The following Python program flips a coin several times, then reports the longest series of heads and tails. I am trying to convert this program into a program that uses functions so it uses basically less code. I am very new to programming and my teacher requested this of us, but I have no idea how to do it. I know I'm supposed to have the function accept 2 parameters: a string or list, and a character to search for. The function should return, as the value of the function, an integer which is the longest sequence of that character in that string. The function shouldn't accept input or output from the user.
import random
print("This program flips a coin several times, \nthen reports the longest
series of heads and tails")
cointoss = int(input("Number of times to flip the coin: "))
varlist = []
i = 0
varstring = ' '
while i < cointoss:
r = random.choice('HT')
varlist.append(r)
varstring = varstring + r
i += 1
print(varstring)
print(varlist)
print("There's this many heads: ",varstring.count("H"))
print("There's this many tails: ",varstring.count("T"))
print("Processing input...")
i = 0
longest_h = 0
longest_t = 0
inarow = 0
prevIn = 0
while i < cointoss:
print(varlist[i])
if varlist[i] == 'H':
prevIn += 1
if prevIn > longest_h:
longest_h = prevIn
print("",longest_h,"")
inarow = 0
if varlist[i] == 'T':
inarow += 1
if inarow > longest_t:
longest_t = inarow
print("",longest_t,"")
prevIn = 0
i += 1
print ("The longest series of heads is: ",longest_h)
print ("The longest series of tails is: ",longest_t)
If this is asking too much, any explanatory help would be really nice instead. All I've got so far is:
def flip (a, b):
flipValue = random.randint
but it's barely anything.
import random
def Main():
numOfFlips=getFlips()
outcome=flipping(numOfFlips)
print(outcome)
def getFlips():
Flips=int(input("Enter number if flips:\n"))
return Flips
def flipping(numOfFlips):
longHeads=[]
longTails=[]
Tails=0
Heads=0
for flips in range(0,numOfFlips):
flipValue=random.randint(1,2)
print(flipValue)
if flipValue==1:
Tails+=1
longHeads.append(Heads) #recording value of Heads before resetting it
Heads=0
else:
Heads+=1
longTails.append(Tails)
Tails=0
longestHeads=max(longHeads) #chooses the greatest length from both lists
longestTails=max(longTails)
return "Longest heads:\t"+str(longestHeads)+"\nLongest tails:\t"+str(longestTails)
Main()
I did not quite understand how your code worked, so I made the code in functions that works just as well, there will probably be ways of improving my code alone but I have moved the code over to functions
First, you need a function that flips a coin x times. This would be one possible implementation, favoring random.choice over random.randint:
def flip(x):
result = []
for _ in range(x):
result.append(random.choice(("h", "t")))
return result
Of course, you could also pass from what exactly we are supposed to take a choice as a parameter.
Next, you need a function that finds the longest sequence of some value in some list:
def longest_series(some_value, some_list):
current, longest = 0, 0
for r in some_list:
if r == some_value:
current += 1
longest = max(current, longest)
else:
current = 0
return longest
And now you can call these in the right order:
# initialize the random number generator, so we get the same result
random.seed(5)
# toss a coin a hundred times
series = flip(100)
# count heads/tails
headflips = longest_series('h', series)
tailflips = longest_series('t', series)
# print the results
print("The longest series of heads is: " + str(headflips))
print("The longest series of tails is: " + str(tailflips))
Output:
>> The longest series of heads is: 8
>> The longest series of heads is: 5
edit: removed the flip implementation with yield, it made the code weird.
Counting the longest run
Let see what you have asked for
I'm supposed to have the function accept 2 parameters: a string or list,
or, generalizing just a bit, a sequence
and a character
again, we'd speak, generically, of an item
to search for. The function should return, as the value of the
function, an integer which is the longest sequence of that character
in that string.
My implementation of the function you are asking for, complete of doc
string, is
def longest_run(i, s):
'Counts the longest run of item "i" in sequence "s".'
c, m = 0, 0
for el in s:
if el==i:
c += 1
elif c:
m = m if m >= c else c
c = 0
return m
We initialize c (current run) and m (maximum run so far) to zero,
then we loop, looking at every element el of the argument sequence s.
