The Theory of Poker - Newbie

[Maximus]

I am reading The Theory of Poker and am trying to understand how he gets the calculations done in Chapter 5, pages 38 to 41. For example, playing 7-stud, you start with 3 spades and have seen seven other cards. The following table shows the effect of the other cards on making the flush.

Number of Spades / Chances of Flush %
0 / 23.6%
1 / 19.6%
2 / 15.8%
3 / 12.3%
4 / 9.1%

I do not understand how this is determined.
I tried myself using this equation:
52 cards minus 3 I have minus 7 on the table. That gives me 42 remaining cards. There are 11 cards I can get to get the flush, thus 31 cards that can beat me. That means I have a 31 to 11 chance of getting a spade. However, that does not equal 23.6% (rather 35%). Obviously, I am not understanding how to calculate this. Any help would be GREATLY appreciated!

Thanks!

[ramjam]

You will need two more spades (not just one) to make a flush.

The table is showing the probability of two or more of your next four cards being spades. Working this out accurately involves some long-handed combinatorics.

For a simplified inaccurate version, think of it like this. If there are ten spades left in the deck, your chance of getting a spade on the next street is 10/42. Your chance of catching another spade immediately after is 9/41. Your chances of getting blanks on the next two streets are 32/40 and 31/39. However rather than catching spades on the next two streets, you could also make a flush by catching them on streets 1 and 3, 1 and 4, 2 and 3, 2 and 4 or 3 and 4. That's six different ways of doing it in total. So the probablility of making a flush is roughly 6.10/42.9/41.32/40.31/39 which is about 20%. To get a more accurate answer you'd also have to factor in the probability that you catch 3 or 4 spades rahter than just 2.

[Mackas]

Cheers Ramjam, I was halfway through trying to explain the same thing in a much worse and less concise way [img]/images/graemlins/confused.gif[/img] when I realised you had answered the problem already. Maybe my head will last another week now without exploding. [img]/images/graemlins/smile.gif[/img]

[BruceZ]

Welcome to the forum Maximus.

See this post for the calculations of flush draws in stud.

Note: C(n,k) = n!/(n-k)!/ k!

= n*(n-1)*(n-2)*...*(n-k+1) / [k*(k-1)*(k-2)*...*1]

Example: C(33,4) = 33*32*31*30 / (4*3*2*1)

[Al Schoonmaker]

David does not try to teach people HOW to perform those calculations. It is not important that you know how to do so.
If you want the general computional principles, you can get them from Petrivs' "Hold'em Odds." Since your question concerns stud, you will have to do some extra work, but you should not have much trouble.
Regards,
Al

[Phishy McFish]

You beat me to my first (what I thought was an obvious) point......

He's asking the question as if 4 of a particular suit would make his flush draw.