Determining Probability of a Streak of 15+ Tails in 1,000 Coin Flips

[kaleidoscope]

Every student in a group of 1,000 is given a penny and is asked to flip it 40 times. What is the probability that someone had a streak of 15 or more tails?

 

[D H]

Welcome to PhysicsForums, kaleidoscope.

This is a homework problem. You need to show some work. We help you do your homework here; we do not do it for you.

 

[kaleidoscope]

My estimations are that p shoud be between 1/3 and 3/4, but I would like to come to a more definitive answer. Fellow forum users, please feel free to add to this thread. Any of your contributions will be greatly appreciated, thanks.

 

[D H]

How did you come up with that answer?

The way to solve this problem is to break it down to a simpler problem and build back up to the original problem.

The first thing to do is to realize that each of the 1000 students is an independent event. Designate the probability that one student will have a streak of 15 tails in 40 rolls as p. The probability that one student will not have a streak of 15 tails in 40 rolls is then 1-p. The probability that all 1000 students will not have such a streak is this number raised to the 1000^th power. The probability that at least one student has such streak is one less this probability:

p_{\text{1000 students}} = 1-(1-p)^{1000}

So all you have to do is figure out what p is. Once again, you do this by breaking it down to a simpler problem and building back up. If the number of tosses was 15, this would be simple. There are 2^{15} possible outcomes, of which only one comprises 15 tails in a row. Thus

p_{15} = \frac{1}{2^{15}}

With 16 tosses, if the first 15 tosses are all tails, the final toss is irrelevant. This makes for two ways to get 15 tails in 16 tosses. There is one more way: heads on the first toss followed by 15 straight tails. Thus

p_{16} = \frac{3}{2^{16}}

You can painstakingly continue with this analysis -- or you could derive a recursive relationship, from which you can derive a general expression. There is a nice simple expression that works up to 30 tosses. However, at 31 tosses, the student could toss 15 straight tails, a heads, and 15 more tails. If you want an exact answer you will need to account for this. If you want an approximate answer, you can simply ignore the very small probability of tossing two sequences of 15 or more tails.

 

[jacksonpeeble]

It's a basic statistics/probability problem. What DH is saying is that we'd appreciate if you would make an effort to solve it yourself, and if you're going about it the wrong way, we'll certainly try to help. I come here with AP Calculus problems all the time, but I follow the suggested format, and show the work I've attempted. If I truly have no idea how to do it, I usually post it in a clump of several problems that I have shown work with. It's not that he doesn't trust you, you just learn better when you attempt it yourself. You could end up right and have the satisfaction of solving it yourself!

It's been a while since I did that sort of problem, but (I may be incorrect) I suggest looking into http://www.google.com/search?q=Permutations+and+Combinations".

Good luck and post back if you need more help!

 

[njama]

D H there is much easier way to do it.

Its by taking a=T...T (15 tails)

Now a ..... (25 dots representing either T or H)


I hope you get it.

 

[D H]

D H there is much easier way to do it.

That approach (not very well described) will give the wrong answer.

Try your approach with a string of at least 2 tails in 5 tosses (you should get a probability of 19/31) or a string of at least 3 tails in 7 tosses (probability = 107/256).

 

[njama]

That approach (not very well described) will give the wrong answer.

Try your approach with a string of at least 2 tails in 5 tosses (you should get a probability of 19/31) or a string of at least 3 tails in 7 tosses (probability = 107/256).

Sorry, but I could not describe further, because the OP did not show any work.

TT... 2³
HTT.. 2²
.HTT. 2²

..HTT
subcases
---------
.HHTT 2¹
HTHTT 2⁰

Anyways its still complicated, but if he try to do it, maybe he will find a pattern that will work.

a=TTTTTTTTTTTTTTT

TTTTTTTTTTTTTTT. 2¹
.TTTTTTTTTTTTTTT
subcases
----------------------------
HTTTTTTTTTTTTTTT 2⁰

At the end, we are doing the same thing

 

[D H]

No, we aren't. This problem is looking for a sequence of 15 or more, not a sequence of exactly 15.

 

[njama]

No, we aren't. This problem is looking for a sequence of 15 or more, not a sequence of exactly 15.

So, do I . I was giving example for 15 tails out of 16 tosses.

And what about:

How do you know that:

The probability that all 1000 students will not have such a streak is this number raised to the 1000th power

?? Why 1000th power?

 

[D H]

So, do I . I was giving example for 15 tails out of 16 tosses.

Try your method on something simpler, such as the probability of getting a streak of 2 or more tails in 5 tosses. Be careful not to double-count TTHTT.

How do you know that:

The probability that all 1000 students will not have such a streak is this number raised to the 1000^th power.

?? Why 1000th power?

To be brutally honest, if you don't know the answer to this question you have no business providing "help" in these threads on probability.