DATA 605 Week05 Discussion

Discrete Probability Distributions (Chapter 1 - Section 1.1) - Question 5

Solution:

There are 3 dice so number of possible outcomes is \(6^{3}\). Out of these possible outcomes, only 1 outcome (three sixes) is in favour of the bet. So the probability of getting at least one time three sixes is equal to the probability 1 minus the probability of not getting three sixes in n trials. Therefore, the required probabillty will be

\(p = f(n)\)

\(p = 1-\left ( \frac{6^{3}-1}{6^{3}} \right )^{n}\)

For any favourable bet, probability of success should be equal to 0.5 or higher. The following code shows the probability for different n from 1 to 160. The code breaks out of the loop as soon as the probability reaches 0.50 or higher:

for (n in 1:160) {
  p = 1-((6^3-1)/6^3)^n
  message(sprintf("%s  %s ", n, round(p, 4)))
  
  if(p >= 0.50) {
    print (n)
    break
  }
}

## 1  0.0046

## 2  0.0092

## 3  0.0138

## 4  0.0184

## 5  0.0229

## 6  0.0275

## 7  0.032

## 8  0.0364

## 9  0.0409

## 10  0.0453

## 11  0.0498

## 12  0.0542

## 13  0.0585

## 14  0.0629

## 15  0.0672

## 16  0.0716

## 17  0.0759

## 18  0.0801

## 19  0.0844

## 20  0.0886

## 21  0.0929

## 22  0.0971

## 23  0.1012

## 24  0.1054

## 25  0.1095

## 26  0.1137

## 27  0.1178

## 28  0.1218

## 29  0.1259

## 30  0.13

## 31  0.134

## 32  0.138

## 33  0.142

## 34  0.146

## 35  0.1499

## 36  0.1538

## 37  0.1578

## 38  0.1617

## 39  0.1655

## 40  0.1694

## 41  0.1733

## 42  0.1771

## 43  0.1809

## 44  0.1847

## 45  0.1885

## 46  0.1922

## 47  0.196

## 48  0.1997

## 49  0.2034

## 50  0.2071

## 51  0.2107

## 52  0.2144

## 53  0.218

## 54  0.2217

## 55  0.2253

## 56  0.2288

## 57  0.2324

## 58  0.236

## 59  0.2395

## 60  0.243

## 61  0.2465

## 62  0.25

## 63  0.2535

## 64  0.2569

## 65  0.2604

## 66  0.2638

## 67  0.2672

## 68  0.2706

## 69  0.274

## 70  0.2773

## 71  0.2807

## 72  0.284

## 73  0.2873

## 74  0.2906

## 75  0.2939

## 76  0.2972

## 77  0.3004

## 78  0.3037

## 79  0.3069

## 80  0.3101

## 81  0.3133

## 82  0.3165

## 83  0.3197

## 84  0.3228

## 85  0.3259

## 86  0.3291

## 87  0.3322

## 88  0.3353

## 89  0.3383

## 90  0.3414

## 91  0.3444

## 92  0.3475

## 93  0.3505

## 94  0.3535

## 95  0.3565

## 96  0.3595

## 97  0.3624

## 98  0.3654

## 99  0.3683

## 100  0.3713

## 101  0.3742

## 102  0.3771

## 103  0.38

## 104  0.3828

## 105  0.3857

## 106  0.3885

## 107  0.3914

## 108  0.3942

## 109  0.397

## 110  0.3998

## 111  0.4025

## 112  0.4053

## 113  0.4081

## 114  0.4108

## 115  0.4135

## 116  0.4163

## 117  0.419

## 118  0.4216

## 119  0.4243

## 120  0.427

## 121  0.4296

## 122  0.4323

## 123  0.4349

## 124  0.4375

## 125  0.4401

## 126  0.4427

## 127  0.4453

## 128  0.4479

## 129  0.4504

## 130  0.453

## 131  0.4555

## 132  0.458

## 133  0.4605

## 134  0.463

## 135  0.4655

## 136  0.468

## 137  0.4705

## 138  0.4729

## 139  0.4753

## 140  0.4778

## 141  0.4802

## 142  0.4826

## 143  0.485

## 144  0.4874

## 145  0.4898

## 146  0.4921

## 147  0.4945

## 148  0.4968

## 149  0.4991

## 150  0.5015

## [1] 150

From the above results we can say that the smallest value of n necessary for a favorable bet that a triple-six will occur is 150.

n_DeMoivre = 216*log(2)
print (n_DeMoivre)

## [1] 149.7198

I would agree with DeMoivre that we can use \(216*log(2)\) to determine the n necessary for a favorable bet.