how to calculate post holecards but preflop in texas hold em poker?
If i have 2 random holecards in my hands, how high are the odds that I will get a pair preflop? I'm assuming there are 5 cards in the middle of the table for anyone that don't know what preflop means.
Including a calculation would be appreciated. thanks
do you know?
how many words do you know
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 translate this math func in c program

calculate pixels to meter different heights
I have data taken at 73 M height, it has an X coordinate and Ycorddinate that are known to me. the problem is the image was taken at a 79meter height and while I try to convert meters to pixels I get an error and the real object is not marked
for example  I hope o mark the red point , but my code marks the yellow one

How do i print out a number triangle in python? Using anything to string is not allowed. Only arithmetic operation
Example
n = 5
Output of each line is int type
1 22 333 4444

How can I optimize the expected value of a function in R?
I have derived a survival function for a system of components (ignore the details of how this system is setup) and I am trying to maximize its expected, or more specifically, maximizing the expected value of the function:
surv_func = function(x,mu) = {(exp((x/(mu))^(1/3))*((1exp((4/3)*x^(3/2)))+exp(((4/3)*x^(3/2)))))*exp((x/(3mu))^(1/3))}
and I am supposed (since the pdf including my tasks gives a hint about it) to use the function
optimize()
and the expected value for a function can be computed with
# Computes expected value of a the function "function" E < integrate(function, 0, Inf)
but my function depends on x and mu. The expected value could (obviously) be computed if the integral had no mu but instead only depended on x. For those interested, the mu comes from the fact that one of the components has a Weibulldistribution with parameters (1/3,mu) and the 3mu comes from that has a Weibulldistribution with parameters (1/3,lambda). In the task there is a constraint mu + lambda = 3, so I tought substituting the lambdaparameter in the second Weibulldistribution with lambda = 3  mu and trying to maximize this problem would yield not only mu, but also lambda.
If I try to, just for the sake of learing about R, compute the expected value using the code below (in the console window), it just gives me the following:
> E < integrate(surv_func,0,Inf) Error in (function (x, mu) : argument "mu" is missing, with no default
I am new to R and seem to be a little bit "slow" at learning. How can I approach this problem?

What's the actual probability of an event occurring predicted by a classification model?
I have a classification model that predicts whether event A will occur or event B. The accuracy of the model is 49%. Say for a test case it predicts that event A will occur with a probability of 72%. So what is the probability that Event A will occur?

Calculate convolution of exponential variables
have question for below convolution problem Let variables a1, a2 and a3 independently follows exponential(1) distributions. Find P(a1<2, a1+a2>2) and P(a1+a2<2, a1+a2+a3>2)

simulating a three person card game, what is the probability that exactly two people
so I wrote some code, but i think it's wrong.
I want to simulate a three person games were exactly two people get three of a kind
code:
from random import shuffle from itertools import product #generating deck suits = ["s","d","h","c"] values = ["1","2","3","4","5","6","7","8","9","10","11","12","13"] deck = list(product(values,suits)) sim = 100000 two_threeOfaKind = 0 for i in range(sim): hand_one = [] three_hand_one = 0 hand_two = [] three_hand_two = 0 hand_three = [] three_hand_three = 0 shuffle(deck) #generating hand for i in range(5): hand_one.append(deck[i]) for i in range(5,11): hand_two.append(deck[i]) for i in range(11,16): hand_three.append(deck[i]) #checking if cards have three of a kind values = sorted(card[0] for card in hand_one) if values.count(values[2]) == 3: three_hand_one = 1 values = sorted(card[0] for card in hand_two) if values.count(values[2]) == 3: three_hand_two = 1 values = sorted(card[0] for card in hand_three) if values.count(values[2]) == 3: three_hand_three = 1 threeofakinds = [three_hand_one,three_hand_two,three_hand_three] if threeofakinds.count(1) == 2: two_threeOfaKind += 1 probtwothree = two_threeOfaKind/sim print(probtwothree)

playing a three person game what is the possibility that exactly two people get three of a kind
Basically i am doing exercises to learn python. The goal is simulating a three person game and finding the possibility that exactly two people get three of a kind. My problem arises when i am removing the cards from the deck that have already been handed out to a player. I get an Error saying " 'NoneType' was no attribute remove, although my deck is a list. (i have no idea if the rest of the code works though)
from random import shuffle, sample from itertools import product #generating deck suits = ["s","d","h","c"] values = ["1","2","3","4","5","6","7","8","9","10","11","12","13"] deck = list(product(values,suits)) sim = 100000 two_three_of_a_kind = 0 for i in range(sim): three_of_a_kind = 0 shuffle(deck) #generating hand hand_one = sample(deck,5) for i in range(5): deck = deck.remove(hand_one[i]) hand_two = sample(deck,5) for i in range(5): deck = deck.remove(hand_two[i]) hand_three = sample(deck,5) hands = [hand_one,hand_two,hand_three] #checking for three of a kind for all three hands for i in range(3): values = sorted(card[0] for card in hands[i]) if values.count(values[2]) == 3: three_of_a_kind += 1 if three_of_a_kind == 2: two_three_of_a_kind += 1 prob_two_three = two_three_of_a_kind/sim print(prob_two_three)

probability of getting three of a kind by drawing 5 cards
so my goal is to try to simulate an actaul deck and draw five cards and check if there is a three of a kind. I have no problem making the deck and drawing five cards, the problem arises when i check for three of a kind
my code:
from random import shuffle, sample from itertools import product #generating deck suits = ["s","d","h","c"] values = ["1","2","3","4","5","6","7","8","9","10","11","12","13"] deck = list(product(values,suits)) sim = 100000 three_of_a_kind = 0 for i in range(sim): shuffle(deck) #generating hand hand = sample(deck,5) #checking for three of a kind if any(hand[0][0] == x[0] for x in hand): three1 += 1 elif any(hand[1][0] == x[0] for x in hand): three2 += 1 elif any(hand[2][0] == x[0] for x in hand): three3 += 1 if three1 == 3 or three2 == 3 or three3 == 3: three_of_a_kind += 1 prob_three = three_of_a_kind/sim print(prob_three)
edit: my deck only had 12 cards and I changed it to 13 but my question has not changed

