In: Math
A group of 20 people have a Russian roulette party. Each person at the party plays (pulls the trigger) three times.
In between the times they re-spin the barrel. What is a box model for the fraction of people who survive the party? What is the expected value and standard error?
If they do not re-spin the barrel, what is a box model for the fraction of survivors? What is the expected value and standard error?
The odds of dying in one trigger is: a one in six chance.
Probability of surviving 3 triggers = (5/6)^3 = 0.5787
Probability of dying = 0.4213
Now, making a probability distribution for number of people surviving with re-spin of barrel:
People suviving | Probability |
1 | 8.5242E-07 |
2 | 1.1124E-05 |
3 | 9.1678E-05 |
4 | 0.00053521 |
5 | 0.00235255 |
6 | 0.00807883 |
7 | 0.02219458 |
8 | 0.04954147 |
9 | 0.0907353 |
10 | 0.13710004 |
11 | 0.17120385 |
12 | 0.17637759 |
13 | 0.14909348 |
14 | 0.10239937 |
15 | 0.05626339 |
16 | 0.02415152 |
17 | 0.00780592 |
18 | 0.00178707 |
19 | 0.0002584 |
20 | 1.7747E-05 |
formula used in excel:
Expected value is sumproduct of above two columns = 11.57
A | B | ||
People suviving | Probability | A*B | A^2*B |
1 | 8.5242E-07 | 8.52416E-07 | 8.52416E-07 |
2 | 1.1124E-05 | 2.22471E-05 | 4.44942E-05 |
3 | 9.1678E-05 | 0.000275033 | 0.000825099 |
4 | 0.00053521 | 0.002140825 | 0.008563298 |
5 | 0.00235255 | 0.011762772 | 0.058813862 |
6 | 0.00807883 | 0.048472963 | 0.290837777 |
7 | 0.02219458 | 0.15536206 | 1.087534423 |
8 | 0.04954147 | 0.396331787 | 3.170654295 |
9 | 0.0907353 | 0.816617693 | 7.349559233 |
10 | 0.13710004 | 1.3710004 | 13.710004 |
11 | 0.17120385 | 1.883242307 | 20.71566538 |
12 | 0.17637759 | 2.116531065 | 25.39837278 |
13 | 0.14909348 | 1.938215261 | 25.19679839 |
14 | 0.10239937 | 1.433591169 | 20.07027637 |
15 | 0.05626339 | 0.843950845 | 12.65926268 |
16 | 0.02415152 | 0.38642438 | 6.182790075 |
17 | 0.00780592 | 0.13270068 | 2.255911557 |
18 | 0.00178707 | 0.032167262 | 0.579010723 |
19 | 0.0002584 | 0.004909533 | 0.093281133 |
20 | 1.7747E-05 | 0.00035494 | 0.007098805 |
11.57407407 | 138.8353052 | ||
Var | 4.87611454 | ||
SD | 2.208192596 |
Standard error = 2.21
Now, if we don't respin the barrel
Probability of surviving 3 triggers = 5/6*4/5*3/4 = 0.5
Now, updated excel calculations:
A | B | ||||
People suviving | Probability | A*B | A^2*B | ||
1 | 1.9073E-05 | 1.90735E-05 | 1.90735E-05 | Survive | 0.5 |
2 | 0.0001812 | 0.000362396 | 0.000724792 | Die | 0.5 |
3 | 0.00108719 | 0.003261566 | 0.009784698 | ||
4 | 0.00462055 | 0.018482208 | 0.073928833 | ||
5 | 0.01478577 | 0.073928833 | 0.369644165 | ||
6 | 0.03696442 | 0.221786499 | 1.330718994 | ||
7 | 0.07392883 | 0.517501831 | 3.622512817 | ||
8 | 0.12013435 | 0.961074829 | 7.688598633 | ||
9 | 0.16017914 | 1.441612244 | 12.97451019 | ||
10 | 0.17619705 | 1.76197052 | 17.6197052 | ||
11 | 0.16017914 | 1.76197052 | 19.38167572 | ||
12 | 0.12013435 | 1.441612244 | 17.29934692 | ||
13 | 0.07392883 | 0.961074829 | 12.49397278 | ||
14 | 0.03696442 | 0.517501831 | 7.245025635 | ||
15 | 0.01478577 | 0.221786499 | 3.326797485 | ||
16 | 0.00462055 | 0.073928833 | 1.182861328 | ||
17 | 0.00108719 | 0.018482208 | 0.31419754 | ||
18 | 0.0001812 | 0.003261566 | 0.058708191 | ||
19 | 1.9073E-05 | 0.000362396 | 0.006885529 | ||
20 | 9.5367E-07 | 1.90735E-05 | 0.00038147 | ||
10 | 105 | ||||
Var | 5 | ||||
SD | 2.236067977 |
Expected value is sumproduct of above two columns = 10
Standard error = 2.236