Probability of getting a straight after drawing 20 cards?

Question

Here is the question I am trying to solve:

"Say you draw a 7 from a deck of 50 cards, what is the probability that after drawing 19 more cards from the deck, you will have the cards required to form a straight using the 7 you drew at the beginning?

Note that the deck is missing two 5s (hence the reason for the deck only having 50 cards to start)."

So far, this is what I've surmised:

I've calculated the odds of getting at least the cards required to form a 3-4-5-6-7 as follows:

4 - 3*$\frac {{45 \choose 19}}{49 \choose 19}$ - $\frac {{47 \choose 19}}{49 \choose 19}$

I arrived at this conclusion because to get at least one of each of the required cards, I need to subtract the complement of getting none of the required cards from 1. Since there are 4 different "sets" of cards to be chosen, and each requires a complement to be calculated, I've subtracted the complements from 4. For the first 3 "sets" (3, 4, and 6), there are 45 other cards to choose from. For the last one (5), there are 47 other cards to choose from.

I'm pretty sure the above makes sense; however, I am not sure where to go from here. I don't need a long explanation, just some help and pointers with how to calculate the chance of getting the cards required for "at least one" of all of the possible straights. To elaborate, say I repeat the process above for all of the possible straights (4-5-6-7-8, 5-6-7-8-9, etc.), would I simply add the odds of obtaining each of them together and then take the average?

Answer 1

Suppose you draw $19$ more cards in one of the ${49\choose19}$ possible ways and don’t have a straight using a $7$. Then your draw must be in one of the following ten sets. Each set contains draws that avoid any straight-with-a-$7$ because of a specific obstacle:

• $E_1$: Draws with no $3$s and no $8$s. Size: ${41\choose19}$.
• $E_2$: Draws with no $4$s and no $8$s. Size: ${41\choose19}$.
• $E_3$: Draws with no $4$s and no $9$s. Size: ${41\choose19}$.
• $E_4$: Draws with no $5$s and no $8$s. Size: ${43\choose19}$.
• $E_5$: Draws with no $5$s and no $9$s. Size: ${43\choose19}$.
• $E_6$: Draws with no $5$s and no $10$s. Size: ${43\choose19}$.
• $E_7$: Draws with no $6$s and no $8$s. Size: ${41\choose19}$.
• $E_8$: Draws with no $6$s and no $9$s. Size: ${41\choose19}$.
• $E_9$: Draws with no $6$s and no $10$s. Size: ${41\choose19}$.
• $E_{10}$: Draws with no $6$s and no Jacks. Size: ${41\choose19}$.

The size counts (row 1) with their tallies (row 2), are summarized in the following array. $$\left[ \begin{array}{cc} {43\choose19} & {41\choose19}\\ 7 & 3 \\ \end{array} \right]$$

More than one of these obstacles can occur (in other words, the sets are not pairwise disjoint), but we can use the inclusion-exclusion principle to find out the number of possible card draws in at least one of these sets. These are the draws with no straight-with-a-$7$. This requires finding the size of the intersection $\cap_{i\in S}E_i$ for any subsets $S$ of $\{1,2,3,4,5,6,7,8,9,10\}$.

First consider subsets $S$ of size $2$. There are $10\choose2$ pairs of two different obstacles. The number of ways to encounter both obstacles simultaneously is ${33\choose19}$, if the pairs involve four card ranks, none of which is a $5$ (ten such pairs), ${35\choose19}$ if the pairs involve four ranks including a $5$ (fifteen pairs), ${37\choose19}$, if the pairs involve three ranks, none of which is a $5$ (eleven pairs), and ${39\choose19}$, if the pairs involve three ranks including a $5$ (nine pairs). This information is summarized as before.

$$\left[\begin{array}{cccc} {39\choose19} & {37\choose19} & {35\choose19} & {33\choose19}\\ 9 & 11 & 15 & 10\\ \end{array}\right]$$

