Thread: Probability Involving a Variation on Poker

1. Probability Involving a Variation on Poker

Hey everyone,

I've been experimenting with probability lately, and I've been creating various games for fun and analyzing the math behind them. On this one though, I'm stuck and have no idea what I'm doing wrong.

The game: Using a random number generator, (1-7) draw 5 numbers to use as a "poker hand." The numbers 1-7 each represent a different color. The colors used are Red, Yellow, Orange, Purple, Blue, Green, White. Each color/number has an equal chance of being drawn.

As such, there are 7 different hands you can draw. 5 of a kind, 4 of a kind, full house, 3 of a kind, 2 pair, 1 pair, and a bust.

5 of a kind = All 5 of same color
4 of a kind = 4 of same color, 1 of another
Full House = 3 of same color, 2 of same color of another
3 of a kind = 3 of same color, 1 of another, 1 of another
2 pair = 2 of same color, 2 of same color of another, 1 of another
1 pair = 2 of same color, 1 of another, 1 of another, 1 of another
Bust = All 5 of different color

I'm trying to find the probability of obtaining each hand. However, I'm doing something wrong, and I'm hoping someone can explain to me what it is. Here is what I've done so far.

Distinguishable hands: (note oak = "of a kind" and #c# is a combination. Like 3c2 means 3 choose 2 = 3)
5 oak = (7c1) = 7
4 oak = (7c1)(6c1) = 42
Full House = (7c1)(6c1) = 42
3 oak = (7c1)(6c2) = 105
2 pair = (7c1)(6c1)(5c1) = 210
1 pair = (7c1)(6c3) = 140
Bust = (7c5) = 21

Possible "Layouts": (for example. a 4 oak could be Y,Y,Y,Y,B or Y,Y,Y,B,Y, etc.)
5 oak = (5c5) = 1
4 oak = (5c4)(1c1) = 5
Full House = (5c3)(2c2) = 10
3 oak = (5c3)(2c1)(1c1) = 20
2 pair = (5c2)(3c2)(1c1) = 30
1 pair = (5c2)(3c1)(2c1)(1c1) = 60
Bust = (5c1)(4c1)(3c1)(2c1)(1c1) = 120

Therefore, the total number of possible ways to get each hand should be equal to the number of distinguishable hands multiplied by the number of possible "layouts."

Possible Hands:
5 oak = 7(1) = 7
4 oak = 42(5) = 210
Full House = 42(10) = 420
3 oak = 105(20) = 2100
2 pair = 210(30) = 6300
1 pair = 140(60) = 8400
bust = 21(120) = 2520
Total = 19957

However, the total number of hands should be 7^5 = 16807. I have calculated a total of 19957 possible hands. I've been trying to figure out what I've been doing wrong for over 2 days now, and I'm just not seeing it.

I'd be thrilled if someone could help me.

Thanks,
SDRS

2. Re: Probability Involving a Variation on Poker

You double-counted when you computed 2 pairs.

The number of distinguishable hands for 2 pairs is (7c2)(5c1) = 105, not 210.

3. The Following User Says Thank You to Mike Z. For This Useful Post:

SDRS (08-21-2012)

4. Re: Probability Involving a Variation on Poker

Thank you so much Mike!

5. Re: Probability Involving a Variation on Poker

Originally Posted by Mike Z.
You double-counted when you computed 2 pairs.
Which 2 pairs are you referring to Mike?

6. Re: Probability Involving a Variation on Poker

I'm pretty sure he's referring to the 2 pair in the distinguishable hands section.

Using (7c2)(5c1) makes sense when I think about it, and using it seems to correct the problem. However, I'm still unsure as to why (7c1)(6c1)(5c1) would not yield the same and instead results in double counting. I've never had a statistics or probability class, so everything I know is mostly what I've figured out myself and/or researched. I'm assuming my knowledge of combinations is wrong in some fundamental way since I am not seeing why these two models can't represent the same situation. If anyone can explain why, I'd be grateful.

Here's my reasoning:
Method 1 - (7c1)(6c1)(5c1)
(7c1) selects a color for the first pair. (6c1) selects a different color for the second pair. (5c1) selects a different color for the "odd one out."

Method 2 - (7c2)(5c1)
(7c2) selects 2 different colors, 1 for each pair. (5c1) selects a different color for the "odd one out."

~~~~~~~~~~~

Looking at it that way, it seems reasonable that either one could be used to model the 2 pair distinguishable hands. However, this is certainly not the case since each method yields a different result.

(5c1) is used either way. It all comes down to the difference between (7c1)(6c1) and (7c2). Clearly different results, but why? (7c1)(6c2) = 42, (7c2) = 21

7. Re: Probability Involving a Variation on Poker

SDRS, in case you didn't notice my question was directed toward Mike Z., mkay.

8. Re: Probability Involving a Variation on Poker

Dragan: "SDRS, in case you didn't notice my question was directed toward Mike Z., mkay."

Why the attitude Dragan? Only my first sentence in that post was directed towards you. The rest was just an extension to my original question.

Oh and I'm sorry, but it's incredibly obvious which 2 pair he was talking about. It's rather clear in my original post that a 2 pair is one of the hands you can get. He said it's not 210, and I computed 210 under the distiguishable hands section. So what other one could it possibly be?

So please, I don't care if you're a "super moderator", cut the arrogance and attitude. Oh and perhaps you should read my original post before responding too, because maybe then you'd actually be able to figure out what Mike was talking about. These forums are here for people to ask for help with math, and that's what I'm doing. I've followed the rules, and I've followed up on the original problem. So I don't need uncalled for attitude from a macho mod.

9. Re: Probability Involving a Variation on Poker

Wash, rinse, and repeat. In other words, I want to see a response from Mike Z.

10. Re: Probability Involving a Variation on Poker

(7c1)(6c1) incorrectly counts every result twice.

Since 42 is not very large, perhaps it would help to actually list all 42 results of (7c1)(6c1), so that you can see the unwanted duplicates. I will label the 7 colors a, b, c, d, e, f, and g:

ab, ac, ad, ae, af, ag
ba, bc, bd, be, bf, bg
ca, cb, cd, ce, cf, cg
da, db, dc, de, df, dg
ea, eb, ec, ed, ef, eg
fa, fb, fc, fd, fe, fg
ga, gb, gc, gd, ge, gf

Every pair of colors is incorrectly counted twice; for example, ab and ba both appear. So the 5-card hand aabbc would be counted as different than the 5-card hand bbaac, when actually they are the same hand and should only be counted once.

Choosing one of 7 colors, and then choosing one of the 6 remaining colors, means that both ab and ba are incorrectly counted. The combination (7c2) avoids that problem since order is not important for combinations.

11. Re: Probability Involving a Variation on Poker

Originally Posted by Mike Z.
You double-counted when you computed 2 pairs. Daniel Negreanu

The number of distinguishable hands for 2 pairs is (7c2)(5c1) = 105, not 210.
Took me a while to see this

12. Re: Probability Involving a Variation on Poker

Originally Posted by Mike Z.
(7c1)(6c1) incorrectly counts every result twice.
Yes, Mike, you are correct for this specific case. However, and more generally, I don’t see how your explanation resolves all of the problems associated with the original post.

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