times\frac{35}{38}\times\frac{34}{37}\times\frac{33}{36}\times\frac{32}{35}$$
$$= \frac{34 \times 33 \times 32}{46 \times 45 \times 44} \approx 0.39$$
so subtracting this from 1 (and multiplying by 100 to get a percentage) means your opponents have a chance of about 61% of having at least one of the three particular cards.
Even if they do have one or two of these cards, it is possible, though less likely, that your partner has an even higher one.