Guest Posted April 14, 2011 Report Share Posted April 14, 2011 I saw this puzzle the other day and after spending all of English class solving it, thought it was so cool that I had to post it... good luck Starting at number zero, you throw a coin. Tails means you go down one (therefore equaling -1) Heads means you go up one (therefore equaling 1) You repeat this procedure infinite times, so here's an example variation: 0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 3, 4, 5, 6, 7, 6, 7......... Question is: how many times on average would 0 have been 'hit' if you threw the coin N times? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2011 Report Share Posted April 14, 2011 I saw this puzzle the other day and after spending all of English class solving it, thought it was so cool that I had to post it... good luck Starting at number zero, you throw a coin. Tails means you go down one (therefore equaling -1) Heads means you go up one (therefore equaling 1) You repeat this procedure infinite times, so here's an example variation: 0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 3, 4, 5, 6, 7, 6, 7......... Question is: how many times on average would 0 have been 'hit' if you threw the coin N times? 1/2N but it can't be that simple Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 14, 2011 Report Share Posted April 14, 2011 (edited) 1/2N but it can't be that simple the answer would be 1/2N example. 0TH=0TH=0TH=0TH=0TH=0-1+1=0 therefore on average every two times the coin is flipped it would result in 0. Edited April 14, 2011 by JDM13 Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 14, 2011 Report Share Posted April 14, 2011 (edited) N/2 (rounded down) would be the max it could be since odd tosses can not equal 0. Just because there is a 50-50 chance in a toss doesn't mean there is a 100% chance N flips will be split 50-50. Take N = 4 There are 16 possible combinations with scoring HHHH : 1 2 3 4 HHHT : 1 2 3 2 HHTH : 1 2 1 2 HHTT : 1 2 1 0 HTHH : 1 0 1 2 HTHT : 1 0 1 0 HTTH : 1 0 -1 0 HTTT : 1 0 -1 -2 THHH : -1 0 1 2 THHT : -1 0 1 0 THTH : -1 0 -1 0 THTT : -1 0 -1 -2 TTHH : -1 -2 -1 0 TTHT : -1 -2 -1 -2 TTTH : -1 -2 -3 -2 TTTT : -1 -2 -3 -4 Out of 64 scorings 14 were zero. If the first 0 is counted then there is 80 scorings with 30 zeros. edit - added code tags So average for N = 4 is 7/32 or 3/8 depending on if the first 0 is counted. Edited April 14, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 14, 2011 Report Share Posted April 14, 2011 (edited) I am getting for even values of N (counting the first 0) 2N + 2N * 1/2 + 2N * 1/2 * 3/4 + 2N * 1/2 * 3/4 * 5/6 + 2N * 1/2 * 3/4 * 5/6 * 7/8 + ... So that number divided by the total count of scorings 2N * (N + 1) 4 : ( 16 + 8 + 6) / (16 * 5) 6 : ( 64 + 32 + 24 + 20) / (64 * 7) 8 : (256 + 128 + 96 + 80 + 70) / (256 * 9) uhm... wait... that is the chance a given scoring will be 0 Edited April 14, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 14, 2011 Report Share Posted April 14, 2011 Average count of zeros 2 = 1.5 4 = 1.875 6 = 2.1875 8 = 2.4609375 So that leads to 1 + 1 * 1/2 + 1 * 1/2 * 3/4 + 1/2 * 3/4 * 5/6 + 1/2 * 3/4 * 5/6 * 7/8 + ... + 1 * 1/2 * ... * (N-1)/N Quote Link to comment Share on other sites More sharing options...
