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TPS_Reports
07-18-2003, 07:26 PM
I have just started reading "The Theory of Poker" and am trying to verify some of the listed probabilities. In particular, p.38 contains a discussion on 7-card stud where your first three cards are spades. There are seven other cards showing (8 player game). The table then lists the probabilities for making the flush if, of the 7 other cards showing, 0,1,2,3,or 4 of them are spades. The probabilities listed are as follows:

# Showing.................Chance for Flush %
0......................... 23.6
1......................... 19.6
2.......................... 15.8
3...........................12.3
4...........................9.1

I think I am able to calculate this if you only received two more cards (10/42)*(9/41) = 5.2%, but I don't know how to do this for 7 card. I would greatly appreciate any help you could give me. Does anyone know of a good source that lists some of the common probability formulas? Thanks again.

"Did you get that memo about the TPS Reports?"

Robk
07-20-2003, 01:04 AM
One way of doing it is using the following formula for drawing without replacement.

Say you have G "good" and B "bad" marbles in a jar making N total marbles (ie G + B = N), and you draw out n of them ( you don't put one back after drawing it). The probability of having g "good" and b "bad" ones (ie g + b = n) is given by P = ((G c g)*(B c b))/(N c n). (where x c y = x!/(y!*(x-y)!). Now we can apply this to the card situation. For example when there are no other spades out we can find the probability of making the flush by finding the probability of missing and subtracting this from one. You miss if you draw either 4 or 3 "bad" cards, ie non spades. So N = 42 (the number of unseen cards), n = 4 (the number you will get), G = 10 (number of spades left), B = 32 (the number of nonspades). Now when b = 4, g = 0, we have P = (10 c 0)*(32 c 4)/(42 c 4) = .3213. And when b = 3, g = 1, P = (10 c 1)*(32 c 3)/(42 c 4) = .4431. Now if you get any more than one "good" card you will have a flush. So the chances of missing are .3213 + .4431 = .7644, ie you'll make your flush with P = 1 - .7644 = .2356. You can find the other probabilities in the same way. There might be a neater way of doing it but this is all I could think of right now.

TPS_Reports
07-20-2003, 07:51 PM
Rob,

Thanks for taking the time, it was very helpful. I thought your method was plenty "neat" and makes intutive sense. Great job, I appreciate it.

On a sidenote, this works the other way for the positive thinkers out there. You can calculate the probabability of getting 4 spades at .1876%, 3 spades at 3.4307%, and 2 spades at 19.94%. These add to our elusive 23.6%.

Although this is very interesting to me, I can't do factorials at a poker table. When they talk about players "knowing the odds" do they have these percentages memorized for different situations? Are there tables out there available with the odds for different situations?

Thanks for helping out a beginner.

TPS_Reports
07-20-2003, 11:16 PM
Now that you've sparked my curiousity, I was trying to figure out some non-poker related problems. This problem is from:

http://probabilitynet.com/problems.htm

there are four red ladybugs, four green lady bugs, and four yellow ladybugs. Upon closer examination I see that one of the red lady bugs has one spot, another has two spots, the third has three spots, and the fourth has four spots. The same is also the case for the green and yellow ladybugs.

All of a sudden, five of these ladybugs fly through my open window and take up residence on my Creeping Charlie. What is the probability that three of them have the same number of spots, and two others also have the same number of spots but a different number of spots than the first three ladybugs have?

How would you think this through and set it up? The answer is 4.5% BTW.

Thanks again.

BruceZ
07-21-2003, 12:23 AM
Well, there are 4 ways to pick the 3 numbers that will be the same, and then there are 3 ways to pick the 2 numbers that will be the same, times 3 ways to pick which of the 3 colors will have the 2 numbers that are the same (one of the 3 colors gets left out). Divide all this by the total number of ways to pick the 5 ladybugs out of 12 total lady bugs which is C(12,5) = 12*11*10*9*8/5*4*3*3*1 = 792. So altogether we have:

4*3*3/792 = 4.5%.

BruceZ
07-21-2003, 12:55 AM
Here are the solutions for all the cases.

There are 42 unseen cards containing 6-10 spades. You must draw at least 2 spades out of your remaining 4 cards to make a flush. It is easiest to add the probabilities of drawing 0 or 1 spade and subtract this from 1.

0 spades out, 10 spades in deck:

1 - [ C(32,4) + C(32,3)*10 ]/C(42,4) = 23.6%

C(32,4) is the number of ways to draw 0 spades, and C(32,3)*10 is the number of ways to draw 3 non-spades and 1 spade. C(42,4) is the number of ways to draw 4 cards out of 42. The others are done the same way.

1 spades out, 9 spades in deck:

1 - [ C(33,4) + C(33,3)*9 ]/C(42,4) = 19.6%

2 spades out, 8 spades in deck:

1 - [ C(34,4) + C(34,3)*8 ]/C(42,4) = 15.8%

3 spades out, 7 spades in deck:

1 - [ C(35,4) + C(35,3)*7 ]/C(42,4) = 12.3%

4 spades out, 6 spades in deck:

1 - [ C(36,4) + C(36,3)*6 ]/C(42,4) = 9.1%