All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Adding to rolls (Posted on 2014-09-15)
Create an infinite sequence of numbers by the following method:

Roll a 6-sided die and record the number.
Roll again, add 1 and record the number.
Roll again, add 2 and record the number.
and so on...

Question:
What proportion of numbers in the sequence, beyond the first 5, are larger than any previous number?

 No Solution Yet Submitted by Jer No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution Comment 1 of 1
Any six consecutive rolls are k+r1, k+1+r2, k+2+r3, k+3+r4, k+4+r5, k+5+r6, where k is the first of the "add" numbers at the point where you're looking and r1 through r6 are the six rolls in the portion we're looking at. The last of these, k+5 + r6 is the one we're testing; it must, with probability 1, be larger than k-1+r0 as the largest this latter can be is k+5.

So we need look only at six consecutive rolls.

If the last is 6 (probability 1/6) then it is certain to be the highest.
If the last is 5 (same probability) it is 5/6 probable of being the highest.
If the last is 4 it is 4/6 probable of being higher than the one before it and 5/6 of being higher than the one before that, for an overall probability of 20/36 = 5/9
If the last is 3 the probability is 3/6 * 4/6 * 5/6 = 5/18
If the last is 2 the probability is 2/6 * 3/6 * 4/6 * 5/6 = 5/54
If the last is 1 the probability is 1/6 * 2/6 * 3/6 * 4/6 * 5/6 = 5/324

The weighted average is thus 899/1944 ~= 0.4624855967.

 Posted by Charlie on 2014-09-15 18:05:27

 Search: Search body:
Forums (0)