import numpy as np
import pandas as pd
from scipy.stats import uniform, norm, expon

Case 1

Branch banks must keep enough money on hand to satisfy customers' cash demands. An armored truck delivers cash to the bank once a week. The bank manager can choose the amount of weekly cash to have delivered. Running out of cash during the week is terrible customer service and the manager wants to avoid this. On the other hand, keeping excessive cash reserves costs the bank profits, since cash is a non-interest earning asset.

The daily demand for cash at this particular bank follows a normal distribution with daily means and std dev summarized in Table 1.

Table 1:

Info	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday	Sunday
Mean (\$1,000s)	175	120	90	60	120	140	65
Std_Dev (\$1,000s)	26	18	13	9	18	21	9

(a)

daily = {'Monday':{'Mean': 175, "Std": 26},
        'Tuesday':{'Mean': 120, "Std": 18},
        'Wednesday':{'Mean': 90, "Std": 13},
        'Thursday':{'Mean': 60, "Std": 9},
        'Friday':{'Mean': 120, "Std": 18},
        'Saturday':{'Mean': 140, "Std": 21},
        'Sunday':{'Mean': 65, "Std": 9}}


df = pd.DataFrame(daily)*1000

begin_cash = 825000
num_trails = 10000

daily_model = pd.DataFrame(columns = df.columns)
daily_model.insert(0,'Begin', '')
daily_model

for t in np.arange(1, num_trails + 1):
    cash_daily = [begin_cash]
    for y in np.arange(1,8):
        cash_daily.append(cash_daily[y-1] - norm.rvs(size = 1, loc = df.iloc[0,y-1], scale = df.iloc[1,y-1]).item())
    
    daily_model.loc[t] = cash_daily       
daily_model
        

	Begin	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday	Sunday
1	825000.0	638283.822689	532629.681394	466886.889368	408371.427039	330003.263185	131500.216712	85732.465277
2	825000.0	642045.994945	518941.574362	433351.965881	372088.082190	254644.549847	156767.601934	80021.096505
3	825000.0	664396.822943	525568.860157	410482.504825	345482.576462	250116.865890	74486.697389	13554.425454
4	825000.0	667781.355873	541694.800709	432606.514755	389341.892504	264704.567488	129633.801852	61024.033772
5	825000.0	657851.011790	535459.416010	452212.940897	386454.686193	291998.088771	172826.523502	111994.092500
...	...	...	...	...	...	...	...	...
9996	825000.0	679413.017089	569128.712065	489701.491199	422660.485143	305947.813465	146174.916163	88952.627227
9997	825000.0	626398.869276	529794.402867	436692.110923	376047.019569	261174.831393	144621.647562	86339.812220
9998	825000.0	696906.348611	571880.761870	459431.469984	386992.446774	306066.601370	143958.405453	92430.473279
9999	825000.0	596569.292631	475361.112516	371748.622718	324638.227006	231945.619725	93073.413837	28440.204460
10000	825000.0	638110.967066	533546.745759	436740.893838	384710.601626	282811.541457	133515.920395	75311.538330

10000 rows × 8 columns

running_out_prob = pd.DataFrame(columns = ['Prob_of_running_out'], index = df.columns)
running_out_prob

	Prob_of_running_out
Monday	NaN
Tuesday	NaN
Wednesday	NaN
Thursday	NaN
Friday	NaN
Saturday	NaN
Sunday	NaN

for k in running_out_prob.index:
    running_out_prob.loc[k,'Prob_of_running_out'] = sum(daily_model[k] < 0)/len(daily_model[k])

running_out_prob

	Prob_of_running_out
Monday	0.0
Tuesday	0.0
Wednesday	0.0
Thursday	0.0
Friday	0.0
Saturday	0.0039
Sunday	0.1152

sum(running_out_prob['Prob_of_running_out'])

0.1191

The total probability of the bank running out of money is around 11.91%.

(b)

Trying some big numbers

• begin_cash = 860,000

begin_cash = 860000
num_trails = 10000

daily_model = pd.DataFrame(columns = df.columns)
daily_model.insert(0,'Begin', '')
for t in np.arange(1, num_trails + 1):
    cash_daily = [begin_cash]
    for y in np.arange(1,8):
        cash_daily.append(cash_daily[y-1] - norm.rvs(size = 1, loc = df.iloc[0,y-1], scale = df.iloc[1,y-1]).item())
    
    daily_model.loc[t] = cash_daily
    
running_out_prob = pd.DataFrame(columns = ['Prob_of_running_out'], index = df.columns)
for k in running_out_prob.index:
    running_out_prob.loc[k,'Prob_of_running_out'] = sum(daily_model[k] < 0)/len(daily_model[k])

running_out_prob

	Prob_of_running_out
Monday	0.0
Tuesday	0.0
Wednesday	0.0
Thursday	0.0
Friday	0.0
Saturday	0.0002
Sunday	0.0265

sum(running_out_prob['Prob_of_running_out'])

0.026699999999999998

The sum of probability of running out is around 2.67% in 10,000 trails.