The logic is straightforward but for elif c: whose block is executed at the end of a run (because c is greater than zero and logically True) but not when the previous item (not the current one) was not equal to i. The savings are small but are savings...
Flipping coins (and more...)
How can we simulate flipping n coins? We abstract the problem and recognize that flipping n coins corresponds to choosing from a collection of possible outcomes (for a coin, either head or tail) for n times.
As it happens, the random module of the standard library has the exact answer to this problem
In [52]: random.choices?
Signature: choices(population, weights=None, *, cum_weights=None, k=1)
Docstring:
Return a k sized list of population elements chosen with replacement.
If the relative weights or cumulative weights are not specified,
the selections are made with equal probability.
File: ~/lib/miniconda3/lib/python3.6/random.py
Type: method
Our implementation, aimed at hiding details, could be
def roll(n, l):
'''Rolls "n" times a dice/coin whose face values are listed in "l".
E.g., roll(2, range(1,21)) -> [12, 4] simulates rolling 2 icosahedron dices.
'''
from random import choices
return choices(l, k=n)
Putting this together
def longest_run(i, s):
'Counts the longest run of item "i" in sequence "s".'
c, m = 0, 0
for el in s:
if el==i:
c += 1
elif c:
m = m if m >= c else c
c = 0
return m
def roll(n, l):
'''Rolls "n" times a dice/coin whose face values are listed in "l".
E.g., roll(2, range(1,21)) -> [12, 4] simulates rolling 2 icosahedron dices.
'''
from random import choices
return choices(l, k=n)
N = 100 # n. of flipped coins
h_or_t = ['h', 't']
random_seq_of_h_or_t = flip(N, h_or_t)
max_h = longest_run('h', random_seq_of_h_or_t)
max_t = longest_run('t', random_seq_of_h_or_t)
Related
I am trying to generate a random sequence of numbers, with each "result" having a chance of {a}48.6%/{b}48.6%/{c}2.8%.
Counting how many times in a sequence of 6 or more {a} occurred, same for {b}.
{c} counts as neutral, meaning that if an {a} sequence is happening, {c} will count as {a}, additionally if a {b} sequence is happening, then {c} will count as {b}.
The thing is that the results seem right, but every "i" iteration seems to give results that are "weighted" either on the {a} side or the {b} side. And I can't seem to figure out why.
I would expect for example to have a result of :
{a:6, b:7, a:8, a:7, b:9} but what I am getting is {a:7, a:9, a:6, a:8} OR {b:7, b:8, b:6} etc.
Any ideas?
import sys
import random
from random import seed
from random import randint
from datetime import datetime
import time
loopRange = 8
flips = 500
median = 0
for j in range(loopRange):
random.seed(datetime.now())
sequenceArray = []
maxN = 0
flag1 = -1
flag2 = -1
for i in range(flips):
number = randint(1, 1000)
if(number <= 486):
flag1 = 0
sequenceArray.append(number)
elif(number > 486 and number <= 972):
flag1 = 1
sequenceArray.append(number)
elif(number > 972):
sequenceArray.append(number)
if(flag1 != -1 and flag2 == -1):
flag2 = flag1
if(flag1 != flag2):
sequenceArray.pop()
if(len(sequenceArray) > maxN):
maxN = len(sequenceArray)
if(len(sequenceArray) >= 6):
print(len(sequenceArray))
# print(sequenceArray)
# print(sequenceArray)
sequenceArray = []
median += maxN
print("Maximum sequence is %d " % maxN)
print("\n")
time.sleep(random.uniform(0.1, 1))
median = float(median/loopRange)
print("\n")
print(median)
I would implement something with two cursors prev (previous) and curr (current) since you need to detect a change between the current and the previous state.
