[Vinish]

Hopefully someone with more knowledge than I have of statistics can help me with a question.

How can I calculate the % chance of having an average of 7.48 hcp over the course of 25 hands? In recent BBO play in a casual side game created by me, I played with three friends and we played 25 hands. I went back and looked at all my hands from that session and counted the hcp in each. My max was 12 which happened only once. I had 11 hcp twice and 10 hcp four times. I even had 0 hcp once. My average for 25 hands was 7.48 hcp. I wonder how you would go about calculating the odds of getting that few an avg. number of hcp over 25 hands. According to website with some bridge statistics, the odds of getting exactly 7 hcp in one hand is 8.03% and the odds of getting 8 hcp in one hand is 8.89%. Graphing and interpolating yields the odds of getting 7.48 hcp as being about 8.5043%. I do not think this is a correct analysis but is there any chance that the odds of averaging exactly 7.48 hcp over 25 hands is (1-0.085043)^25 which would be (0.914957)^25 = 0.108 or 10.8% of the time?

Thanks for any help you can offer.

G'day,

Vinish

[AL78]

Vinish, on 2021-February-18, 10:07, said:

Hopefully someone with more knowledge than I have of statistics can help me with a question.

How can I calculate the % chance of having an average of 7.48 hcp over the course of 25 hands? In recent BBO play in a casual side game created by me, I played with three friends and we played 25 hands. I went back and looked at all my hands from that session and counted the hcp in each. My max was 12 which happened only once. I had 11 hcp twice and 10 hcp four times. I even had 0 hcp once. My average for 25 hands was 7.48 hcp. I wonder how you would go about calculating the odds of getting that few an avg. number of hcp over 25 hands. According to website with some bridge statistics, the odds of getting exactly 7 hcp in one hand is 8.03% and the odds of getting 8 hcp in one hand is 8.89%. Graphing and interpolating yields the odds of getting 7.48 hcp as being about 8.5043%. I do not think this is a correct analysis but is there any chance that the odds of averaging exactly 7.48 hcp over 25 hands is (1-0.085043)^25 which would be (0.914957)^25 = 0.108 or 10.8% of the time?

Thanks for any help you can offer.

G'day,

Vinish

The number of possible bridge hands is finite so it theoretically is possible to calculate exactly the probability of getting less than or equal to a HCP average over some number of boards. You can get an approximation by randomly dealing a large number of hands and working out the mean and standard deviation of the mean HCP of 25 hands. Although the mean is bounded, the central limit theorem states that the distribution of the means approaches a normal distribution for large sample sizes, so you can compute the mean and standard deviation of the mean HCP of a large number of groups of 25 hands, assume this closely resembles the full population distribution, and use the normal cumulative distribution function to calculate the probability of exceedance. It is recommended you use a computer to generate the random deals.

[smerriman]

The probability you averaged 7.48 hcp over your last 25 hands is 100%

For the probability you average 7.48 hcp over your *next* 25 hands, you can estimate it with a normal distribution - for one hand, the mean is 10 and standard deviation is 4.1302.

Over 25 hands, the mean is 10 and standard deviation is 4.1302/sqrt(25) = 0.8620. Plugging that into a normal distribution gives a probability 0.1%.

There have been numerous posters who confuse the 0.1% with the 100% though, and think this means something was 'wrong' with the deal generation - while rare, over a huge population, these 0.1% still can and will happen regularly.

[Vinish]

smerriman, on 2021-February-18, 13:22, said:

<snip>
There have been numerous posters who confuse the 0.1% with the 100% though, and think this means something was 'wrong' with the deal generation - while rare, over a huge population, these 0.1% still can and will happen regularly.

In fact, it will happen precisely 0.1% of the time which is about 1 in 1000 25 board sessions. My team still beat the other team handily. As all duplicate players know - it is not the cards you are dealt, it is what you do with the cards you are dealt.

