View Full Version : Average 5stud Lowball Hand

Al Mirpuri
11-08-2003, 11:02 PM
What would the average 5stud Lowball hand be using the Bicycle (A2345) for low and it being seven handed. What if with the same conditions the Bug was added?

This is directed primarily at BruceZ if he is about as he seems to be a mathematical wiz. However, all should feel that they can post in response.


11-13-2003, 03:04 AM
What would the average 5stud Lowball hand be using the Bicycle (A2345) for low and it being seven handed.

If you list out all the hands, including different suits, then the one in the middle is KQJ87. I assume this is what you want. It's actually the median hand.

Here's the spreadsheet that solves this:

<font class="small">Code:</font><hr /><pre>
1317888 np 1 1294336
1098240 1p 2 1272832 1295360
123552 2p 3 1149952 1274880 1296384
54912 set 4 812032 1153024 1277952 1297408
3744 fh 5 1024 816128 1159168 1282048 1298432
624 quads 6 6144 826368 1169408 1287168 1299456
2598960 total 7 21504 846848 1184768 1293312 <font color="red">1300480</font>
2598960 C(52,5) 8 57344 882688 1206272 <font color="red">1300480</font>
<font color="red">1299480</font> C(52,5)/2 9 129024 940032 1234944 1308672
10 258048 1026048 1271808 1317888
11 473088 1148928 <font color="red">1317888</font>
12 811008 <font color="red">1317888 </font>
13 <font color="red">1317888 </font>

</pre><hr />

The first column counts each type of hand.

np: 52*48*44*40*36/5!

1p: 13*6*48*44*40/3!

2p: 13*6*12*6/2*44

set: 13*4*48*44/2

fh: 13*4*12*6

quads: 13*48

The median hand is hand number 1299480 or 1299481. These are both the same hand with different suits, and we see that the median hand is a no-pair. Then we count each no-pair hand from best to worst, and we find that hand 1299480 is a K-high. Then in the next column we count each K-high hand, and we find that hand 1299480 has the second card a Q. We continue like this for each card. The last column starts counting from 1293312, the first KQJ8x. Each column accumulates 4^5*C(n,k) where n are the number of combinations for the remaining cards (number in 3rd column minus 1 = 2 through 12) and k is the number of remaining cards (4,3,2,1,0). 4^5 counts all combinations of suits.

I haven't done the joker case yet. I probably won't be able to do that one without sending you a bill. /images/graemlins/laugh.gif

Phat Mack
11-13-2003, 10:21 AM
I gave this problem to Poker Probe. Loaded seven players, no joker, 1,000,000 hands. Here are the stats of one player. (All players had nearly the same stats. Totals can be estimated by multiplying by 7.)
<font class="small">Code:</font><hr /><pre> 5 425 0.30% 0.30%
6 1921 1.35% 1.64%
7 5886 4.12% 5.77%
8 12755 8.94% 14.70%
9 23142 16.21% 30.92%
T 32973 23.10% 54.02%
J 34159 23.93% 77.95%
Q 22485 15.75% 93.70%
K 7812 5.47% 99.18%
Pr 1177 0.82% 100.00%
2 Pair 0 0.00% 100.00%
Trips 0 0.00% 100.00%
Worse 0 0.00% 100.00%
</pre><hr />

The same, but with joker:

<font class="small">Code:</font><hr /><pre>
5 814 0.57% 0.57%
6 2654 1.86% 2.43%
7 6809 4.77% 7.20%
8 14204 9.95% 17.14%
9 24356 17.06% 34.20%
T 33086 23.17% 57.37%
J 32697 22.90% 80.27%
Q 20810 14.57% 94.84%
K 6496 4.55% 99.39%
Pr 869 0.61% 100.00%
2 Pair 0 0.00% 100.00%
Trips 0 0.00% 100.00%
Worse 0 0.00% 100.00%
</pre><hr />

Poker Probe generates some nice reports. I ran these sims on an old un-networked 386 box in my basement. If you are really interested, I could get the data for all the players to a modern machine and send it to you. But I think this answers your questions about 7-handed games.

11-13-2003, 11:32 AM
I was able to do the problem with a joker after all. It came out to KQT75.

The solution is complicated, but it was possible to modify the previous spreadsheet in a systematic way to get the answer. The new spreadsheet is shown below. I'll explain how it was modified. First, here is a typo fix for the previous explanation:

Each column accumulates 4^5*C(n,k) where n are the number of <font color="red">combinations</font> for the remaining cards (number in 3rd column minus 1 = 2 through 12) and k is the number of remaining cards (4,3,2,1,0). 4^5 counts all combinations of suits.

That should say, n is the number of <font color="blue">ranks</font> for the remaining cards. For example, if we have K-high, then there are 12 possible ranks for the remaining 4 cards.