I just write the code of the inner loop on i since the external loop adds complexity without focusing on the source of the problem. You can then include this in your piece of code. It seems to work for me, but I am not sure to understand perfectly how you want to manage all the behaviours (especially at start).
prev = -1
curr = -1
seq = []
maxN = 0
for i in range(flips):
number = randint(1, 1000)
if number<=486:
curr = 0 # case 'a'
elif number<=972:
curr = 1 # case 'b'
else:
curr = 2 # case 'c', the joker
if (prev==-1) and (curr==2):
# If we start with the joker, don't do anything
# You can add code here to change this behavior
pass
else:
if (prev==-1):
# At start, it is like the previous element is the current one
prev=curr
if (prev==curr) or (curr==2):
# We continue the sequence
seq.append(number)
else:
# Break the sequence
maxN = max(maxN,len(seq)) # save maximum length
if len(seq)>=6:
print("%u: %r"%(prev,seq))
seq = [] # reset sequence
seq.append(number) # don't forget to append the new number! It breaks the sequence, but starts a new one
prev=curr # We switch case 'a' for 'b', or 'b' for 'a'
I am trying to solve the usaco problem combination lock where you are given a two lock combinations. The locks have a margin of error of +- 2 so if you had a combination lock of 1-3-5, the combination 3-1-7 would still solve it.
You are also given a dial. For example, the dial starts at 1 and ends at the given number. So if the dial was 50, it would start at 1 and end at 50. Since the beginning of the dial is adjacent to the end of the dial, the combination 49-1-3 would also solve the combination lock of 1-3-5.
In this program, you have to output the number of distinct solutions to the two lock combinations. For the record, the combination 3-2-1 and 1-2-3 are considered distinct, but the combination 2-2-2 and 2-2-2 is not.
I have tried creating two functions, one to check whether three numbers match the constraints of the first combination lock and another to check whether three numbers match the constraints of the second combination lock.
a,b,c = 1,2,3
d,e,f = 5,6,7
dial = 50
def check(i,j,k):
i = (i+dial) % dial
j = (j+dial) % dial
k = (k+dial) % dial
if abs(a-i) <= 2 and abs(b-j) <= 2 and abs(c-k) <= 2:
return True
return False
def check1(i,j,k):
i = (i+dial) % dial
j = (j+dial) % dial
k = (k+dial) % dial
if abs(d-i) <= 2 and abs(e-j) <= 2 and abs(f-k) <= 2:
return True
return False
res = []
count = 0
for i in range(1,dial+1):
for j in range(1,dial+1):
for k in range(1,dial+1):
if check(i,j,k):
count += 1
res.append([i,j,k])
if check1(i,j,k):
count += 1
res.append([i,j,k])
print(sorted(res))
print(count)
The dial is 50 and the first combination is 1-2-3 and the second combination is 5-6-7.
The program should output 249 as the count, but it instead outputs 225. I am not really sure why this is happening. I have added the array for display purposes only. Any help would be greatly appreciated!
You're going to a lot of trouble to solve this by brute force.
First of all, your two check routines have identical functionality: just call the same routine for both combinations, giving the correct combination as a second set of parameters.
The critical logic problem is handling the dial wrap-around: you miss picking up the adjacent numbers. Run 49 through your check against a correct value of 1:
# using a=1, i=49
i = (1+50)%50 # i = 1
...
if abs(1-49) <= 2 ... # abs(1-49) is 48. You need it to show up as 2.
Instead, you can check each end of the dial:
a_diff = abs(i-a)
if a_diff <=2 or a_diff >= (dial-2) ...
Another way is to start by making a list of acceptable values:
a_vals = [(a-oops) % dial] for oops in range(-2, 3)]
... but note that you have to change the 0 value to dial. For instance, for a value of 1, you want a list of [49, 50, 1, 2, 3]
With this done, you can check like this:
if i in a_vals and j in b_vals and k in c_vals:
...
If you want to upgrade to the itertools package, you can simply generate all desired combinations:
combo = set(itertools.product(a_list, b_list_c_list) )
Do that for both given combinations and take the union of the two sets. The length of the union is the desired answer.
