# 3 cards of 52, Making Straight or better with 2 imaginary wild Jokers

Random shuffled 52-card deck, 3 cards are dealt.

Find a probability of making 5-card Straight or better with imaginary 2 wild Jokers.

In order to make Straight or better, we need the three cards to represent either of the following:

1. Three of a kind,
2. A pair,
3. 3 suited cards,
4. 3 cards to a straight

Total combinations: C(52,3) = 22,100

• Three of a kind: Choose 1 of 13 ranks, choose 3 out of the 4 same rank of cards: C(4,3)=4 N= C(13,1)*C(4,3) = 52.

• A pair: Choose 1 of 13 ranks, choose 2 out of the 4 same rank of cards: C(4,2)=6 For the remaining 3rd card, we can choose any of the other ranks' 48 cards. N= C(13,1)*C(4,2)*48 = 3744

• 3 suited cards: Choose 3 of the 13 from a suit: C(13,3)=286, times any of the 4 suits. N=286*4=1144

• 3 to a straight: We take the smallest card of each comb for the reference point. If the smallest card is 1,2,3,4,5,6,7,8,9,10, the possible straight combinations are 6 for each card.

• n, n+1, n+2

• n, n+1, n+3
• n, n+1, n+4
• n, n+2, n+3
• n, n+2, n+4
• n, n+3, n+4

If the smallest card is J, such combinations are only 3:

• JQK
• JQA
• JKA

and in case of Q, there is only one combination:

• QKA

Total: 64 triplets.

We multiply 64 by the suits each card can take, 4*4*4=64, but need to subtract the all-same-suit variants (to exclude flushes): one for each suit (total of 4).

N= 64*(64-4) = 3840

N_Total: 52 + 3744 + 1144 + 3840 = 8780

P_Total: 8780 / 22100 = 0.3973

Since I didn't calculate the rest of the hands to make sure all probabilities add up to 100%, I would like to ask for verification of my method and calculations, please. Method being the most important part of my question.

• Both the method and the numbers appear to be correct. Commented Apr 28, 2015 at 18:46
• Do you need to calculate having three of a kind or better? If you have 3 of a kind you have a pair and that's enough, isn't it. Commented Mar 24, 2016 at 21:16
• I don't see any compensation for double counting (maybe I just missed) it. If you have 3 of a kind then you have a pair. And there is overlap on suited pair cards and potential straight. Example 5,6 and 9 of diamonds I both a straight and a flush. Commented Mar 24, 2016 at 21:20
• Oh, I see where you compensated for double counting. (I think it'd be easier to just do pairs.) Commented Mar 24, 2016 at 21:56
• @fleablood - You are correct, it was easier to do with the pairs. I was just making sure I was not missing flops of any configuration. Besides, it was testing my abilities for enumeration. Commented Mar 26, 2016 at 2:57

Note that to count the $3$ to a straight, you could have said that there are $\binom42$, $\binom32$ and $\binom22$ possibilities, respectively, since you can choose any $2$ of the up to $4$ cards above the lowest.