With a joker, there are C(53,5)hands, and the median hand is hand number C(53,5)/2 = 1434843. The number of no-pair hands is

52*48*44*40*36/5! + 52*48*44*40/4! = 1500928

This is no-pair hands with a joker added to no-pair hands without a joker. Since this number is greater than the median, we know the median hand is again a no-pair hand.

For the problem without a joker, we said that we are counting hands by accumulating terms of the form 4^5*C(n,k). We must now increase some of these terms to count the additional ways that some hands can be made with a joker. For example, the first number in each column reading across corresponds to the hands 5432A, K432A, KQ32A, KQT2A, and KQT7A. These hands can be made by substituting the joker for any of the low cards. Here is how to count each of these hands:

5432A: 4^5 + 5*4^4 since 5 cards can be the joker

K432A: 4^5 + 4*4^4 since 4 cards can be the joker

KQ32A: 4^5 + 3*4^4 since 3 cards can be the joker

KQT2A: 4^5 + 2*4^4 since 2 cards can be the joker

KQT7A: 4^5 + 1*4^4 since 1 card can be the joker

The 4^5 counts the naturals since 5 cards can each be one of 4 suits, and the second term counts hands made with the joker.

The joker is always the lowest card not already in the hand. This means not all hands can be made with a joker. Only hands with an A can made with a joker, since if there is no A, then the joker becomes the A. The hands with no A are counted as in the previous problem, except now the terms become 4^5*C(n-1,k) where the n-1 accounts for the fact that we can't choose an A. The following are the remaining general cases and how they are counted. Note the similarities to the examples above. The ' symbol indicates NOT. For example, A2' means the hand contains an A but not a 2.

no A: 4^5*C(n-1,k)

A2': (4^5 + 1*4^4)*C(n-2,k-1)

A23': (4^5 + 2*4^4)*C(n-3,k-2)

A234': (4^5 + 3*4^4)*C(n-4,k-3)

A2345': (4^5 + 4*4^4)

For each entry in a column beginning with the second entry, we must add together the appropriate terms above (the first entry is computed as described previously). For example, in the first column, we would add all 5 of the above terms since all of these hands are possible. In the second column, we would only add the first 4, since 1 card is fixed (K), and the second card is a 5 or greater, so the last case is not possible. Column 3 uses the first 3 terms, column 2 uses the first 2, and the last column uses only the first term. As we move down each column, n increases since we have more cards to choose from, and as we move across, k decreases since we hae fewer remaining cards to fill.

This is a hard thing to explain, so if you have any questions, don't hesitate to ask. It actually took me a lot longer to write this explanation than it did to do the problem. If you want the Excel spreadsheets that do this, PM me. Along with the above, they should make clearer what is going on. They could also be modified for more research into hand rankings.

Note that in both solutions, each column sums to the number of hands counted in the preceding column, up to the card in question. This provides a check, and gives us confidence in the answer. Also, in both cases, the columns which counted up all the no-pair hands, which were the first 4 columns in the first problem, and the first 3 columns in the second problem, summed to the same number as the total number of no-pair hands computed independently, as they should.

<font class="small">Code:</font><hr /><pre>
X X.... KX... KQX.. KQTX. KQT7X

1 1431296
2 1414656 1432320
3 1319680 1416960 1433344
4 939776 1323520 1420288 1434368
5 2304 945408 1330688 1424640 1435392
6 9984 958208 1342208 1430016 <font color="red">1436416</font>
7 30464 982528 1359104 <font color="red">1436416 </font>
8 75264 1023744 1382400 1443840
9 161280 1088256 1413120 1452288
10 311808 1183488 <font color="red">1452288</font>
11 557568 1317888 1500928
12 937728 <font color="red">1500928</font>
13 <font color="red">1500928</font>
</pre><hr />

11-13-2003, 01:46 PM
Are you simulating 5 stud/draw, or razz? Also, are you simulating actual play with some strategy, rather than just taking all hands to the river? Or just looking at winning hands? It must be one of these since nobody made 2-pr or worse in a million hands. You're showing a median at T high, while I have calculated it at K high above considering all hands.

Did the original poster want the average winning hand with 7 players? If so, I wish I'd known that before I wasted time. If you just want the average hand for all players, then the number of players doesn't matter.

I just noticed that the numbers I have in my column for 9-high or better through bicycle agree exactly with the numbers in Super System p. 589, table XXXII, for both a joker and no joker. With the various checks, I believe the average hands I was computing are solid.

Phat Mack
11-13-2003, 06:07 PM
It's 5 stud w/ all hands played to the river. These are the winning hands for one of the seven players. Since Al specified seven players, I assumed he wanted the winning hand, but I may be wrong...