The odds of getting 5 consecutive heads are 1/32. Realize, if your first 4 flips are "HHHT", not only does the first of 1000 flips fail, but the next 3 do as well. For this reason, you have to figure out the average length per trial.
How many "chances" do you get? On failed chances, you have these sequences:
T (1/2)
HT (1/4)
HHT (1/8)
HHHT (1/16)
HHHHT (1/32)
The remainder of successful sequences begin with HHHHH. For this question, I'm treating all such sequences as length 6, although you could compute an accurate length for this yourself.
The average length of each "trial" is 1(0.5) + 2 (0.25) + 3 (0.125) + 4 (1/16) + 5 (1/32) + 6 (1/32)
or about 2.0. You have about 500 "trials" to "succeed".
If each trial succeeds 1/32 of the time, and you have 500 trials, the odds of succeeding in at least one trial is 1 - (1-1/32) ^ 500
The second term (1-1/32) ^ 500 is the odds of failing 500 consecutive times. This is close to 1.3x10e-7. It is very likely that you'll succeed once.
A lot of people misunderstand sequence testing. Another example would be using a streak of 32 flips. An unsophisticated bettor might assume that with a 1/32 chance, and 32 flips (or even 36 to finish his last "run"), he should get 1 success.
1-(1-1/32)^16 = 0.4
A more clever bettor, who does not appreciate sequence lengths might assume his odds are
1 - (1-1/32)^32 = 0.64
Either way, if you don't understand sequencing, and believe you should get 1 or .64 successes when the actual number is closer to 0.4, you're probably going to end up broke.
How many "chances" do you get? On failed chances, you have these sequences:
T (1/2)
HT (1/4)
HHT (1/8)
HHHT (1/16)
HHHHT (1/32)
The remainder of successful sequences begin with HHHHH. For this question, I'm treating all such sequences as length 6, although you could compute an accurate length for this yourself.
The average length of each "trial" is 1(0.5) + 2 (0.25) + 3 (0.125) + 4 (1/16) + 5 (1/32) + 6 (1/32)
or about 2.0. You have about 500 "trials" to "succeed".
If each trial succeeds 1/32 of the time, and you have 500 trials, the odds of succeeding in at least one trial is 1 - (1-1/32) ^ 500
The second term (1-1/32) ^ 500 is the odds of failing 500 consecutive times. This is close to 1.3x10e-7. It is very likely that you'll succeed once.
A lot of people misunderstand sequence testing. Another example would be using a streak of 32 flips. An unsophisticated bettor might assume that with a 1/32 chance, and 32 flips (or even 36 to finish his last "run"), he should get 1 success.
1-(1-1/32)^16 = 0.4
A more clever bettor, who does not appreciate sequence lengths might assume his odds are
1 - (1-1/32)^32 = 0.64
Either way, if you don't understand sequencing, and believe you should get 1 or .64 successes when the actual number is closer to 0.4, you're probably going to end up broke.
OMT,
Keep in mind that these probabilities are only useful if you define the event that you are going to observe BEFOREHAND!!!
If you notice that a pitcher has gone 4-0 and you're thinking of betting on his next outing, then using the 1/32 odds here will mislead you. You need to have asked the question when he was 0-0. Here you are interested in the conditional probability that he goes 5-0 GIVEN that he has already gone 4-0 and the bad news here is that it is still 50%. This is known as the memoryless property of random events.
Keep in mind that these probabilities are only useful if you define the event that you are going to observe BEFOREHAND!!!
If you notice that a pitcher has gone 4-0 and you're thinking of betting on his next outing, then using the 1/32 odds here will mislead you. You need to have asked the question when he was 0-0. Here you are interested in the conditional probability that he goes 5-0 GIVEN that he has already gone 4-0 and the bad news here is that it is still 50%. This is known as the memoryless property of random events.
