card-games
What would the formula be to calculate the odds be of pulling all four 2's, all four 3's and at least 2 of the 4 4's from a standard 52 card deck. No order is necessary as long as these ten cards are pulled first.
A second question is what is the formula or odds of having any three of the four 8's and any 2 of the 4 7's remaining as the last five cards in a standard deck of cards.
Imagine if you will for a moment that for each specifically named card, you have a unique id number on it. I.e. you have an Ornithopter1, an Ornithopter2, an Ornithopter3, etc...
As there are 60 cards total in the deck and your handsize is 7 cards, the number of different starting hands is $\binom{60}{7}$.
Our ideal hand contains at least one of each of our five card categories, abbreviated hereout as $A,B,C,D,E$, which have 6,4,4,4, and 17 copies available respectively. There are then 25 other cards (not of the aforementioned types) in the deck.
We can approach the counting required directly or via inclusion-exclusion. Both methods at first glance will require similar amounts of work, so I will go directly.
$\begin{array}{c|c|c|c|c|l} ~A&B&C&D&E&\text{total hands}\\ \hline 1&1&1&1&1&6\cdot 4\cdot 4\cdot 4\cdot 17\cdot \binom{25}{2}\\ 2&1&1&1&1&\binom{6}{2}\cdot 4\cdot 4\cdot 4\cdot 17\cdot 25\\ 1&2&1&1&1&6\cdot \binom{4}{2}\cdot 4\cdot 4\cdot 17\cdot 25\\ 1&1&2&1&1&6\cdot 4\cdot\binom{4}{2}\cdot 4\cdot 17\cdot 25\\ 1&1&1&2&1&6\cdot 4\cdot 4\cdot \binom{4}{2}\cdot 17\cdot 25\\ 1&1&1&1&2&6\cdot 4\cdot 4\cdot 4\cdot \binom{17}{2}\cdot 25\\ \color{red}{2}&\color{red}{2}&\color{red}{1}&\color{red}{1}&\color{red}{1}&\color{red}{\binom{6}{2}\cdot \binom{4}{2}\cdot 4\cdot 4\cdot 17}\\ 2&1&2&1&1&\binom{6}{2}\cdot4\cdot\binom{4}{2}\cdot4\cdot17\\ \vdots\\ 3&1&1&1&1&\binom{6}{3}\cdot4\cdot4\cdot4\cdot17\\ \vdots\\ \end{array}$
6*4^3*17*25*12 + 15*4^3*17*25 + 6*6*4^2*17*25*3 + 6*4^3*17*8*25 + 15*6*4^2*17*3 + 6^3*4*17*3 + 6^3*4*17*3 + 6^2*4^2*17*8*3 + 20*4^3*17+6*4^3*17*3 + 6*4^3*17*8*5 = 4479840 + 625600 = 5105440
The probability then is: $$\frac{5105440}{\binom{60}{7}}\approx 0.0132$$
or about one in seventy-five starting hands.
How to calculate one of the rows:
Taking the red row for example, we are in the situation where there are 2 of card A, two of card B, one of card C, one of card D, one of card E, and 0 cards of an unlisted type. By multiplication principle, pick which specific copies of card A you use: $\binom{6}{2}$ number of choices, pick which specific copies of card B you use: $\binom{4}{2}$ number of choices, pick which specific copy of card C you use: $\binom{4}{1}=4$ number of choices, etc... for a total number of $\binom{6}{2}\binom{4}{2}\cdot 4\cdot 4\cdot 17$ different hands in this specific case.
Note: I did not account for the cases where you have no lands and two Mox Opal yet, but those can be rather easily added in if you feel like stretching your muscles a bit. Just add the cases where $D\geq 2$ and $E=0$ and $A,B,C\geq 1$. This also does not take into account mulligan rules where you might be allowed to ditch a starting hand if it was particularly bad.
In particular with $A=B=C=1, D=2$ and $E=0$ it will add another $6\cdot4\cdot4\cdot\binom{4}{2}\cdot\binom{25}{2} = 172800$ additional "acceptable" hands, bumping the probability up a little higher. The other cases will add a rather insignificant amount in comparison.
It'll be a substantial amount of work, but it's doable. My intuition says the probability should be pretty high - let's see how that works out.
Final result: Approximately $57.8\%$. Somewhat lower than I would have initially guessed, but still quite large.