The number of ways to encounter three of these obstacles simultaneously is ${49-4a-2b\choose19}\big{/}{49\choose19}$, where $a$ is the number of different non-$5$ ranks involved and $b$ is $1$ or $0$, depending on whether rank $5$ is involved. The possible values of $49-4a-2b$ among the size-$3$ subsets of obstacles, with the number of $3$-subsets that give that value, is $$\left[ \begin{array}{cccccc} 35\choose19 & 33\choose19 & 31\choose19 & 29\choose19 & 27\choose19 & 25\choose19 \\ 28 & 17 & 44 & 16 & 13 & 2 \\ \end{array} \right]$$

In fact, this approach works for any sized subset. In order to apply the inclusion-exclusion principle, we need to find out how many $k$-subsets of obstacles there are with each possible value of $49-4a-2b$. For four obstacles, this is the summary.

$$\left[ \begin{array}{ccccccccc} 35\choose19 & 33\choose19 & 31\choose19 & 29\choose19 & 27\choose19 & 25\choose19 & 23\choose19 & 21\choose19 & 19\choose19 \\ 4 & 1 & 66 & 20 & 83 & 13 & 21 & 1 & 1 \\ \end{array} \right]$$

Briefly, the inclusion-exclusion principle states that the number of card draws encountering at least one of the $E_1$ is the sum of the $|E_i|$, minus the sum of the $|E_i\cap E_j|$ ($i\ne j$), plus the sum of the $|E_i\cap E_j\cap E_k|$ ($i,j,k$ distinct), and so on.

Note that the dot product of the two rows of the summary array equals one term in the inclusion-exclusion formula.

For five through ten obstacles simultaneously the summaries are (now just presenting the top number of the binomial coefficient that gives the count)

$$\left[ \begin{array}{ccccccc} 31 & 29 & 27 & 25 & 23 & 21 & 19 \\ 21 & 3 & 115 & 14 & 84 & 4 & 11 \\ \end{array} \right]$$

$$ \left[ \begin{array}{cccccc} 31 & 27 & 25 & 23 & 21 & 19 \\ 2 & 56 & 3 & 112 & 4 & 33 \\ \end{array} \right], \left[ \begin{array}{cccc} 27 & 23 & 21 & 19 \\ 12 & 65 & 1 & 42 \\ \end{array} \right] $$

$$ \left[ \begin{array}{ccc} 27 & 23 & 19 \\ 1 & 18 & 26 \\ \end{array} \right], \left[ \begin{array}{cc} 23 & 19 \\ 2 & 8 \\ \end{array} \right], \left[ \begin{array}{c} 19 \\ 1 \\ \end{array} \right] $$

The alternating sum of dot products of these arrays (restoring the binomial coefficients in the top rows) is $3340356480525$. This is the number of $19$-card draws that lead to no straight-with-a-$7$. There are ${49\choose19}=18851684897584$ possible draws, so the number that yield a straight-with-a-7 is therefore $18851684897584-3340356480525=15511328417059$, and the probability of a straight-with-a-7 is $$\frac{15511328417059}{49\choose19}=\frac{15511328417059}{18851684897584}\approx0.82281.$$

Mathematica did the heavy lifting.

1/Binomial[49,19] (Binomial[49,19]-Total[Map[#[[1]].#[[2]]&,Table[Transpose[Tally[Map[(-1)^(k+1)Binomial[49-4*Length[#]+If[MemberQ[#,5],2,0],19]&,Apply[Union,Subsets[Flatten[Table[{i,j},{i,3,6},{j,8,i+5}],1],{k}],1]]]],{k,1,10}]]])//N

Also, for what it’s worth, this simulation gives close to $82\%$ each time I run it.

    SeedRandom[Method -> "MersenneTwister"]
deck = {1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6,
    6, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 
   11, 12, 12, 12, 12, 13, 13, 13, 13};
c = Table[Append[RandomSample[deck, 19], 7], {i, 1, 10000}];
Select[Tally[
      Table[Max[
        Table[Length[
          Intersection[Table[n + i, {i, 0, 4}], c[[i]]]], {n, 3, 
          7}]], {i, 1, 10000}]], #[[1]] == 5 &][[1]][[2]]/10000 // N