Probability of drawing certain hand, incorrect answer, but why?

Question

So I am drawing $5$ cards from a standard deck of $52%$

I want to find the probability that I draw $5$ consecutive cards of same suit with no card looping, and the ace is card $1$.

So the consecutive cards can start on any of $A,2,3,4,5,6,7,8,9$ but can't start on $10$ since that would not complete, e.g $10,J,Q,K$, and then it can't loop so doesn't work.

So we should have $\frac{9*4}{52}$ starting cards, being the four suits and the $9$ possible numbers. Then we should have four appropriate options, being any of the four cards that needs to be grabbed, out of $51$ remained, and then three cards out of $50$

Giving me: $$\frac{9*4}{52}\cdot\frac{4}{51}\cdot\frac{3}{50}\cdot\frac{2}{49}\cdot\frac{1}{48}=2.77*10^{-6}$$

What do you mean with "no card looping"? And all drawn cards should be of the same suit? So I guess after the first card has been drawn, there is only one option for the next card and not four? — SPK.z, Oct 16 '14 at 06:34
@SPK I mean you can't go $Q,K,A,2,3$ so you can't jump from $13$ to $1$, and my logic is that if you think of the $5$ consecutive that you get, you could have picked any of those on the second turn. — Wow you have a good name, Oct 16 '14 at 06:36
Do you have to draw the cards in consecutive order (i.e. first A, then 2, then 3 ...) ? That's what I understood. — Alexandre Halm, Oct 16 '14 at 06:37
@AlexH. No, all five are given, and I pick up the five together. — Wow you have a good name, Oct 16 '14 at 06:39
If I understand correctly, there are $4\cdot 9$ "good" hands. Divide by $\binom{52}{5}$. — André Nicolas, Oct 16 '14 at 06:45
@AlexH. I was asking what was wrong with my method, so unless someone got the same wrong answer as me, it isn't a duplicate — Wow you have a good name, Oct 16 '14 at 11:53

score 1 · Answer 1 · answered Oct 16 '14 at 06:37

1

You ask the probability of first drawing the lowest card of your straight flush, then drawing any of the remaining four card of the tsraight flush, then drawing any of the remaing three cards, ... It is, however, not given that the lowest card will be the first you draw.

answered Oct 16 '14 at 06:37

Hagen von Eitzen

374,180

How can I get rid of that issue? Setting it up so I assume I draw the middle card? and staying two cards away from the edges? – Wow you have a good name Oct 16 '14 at 06:40
@AlgebraNo, the assumption that you draw any specific of the five cards first is nnot justified (though you might add all 5 variants. Just note that there are $4\cdot9$ straight flush hands out of $52\choose 5$ hands in total. – Hagen von Eitzen Oct 16 '14 at 14:24

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

0

Given that order doesn't matter, your universe is all unordered subsets of $\{1,...,52\}$ of size $5$ (so card = $\binom{52}{5}$) and your event is of size $4 \times 9$ (each satisfactory draw is uniquely determined by the choice of a card below $9$ in each of 4 colors).

I get $\frac{4\times 9 }{\binom{52}{5}} = 1.38 \times 10^{-5}$.

Btw, this question is a duplicate: link

edited Apr 13 '17 at 12:21

Community

1

answered Oct 16 '14 at 06:36

Alexandre Halm

3,714

score 0 · Answer 3 · answered Oct 16 '14 at 06:51

I want to find the probability that I draw 5 consecutive cards of same suit with no card looping, and the ace is card 1.

@AlexH. No, all five are given, and I pick up the five together. – Algebra

So you use the face and suit of the lowest card in the 5 as the identity of the hand. There are $9\times 4$ ways to pick 5 consecutive cards of the same suit when order does not matter. There are ${52\choose 5}$ ways to select any 5 cards when order doesn't matter. So the probability is:

$$\frac{9\cdot 4\cdot 5!\cdot 47!}{52!}=\frac{3}{216580}$$

Probability of drawing certain hand, incorrect answer, but why?

3 Answers3