I see the follow-up isn't obvious -- at least, it's not appearing in the comments.
You have 5*5*5 solutions for each combination; start with 250 as your total.
Compute the sizes of the overlap sets: the numbers in each triple that can serve for each combination. For your given problem, those are [3],[4],[5]
The product of those set sizes is the quantity of overlap: 1*1*1 in this case.
The overlapping solutions got double-counted, so simply subtract the extra from 250, giving the answer of 249.
For example, given 1-2-3 and 49-6-6, you would get sets
{49, 50, 1}
{4}
{4, 5}
The sizes are 3, 1, 2; the product of those numbers is 6, so your answer is 250-6 = 244
Final note: If you're careful with your modular arithmetic, you can directly compute the set sizes without building the sets, making the program very short.
Here is one approach to a semi-brute-force solution:
import itertools
#The following code assumes 0-based combinations,
#represented as tuples of numbers in the range 0 to dial - 1.
#A simple wrapper function can be used to make the
#code apply to 1-based combos.
#The following function finds all combos which open lock with a given combo:
def combos(combo,tol,dial):
valids = []
for p in itertools.product(range(-tol,1+tol),repeat = 3):
valids.append(tuple((x+i)%dial for x,i in zip(combo,p)))
return valids
#The following finds all combos for a given iterable of target combos:
def all_combos(targets,tol,dial):
return set(combo for target in targets for combo in combos(target,tol,dial))
For example, len(all_combos([(0,1,2),(4,5,6)],2,50)) evaluate to 249.
The correct code for what you are trying to do is the following:
dial = 50
a = 1
b = 2
c = 3
d = 5
e = 6
f = 7
def check(i,j,k):
if (abs(a-i) <= 2 or (dial-abs(a-i)) <= 2) and \
(abs(b-j) <= 2 or (dial-abs(b-j)) <= 2) and \
(abs(c-k) <= 2 or (dial-abs(c-k)) <= 2):
return True
return False
def check1(i,j,k):
if (abs(d-i) <= 2 or (dial-abs(d-i)) <= 2) and \
(abs(e-j) <= 2 or (dial-abs(e-j)) <= 2) and \
(abs(f-k) <= 2 or (dial-abs(f-k)) <= 2):
return True
return False
res = []
count = 0
for i in range(1,dial+1):
for j in range(1,dial+1):
for k in range(1,dial+1):
if check(i,j,k):
count += 1
res.append([i,j,k])
elif check1(i,j,k):
count += 1
res.append([i,j,k])
print(sorted(res))
print(count)
And the result is 249, the total combinations are 2*(5**3) = 250, but we have the duplicates: [3, 4, 5]
I'm trying to write Python code to see how many coin tosses, on average, are required to get a sequences of N heads in a row.
The thing that I'm puzzled by is that the answers produced by my code don't match ones that are given online, e.g. here (and many other places) https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads
According to that, the expected number of tosses that I should need to get various numbers of heads in a row are: E(1) = 2, E(2) = 6, E(3) = 14, E(4) = 30, E(5) = 62. But I don't get those answers! For example, I get E(3) = 8, instead of 14. The code below runs to give that answer, but you can change n to test for other target numbers of heads in a row.
What is going wrong? Presumably there is some error in the logic of my code, but I confess that I can't figure out what it is.
You can see, run and make modified copies of my code here: https://trinket.io/python/17154b2cbd
Below is the code itself, outside of that runnable trinket.io page. Any help figuring out what's wrong with it would be greatly appreciated!
Many thanks,
Raj
P.S. The closest related question that I could find was this one: Monte-Carlo Simulation of expected tosses for two consecutive heads in python
However, as far as I can see, the code in that question does not actually test for two consecutive heads, but instead tests for a sequence that starts with a head and then at some later, possibly non-consecutive, time gets another head.