And now, the calculation. I'll sort by hand type.
Many of a kind. We count all of the ten-card stacks in which it's possible to make 4 of a kind, full house, or 3 of a kind.
To get four of a kind, we choose the denomination and then choose six other cards, for $13\cdot \binom{48}{6}$ ways. Some of these double up with two denominations each having four; that happens in $\binom{13}{2}\cdot \binom{44}{2}$ ways, which we subtract off for their double-counting.
We won't separately track full house hands; every hand that could be a full house is already three of a kind, so these add nothing to the count.
To get three of a kind, choose a denomination, then choose three cards from that denomination and seven cards of other denominations. That's $13\cdot \binom{4}{3}\cdot\binom{48}{7}$ ways. We can also have three of two kinds in $\binom{13}{2}\cdot \binom{4}{3}^2\cdot\binom{44}{4}$ ways, or three of three kinds in $\binom{13}{3}\cdot\binom{4}{3}^3\cdot\binom{40}{1}$ ways. Counting by inclusion-exclusion, we subtract the "three of two kinds" options and add back the "three of three kinds" options. It's also possible to get three and four in $13\cdot 12\cdot\binom{4}{3}\cdot\binom{44}{3}$ ways, or three, three, and four in $13\cdot \binom{12}{2}\cdot\binom{4}{3}^2$ ways. We subtract the former since they're already counted as four of a kind hands, and add the latter back by inclusion-exclusion.
So far, with just counting hands of four of a kind, full house, or three of a kind, we're at \begin{align*}M &= 13\cdot \binom{48}{6} - \binom{13}{2}\cdot \binom{44}{2} + 13\cdot \binom{4}{3}\cdot\binom{48}{7} - \binom{13}{2}\cdot \binom{4}{3}^2\cdot\binom{44}{4}\\ &\quad + \binom{13}{3}\cdot\binom{4}{3}^3\cdot\binom{40}{1} - 13\cdot 12\cdot\binom{4}{3}\cdot\binom{44}{3} + 13\cdot \binom{12}{2}\cdot\binom{4}{3}^2\end{align*} ways out of the $N=\binom{52}{10}$ ways to choose ten cards. Computing it, $N\approx 1.582\cdot 10^{10}$ and $M\approx 3.811\cdot 10^9$, for a probability $\frac MN$ of about $24.1\%$.
Straights. Our count will exclude all "many of a kind" hands.
To get at straights, we look at the list of denominations among the cards we've drawn. We can have one, two, three, or four cards per denomination - but we've already counted everything with three or more of a kind, so we'll just look at ones and twos. The options:
Adding these up, straights that aren't also 3 or more of a kind account for \begin{align*}S &= 9\cdot 6^5 + (9\cdot 8-8)\cdot\binom{6}{2}\cdot 6^4\cdot 4^2 + \left(9\cdot \binom{8}{2}-8\cdot 7\right)\cdot \binom{7}{4}\cdot 6^3\cdot 4^4\\ &\quad +\left(9\cdot\binom{8}{3}-8\cdot\binom{7}{2}\right)\cdot \binom{8}{6}\cdot 6^2\cdot 4^6 + \left(9\cdot\binom{8}{4}-8\cdot\binom{7}{3}\right)\cdot \binom{9}{8}\cdot 6\cdot 4^8\\ &\quad +\left(9\cdot\binom{8}{5} - 8\cdot\binom{7}{4} - 6\right)\cdot 4^{10}\end{align*} Computing that, it's about $3.253\cdot 10^9$, for an added probability $\frac{S}{N}\approx 20.6\%$. We're up to nearly half.
Since straight flushes, including the royal variety, are already counted as both straights and flushes, we won't be treating them separately. Instead, we'll look at stacks that contain both a straight and a flush - even if those aren't the same five-card hand. That can wait, though.
Flushes. We sort by the length of the long suit, then account for the elements outside that suit. The elements off-suit may cause us to double-count with three or more of a kind, so we'll subtract them.
