**Question:** An unbiased die is successively rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a six and a five. Find

- E[X]
- E[X|Y = 1]
- E[X|Y = 5]

**Analytical Solution** Let us solve this part by part

- E[X]
- Intuitively this can be solved in a single step. Since the probability obtaining any of the numbers is independent and equal to
^{1}⁄_{6}this means that the E[X] is simply the reciprocal of^{1}⁄_{6}which is 6 - Formally we can start modelling this as E[X] = E[E[X|T]] where T is the variable on which we wish to condition the variable X on. We can condition X on the outcome of the first roll.
- ⇒ E[X] = E[X|T=6] P{T=6} + E[X|T≠6] P{T≠6}
- Let us examine a few things before getting to the next step
- E[X|T=6] is simply 1 since the first roll is itself 6
- P{T=6} =
^{1}⁄_{6}

- E[X|T≠6] would be E[X]+1 since we have already rolled the dice once and still trying to determine the value of E[X]
- P{T≠6} = 5⁄
_{6}

- E[X] =
^{1}⁄_{6}+ (E[X]+1)^{5}⁄_{6} - ⇒ E[X] = 6

- ⇒ E[X] = E[X|T=6] P{T=6} + E[X|T≠6] P{T≠6}

- Intuitively this can be solved in a single step. Since the probability obtaining any of the numbers is independent and equal to
- E[X|Y=1]
- This is a straightforward item
- E[X|Y=1] = E[X|T≠6] = 1+E[X] = 1+6 = 7

- E[X|Y=5]
- This might look straightforward but there is a rather a twist to it. Let us examine this step by step
- Firstly we know that the 5
^{th}one is the number 5 (OR not 6) - Also, this means that we know partial information about the so called random system
- this knowledge means that our calculation of the expected value
- this knowledge means that the there are only five possibilities in all the rolls except the fifth one

- E[X|Y=5] = 1(
^{1}⁄5) + 2(^{1}⁄5)(^{4}⁄5)^{1}+ 3(^{1}⁄5)(^{4}⁄5)^{2}+ 4(^{1}⁄5)(^{4}⁄5)^{3}+ 6(^{1}⁄6)(^{4}⁄5)^{4}+ 7(^{1}⁄6)(^{5}⁄6)(^{4}⁄5)^{4}+ . . .

**Simulation Solution**

Since the outcomes are small here, I have simulated the sections 1 and 3 (since the section 2 is simple enough to not need any). The code is also provided just below, so you could run the code yourself in your Mathematica Notebook.

**Observations**

- The plots are not of the actual outcomes. Plotting the actual outcomes would mean that we would only see integral bins in the histogram
- Vital things to note of this operation
- There is loss of the data in the frequency (or counts) realm
- There is additional information created that shows how the batches of set sizes are behaving

- Vital things to note of this operation
- Here is what I have done to eliminate the non integral-ity
- I took the list and partitioned it into non overlapping sublists (or batches) of
- 1000 in the first case &
- 100 in the second case

- Then, I calculated the means of these partitioned list and plotted them as histograms

- I took the list and partitioned it into non overlapping sublists (or batches) of

**Code**

I have coded this problem in both Python and Mathematica. I have pasted them as two distinct blocks of code. It should be fairly simple to tell them apart.

from mathematica.random_functions import RandomChoice, RandomInteger _masterList = [] for r in range(0, 1000000): _list = [] _i = 1 while True: if _i == 5: _list.append(5) elif _i < 6 and _i != 5: _rand = RandomChoice([1, 2, 3, 4, 6]) _list.append(_rand) elif _i >= 6: _rand = RandomInteger(1, 6) if _rand != 6: _list.append(_rand) elif _rand == 6: _list.append(_rand) break _i = _i + 1 _masterList.append(_list.index(6) + 1) _file = open('FilePath.txt', 'w') for r in _masterList: _file.write(r.__str__() + '\n')

Histogram[ Parallelize[ Mean[Differences[ Position[RandomInteger[{1, 6}, 1000], 6] // Flatten]] & /@ Range[10000]], {0.02}, "Probability", ImageSize -> 800, Frame -> True, PlotLabel -> "Mean[X]"] Module[{data = Partition[ ToExpression[#] & /@ StringSplit[Import[FileNames["*.txt", NotebookDirectory[]][[1]]], "\n"], 100]}, Histogram[Mean[#] & /@ data, {0.02}, ImageSize -> 800, Frame -> True, PlotLabel -> "Mean[X|Y=5]"]]

End of the post 😉

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