Question of luck!

Probability Level 1

You roll two dices. Find the expected value of the sum you'll get.

Bonus: Can you Generalise it to n number of dice ?

Difficulty: $\dagger \color{grey}{}\dagger \dagger \dagger \color{grey}{\dagger}$

9 11 12 8 5 7 6 3

5 solutions

Mahdi Raza
Jun 30, 2020

[Bonus]

Average face is:

$\dfrac{1}{6} \times (1 + 2 + 3 + 4 + 5 + 6) = 3.5$

This is rolled $n$ times to give:

$\boxed{3.5 \times n}$

We can create square-grids in 2D (for two rolls), cube-grids in 3D (for three rolls) and so on... But to visualize $n$ dice-rolls it gets cumbersome and messy to see this beyond 2D.

Excellent work @Mahdi Raza . Thanku for sharing your solution.

Aryan Sanghi - 11 months, 2 weeks ago

Thanks, glad you like it! Quite simply actually!

Mahdi Raza - 11 months, 2 weeks ago

As I said it's an easy problem.

My solution was Expected value = ((lowest value) + (highest value))/2

Expected value = (n + 6n)/2

Expected value = 3.5n

But yours is an elegant one. :)

Aryan Sanghi - 11 months, 2 weeks ago

@Mahdi Raza Even I thought similarly by thinking of the average face, but as you have written the solution, I don't want to write again. Also, a pattern arises if you plot the sample spaces. @Aryan Sanghi great question!

Siddharth Chakravarty - 11 months, 2 weeks ago

Thanku @Siddharth Chakravarty . :)

Aryan Sanghi - 11 months, 2 weeks ago

You are right! I have a lot of die and time, so I tested it with $1,2,3\cdots100$ dice. The avarage sums with 10,000,000 rolls are:

1	3.5000013
2	6.9989863
3	10.4998481
4	13.9971883
5	17.5013234
6	20.9986741
7	24.4981529
8	27.9999454
9	31.5005634
10	34.9972343
11	38.5002841
12	42.16247
13	45.4980172
14	48.9981100
15	52.4980437
16	55.9994406
17	59.4982269
18	62.9968936
19	66.4947381
20	69.9968198
21	73.4973500
22	76.9961685
23	80.4979282
24	84.19010
25	87.4952486
26	90.9986911
27	94.4976375
28	97.9919740
29	101.4956757
30	104.9940632
31	108.4983679
32	111.9977235
33	115.4972338
34	118.9973354
35	122.4920760
36	125.9942138
37	129.4956380
38	132.9968175
39	136.4984199
40	139.9950933
41	143.4914077
42	146.9943447
43	150.4989645
44	153.9929684
45	157.4961524
46	160.9889031
47	164.4948546
48	167.9979304
49	171.4962770
50	174.9896285
51	178.4913413
52	181.9985439
53	185.4959921
54	188.9874608
55	192.4917256
56	195.9977068
57	199.4945960
58	202.9884386
59	206.4911641
60	209.9979973
61	213.4920127
62	216.9874371
63	220.4982939
64	223.9921324
65	227.4882880
66	230.9920185
67	234.4941657
68	237.9898166
69	241.4896067
70	244.9936467
71	248.4904991
72	251.9895294
73	255.4940553
74	258.9898544
75	262.4895043
76	265.9958375
77	269.4868072
78	272.9924676
79	276.4925781
80	279.9838065
81	283.4940814
82	286.9893526
83	290.4871943
84	293.9948063
85	297.4856383
86	300.9929921
87	304.4891023
88	307.9846959
89	311.4956685
90	314.9830647
91	318.4934824
92	321.9854683
93	325.4913286
94	328.9897604
95	332.4826614
96	335.9931466
97	339.4903680
98	342.9859449
99	346.4901426
100	349.9809587

A Former Brilliant Member - 11 months, 2 weeks ago

Excellent job @Páll Márton

Aryan Sanghi - 11 months, 2 weeks ago

Did u code this? I mean how did you get decimal values. The formula 3.5n suggests that the sum is close to it.

Siddharth Chakravarty - 11 months, 2 weeks ago

For you and @Aryan Sanghi :

#include <iostream>
#include <cstdlib>
#include <ctime>
#include <fstream>
using namespace std;

int main()
{
    //save to the bin.txt file
    ofstream fout("bin.txt");
    srand(time(0));
    for(int n=1;n<101;n++){
        long long int sum=0;
        for(long long int number=0;number<10000000;number++){
            for(int x=0;x<n;x++)
                sum+=rand()%6+1;
        }
        fout << n << "\t" << int(sum/10000000) << "." << int(sum%10000000)<<endl;
    }
    fout.close();
    return 0;
}

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – Excellent code. But, why don't you use <bits/stdc++.h>, you won't need to include so many files then. Just one file and it's done.