# Click here to run and/or modify this code:
# https://trinket.io/python/17154b2cbd
import random
# n is the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 3
possible_tosses = [ 'h', 't' ]
num_trials = 1000
target_seq = ['h' for i in range(0,n)]
toss_sequence = []
seq_lengths_rec = []
for trial_num in range(0,num_trials):
if (trial_num % 100) == 0:
print 'Trial num', trial_num, 'out of', num_trials
# (The free version of trinket.io uses Python2)
target_reached = 0
toss_num = 0
while target_reached == 0:
toss_num += 1
random.shuffle(possible_tosses)
this_toss = possible_tosses[0]
#print([toss_num, this_toss])
toss_sequence.append(this_toss)
last_n_tosses = toss_sequence[-n:]
#print(last_n_tosses)
if last_n_tosses == target_seq:
#print('Reached target at toss', toss_num)
target_reached = 1
seq_lengths_rec.append(toss_num)
print 'Average', sum(seq_lengths_rec) / len(seq_lengths_rec)
You don't re-initialize toss_sequence for each experiment, so you start every experiment with a pre-existing sequence of heads, having a 1 in 2 chance of hitting the target sequence on the first try of each new experiment.
Initializing toss_sequence inside the outer loop will solve your problem:
import random
# n is the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 4
possible_tosses = [ 'h', 't' ]
num_trials = 1000
target_seq = ['h' for i in range(0,n)]
seq_lengths_rec = []
for trial_num in range(0,num_trials):
if (trial_num % 100) == 0:
print('Trial num {} out of {}'.format(trial_num, num_trials))
# (The free version of trinket.io uses Python2)
target_reached = 0
toss_num = 0
toss_sequence = []
while target_reached == 0:
toss_num += 1
random.shuffle(possible_tosses)
this_toss = possible_tosses[0]
#print([toss_num, this_toss])
toss_sequence.append(this_toss)
last_n_tosses = toss_sequence[-n:]
#print(last_n_tosses)
if last_n_tosses == target_seq:
#print('Reached target at toss', toss_num)
target_reached = 1
seq_lengths_rec.append(toss_num)
print(sum(seq_lengths_rec) / len(seq_lengths_rec))
You can simplify your code a bit, and make it less error-prone:
import random
# n is the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 3
possible_tosses = [ 'h', 't' ]
num_trials = 1000
seq_lengths_rec = []
for trial_num in range(0, num_trials):
if (trial_num % 100) == 0:
print('Trial num {} out of {}'.format(trial_num, num_trials))
# (The free version of trinket.io uses Python2)
heads_counter = 0
toss_counter = 0
while heads_counter < n:
toss_counter += 1
this_toss = random.choice(possible_tosses)
if this_toss == 'h':
heads_counter += 1
else:
heads_counter = 0
seq_lengths_rec.append(toss_counter)
print(sum(seq_lengths_rec) / len(seq_lengths_rec))
We cam eliminate one additional loop by running each experiment long enough (ideally infinite) number of times, e.g., each time toss a coin n=1000 times. Now, it is likely that the sequence of 5 heads will appear in each such trial. If it does appear, we can call the trial as an effective trial, otherwise we can reject the trial.
In the end, we can take an average of number of tosses needed w.r.t. the number of effective trials (by LLN it will approximate the expected number of tosses). Consider the following code:
N = 100000 # total number of trials
n = 1000 # long enough sequence of tosses
k = 5 # k heads in a row
ntosses = []
pat = ''.join(['1']*k)
effective_trials = 0
for i in range(N): # num of trials
seq = ''.join(map(str,random.choices(range(2),k=n))) # toss a coin n times (long enough times)
if pat in seq:
ntosses.append(seq.index(pat) + k)
effective_trials += 1
print(effective_trials, sum(ntosses) / effective_trials)
# 100000 62.19919
Notice that the result may not be correct if n is small, since it tries to approximate infinite number of coin tosses (to find expected number of tosses to obtain 5 heads in a row, n=1000 is okay since actual expected value is 62).
