Comments on: What are the odds of 5 or more heads or tails in a row out of 20 coin tosses? http://www.24gambling.net/casino/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses-10241/ Just another WordPress weblog Fri, 01 Apr 2011 22:58:33 +0000 hourly 1 http://wordpress.org/?v=3.2-RC1 By: ☮ Vašek http://www.24gambling.net/casino/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses-10241/comment-page-1/#comment-18984 ☮ Vašek Tue, 18 May 2010 22:40:26 +0000 http://www.24gambling.net/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses#comment-18984 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<k tosses and then changed. The rest behaves exactly as the definition of a(k, n-l) goes: for each positive case of the couple (k, n-l), we can fill the appropriate "history" only one way. So we just sum a(k, n-l) for 1 ≤ l < k. This gives a recurrence relation: 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, 8) = 2*2^3 + 16 + 6 + 2 + 0 = 16 + 16 + 6 + 2 = 40 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%. 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:
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, 8) = 2*2^3 + 16 + 6 + 2 + 0 = 16 + 16 + 6 + 2 = 40
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%.

]]> By: Website http://www.24gambling.net/casino/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses-10241/comment-page-1/#comment-18983 Website Tue, 18 May 2010 22:19:07 +0000 http://www.24gambling.net/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses#comment-18983 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 ------------ 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. 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

————
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.

]]> By: Ravindra P http://www.24gambling.net/casino/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses-10241/comment-page-1/#comment-18982 Ravindra P Tue, 18 May 2010 21:30:55 +0000 http://www.24gambling.net/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses#comment-18982 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. 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.

]]> By: DDs wife http://www.24gambling.net/casino/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses-10241/comment-page-1/#comment-18981 DDs wife Tue, 18 May 2010 20:57:19 +0000 http://www.24gambling.net/what-are-the-odds-of-5-or-more-heads-or-tails-in-a-row-out-of-20-coin-tosses#comment-18981 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 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

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