What are the odds of 5 or more heads or tails in a row out of 20 coin tosses?
I think that this problem is very difficult partly because it does not have a closed form algebraic solution. I was able to solve it with a Markov transition matrix or a recursive solution. Here is the exact answer using 10 coin tosses and different size streaks from 2 to 10. The extreme cases of 2 and 10 are fairly obvious.
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Streak length , probability of a streak at least that length out of 10 coin tosses.
2, 1022/1024
3, 846/1024
4, 476/1024
5, 222/1024
6, 96/1024
7, 40/1024
8, 16/1024
9, 6/1024
10, 2/1024
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You can check your technique using these answers.
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The wizard of odds published my paper about this subject on Thursday in his column.
http://wizardofodds.com/askthewizard/232
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I would like opinions. Is the paper interesting? Do the results suprise you? Do you think people totally understimate the commonality of streaks? Do you think casinos capitalize on this belief? Can you get the results yourself ? If you think you can answer the question I would prefer it n the form the answer is m/2^20 and the value of m= ____, however if you give the answer as a percentage that is fine.
This is a challenge question. I do know the answer. I want people who enjoy mathematics to read my paper.
Ravindra, it’s not nearly that easy that it would be a power of 2. Try your technique on the easier problem and you won’t get the answer.
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Website: I suppose you could check out all million plus combinations by computer, but it would be a major problem if I ask you for the probability of a run of 10 in a row out of 1000 coin tosses. It would take some serious computer power.
As a hint, the answer for 5 out of 20 must be larger than the answer for 5 out of 10.
website: The answer for 5 out of 20 must be larger than for 5 out of 10 (220/1024), so the answer must be larger than 20%. If the answer is in the form m/2^20 ,then m must be larger than 200,000. Your answer of 564 is not in the correct ballpark. You should read the paper.
The probability of 5 heads or tails out of 5 coin tosses is 5/32=6.250% . So the solution at an absolute minimum has to be more than 6.250%.
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What if you consider the probability of series like HTHHT…T, so that there are no more than 4 in a row and then you substract it from 1?
Maybe it’s easy that way… I will think about your problem but maybe this suggestion helps.
Ana
5 H in a row = 1/2^5; 6 H in a row = 1/2^6; … 20 H in a row = 1/2^20.
=> P(5 or more in a row) = 1/2^5*(1 + 1/2 + 1/2^2 + … + 1/2^15)
= (1/2^5)*([1 - 1/2^15]/[1 - 1/2]) = (1/16)(32767/32768)
= 32767/524288.
Paco:
I think that an exact solution for 20 can be laboriously determined. I know that a computer program can easily be written which will check all 2^20 possible flips out will give the exact answer too. (I am tempted to do this) I did figure out the odds for 20 heads to 11 heads (or tails) and those odds were 110/2^20, but as you can guess the odds for 10 or less heads are different because we can get multiple streaks of these runs occurring.
Because of the amount of time to compute the solutions for 5 to 9 by hand is very timeconsuming, I will give up there for now. I do know that they can be worked out also by hand.
Okay, this problem got to me, I worked it entirely out by hand
5 heads
1 streak = 16 cases
2 streaks = 55 cases
3 streaks = 20 cases
6 heads
1 streak = 15 cases
2 streaks = 36 cases
3 streaks = 1 case
7 heads
1 streak = 14 cases
2 streaks = 21 cases
8 heads
1 streak = 13 cases
2 streaks = 10 cases
9 heads
1 streak = 12 cases
2 streaks = 3 cases
10 heads
1 streak = 11 cases
11 heads
1 steak = 10 cases
12 heads
1 streak = 9 cases
13 heads
1 streak = 8 cases
14 heads
1 streak = 7 cases
15 heads
1 streak = 6 cases
16 heads
1 streak = 5 cases
17 heads
1 steak = 4 cases
18 heads
1 streak = 3 cases
19 heads
1 streak = 2 cases
20 heads
1 streak = 1 cases
Totaling all these
we have for 1 cases
sum of 1..16 = 136
we have for 2 cases
sum of 55 + 36 + 21 + 10 + 3 (which are the summations of 2n for n=1..5) = 125
and for 3 cases
sum of 20 + 1 = 21
So for heads only we have 136 + 125 + 21 = 282
And we have the same exact cases for tails = 282
So the total odds on the chances of 5 or more heads or tails in a run of 20 tosses is 564/2^20
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and after having seen your answer.. and looking at the behavior of the n-cases, I think the pattern is discernable, hence can be worked out as multiple summations.
I got 480670 / 2^20, or 45.84%.
Let a(k,n) denote the number of possibilities of getting a streak of k or more in n tosses. For k>n, let a(k,n) be trivially zero.
Let n ≥ k.
The streak can begin at the very first toss, which generates 2*2^(n-k) positive cases.
If the streak does not begin at the first toss, it means that the first run is shorter than k equal tosses. In other words, the result was the same for l
This gives a recurrence relation:
= 2*2^3 + 16 + 6 + 2 + 0 = 16 + 16 + 6 + 2 = 40
a(k, n) = 2*2^(n-k) + Σ [l=1 to k-1] a(k, n-l),
which can be solved in closed form only for a few small k's; for k>5, it presumably leads to insolvable polynomials (of degree ≥ 5). This also excludes finding a closed form for all a(k, n) unless we are extremely lucky to get a sequence of exceptional cases of solvable polynomials. For k=5, it is possible to express a(5, n) analytically but it would take quite some time, so I will calculate the terms explicitly.
a(5, 1) = a(5, 2) = a(5, 3) = a(5, 4) = 0
a(5, 5) = 2*2^0 + 0 + 0 + 0 + 0 = 2
a(5, 6) = 2*2^1 + 2 + 0 + 0 + 0 = 4 + 2 = 6
a(5, 7) = 2*2^2 + 6 + 2 + 0 + 0 = 8 + 6 + 2 = 16
a(5,
a(5, 9) = … = 96
a(5, 10) = … = 222 <== We can check with your result cited above
a(5, 11) = 502
…
a(5, 19) = 229664
a(5, 20) = 480670 <== For 20 tosses.
The probability = number of positive cases / number of all cases = a(5, 20) / 2^20 = 480670 / 1048576 = 45.84%.