[hrothgar]

sim_length = 1000000

foo_aces <- rep(4,4)
foo_kings <- rep(3,4)
foo_queens <- rep(2,4)
foo_jacks <- rep(1,4)
foo_crap <- rep(0, 36)

foo <- c(foo_aces, foo_kings,foo_queens,foo_jacks,foo_crap)


storage <- matrix(0, sim_length)

for (i in 1:sim_length){

storage[i] <- sum(sample(foo,13, replace = FALSE))

}

mean(storage)
sd(storage)

bar <- table(storage)
bar <- as.data.frame(bar)

[smerriman]

Vinish, on 2021-February-18, 13:37, said:

In fact, it will happen precisely 0.1% of the time which is about 1 in 1000 25 board sessions.

From your perspective, yes. From the forum's perspective, consider that there are 45000 people online right now - 45 of them are experiencing streaks at least as bad as yours and wondering why the dealer hates them They tend to notice and post it about it on the forum much more than the 45 people on the other end of the spectrum who have been averaging 12.5+!

[PeterAlan]

The chances of <= 187 (= 7.48 x 25) HCP in 25 independent hands is 1 in 1045.160.

The HCP distribution of a hand is not normal, obviously, as it has a range of 0-37 and a mean of 10, but it has a long and very flat tail, and so, as you see above, gives a fairly reasonable approximation when applied to this sort of problem.

A while ago I built myself a simple program that calculates exactly the statistic you want, based on the full 0-37 HCP distribution, and that's the basis of the answer I've given you above.

[helene_t]

According to Richard Pavlicek (http://www.rpbridge.net/8j17.htm#1), the standard deviation of the HCPs in a single bridge hand is 4.1302, so the standard deviation of the average in n bridge hands is 4.1302/sqrt(n). Since the mean is 10, you can for example enter this formula in a spreadsheet:

NORM.DIST(7.48,10,4.1302/SQRT(25),TRUE)

This will give you a probability of 0.001142 of getting less than or equal to 7.48 on average in 25 hands, in other words a bit more than once every thousand times you play 25 hands.

PeterAllan's number is more accurate since I have here assumed that the HCPs are normal distributed. In reality they are a bit right-skewed as the range goes further to the right (up to 37) than to the left (down to 0).

[PeterAlan]

Vinish, on 2021-February-18, 13:37, said:

My team still beat the other team handily. As all duplicate players know - it is not the cards you are dealt, it is what you do with the cards you are dealt.

Congratulations. But you also have a partner; what were their HCP?

Incidentally, whilst the HCP distribution of a single hand is not normal, the HCP distribution of a pair of hands from the same deal is pretty close to normal. (It is obviously symmetrical about its mean of 20: for every pair of hands totalling N HCP there is another pair of hands with 40-N HCP, namely the other two.)

[AL78]

PeterAlan, on 2021-February-18, 19:19, said:

The HCP distribution of a hand is not normal, obviously, as it has a range of 0-37 and a mean of 10, but it has a long and very flat tail, and so, as you see above, gives a fairly reasonable approximation when applied to this sort of problem.

The question was not about the probability of picking up one hand with a specific HCP, it was about having a mean of a specific HCP (or less) over a session with 25 hands. We are interested in the distribution of means of 25 randomly sampled hands, which will approach normal as the sample size gets very large. You see the same in the UK national lottery, the distribution of the mean sum of the six randoms numbers selected is extremely close to normal if evaluated over all possible random samples, even though the distribution is bounded, so statistical inferences which assume normality of the distribution can still be used.

https://en.wikipedia...l_limit_theorem

[AL78]

Vinish, on 2021-February-18, 13:37, said:

In fact, it will happen precisely 0.1% of the time which is about 1 in 1000 25 board sessions. My team still beat the other team handily. As all duplicate players know - it is not the cards you are dealt, it is what you do with the cards you are dealt.

It is also what other people do with the cards you are dealt (at duplicate), which you have no influence over. If your opponents punt a slam on a hook and a 3-2 trumpm break which no-one else bids, you are getting a bottom or a double figure negative IMP score whether you like it or not.

[AL78]

This reminds me of a friend of mine who averaged 7.17 HCP over 18 boards. She declared twice, was dummy twice, and defended 14 out of 18 hands. This is the lowest HCP average I have seen over a session (even lower than my worst).

[PeterAlan]

AL78, on 2021-February-19, 04:16, said:

The question was not about the probability of picking up one hand with a specific HCP, it was about having a mean of a specific HCP (or less) over a session with 25 hands.