I'm trying to build a function to do internal metric conversion on a wavelength to frequency conversion program and have been having a hard time getting it to behave properly. It is super slow and will not assign the correct labels to the output. If anyone can help with either a different method of computing this or a reason on why this is happening and any fixes that I cond do that would be amazing!
def convert_SI_l(n):
if n in range( int(1e-12),int(9e-11)):
return n/0.000000000001, 'pm'
else:
if n in range(int(1e-10),int(9e-8)):
return n/0.000000001 , 'nm'
else:
if n in range(int(1e-7),int(9e-5)):
return n/0.000001, 'um'
else:
if n in range(int(1e-4),int(9e-3)):
return n/0.001, 'mm'
else:
if n in range(int(0.01), int(0.99)):
return n/0.01, 'cm'
else:
if n in range(1,999):
return n/1000, 'm'
else:
if n in range(1000,299792459):
return n/1000, 'km'
else:
return n , 'm'
def convert_SI_f(n):
if n in range( 1,999):
return n, 'Hz'
else:
if n in range(1000,999999):
return n/1000 , 'kHz'
else:
if n in range(int(1e6),999999999):
return n/1e6, 'MHz'
else:
if n in range(int(1e9),int(1e13)):
return n/1e9, 'GHz'
else:
return n, 'Hz'
c=299792458
i=input("Are we starting with a frequency or a wavelength? ( F / L ): ")
#Error statements
if i.lower() == ("f"):
True
else:
if not i.lower() == ("l"):
print ("Error invalid input")
#Cases
if i.lower() == ("f"):
f = float(input("Please input frequency (in Hz): "))
size_l = c/f
print(convert_SI_l(size_l))
if i.lower() == ("l"):
l = float(input("Please input wavelength (in meters): "))
size_f = ( l/c)
print(convert_SI_f(size_f))
You are using range() in a way that is close to how it is used in natural language, to express a contiguous segment of the real number line, as in in the range 4.5 to 5.25. But range() doesn't mean that in Python. It means a bunch of integers. So your floating-point values, even if they are in the range you specify, will not occur in the bunch of integers that the range() function generates.
Your first test is
if n in range( int(1e-12),int(9e-11)):
and I am guessing you wrote it like this because what you actually wanted was range(1e-12, 9e-11) but you got TypeError: 'float' object cannot be interpreted as an integer.
But if you do this at the interpreter prompt
>>> range(int(1e-12),int(9e-11))
range(0, 0)
>>> list(range(int(1e-12),int(9e-11)))
[]
you will see it means something quite different to what you obviously expect.
To test if a floating-point number falls in a given range do
if lower-bound <= mynumber <= upper-bound:
You don't need ranges and your logic will be more robust if you base it on fixed threshold points that delimit the unit magnitude. This would typically be a unit of one in the given scale.
Here's a generalized approach to all unit scale determination:
SI_Length = [ (1/1000000000000,"pm"),
(1/1000000000, "nm"),
(1/1000000, "um"),
(1/1000, "mm"),
(1/100, "cm"),
(1, "m"),
(1000, "km") ]
SI_Frequency = [ (1, "Hz"), (1000,"kHz"), (1000000,"MHz"), (1000000000,"GHz")]
def convert(n,units):
useFactor,useName = units[0]
for factor,name in units:
if n >= factor : useFactor,useName = factor,name
return (n/useFactor,useName)
print(convert(0.0035,SI_Length)) # 3.5 mm
print(convert(12332.55,SI_Frequency)) # 12.33255 kHz
Each unit array must be in order of smallest to largest multiplier.
EDIT: Actually, range is a function which is generally used in itaration to generate numbers. So, when you write if n in range(min_value, max_value), this function generates all integers until it finds a match or reach the max_value.
The range type represents an immutable sequence of numbers and is commonly used for looping a specific number of times in for loops.
Instead of writing:
if n in range(int(1e-10),int(9e-8)):
return n/0.000000001 , 'nm'
you should write:
if 1e-10 <= n < 9e-8:
return n/0.000000001 , 'nm'
Also keep in mind that range only works on integers, not float.
