Zodd
Zodd
Joined: Jan 28, 2016
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February 14th, 2017 at 2:43:13 PM permalink
(1) I was wondering what the odds were that a flop *could* be a straight (for someone/anyone, not given a specific set of hole cards). Including straight flush.

So, any flop like AKQ, AKJ, AKT, ..., 32A.

My thought is to take the number of 5 card straights, 10,240, and multiply by the number of ways to choose 3 cards from 5, 10, to get (10240)(10) = 102,400. Which would be 3.9%.

Is that right? I figured I'd post here since people are more up to speed on how to calculate these odds.

(2) What about the probability that a flop could be a straight draw (2 cards in a possible straight)? I calculated 11.8%. Is that right?
Romes
Romes
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February 27th, 2017 at 12:14:47 PM permalink
Hi General Zodd.

You're more/less just asking statistics questions, but we can relate them to poker. The probability, P(straight), isn't all that hard if you break it down one card at a time.

P(Straight) = P(1st card) * P(2nd card) * P(3rd card)

P(1st card) is what is the probability of getting a card that could make a straight. Well, since we can start at the top or bottom it doesn't matter whether it's and Ace, or 2 or anything in between even. EVERY card on this first card will give us a potential for a 3 card straight. So our odds (assuming we don't know other card information) is 52/52 for this first card to complete a 3 card straight on the flop.

P(1st card) = 52/52 = 1

P(2nd card) is what is the probability of getting a card within ONE of the first card? Though I suspect some special cases such as 1st card K, 2nd card A, because that would break the chain (as we're not counting wrap around K-A-2 as a straight) and from your question it appeared it needs to be sequential, so K-A-Q does not count either. For now, because I don't want to spend all day on this, I'm going to ignore the two one off cases of K-A and 2-A. So what are the odds we get within 1 card of the first card? Well, each card on either side of our original card yields us 4 chances, so for both sides we have 8/51.

P(2nd card) = 8/51 = .1569

P(3rd card) is what is the probability, given we have a 2 card straight, the 3rd card will complete it? That should also be simple... At this point we have 2 cards in sequential order and there's only one possible card to complete it (but 4 of them in the deck). An example here would be if we have 10-9 so far, we HAVE to draw an 8 to complete the straight, because again your question seemed to be sequential, so 10-9-J would not count. Thus, only an 8 would make a straight, and for other situations the same would follow that only 1 card (time 4 in the deck) will complete the straight... so we have 4/50

P(3rd card) = 4/50 = .08

In this scenario we need P(1 card) AND P(2 card) AND P(3 card)... when things are "AND" they are multiplicative... When they're "OR" they're additive. For example if you asked "What are the odds of drawing an Ace OR 9 OR 3?" Then it would be P(A) + P(9) + P(3).

Thus, P(Straight) = P(1st card) * P(2nd card) * P(3rd card) = 1 * .1569 * .08 = .01255 = 1.26%***


***don't forget I ignored those two one off cases of K then drawing A, which would break the chain as well as 2 then drawing an A. This should be rather close though.

Edit
Reviewing now it looks like you want to know what flops would give a "potential" straight, which would be any 3 cards within 5 ranks of one another...

P(Potential Straight) = P(1st card) + P(2nd card within 4 ranks) + P(3rd card within 4 ranks of 1st card)

P(1st card) = 52/52 = 1

P(2nd card within 4 ranks) = 32/51 = .6275 ...again I'm ignoring end cases such as 1st card 3... then you can only draw A,2,4,5,6,7... where as with a 1st card 7 you can draw 6,5,4,3,8,9,10,J.

P(3rd card within 4 ranks of 1st card) = 24/50 = .48... you have to take one possible card away for the 2nd card pair and 1 card away because the direction (higher or lower) than the first card will limit the end of the straight. Example: first card 7... 2nd card 8... 3rd card can not be 8 (pair) and can not be 3 (no straight then).

P(Potential Straight) = 1 * .6275 * .48 = .3012... or 30.12% chance of a flop that could lead to a "potential straight" for someone.***


***Again I left off the end cases, which would affect the math here a little bit considering anything A-2-3 and Q-K-A will all have slightly less odds. So I overvalued the final number, but it shouldn't be by "that" much... maybe a few %.

For someone to flop a straight draw you'd have to consider both hole cards. The flop could come 10-6-9, and they'd have a straight draw with either a 7 or an 8... similarly the flop could come A-6-9 and if the person held 8-7 they'd have flopped a straight draw. Just giving some thought, I don't have time to work that one out =P.
Playing it correctly means you've already won.
charliepatrick
charliepatrick
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February 27th, 2017 at 1:23:20 PM permalink
You need to look for an approach. Assuming the question is if we are to draw 3 cards what are the chances of it being possible for someone to have two cards that form a 5-card straight, then one way might be
(i) how many three card draws are there
(ii) how many of these are made up of three different cards
(iii) how many of these satisfy the requirement of being three of five cards in any given straight (warning make sure you don't double count).

btw I didn't understand the second part - it doesn't matter what the first three cards are, it's always possible for someone to make a straight by the time you add the turn and river (e.g. AAA K Q). If there are quads or a full house then it becomes impossible.
ChesterDog
ChesterDog
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Thanks for this post from:
Zodd
February 27th, 2017 at 1:40:53 PM permalink
Quote: Zodd

...So, any flop like AKQ, AKJ, AKT, ..., 32A...



I would count them by breaking them into three categories depending on the difference (either 2, 3, or 4) between the lowest card and the highest card. Examples: For A23, the difference is 2; for A24, the difference is 3; and for A25, then difference is 4.

For each lowest card and highest card combination, if the difference is 3, there can be two different middle cards; and when the difference is 4, there can be three different middle cards.

For a difference of 2, the combinations range from A23 up to QKA, for 12 combinations.

For a difference of 3, the combinations range from A24 (or A34) up to JQA (or JKA) for 11 times 2 = 22 combinations.

And for a difference of 4, the combinations range from A25 (or A34 or A45) up to TJA (or TQA or TKA) for 10 times 3 = 30 combinations.

The total is 12 + 22 + 30 = 64 combinations. Each of the three cards can be any of the four suits, so the final number of ways is 64 * 4^3 = 4096. Dividing this by combin(52,3) makes a probability of 18.53%.
Zodd
Zodd
Joined: Jan 28, 2016
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March 2nd, 2017 at 12:09:32 PM permalink
Yeah, that's exactly what I was going for - I see it now. The 18.53% figure is right. That's a nice elegant/simple way to work it out. I'm surprised it's that high.

As for flops that could be a straight draw, this is flops like 23K,, 24K, 25K, 26K, .... flops with 2 (or 3) cards where 2 hole cards could complete a straight draw (open ended or gutshot). As I looked at it though I realized I'm not that interested because over 90% of all flops could be that. All you need is 2 cards with a distance of 4 or less between them, so you'd need a flop like 27Q or 38K that couldn't be a straight draw. So nevermind that!

Thanks for all the thoughts.

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