• begin_cash = 865,000

begin_cash = 865000
num_trails = 10000

daily_model = pd.DataFrame(columns = df.columns)
daily_model.insert(0,'Begin', '')
for t in np.arange(1, num_trails + 1):
    cash_daily = [begin_cash]
    for y in np.arange(1,8):
        cash_daily.append(cash_daily[y-1] - norm.rvs(size = 1, loc = df.iloc[0,y-1], scale = df.iloc[1,y-1]).item())
    
    daily_model.loc[t] = cash_daily
    
running_out_prob = pd.DataFrame(columns = ['Prob_of_running_out'], index = df.columns)
for k in running_out_prob.index:
    running_out_prob.loc[k,'Prob_of_running_out'] = sum(daily_model[k] < 0)/len(daily_model[k])

running_out_prob

	Prob_of_running_out
Monday	0.0
Tuesday	0.0
Wednesday	0.0
Thursday	0.0
Friday	0.0
Saturday	0.0005
Sunday	0.0212

sum(running_out_prob['Prob_of_running_out'])

0.0217

The sum of probability of running out is around 2.17% in 10,000 trails.

• begin_cash = 870,000

begin_cash = 870000
num_trails = 10000

daily_model = pd.DataFrame(columns = df.columns)
daily_model.insert(0,'Begin', '')
for t in np.arange(1, num_trails + 1):
    cash_daily = [begin_cash]
    for y in np.arange(1,8):
        cash_daily.append(cash_daily[y-1] - norm.rvs(size = 1, loc = df.iloc[0,y-1], scale = df.iloc[1,y-1]).item())
    
    daily_model.loc[t] = cash_daily
    
running_out_prob = pd.DataFrame(columns = ['Prob_of_running_out'], index = df.columns)
for k in running_out_prob.index:
    running_out_prob.loc[k,'Prob_of_running_out'] = sum(daily_model[k] < 0)/len(daily_model[k])

sum(running_out_prob['Prob_of_running_out'])

0.012799999999999999

The sum of probability of running out is around 1.28% in 10,000 trails. Therefore, the range of amount of money needed at the start of the week to ensure that there is at most a 2.0% probability of running out of money is in (865,000,870,000). Let's find a more accurate number!!!

begin_cash = 865000
num_trails = 10000


begin_cash_range = np.arange(865000,870000,500)
running_out_prob_b = pd.DataFrame(columns = ['Prob_of_running_out'], index = begin_cash_range)

for i in begin_cash_range:
    daily_model = pd.DataFrame(columns = df.columns)
    daily_model.insert(0,'Begin', '')
    for t in np.arange(1, num_trails + 1):
        cash_daily = [begin_cash]
        for y in np.arange(1,8):
            cash_daily.append(cash_daily[y-1] - norm.rvs(size = 1, loc = df.iloc[0,y-1], scale = df.iloc[1,y-1]).item())
    
        daily_model.loc[t] = cash_daily
    
    running_out_prob = pd.DataFrame(columns = ['Prob_of_running_out'], index = df.columns)
    for k in running_out_prob.index:
        running_out_prob.loc[k,'Prob_of_running_out'] = sum(daily_model[k] < 0)/len(daily_model[k])
    running_out_prob_b.loc[i,'Prob_of_running_out'] = sum(running_out_prob['Prob_of_running_out'])
    
running_out_prob_b

	Prob_of_running_out
865000	0.0189
865500	0.0206
866000	0.0184
866500	0.0198
867000	0.0202
867500	0.0221
868000	0.0198
868500	0.0188
869000	0.0175
869500	0.0198

The amount of money needed at the start of the week to ensure that there is at most a 2.0% probability of running out of money is around 867,000.

Case 2

Winning the pass bet in the dice game Craps follows the following rules:

• Player rolls the two dice (the “come out” roll).
• If the first roll (total of both dice) is a 7 or 11, the pass bet wins.
• If the first roll is a 2, 3 or 12, the pass bet loses.
• If any other total is rolled, the player rolls again repeated until one of the following…
  • If the player rolls the same total as the first (come out) roll, the pass bet wins.
  • If the player rolls a 7, the pass bet loses.

import random

a.

def diceRoll(num_trials):
    min = 1
    max = 6
    trails_num_dolls = []
    trails_results = []
    
    for i in range(num_trials):
        first_roll_1 = random.randint(min, max)
        first_roll_2 = random.randint(min, max)
        first_roll_sum = first_roll_1 + first_roll_2
        num_dolls = 1
        if first_roll_sum == 7 or first_roll_sum == 11:
            result = 'Win'
        else:
            if first_roll_sum == 2 or first_roll_sum == 3 or first_roll_sum == 12:
                result = 'Lose'
            else:
                while num_dolls >= 1:
                    new_roll_1 = random.randint(min, max)
                    new_roll_2 = random.randint(min, max)
                    new_roll_sum = new_roll_1 + new_roll_2
                    num_dolls += 1
                    if new_roll_sum == 7:
                        result = 'Lose'
                        break
                    else:
                        if new_roll_sum == first_roll_sum:
                            result = 'Win'""
                            break
        trails_num_dolls.append(num_dolls)
        trails_results.append(result)
    return trails_num_dolls, trails_results

trails_num_dolls, trails_results = diceRoll(1)
print('number of dice rolls = ', trails_num_dolls, 'result = ', trails_results)

number of dice rolls =  [8] result =  ['Lose']

b.

sim = diceRoll(2000)
avg_num_dice_roll = np.average(sim[0])
avg_num_dice_roll

3.295

avg_winnings = sum(map(lambda x : x == 'Win', sim[1]))/len(sim[1])
avg_winnings

0.4775

The number of trials is 2000, we get the average number of dice rolls in a Craps game is 3.2455 and the estimated probability of winning is 0.4835.