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Using a standard $52$ card deck, if you are given the $5$ and $7$ of hearts from it, what is the probability that you end up with a straight if $3$ additional cards from that same deck are given to you?

I was trying to set this up and got a little confused as to how to do this problem because if the next card is a $3$ or $9$ of any suit, then there are $2$ possible ranks for the following card and if the next card is a $4$ or $8$ of any suit, then there are $3$ possible ranks for the following card. Finally, if the next card is a $6$ of any suit, then there are $4$ possible ranks for the following card.

I'm not sure how to compute this because the probability changes for different cards. Do I add them all up?

Edit:
My attempt at a solution:

$3\cdot\frac{({4\choose 1}(3\cdot2\cdot 1) - 1)}{{50 \choose 3 }} \approx 0.35\%$

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    $\begingroup$ One "tipoff" that $0.35$% is wrong is look up the probability of a straight for a $5$ card hand. It is about $0.39$%. So if we are given $2$ cards that are already part of a $5$ card straight (vs. any $2$ cards), the probability should go up, not down. $\endgroup$
    – David
    Commented Nov 17, 2014 at 11:54

3 Answers 3

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I think a simpler way to proceed would be to observe that if you have a straight, it must be one of

  • 5–6–7–8–9
  • 4–5–6–7–8
  • 3–4–5–6–7

You can calculate the probabilities for each of these three separately (each is straightforward since you know just what three cards you need) and then, since they are disjoint, add them up.

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  • $\begingroup$ So it would just be $3 * \frac{{4 \choose 1}(3 * 2*1)}{{50 \choose 3}}$ ? (this gives 0.36%) $\endgroup$
    – lschlessinger
    Commented Nov 17, 2014 at 5:26
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    $\begingroup$ You will have to think about the cases where the 3 cards are all hearts. Then you would have a straight flush. $\endgroup$
    – bobbym
    Commented Nov 17, 2014 at 5:31
  • $\begingroup$ I'm not sure where your $3\cdot 2\cdot 1$ is coming from. In each case, there are 12 cards you can draw that will help you complete a straight. $\endgroup$
    – MJD
    Commented Nov 17, 2014 at 5:34
  • $\begingroup$ @MJD The 3*2*1 comes from the fact that in each of the three possible straights there are three ranks that can be chosen. After seeing the third card, there are two ranks, etc. I have a factor of 4 for each suit and a factor of 3 because of the 2 possible straights. $\endgroup$
    – lschlessinger
    Commented Nov 17, 2014 at 5:42
  • $\begingroup$ @bobbym So the I would just subtract 1 in the numerator correct? $\endgroup$
    – lschlessinger
    Commented Nov 17, 2014 at 5:43
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We definitely need the $6$ to make a straight. There are $4$ of those cards. For the last $2$ cards, we need either a $8$, $9$ combo, a $4$, $8$ combo, or a $3$, $4$ combo.

There are $16$ choices for each needed combo.

We have to subtract out the $3$ cases where we get straight flushes.

So the answer is: $$\frac {4*(16+16+16)-3} {50 \choose 3} = \frac {189} {19,600} \approx 0.00964$$

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    $\begingroup$ Thanks! This makes a lot of sense. I believe the ${50 \choose 3}$ should be equal to $19,600$ not $19,200$ though. $\endgroup$
    – lschlessinger
    Commented Nov 17, 2014 at 23:53
  • $\begingroup$ Also, doesn't this imply that a 6 has to come first? $\endgroup$
    – lschlessinger
    Commented Nov 18, 2014 at 1:37
  • $\begingroup$ We absolutely need the $6$ so no it doesn't imply we need to pick it first. The expression $4$ * ($16$ + $16$ + $16$) is the same as $64$ + $64$ + $64$ which is the number of ways to pick the triples we need, namely ($6,8,9$) or ($4,6,8$) or ($3,4,6$) so as you can see, no order is implied by the $6$. $\endgroup$
    – David
    Commented Nov 18, 2014 at 21:17
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This should be self explanatory considering the comments made above.

$$P(straight) = 3\frac{ (12\cdot 8\cdot 4-6)}{50\cdot 49\cdot 48} = \frac{27}{2800}$$

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