A group of 20 people have a Russian roulette party. Each person at the party plays...

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?

Solutions

Expert Solution

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

milcah answered 2 days ago

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