Very difficult problem. I would greatly appreciate more help. What is the expected value of (1+1)^4?

Question

We usually think of addition as an operation applied to the rational numbers or the real numbers. Now suppose we defined addition another way, using a classic model organism: the nematode. To compute the sum of x and y, you combine groups of x and y nematodes and leave them for 24 hours. When you come back, you count up how many you have — and that’s the sum.

It turns out that, over the course of 24 hours, the nematodes pair up, and each pair has one offspring 50 percent of the time. (If you have an odd number of nematodes, they will still pair up, but one will be left out.) So if you want to compute 1+1, half the time you’ll get 2 and half the time you’ll get 3. If you compute 2+2, 25 percent of the time you get 4, 50 percent of the time you’ll get 5, and 25 percent of the time you’ll get 6.

While we’re at it, let’s define exponentiation for sums of nematodes. Raising a sum to a power means leaving that sum of nematodes for the number of days specified by the exponent. With this number system, what is the expected value of (1+1)⁴?

As N gets larger and larger, what does the expected value of (1+1)^N approach?

Answer 1

possible outcomes expected value

(1+1)^1 2 or 3 2*0.5 + 3*0.5 = 1+1.5 = 2.5

(1+1)^2 2+2=4 or 4 * 0.25 + 5*0.5 + 06*0.25 =

2+3=5 or 1+2.5+1.5 = 5

3+2=5 or

3+3=6

(1+1)^3 2+2+2 = 6 or 6 * 1/8 + 7*3/8 + 8*3/8 + 9*1/8 =

2+2+3 = 7 6/8 + 21/8 + 24/8 + 9/8 =60/8 = 30/4 =

2+3+2 = 7 15/2 = 7.5

2+3+3 = 8

3+2+2 = 7

3+2+3 = 8

3+3+2 = 8

3+3+3 = 9

(1+1)^4 2+2+2+2 = 8

2+2+2+3 = 9

2+2+3+2 = 9 8* 1/16 + 9* 4/16 + 10 * 6/16 + 11 * 4/16 + 12 * 1/16 =

2+2+3+3 = 10 8/16 + 36/16 + 60/16 + 44/16 + 12/16 = 160/16 = 10

2+3+2+2 = 9

2+3+2+3 = 10

2+3+3+2 = 10

2+3+3+3 = 11

3+2+2+2 = 9

3+2+2+3= 10

3+2+3+2 = 10

3+2+3+3 = 11

3+3+2+2 = 10

3+3+2+3 = 11

3+3+3+2 = 11

3+3+3+3 = 12

(1+1)^5 there are 32 outcomes 10 * 1/32 + 11*6/32 + 12* 9/32 + 13* 9/32+ 14 * 6/32+ 15 * 1/32=

ranging from 10 to 15 (10 +66+108 + 117 + 84 + 15)/32 = 400/32 = 50/4 = 25/2 = 12.5

So the expected value is the MEDIAN/Mean of the smallest outcome and the largest outcome.

As N grows infinitely large, the distribution becomes Gaussian Normal and the expected

value shall approach the mean=median.

That boils down to finding the average of 2N and 3N , which is (5/2)N

Proceeding by induction

Suppose that after N days, the expected value is (5/2)N...

Waiting an extra day (N+1), then the smallest outcome is 2(N+1) and hte largest outcome is 3(n+1)

the average is [2(n+1)+3(n+1)]/2 = (5n+5)/2 = 5(n+1)/2 = (5/2)(N+1)