probability of making a straight?

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durexlw

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Let's say I have a 3 and a 2... I take 3 cards of the remaining 50. What is the chance of drawing a straight (5 cards that follow)?

I'd be tempted to say:
(4*4*4 * 3) / 50C3 = 0.0098 or 0.98%

Let's say I have a 3 and a 2... I take 3 cards of the remaining 50. What is the chance of having a straight-draw (4 cards that follow)?

I'd be tempted to say:
(4*4*44 * 3) / 50C3 = 0.1078 or about 10%

[Question:]
-> Are the above calculations right?
-> However about straights: what confuses me here is: Let's say I have J-10, then I have 4 possible straights (10-J-Q-K-A; 9-10-J-Q-K; 8-9-10-J-Q and 7-8-9-10-J). If I have J-9 however, I have only 3 possible straights (7-8-9-10-J; 8-9-10-J-Q; 9-10-J-Q-K) so my odds would drop... but this does not seem to be included in the calculations I suggested... Do you happen to know what I am overlooking here?
 
durexlw said:
Let's say I have a 3 and a 2... I take 3 cards of the remaining 50. What is the chance of drawing a straight (5 cards that follow)?
Click to expand...
Straight can be 1-2-3-4-5 OR 2-3-4-5-6 : realise that?

So 1st card has 16/50 chance of being within range : see that?

Btw, language like "tempted to say" is a NO-NO in mathematics :wink:
 
Denis said:
durexlw said:
Let's say I have a 3 and a 2... I take 3 cards of the remaining 50. What is the chance of drawing a straight (5 cards that follow)?
Click to expand...
Straight can be 1-2-3-4-5 OR 2-3-4-5-6 : realise that?

So 1st card has 16/50 chance of being within range : see that?
Click to expand...
I realize the first part, but I don't see how the first card has a 16/50 chance of being within range... can you elaborate on this?
 
durexlw said:
Denis said:
durexlw said:
Let's say I have a 3 and a 2... I take 3 cards of the remaining 50. What is the chance of drawing a straight (5 cards that follow)?
Click to expand...
Straight can be 1-2-3-4-5 OR 2-3-4-5-6 : realise that?

So 1st card has 16/50 chance of being within range : see that?
Click to expand...
I realize the first part, but I don't see how the first card has a 16/50 chance of being within range... can you elaborate on this?
Click to expand...

As a first card it can be one of - 1 - 4 - 5 or 6 - there are 16 of those (four numbers with four different colors)
 
Hello, durexlw!

Let's say I have a 3 and a 2.
I take 3 cards of the remaining 50.
What is the probability of drawing a straight (5 cards in squence)?
Click to expand...

\(\displaystyle \text{There are: }\:{50\choose3} \,=\,19,\!600\text{ possible outcomes.}\)


\(\displaystyle \text{We can draw Ace-Four-Five.}\)
\(\displaystyle \text{There are 4 choices for each card: }\:4^3 \,=\,64\text{ ways.}\)

\(\displaystyle \text{We can draw Four-Five-Six . . . and there are: }\:4^3\,=\,64\text{ ways.}\)

\(\displaystyle \text{Hence, there are: }\:64 + 64 \:=\:128\text{ ways to get a straight.}\)


\(\displaystyle \text{Therefore: }\:p(\text{straight}) \:=\:\frac{128}{19,\!600} \:=\:\frac{8}{1225}\)

 
Ok durexlw!
Show us you "understood" Soroban; do this one: start off cards are 4,5
 
Denis said:
Ok durexlw!
Show us you "understood" Soroban; do this one: start off cards are 4,5
Click to expand...
Fair enough... correct me if I'm wrong:

Hold Cards: 4-5
Possible straights:
1) A-2-3-4-5 = 4*4*4 = 64
2) 2-3-4-5-6 = 4*4*4 = 64
3) 3-4-5-6-7 = 4*4*4 = 64
4) 4-5-6-7-8 = 4*4*4 = 64

So for a 4-5 hold card, there is a 64*4/50C3 chance on hitting a straight = 256/19600 or a 1.3% chance.

-

Dennis, could you verify this:
Let's say I have a 4 and a 5... I take 3 cards of the remaining 50. What is
the chance of having a straight-draw (4 cards that follow)?

Hold Cards: 4-5
Possible straights:
1) A-2-3-4-5 = 4*4*(48-4) = 704
2) 2-3-4-5-6 = 4*4*(48-4) = 704
3) 3-4-5-6-7 = 4*4*(48-4) = 704
4) 4-5-6-7-8 = 4*4*(48-4) = 704

So for a 4-5 hold card, there is a 704*4/50C3 chance on hitting a straight-draw = 2816/19600 or a 14.37% chance... is that right?
 
durexlw said:
> So for a 4-5 hold card, there is a 64*4/50C3 chance on hitting a straight = 256/19600 or a 1.3% chance.
Agree; 256/19600 = 16/1225 (twice compared to 2,3)

> Let's say I have a 4 and a 5... I take 3 cards of the remaining 50.
> What is the chance of having a straight-draw (4 cards that follow)?
Not sure what that means; 23(45) and 3(45)6 and (45)67, so other card can be any of 48?
If so, I make it 1792/19600 = 16/175 = ~9.14%
Not sure how to explain it (I ain't an "expert" gambler like Soroban!)
Perhaps he'll step in and set us straight :wink:
Click to expand...
 
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