pljr Posted May 22, 2019 Report Share Posted May 22, 2019 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 Quote Link to comment Share on other sites More sharing options...
DozyDom Posted May 22, 2019 Report Share Posted May 22, 2019 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. 1 Quote Link to comment Share on other sites More sharing options...
shyams Posted May 25, 2019 Report Share Posted May 25, 2019 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? ThanksThere is a lot of useful statistical analysis on Richard Pavlicek's website (www.rpbridge.net). Hopefully, you will find what you're looking for. Quote Link to comment Share on other sites More sharing options...
hrothgar Posted May 25, 2019 Report Share Posted May 25, 2019 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 = 1000000storage <- 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) Quote Link to comment Share on other sites More sharing options...
Cascade Posted May 26, 2019 Report Share Posted May 26, 2019 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 Quote Link to comment Share on other sites More sharing options...
Cascade Posted May 26, 2019 Report Share Posted May 26, 2019 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 Quote Link to comment Share on other sites More sharing options...
hrothgar Posted May 26, 2019 Report Share Posted May 26, 2019 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 ;-) Quote Link to comment Share on other sites More sharing options...
Cascade Posted May 26, 2019 Report Share Posted May 26, 2019 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. Quote Link to comment Share on other sites More sharing options...
hrothgar Posted May 26, 2019 Report Share Posted May 26, 2019 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... Quote Link to comment Share on other sites More sharing options...
Cascade Posted May 27, 2019 Report Share Posted May 27, 2019 Sorry over my head first thing in the morning. I did think it was a very odd comment. What was your scripting language? Quote Link to comment Share on other sites More sharing options...
hrothgar Posted May 27, 2019 Report Share Posted May 27, 2019 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 Quote Link to comment Share on other sites More sharing options...
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