[pljr]

Can someone please direct me to the data on the frequency of a two suited fit (8+ cards).

I have looked around but not found it.


If I had to guess, I would think it was 1 in 50 hands, whilst a 1 suited fit would be 8/10?

Thanks

[DozyDom]

Your partnership will have two 8-card fits in about one in ten deals, and a double fit where at least one of the fits is 9 cards or better in about 8% of deals. So all in all that's about 18%, or one every five and a half games.

I'm guessing, given that, that the OP is a little mistaken and you meant 1 in 5 hands, which is pretty much right.

[shyams]

pljr, on 2019-May-22, 01:13, said:

Can someone please direct me to the data on the frequency of a two suited fit (8+ cards).

I have looked around but not found it.


If I had to guess, I would think it was 1 in 50 hands, whilst a 1 suited fit would be 8/10?

Thanks

There is a lot of useful statistical analysis on Richard Pavlicek's website (www.rpbridge.net). Hopefully, you will find what you're looking for.

[hrothgar]

Was bored, so I wrote a quick script to estimate this

From the looks of things, the frequency is about 20.2%

I've attached the code because - as always - I am more than capable of screwing this up

sim_length = 1000000
storage <- matrix(0, sim_length, 4)

foo_S <- rep("S", 13)
foo_H <- rep("H", 13)
foo_D <- rep("D", 13)
foo_C <- rep("C", 13)
foo <- c(foo_S, foo_H, foo_D, foo_C)

for (i in 1:sim_length){

bar <- sample(foo,26)
bar2 <-table(bar)
bar3 <- sort(as.vector(bar2))

if (length(bar3) == 3) {
bar3 <- c(0,bar3)
}

storage[i,] <- bar3

}

table(storage[,3] > 7 & storage[,4] > 7)

[Cascade]

It is relatively straight forward to calculate these numbers exactly.

There are 104 distributions between two hands ranging from the most balanced 7-7-6-6 to the most unbalanced 13-13-0-0.

The probabilities (or frequencies) can be calculated using elementary combinations.

For example the 7-7-6-6 distribution has a frequency calculated as combin(13,7)*combin(13,7)*combin(13,6)*combin(13,6) where combin is the excel spreadsheet function which is the combination 13 things choose 7 for example. Here there are 13 cards in each suit and we want one side to have any 7 of those 13 cards. We multiple the frequencies for each suit to get the total number for a particular distribution.

Now in the case of 7-7-6-6 there are six different permutations for the two seven card suits and two six card suits. That is combin(4,2).

To get a probability we divide by combin(52,26) the number of ways of choosing 26 cards out of the 52 in the deck.

Once you do these calculations for all 104 distributions you can collate the distributions with 0, 1, 2 or 3 eight card or longer fits.

The numbers are

No eight card fit 0.157362521 - a bit more often than 1 in 6

One eight card fit 0.640981776 - a bit more often than 2 in 3

Two eight card fits 0.199914617 - almost exactly 1 in 5

Three eight card fits 0.001741085 - around 1 in 574

The probabilities for all 104 distributions follow:

