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Die Toss (Posted on 2003-07-24) Difficulty: 2 of 5
A die is tossed until the sum of the rolls is 15 or greater.

What is the most likely total? What is the the least likely total which can occur?

See The Solution Submitted by Brian Smith    
Rating: 4.3750 (8 votes)

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Solution The probabilities | Comment 5 of 11 |
Let p(n;g) be the probability that n will be the first number at or greater than g.

If 0<=g-n<=5, p(n;g) = p(n;g-1) + p(g-1;g-1)/6, as the event can be achieved if either
1) n is the first number achieved at or higher than g-1, assuring the same for g+1 or
2) The first number achieved at or higher than g-1 was g-1 itself, and then an (n-g) was rolled, the latter with a likelihood of 1/6.

If we start with choosing g = 1, p(6;1) through p(6;1) are all 1/6.

Thereafter we just need to use the recursion formula above. In the below table, the numbers listed start, for each g, at n=g, rather than 1 each time, as that at n=g is the earliest non-zero value:
1 .166667 .166667 .166667 .166667 .166667 .166667
2 .194444 .194444 .194444 .194444 .194444 .027778
3 .226852 .226852 .226852 .226852 .060185 .032407
4 .264660 .264660 .264660 .097994 .070216 .037809
5 .308771 .308771 .142104 .114326 .081919 .044110
6 .360232 .193566 .165788 .133380 .095572 .051462
7 .253604 .225827 .193419 .155611 .111500 .060039
8 .268094 .235687 .197878 .153768 .102306 .042267
9 .280369 .242560 .198450 .146988 .086950 .044682
10 .289288 .245178 .193717 .133678 .091410 .046728
11 .293393 .241931 .181893 .139625 .094943 .048215
12 .290830 .230791 .188524 .143842 .097114 .048899
13 .279263 .236996 .192313 .145585 .097371 .048472
14 .283540 .238857 .192129 .143914 .095016 .046544
15 .286114 .239386 .191171 .142272 .093800 .047257

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So 15 has about a 28.6% probability of being the first number at or above 15 to be reached, while 20 has only a 4.7% chance.

A computer simulation of 1,000,000 trials of this procedure resulted in a verification of the above probabilities, with the results
285502 239261 191102 142391 94015 47729

The probabilities in the columns above do approach a limit as n gets higher. That limit can be determined from the following:

If a is p(n;n), and b is p(n+1;n), and c is p(n+2;n), ..., f is p(n+5;n), (that is, the limit as n increases without bound) then from the above relations:

a=b+a/6
b=c+a/6
c=d+a/6
d=e+a/6
e=f+a/6
f=a/6

from which
b=5a/6; c=4a/6; d=3a/6; e=2a/6 and f=a/6; and of course a=6a/6, so they are in arithmetic progression,and as they add up to 1, they are 6/21, 5/21,...,1/21. This agrees with the common-sense notion that there are 6 ways of getting to n (from n-6+6 to n-1+1), but one fewer for each of the remaining numbers.

  Posted by Charlie on 2003-07-24 12:45:59
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