program doesn't automatically run the statements in a nested if else function?
install.packages("lmomco") install.packages("extRemes") library("lmomco") library("extRemes") distribution < tolower('lpe3') transformed < FALSE # add log Pearson Type 3 to list of distributions supported # by lmomco package base.dist < c(distribution, dist.list()) if( any(distribution %in% base.dist) ) { # log transform series if( distribution == 'lp3' ) { series < log10(Potomac$Flow) transformed < TRUE distribution < 'pe3' } } # compute Lmoments samLmom < lmom.ub(series) # estimate distribution parameter distPar < lmom2par(samLmom, type = distribution) distPar
The code doesn't run the nested if/else function for the log transform series. I want to transform the series into a logseries for the log pearson typeIII distribution.

TensorFlow Probability: Different log probabilities for Sequential vs Named JointDistributions?
I'm fairly new to Bayesian estimation with TensorFlow. I was trying to set up a very simple regression of height on weight (using McElreath's Howell data) to familiarize myself with the machinery in TensorFlow Probability, but I am running into something I don't understand. I presumed that defining a model with
JointDistributionSequentialAutoBatched
would yield a model that was identical to one defined byJointDistributionNamedAutoBatched
, but the latter would just yield some nice handles for getting at parameters.ht_wt: tfd.JointDistributionSequentialAutoBatched = tfd.JointDistributionSequentialAutoBatched([ tfd.Normal(loc=tf.cast(0., dtype=tf.float64), scale=0.2), tfd.Normal(loc=tf.cast(0., dtype=tf.float64), scale=1.), tfd.Uniform(low=tf.cast(1., dtype=tf.float64), scale=10.), lambda beta0, beta1, sigma: tfd.Independent(tfd.Normal( loc=beta0 + beta1 * weight, # weight is a tf.Tensor of weights from Howell sigma=sigma )) ], name="Height vs Weight") ht_wt_named: tfd.JointDistributionNamedAutoBatched = tfd.JointDistributionNamedAutoBatched([ beta0=tfd.Normal(loc=tf.cast(0., dtype=tf.float64), scale=0.2), beta1=tfd.Normal(loc=tf.cast(0., dtype=tf.float64), scale=1.), sigma=tfd.Uniform(low=tf.cast(1., dtype=tf.float64), scale=10.), x=lambda beta0, beta1, sigma: tfd.Independent(tfd.Normal( loc=beta0 + beta1 * weight, # weight is a tf.Tensor of weights from Howell sigma=sigma )) ], name="Height vs Weight (Named)")
However, when I look at the distribution of logged probabilities across different parameter values, I get inconsistently different values. For example...
beta0 = 1., beta1 = 1., sigma = 0.5 yields: Sequential = 19163.0633 Named = inf
beta0 = 1., beta1 = 1., sigma = 1.0 yields: Sequential = 5108.2755 Named = 5108.2755
beta0 = 1., beta1 = 1., sigma = 1.5 yields: Sequential = 2616.9668 Named = 2601.3418
I get different values when I vary the
beta
as well. Have I missed something about underlying differences betweenSequential
andNamed
JointDistributions
? 
how to figure out the probability of elevators and people?
I did work out a distribution, but the sum of all the probabilities is not 1! So it is obviously wrong. Please point out what I did wrong:
**P**(numbers of elevators used is $i$)=$\frac{(C_10^i)(P_12^i)(i^(12i)}{10^12}$ enter code here

How to obtain the witness function of the Stein operator
With the Stein operator:
We can define the Kernel Stein discrepancy as:
I found a nice explanation of the witness function modified by the Stein operator at: https://slideslive.com/38917868/relativegoodnessoffittestsformodelswithlatentvariables, where this function is explaining how the two distributions are different:
I tried to reproduce this figure as an explanation of the witness function, but I can't quite understand how did he calculated the witness function g* to code it up:
import torch import matplotlib.pyplot as plt p = torch.distributions.normal.Normal(torch.tensor([0.0]), torch.tensor([1.0])) q = torch.distributions.normal.Normal(torch.tensor([1.0]), torch.tensor([1.0])) x = torch.arange(4, 4, .1) fig = plt.figure(1) ax = plt.gca() ax.plot(x, torch.exp(p.log_prob(x)), 'red') ax.plot(x, torch.exp(q.log_prob(x)), 'blue') ax.grid(True) ax.spines['left'].set_position('zero') ax.spines['right'].set_color('none') ax.spines['bottom'].set_position('zero') ax.spines['top'].set_color('none') ax.set_xlim(4, 4); ax.set_ylim(.7, .5) """How to compute the $g^*$ (witness function) to plot it."""
Can you please help to explain how did he compute the witness function g* in green line, and to simply code it up !

Conditional ProbabilityBaye's theorem
Everyone in this globe has done at least one PCR test in the last two years to check if he/she has SARSCOV2 virus. It is scientifically proved that the PCR tests are not 100% accurate. Assume that the PCR test has an accuracy of 98.5% of detecti:g the disease, and 55% of people have the SARSCov2 virus. If a patient has done a PCR test and tested positive, what is the probability that the patient has the SARSCov2 virus.
Can anyone explain to me exactly in details how to solve this problem?
It’s probability theory and statistics (Baye’s Theorem)