```Question 212403
One examiner has a pile of 20 CVs to analyze and one coin. If he gets heads then he moves on to the next CV in line (ex: if on the 1st toss he gets heads then he moves on to CV no. 2). Otherwise if he gets tails then he moves on to the previous CV (ex: if on the 1st toss he gets tails then he moves on to CV no. 20). Knowing that your CV is no. 15 then what is the probability of being on your CV on the 40th toss?
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The questions as best I can translate, is that you want to know the probability that the examiner will be on CV number 15 on the 40th toss.
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It doesn't matter if he reached CV number 15 on any other toss.  You want to know the probability of him being on CV number 15 on the 40th toss only.
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That's what I think is being asked and that's what I'll try to solve.
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In order for him to land on CV number 15, he has to net 14 heads or he has to net 6 tails.
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Net heads means the number of heads has to be greater than the number of tails by 14.
Net tails means the number of tails has to be greater than the number of heads by 6.
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I'm not really sure there's a simple way to do this because I don't know the magic formula, so I'll do it the hard way.
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The possible combinations and the number of heads and tails he gets and the number of net heads this generates and the number he lands on are as follows:
40/0/40/1
39/1/38/19
38/2/36/17
37/3/34/15
36/4/32/13
35/5/30/11
34/6/28/9
33/7/26/7
32/8/24/5
31/9/22/3
30/10/20/1
29/11/18/19
28/12/16/17
27/13/14/15
26/14/12/13
25/15/10/11
24/16/8/9
23/17/6/7
22/18/4/5
21/19/2/3
20/20/0/1
19/21/-2/19
18/22/-4/17
17/23/-6/15
16/24/-8/13
15/25/-10/11
14/26/-12/9
13/27/-14/7
12/28/-16/5
11/29/-18/3
10/30/-20/1
9/31/-22/19
8/32/-24/17
7/33/-26/15
6/34/-28/13
5/35/-30/11
4/36/-32/9
3/37/-34/7
2/38/-36/5
1/39/-38/3
0/40/-40/1
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We have 40 tosses.
The possible number of combinations we can get on 40 tosses are 41.
15 can come out 4 times out of the 41 possible combinations so the probability of getting a 15 would be 4/41.
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The patterns are different with different number of CV and number of tosses.
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I tested with 4 numbers and 8 tosses.
The probability of landing on 1,2,3,4 was totally different.
the probability of landing on 2 or 4 was zero.
the probability of landing on 1 or 3 was .5 each.
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I couldn't find a pattern other than the net heads or tails was raised or reduced by 2.  8h = net 8 heads, 7h1t = net 6 heads, etc.
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Could I have derived a formula from the net heads or tails required and somehow translated that to the numbers that would be derived.  Probably, but I was unable to do so.
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You will notice that the probability of getting an even numbered CV on exactly 40 tosses was 0.
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All the CV numbers landed on were odd only.
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I looked at when I landed on 15 and found that I landed on 15 when the net heads were 14 and 34 and -6 and -26.
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Heads of -6 and -26 corresponded to tails of 6 and 26.
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If I had known this when I started, I could have analyzed it as follows:
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I can land on heads when the net heads is 14 or when the net tails is 6.
If I add 20 (number of numbers in the set), then I can land on 14 or 34 with heads, and I can land on 6 or 26 with tails.  Since the number of tosses is 40 I can't go beyond that.  This means I can land on 15 4 times.  With 40 tosses the number of possible combinations is 41.  The probability has to be 4 / 41.
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Easy to do once you go through the grunt work of figuring it all out, but a lot harder to do if you are exposed to it for the first time and have no clue as to what formula to use.
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Interesting problem.
I'll be even more interested to see if I got it right, or even came close to being on the right track.
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