[durexlw]

Could someone verify the following:

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?

[ebaines]

Could someone verify the following:

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%

This is incorrect - it looks like you are trying to capture the probability of drawing 4,5, and 6, and are overlooking the possibility of A, 4,5. Also, to draw 4,5, 6 the probability is 4*4*4*3!/P(50,3)

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%

The denominator should be P(50,3), not C(50,3). And again, this assumes that if starting with 2 and 3 that you need to draw 4, 5 and something else, and ignores that drawing an Ace, 4 and something else is also a winning hand.

-> 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?

You are correct that the odds of drawing a straight when you start with some cards is different than if you start with other cards. If you start with 2 and 3, there are only 2 ways to make a straight (by drawing either A-4-5, or 4-5-6), whereas if you start with 5 and 6 there are four ways (2-3-4, 3-4-7, 4-7-8, and 7-8-9).

[durexlw]

This is incorrect - it looks like you are trying to capture the probability of drawing 4,5, and 6, and are overlooking the possibility of A, 4,5. Also, to draw 4,5, 6 the probability is 4*4*4*3!/P(50,3)

-snip-

The denominator should be P(50,3), not C(50,3). And again, this assumes that if starting with 2 and 3 that you need to draw 4, 5 and something else, and ignores that drawing an Ace, 4 and something else is also a winning hand.

Sir, thanks for your reply. Would you be willing to explain why the denominator should be P(50,3) and not C(50,3)? If I recall correct 'P' is used when the order is important. To quote an article:

When you play a game of cards, with a 'hand of cards' you pick up the collection and change the order to suit what you are going to do next. The 'deal' of the cards has an order - but by the time you play, it is a COMBINATION - the order your received the cards does not matter.

I'm sensing you have a good reason for saying it should be a permutation, so could you explain why you believe it should be P in stead of C?

Also, I'm having a hard time figuring out why the factor '3!' is used. Could you explain this part?

You are correct that the odds of drawing a straight when you start with some cards is different than if you start with other cards. If you start with 2 and 3, there are only 2 ways to make a straight (by drawing either A-4-5, or 4-5-6), whereas if you start with 5 and 6 there are four ways (2-3-4, 3-4-7, 4-7-8, and 7-8-9).

I agree... I guess my question is: how does this influence the calculation. Or differently put: how can I take this into account in the calculation?

[ebaines]

Would you be willing to explain why the denominator should be P(50,3) and not C(50,3)? If I recall correct 'P' is used when the order is important.

Also, I'm having a hard time figuring out why the factor '3!' is used. Could you explain this part?

Perhaps it would been clearer if I had written it this way: 12/50 * 8/49 * 4/48. This gives the same result as what I had before, but is probably easier to follow.

I agree... I guess my question is: how does this influence the calculation. Or differently put: how can I take this into account in the calculation?

You are going to have to weight the calculations based on the probability of starting with:
A,2 - has one way to win
2,3 - has 2 ways
3,4 - has 3 ways
anything from 4,5 to 10,J - has 4 ways
J,Q - has 3 ways
Q,K, - has 2 ways, and
K,A- has 1 way

(I am assuming here that ace can be treated as either high or low -I'm not a poker player, so this may or may not be according to Hoyle)