Easiest to Hardest Heads in a Row
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2 members have voted
April 10th, 2016 at 5:15:01 PM
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a) What is the probability of getting 5 Heads in a row in 5 flips?
b) What is the probability of getting at least 6 Heads in a row in 8 flips?
at least 6 means it could be 6 or 7 or 8.
c) What is the probability of getting at least 7 Heads in a row in 13 flips?
at least 7 means it could be 7 or 8 or 9 or 10 or 11 or 12 or 13.
now rank them from easiest to hardest (highest probability of success to lowest)
if you have come this far or have a guess
vote
This was an extra credit math question in Tobe math class
("TOBE, I DID NOT USE YOUR REAL NAME") and was also a class exercise
Sally
b) What is the probability of getting at least 6 Heads in a row in 8 flips?
at least 6 means it could be 6 or 7 or 8.
c) What is the probability of getting at least 7 Heads in a row in 13 flips?
at least 7 means it could be 7 or 8 or 9 or 10 or 11 or 12 or 13.
now rank them from easiest to hardest (highest probability of success to lowest)
if you have come this far or have a guess
vote
This was an extra credit math question in Tobe math class
("TOBE, I DID NOT USE YOUR REAL NAME") and was also a class exercise
Sally
I Heart Vi Hart
April 10th, 2016 at 5:55:46 PM
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Damn, I can't flip heads. Too bad the question wasn't for tails. I will practice my heads flipping tonight.
Expect the worst and you will never be disappointed.
I AM NOT PART OF GWAE RADIO SHOW
April 10th, 2016 at 6:32:25 PM
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The only way to get 5 heads in a row in 5 flips is if all 5 are heads; this is 1/32
At least 6 in a row in 8 flips is the sum of:
(a) The first 6 being heads = 1/64
(b) #1 being tails, then 2-7 being heads = 1/2 x 1/64 = 1/128
(c) #2 being tails (#1 can be either), then 3-8 being heads = 1/2 x 1/64 = 1/128
The sum = 1/32
At least 7 in a row in 13 flips is the sum of:
(a) The first 7 being heads = 1/128
(b) #1 being tails, then 2-8 being heads = 1/2 x 1/64 = 1/256
(c) #2 being tails (#1 can be either), then 3-9 being heads = 1/2 x 1/64 = 1/256
(d) #3 being tails (1-2 can be either), then 4-10 being heads = 1/2 x 1/64 = 1/256
(e) #4 being tails (1-3 can be either), then 5-11 being heads = 1/2 x 1/64 = 1/256
(f) #5 being tails (1-4 can be either), then 6-12 being heads = 1/2 x 1/64 = 1/256
(g) #6 being tails (1-5 can be either), then 7-13 being heads = 1/2 x 1/64 = 1/256
Note that no group includes having 6 in a row earlier in the set, so no set of flips is counted twice
(had it said 14 flips, then counting #7 as tails and 8-14 as heads would have to exclude sets where 1-6 were all heads, as they were counted in (a))
The sum = 1/32
All three have the same probability