Aryan Sanghi - 11 months, 2 weeks ago

@Aryan Sanghi – I hate the .h headers :) But sometimes they are necessary.

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – Why? Did something wrong happen with them?

Aryan Sanghi - 11 months, 2 weeks ago

@Aryan Sanghi – No. Just .h is seems outdated.

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – But, after sometime, you'll need them desperately like when you design your own games/apps. You can't write a long code in one file, so you'll have to write your code in multiple files and have to save them with .h extension and have to include them. Also, many important header files are .h like <graphics.h>.

So, I think you should start liking them. Just a suggestion.

Aryan Sanghi - 11 months, 2 weeks ago

@Aryan Sanghi – And the C++ is outdated too :)

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – Actually, it's the language which is used mos after python. It is the one which is used to write core code of 90% files. So, you say it to be outdated.

Aryan Sanghi - 11 months, 2 weeks ago

@Aryan Sanghi – I wrote a code, but I used brute-force, so it was a very slow code, so I had to use % and pow, but we can't in one argument use both!!! So I wrote a 200-lines code to get the product of two vectors, and another code to get the % of two vectors. But the code was too long...(memory and time limit)

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – Actually I'll send you a link. It evaluates power in $log n$ time complexity. That is why I say you to start Competitive programming, it teaches how to write efficient codes.

Considering time and memory, I'll try to write a efficient code when as soon as I get time and will post the code.

Aryan Sanghi - 11 months, 2 weeks ago

@A Former Brilliant Member – Here is the link to power in $log n$ time complexity.

Aryan Sanghi - 11 months, 2 weeks ago

@Aryan Sanghi – Ok, but what will be if x=157986453215668516535400653515305403378306835468873210? :) I used big numbers, so I should use vectors.

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – No, you could convert it to string and then use a variant of the link

Aryan Sanghi - 11 months, 2 weeks ago

@Aryan Sanghi – What about $x>10^{500}$ ?

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – Still the same. It is valid till 10 $^{1000000}$

Aryan Sanghi - 11 months, 2 weeks ago

@Aryan Sanghi – Oh! But we can do with a string only those things, what we can do with a vector. And in a vector we can use integers.

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – Yes, I know about vectors

Use string data type, it's a vector of characters

Aryan Sanghi - 11 months, 2 weeks ago

@A Former Brilliant Member – There was a little typo in the 12th line :) for(int n=100;n<101;n++){ instead of for(int n=1;n<101;n++){

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – I am learning to code, but what you have done is great! Hats off.

Siddharth Chakravarty - 11 months, 2 weeks ago

A C++ code?

Aryan Sanghi - 11 months, 2 weeks ago

Aryan Sanghi
Jun 30, 2020

Here is a table of outcomes

Sum	2	3	4	5	6	$\color{#3D99F6}{7}$	8	9	10	11	12
Frequency	1	2	3	4	5	$\color{#3D99F6}{6}$	5	4	3	2	1

Short answer As 7 is the symmetrical term, it is the answer.

Average face is:

$\dfrac{1}{6} \times (1 + 2 + 3 + 4 + 5 + 6) = 3.5$

This is rolled $n$ times to give:

$\boxed{3.5 \times n}$

Mahdi Raza - 11 months, 2 weeks ago

Did you try my question Unexpected Probability problem!! 2

Aryan Sanghi - 11 months, 2 weeks ago

Not yet, will do !

Mahdi Raza - 11 months, 2 weeks ago

Yeetsquad 747
Jul 1, 2020

Here is a chart with all of the possible numbers you can roll. As you can see, 7 is the most likely to happen.

Actually, expected value is not the most likely number. It's the average of all values that you will get if you roll the dice infinitely.

Aryan Sanghi - 11 months, 2 weeks ago

That is intresting

Yeetsquad 747 - 11 months, 2 weeks ago

Lucas Tsui
Oct 23, 2020

If a is the lowest number you can get on a die, b is the highest, and c is the number or dice, the equation is (a+b/2)*c.

Lâm Lê
Jul 15, 2020

'Dices' is not a word. It is supposed to be 'dice'.

Question of luck!

Difficulty: † † † † † \dagger \color{grey}{}\dagger \dagger \dagger \color{grey}{\dagger} † † † † †

5 solutions

[Bonus]

0 pending reports

Difficulty: $\dagger \color{grey}{}\dagger \dagger \dagger \color{grey}{\dagger}$