Adding these up, flushes that aren't also 3 or more of a kind account for \begin{align*}F &= 4\binom{13}{9}\cdot 39 + 4\binom{13}{8}\left(\binom{39}{2}-8\binom{3}{2}\right) + 4\binom{13}{7}\left(\binom{39}{3}-13-7\binom{3}{2}36\right)\\ & +4\binom{13}{10}+ 4\binom{13}{6}\left(\binom{39}{4}-13\binom{36}{1}-6 \binom{3}{2} \binom{36}{2}+\binom{6}{2}\binom{3}{2}^2\right) - \binom{4}{2}\binom{13}{5}^2\\ & + 4\binom{13}{5}\left(\binom{39}{5}-13\binom{36}{2}-5\binom{3}{2}\binom{36}{3}+5\binom{3}{2} 12+\binom{5}{2}\binom{3}{2}^2 33\right) \end{align*} Computing that, it's about $2.922\cdot 10^9$, for an added probability of about $18.5\%$. We're up past $60\%$ overall.
Straights and flushes. Unfortunately, we still have some overcounting to deal with. It's possible to have both a straight and a flush in the same stack, and it's possible with any level of overlap. We'll work with the straight count from earlier to find the doubles. Sorting in the same way:
Adding these up, stacks with both straights and flushes but not three or more of a kind come to \begin{align*}X &= 9\cdot \left(4\cdot 3^5 - \binom{4}{2}\right) + (9\cdot 8-8)\cdot \binom{6}{2}\cdot 4\left(5\cdot 3^4 + 2\cdot 3^5 - 3\right)\\ & + \left(9\cdot 28-8\cdot 7\right)\cdot\binom{7}{4}\cdot 4\left(7\cdot 3^3+16\cdot 3^4+6\cdot 3^5 - 9\right)\\ & + \left(9\cdot 56-8\cdot 28\right)\cdot\binom{8}{6}\cdot 4\left(4\cdot 3^2+24\cdot 3^3+45\cdot 3^4+20\cdot 3^5-30\right)\\ & + \left(9\cdot 70-8\cdot 35\right)\cdot\binom{9}{8}\cdot 4\left(2\cdot 3+16\cdot 3^2+56\cdot 3^3+112\cdot 3^4+70\cdot 3^5 - 105\right)\\ & + \left(9\binom{8}{5} - 8\binom{7}{4} - 6\right) 4 \left(1+10\cdot 3+45\cdot 3^2+120\cdot 3^3+210\cdot 3^4+252\cdot 3^5 - 378\right)\end{align*} Computing that, it's about $0.848\cdot 10^9$, for a probability of about $5.4\%$.
The overall probability we seek is $\dfrac{M+S+F-X}{N} = \dfrac{9139533260}{15820024220}\approx 57.8\%$.
Best Answer
Firstly I'd like to clear up any confusion regarding the term "odds" and how I'll be using it in my answer. If I was to roll a standard fair dice I would have 1 chance in 6 (number of successful outcomes verses the number of all possible outcomes) of rolling a specified number.
If a fair bookmaker (there's an oxymoron) was to take bets on this game he would give odds of 5 to 1, you would (on average) lose 5 times at 1 dollar per time and win 1 time at $5 (the odds the bookmaker gave you ) for every 6 games, leaving you square. Similarly in a standard deck of cards you have 1 chance in 52 of picking the ace of hearts, but the bookmakers odds would be 51 to 1, and so on for various other games of chance. This seeming discrepancy of "1" is only due to the fact that the bookmaker doesn't include your ORIGINAL bet in the odds otherwise it would be the same as the number of chances, either way as the chances increase the discrepancy becomes more insignificant.
Just to be clear, I will be using the number of "CHANCES" as opposes to "ODDS", if you wish to convert, just deduct 1, but in this case with the numbers involved this becomes overwhelmingly insignificant.
Q1
Number of ways 10 cards can be picked from a deck (order unnecessary) = number of ways all four 2's, all four 3's and at least 2 of the 4 4's can be arranged just comes down to how many ways to pick just two of the 4,s (again order excluded)=$\binom {4}{2}$=6
The chances of this happening in the first 10 cards is $\frac{15820024220}{6}\approx2,636,670,703 $
Q2
Number of arrangements (order excluded) of the last 5 cards in a deck = $\binom {52}{5}$=2598960
Number of ways of picking 3 of the 8's = same as the number of ways of excluding 1 of the 8's = 4
Number of ways of picking 2 of the 7's =$\binom {4}{2}$=6
Therefore there are 4x6 = 24 ways of picking the three 8's and two 7's
So the chances of this occurring = $\frac{2598960}{24} =108290$