0 plainglazed Posted April 15, 2011 Report Share Posted April 15, 2011 nice solve curr3nt. really enjoyed this puzzle magician! and appreciate your hard work in English class. thanks for posting. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 15, 2011 Report Share Posted April 15, 2011 Didn't prove it though. I just gave an educated guess based upon evidence from a few scenerios. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 15, 2011 Report Share Posted April 15, 2011 (edited) Start with 0 This is where the 1 comes from Next is 2 HH - 1 +0 leaving score of 1 HT - 1 +1 TH - 1 +1 TT - 1 +0 leaving score of -1 Here is 1/2 average being added Roll 2 more times to get to 4 rolls HH +1 to average if previous score was -2 HT +1 to average if previous score was 0 TH +1 to average if previous score was 0 TT +1 to average if previous score was 2 So from the result of 2 we get the following for 4 HH - 1 +1, 2 +0 leaving a score of 2, 1 +0 with a score of 4 HT - 2 +1, 2 +0 TH - 2 +1, 2 +0 TT - 1 +1, 2 +0 leaving a score of -2, 1 +0 with a score of -4 Of the half that were 0 from 2 rolls, half are +1 to the average. 1/2 * 1/2 Of the half that were +/-2 from 2 rolls, 1/4 are +1 to the average. 1/2 * 1/4 (1/2 * 1/2) + (1/2 * 1/4) = 1/2 * (1/2 + 1/4) = 1/2 * 3/4 Roll 2 more times to get to 6 rolls HH +1 if previous score was -2 HT +1 if previous score was 0 TH +1 if previous score was 0 TT +1 if previous score was 2 From the results of the 4 rolls Of the (1/2 * 1/2) that were 0, half are +1. 1/2 * 1/2 * 1/2 Of the (1/2 * 1/2) that were +/-2, 1/4 are +1. 1/2 * 1/2 * 1/4 Of the (1/2 * 1/4) that were 0, half are +1. 1/2 * 1/4 * 1/2 Of the (1/2 * 1/2) that were +/-2, 1/4 are +1. 1/2 * 1/2 * 1/4 Of the (1/2 * 1/4) that were +/-4, 0 are +1 but 1/4 are +/-2 now. (1/2 * 1/2 * 1/2) + (1/2 * 1/2 * 1/4) + (1/2 * 1/2 * 1/4) + (1/2 * 1/2 * 1/4) = 1/2 * ( (1/2 * 1/2) + (1/2 * 1/4) + (1/2 * 1/4) + (1/2 * 1/4) ) = 1/2 * 1/4 * (1 + 1/2 + 1/2 + 1/2) = 1/2 * 1/4 * 5/2 ( * 3 / 3 ) = 1/2 * (1 * 3)/4 * 5/(2 * 3) = 1/2 * 3/4 * 5/6 Each 2 rolls added a smaller and smaller percentage of the scores are in the +2, 0 or -2 range that can be turned into 0 and adding to the average 0's. I can continue demonstrating...but how to prove that each additional two roll adds 1/2 * 3/4 * ... * (N-1)/N to the average? Edited April 15, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 15, 2011 Report Share Posted April 15, 2011 but not sure it adds. that is, it's not 1 +1 *1/2 +1* 1/2 *3/4 +1* 1/2 *3/4 *5/6 it's simply 1 *1/2 *3/4 *5/6 ... *(n/2 -1)/(n/2)) that is if you have 32 flips, you would have 1 *1/2 *3/4 *5/6 *7/8 *9/10 *11/12 *13/14 *15/16 = 0.1964 percent zeros on average, or 32*.1964 = 6.28 zeros on average. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 15, 2011 Report Share Posted April 15, 2011 edit: sorry didn't get it in time but not sure it adds. that is, it's not 1 +1 *1/2 +1* 1/2 *3/4 +1* 1/2 *3/4 *5/6 it's simply 1 *1/2 *3/4 *5/6 ... *(n -1)/n) that is if you have 32 flips, you would have 1 *1/2 *3/4 *5/6 *7/8 *9/10 *11/12 *13/14 *15/16 *17/18 *19/20 *21/22 *23/24 *25/26 *27/28 *29/30 *31/32 = 0.1399 percent zeros on average, or 32*.1399 = 4.47 zeros on average. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 15, 2011 Report Share Posted April 15, 2011 I actually used a construction very similar to Pascal's triangle for this problem. note: I didn't count the starting zero. it looked something like this: -5 -4 -3 -2 -1 0 1 2 3 4 5 0 1 0 1 01 0 1 122 001 0 I started at positive 1 to skip a step because whichever the first coin flip is is actually irrelevant. The values only go up by one and only when at 0. the number of possible outcomes that lead to 0 at each even step is (2n) C n From this, the solution can be defined recursively: a0=1, an=4*an-2+(2n) C n and the expected number of zeros is an/2n Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 15, 2011 Report Share Posted April 15, 2011 Question is: how many times on average would 0 have been 'hit' if you threw the coin N times? I took this meaning how many times there would be a scoring of 0 at any point of time during the tosses, not the final scoring of 0. If you want finally scoring then use what I added to the total at that toss. In which case we agree. Quote Link to comment Share on other sites More sharing options...
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Guest
I saw this puzzle the other day and after spending all of English class solving it, thought it was so cool that I had to post it... good luck
Starting at number zero, you throw a coin.
Tails means you go down one (therefore equaling -1)
Heads means you go up one (therefore equaling 1)
You repeat this procedure infinite times, so here's an example variation:
0, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 3, 4, 5, 6, 7, 6, 7.........
Question is: how many times on average would 0 have been 'hit' if you threw the coin N times?
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