7	7	6	6	0.104908347
7	7	7	5	0.052454174
8	6	6	6	0.052454174
8	7	6	5	0.236043782
8	7	7	4	0.065567717
8	8	5	5	0.033193657
8	8	6	4	0.049175788
8	8	7	3	0.019670315
8	8	8	2	0.001341158
9	6	6	5	0.065567717
9	7	5	5	0.049175788
9	7	6	4	0.072853019
9	7	7	3	0.014570604
9	8	5	4	0.040979823
9	8	6	3	0.021855906
9	8	7	2	0.005960702
9	8	8	1	0.000372544
9	9	4	4	0.003162023
9	9	5	3	0.004553314
9	9	6	2	0.00165575
9	9	7	1	0.000275958
9	9	8	0	1.59207E-05
10	6	5	5	0.019670315
10	6	6	4	0.014570604
10	7	5	4	0.021855906
10	7	6	3	0.011656483
10	7	7	2	0.00158952
10	8	4	4	0.004553314
10	8	5	3	0.006556772
10	8	6	2	0.002384281
10	8	7	1	0.00039738
10	8	8	0	1.14629E-05
10	9	4	3	0.002023695
10	9	5	2	0.00099345
10	9	6	1	0.000220767
10	9	7	0	1.69821E-05
10	10	3	3	8.09478E-05
10	10	4	2	0.000110383
10	10	5	1	3.3115E-05
10	10	6	0	3.39641E-06
11	5	5	5	0.001341158
11	6	5	4	0.005960702
11	6	6	3	0.00158952
11	7	4	4	0.00165575
11	7	5	3	0.002384281
11	7	6	2	0.000867011
11	7	7	1	7.22509E-05
11	8	4	3	0.00099345
11	8	5	2	0.000487694
11	8	6	1	0.000108376
11	8	7	0	8.33665E-06
11	9	3	3	0.000110383
11	9	4	2	0.000150523
11	9	5	1	4.51568E-05
11	9	6	0	4.63147E-06
11	10	3	2	2.40836E-05
11	10	4	1	1.00349E-05
11	10	5	0	1.38944E-06
11	11	2	2	4.47836E-07
11	11	3	1	5.47356E-07
11	11	4	0	1.05261E-07
12	5	5	4	0.000372544
12	6	4	4	0.000275958
12	6	5	3	0.00039738
12	6	6	2	7.22509E-05
12	7	4	3	0.000220767
12	7	5	2	0.000108376
12	7	6	1	2.40836E-05
12	7	7	0	9.26294E-07
12	8	3	3	3.3115E-05
12	8	4	2	4.51568E-05
12	8	5	1	1.3547E-05
12	8	6	0	1.38944E-06
12	9	3	2	1.00349E-05
12	9	4	1	4.18119E-06
12	9	5	0	5.78934E-07
12	10	2	2	5.47356E-07
12	10	3	1	6.6899E-07
12	10	4	0	1.28652E-07
12	11	2	1	4.97596E-08
12	11	3	0	1.40348E-08
12	12	1	1	3.45553E-10
12	12	2	0	3.18972E-10
13	5	4	4	1.59207E-05
13	5	5	3	1.14629E-05
13	6	4	3	1.69821E-05
13	6	5	2	8.33665E-06
13	6	6	1	9.26294E-07
13	7	3	3	3.39641E-06
13	7	4	2	4.63147E-06
13	7	5	1	1.38944E-06
13	7	6	0	1.42507E-07
13	8	3	2	1.38944E-06
13	8	4	1	5.78934E-07
13	8	5	0	8.01601E-08
13	9	2	2	1.05261E-07
13	9	3	1	1.28652E-07
13	9	4	0	2.47408E-08
13	10	2	1	1.40348E-08
13	10	3	0	3.95852E-09
13	11	1	1	3.18972E-10
13	11	2	0	2.94435E-10
13	12	1	0	8.17876E-12
13	13	0	0	1.20988E-14

[Cascade]

These are the numbers for each suit length and the probability that there is a hand with 0, 1, 2, or 3 suits with at least that many cards.

	7	        8	        9	        10	        11	        12	        13
0	0	        0.157362521	0.614809111	0.89580818	0.982536954	0.998352787	0.999934486
1	0.162325951	0.640981776	0.371681401	0.103925941	0.017461881	0.001647213	6.55142E-05
2	0.675348097	0.199914617	0.013509488	0.000265879	1.16553E-06	6.72715E-10	1.20988E-14
3	0.162325951	0.001741085	0	        0	        0	        0	        0

[hrothgar]

Thanks for posting the numbers Wayne

Looks like your estimate (20.175%) is almost identical to my monte carlo (20.2%)

I suspect that the difference is that I allow for 4+ eight card fits so my estimate is slightly higher ;-)

[Cascade]

hrothgar, on 2019-May-26, 05:53, said:

Thanks for posting the numbers Wayne

Looks like your estimate (20.175%) is almost identical to my monte carlo (20.2%)

I suspect that the difference is that I allow for 4+ eight card fits so my estimate is slightly higher ;-)

You can't have four eight card fits between two hands. Since 8 times 4 is 32 and there are only 26 cards.

Maybe i am not understanding what you mean but that is why I did not calculate four or more eight card fits.

[hrothgar]

Cascade, on 2019-May-26, 15:01, said:

You can't have four eight card fits between two hands. Since 8 times 4 is 32 and there are only 26 cards.

Maybe i am not understanding what you mean but that is why I did not calculate four or more eight card fits.

The use of the ;-) was meant to indicate a joke...

[Cascade]

Sorry over my head first thing in the morning. I did think it was a very odd comment.

What was your scripting language?

[hrothgar]

Cascade, on 2019-May-26, 18:12, said:

What was your scripting language?

R

Not sure why, but I find it amusing to use my stats packages for this stuff rather than Dealer or some such