I was well aware of that, and that was the question I answered, since I had the tool to hand to do it without approximation.

My comment about the normal distribution was badly worded; I was just trying to say that the approach that you and smerriman adopted was giving good results: I should have been clearer about that, and separated it from the first remark. I was aware of the central limit theorem, and if I weren't your first post would have educated me.

As the wiki article says "The central limit theorem gives only an asymptotic distribution. As an approximation for a finite number of observations, it provides a reasonable approximation only when close to the peak of the normal distribution; it requires a very large number of observations to stretch into the tails.", but it works pretty well here. I suspect, without any further analysis, that this is largely due to (1) the value in question (74.8% of the mean) not being that far along the lower tail of the cumulative distribution, and (2) the HCP distribution of a single hand being reasonably well-behaved with only a small proportion of hands in the tail above 20 or so HCP. You may well have a better insight. In particular, I haven't looked at how well, in this application, the CLT approximation improves with the number of observations.

[AL78]

The probability distribution for the HCP for a single hand is a discrete distribution, the density function is given in this table:

https://www.bridgebu...stributions.php

From this table, I think there are statistical formulae which can be used to calculate the probability of getting an average HCP of X over 25 deals, but they escape me at present.

An average HCP of 7.48 over 25 boards is equivalent to a sum of 25 x 7.48 = 187 HCP. It should be possible to work out how many ways there are to combine 25 hands to get less than or equal to a cumulative total of 187 HCP, then work out the probability for each combination, finally summing them all.

[PeterAlan]

AL78, on 2021-February-19, 11:26, said:

An average HCP of 7.48 over 25 boards is equivalent to a sum of 25 x 7.48 = 187 HCP. It should be possible to work out how many ways there are to combine 25 hands to get less than or equal to a cumulative total of 187 HCP, then work out the probability for each combination, finally summing them all.

That is exactly what my program did: it has the HCP distribution of a single hand as its raw data, and does precisely the calculation you describe.

Rather than the rounded probability numbers in your link, the easiest source of raw data (for me, since I already had it) was to start with the 635,013,559,600 possible ways of dealing a 13-card hand, tabulated by the number (2,310,7889,600) with 0 HCP, etc through to (4) 37 HCP.

If anyone wants the parameters to slightly more precision, its standard deviation is 4.13023287. I calculated its third (skewness) and fourth (kurtosis) standardised moments to be 0.24044733 and 2.87800178 respectively.

[Vinish]

Ok, here is a related question. In that same sequence of 25 hands, I had 12 hcp exactly once and all other 24 hands were less than 12 hcp. What is the percent chance of having 25 hands with no hcp higher than 12?

[hrothgar]

Vinish, on 2021-February-19, 18:36, said:

Ok, here is a related question. In that same sequence of 25 hands, I had 12 hcp exactly once and all other 24 hands were less than 12 hcp. What is the percent chance of having 25 hands with no hcp higher than 12?

The cumulative probability that a hand has no more than 12 HCP is 0.732097

The odds that this did not happen is (1 - 0.732097)

The odds that this happened 25 times in a row is (1- 0.732097) ^ 25

[smerriman]

hrothgar, on 2021-February-19, 19:31, said:

The odds that this happened 25 times in a row is (1- 0.732097) ^ 25

That's the probability of having *more* than 12 HCP 25 times in a row. For <= 12, it's just 0.732097^25 = 0.4%.

[AL78]

hrothgar, on 2021-February-19, 19:31, said:

The cumulative probability that a hand has no more than 12 HCP is 0.732097

The odds that this did not happen is (1 - 0.732097)

The odds that this happened 25 times in a row is (1- 0.732097) ^ 25

Surely it is 0.732097^25, the probability of the event happening 25 consecutive times. What you have is the probability of getting more than 12 HCP 25 consecutive times.

[AL78]

smerriman, on 2021-February-19, 20:05, said:

That's the probability of having *more* than 12 HCP 25 times in a row. For <= 12, it's just 0.732097^25 = 0.4%.

You are a factor of 10 out, I make it 0.04%. It is more likely that I would have thought, about the same magnitude as picking up a Yarborough or an Acol 2♣ opening.