More EDIT:
For your specific use case, you can define dictionary of *(value, symbol) pairs, like below:
import collections
symbols = collections.OrderedDict(
[(1e-12, u'p'),
(1e-9, u'n'),
(1e-6, u'μ'),
(1e-3, u'm'),
(1e-2, u'c'),
(1e-1, u'd'),
(1e0, u''),
(1e1, u'da'),
(1e2, u'h'),
(1e3, u'k'),
(1e6, u'M'),
(1e9, u'G'),
(1e12, u'T')])
The use the bisect.bisect function to find the "insertion" point of your value in that ordered collection. This insertion point can be used to get the simplified value and the SI symbol to use.
For instance:
import bisect
def convert_to_si(value):
if value < 0:
value, symbol = convert_to_si(-value)
return -value, symbol
elif value > 0:
orders = list(symbols.keys())
order_index = bisect.bisect(orders, value / 10.0)
order = orders[min(order_index, len(orders) - 1)]
return value / order, symbols[order]
else:
return value, u""
Demonstration:
for value in [1e-12, 3.14e-11, 0, 2, 20, 3e+9]:
print(*convert_to_si(value), sep="")
You get:
1.0p
0.0314n
0
2.0
2.0da
3.0G
You can adapt this function to your needs…
I was trying to calculate the expected value for the longest consecutive heads streak in 200 coin flips, using python. I came up with a code which I think does the job right but it's just not efficient because of the amount of calculations and data storage it requires, and I was wondering if someone could help me out with this, making it faster and more efficient (I took only one course of python programming in last semester without any previous knowledge of the subject).
My code was
import numpy as np
from itertools import permutations
counter = 0
sett = 0
rle = []
matrix = np.zeros(200)
for i in range (0,200):
matrix[i] = 1
for j in permutations(matrix):
for k in j:
if k == 1:
counter += 1
else:
if counter > sett:
sett == counter
counter == 0
rle.append(sett)
After finding rle, I'd iterate over it to get how many streaks of which length there are, and their sum divided by 2^200 would give me the expected value I'm looking for.
Thanks in advance for help, much appreciated!
You don't have to try all the permutations (in fact you cannot), but you can do a simple Monte Carlo style simulation. Repeat the 200 coin flips many times. Average the lengths of longest streaks you get and this will be a good approximation of the expected value.
def oneTrial (noOfCoinFlips):
s = numpy.random.binomial(1, 0.5, noOfCoinFlips)
maxCount = 0
count = 0
for x in s:
if x == 1:
count += 1
if x == 0:
count = 0
maxCount = max(maxCount, count)
return maxCount
numpy.mean([oneTrial(200) for x in range(10000)])
Output: 6.9843
Also see this thread for exact computation without using Python simulation.
This is an answer to a slightly different question. But, as I had invested an hour and half of my time into it, I didn't wanna scrape it off.
Let E(k) denote a k head streak, i.e., you get k consecutive heads from the first toss onwards.
E(0): T { another 199 tosses that we do not care about }
E(1): H T { another 198 tosses... }
.
.
E(198): { 198 heads } T H
E(199): { 199 heads } T
E(200): { 200 heads }
Note that P(0) = 0.5, which is P(tails in first toss)
whereas P(1) = 0.25 , i.e., P(heads in first toss and tails in the second)
P(0) = 2**-1
P(1) = 2**-2
.
.
.
P(198) = 2**-199
P(199) = 2**-200
P(200) = 2**-200 #same as P(199)
Which means if you toss a coin 2**200 times, you'd get
E(0) 2**199 times
E(1) 2**198 times
.
.
E(198) 2**1 times
E(199) 2**0 times and
E(200) 2**0 times.
Thus, the expected value reduces to
(0*(2**199) + 1*(2**198) + 2*(2**197) + ... + 198*(2**1) + 199*(2**0) + 200*(2**0))/2**200
This number is virtually equal to 1.
Expected_value = 1 - 2**-200
How I got the difference.
>>> diff = 2**200 - sum([ k*(2**(199-k)) for k in range(200)], 200*(2**0))
>>> diff
1
This can be generalized to n tosses as
f(n) = 1 - 2**(-n)