February 2nd, 2014 at 9:04:47 AM
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I see where covers.com has a streak survivor contest. The first cash prize is for a win streak of 17 games. Here is my question.......if someone can hit 57% winners, how many selections would need to be made in order to have a streak of 17 wins.

I appreciate the help. Thanks!

I appreciate the help. Thanks!

DO NOT blindly accept what has been spoken.
DO NOT blindly accept what has been written.
Think. Assess. Lead. DO NOT blindly follow.

February 2nd, 2014 at 9:07:08 AM
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p^n:

.57^17 = 0.00007077

1 in 14129

.57^17 = 0.00007077

1 in 14129

Wisdom is the quality that keeps you out of situations where you would otherwise need it

February 2nd, 2014 at 9:09:34 AM
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dwheatley, thanks so much!

DO NOT blindly accept what has been spoken.
DO NOT blindly accept what has been written.
Think. Assess. Lead. DO NOT blindly follow.

February 2nd, 2014 at 9:09:55 AM
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oops, rereading your question that's not what you asked. ..

If you want to make one long string of selections and guarantee you get a 17 game streak, that's not possible. If you want a to have a certain probability of hitting 17, you can use a streak calculator. How many games will you bet on?

If you can make many different 17 game selections and you want one of them to be 17 correct games, you need to pick all 2^17 results to get it right.

If you want to make one long string of selections and guarantee you get a 17 game streak, that's not possible. If you want a to have a certain probability of hitting 17, you can use a streak calculator. How many games will you bet on?

If you can make many different 17 game selections and you want one of them to be 17 correct games, you need to pick all 2^17 results to get it right.

Wisdom is the quality that keeps you out of situations where you would otherwise need it

February 2nd, 2014 at 9:32:13 AM
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dwheatley, let me try to re-phrase. I can hit 57%. I want a high probability of winning 17 games in a row. I'll accept your definition of "high probability".

How many selections would need to be made?

How many selections would need to be made?

DO NOT blindly accept what has been spoken.
DO NOT blindly accept what has been written.
Think. Assess. Lead. DO NOT blindly follow.

February 2nd, 2014 at 11:56:59 AM
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I have an online streak calculator I use. One way of answering your question is:

If you bet 2,100,000 games in your lifetime (and are correct on 57% of them), there is a ~50% chance you will hit a streak of 17 wins.

If you bet 2,100,000 games in your lifetime (and are correct on 57% of them), there is a ~50% chance you will hit a streak of 17 wins.

Wisdom is the quality that keeps you out of situations where you would otherwise need it

February 3rd, 2014 at 5:21:15 AM
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Quote:dwheatleyI have an online streak calculator I use. One way of answering your question is:

If you bet 2,100,000 games in your lifetime (and are correct on 57% of them), there is a ~50% chance you will hit a streak of 17 wins.

dwheatley, wow. I would have never suspected that in excess of 2 million picks would have to be made. Thanks again.

February 3rd, 2014 at 7:01:12 AM
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Precision, please: what do you call selections? Is it one 'prediction' per game and you ask how many successive games (a) until a 17-string or (b) number predefined and max the P of having a 17-string? Or is it a specific group of N games predetermined by the bookmaker, and you ask how many 'set of predictions' have to be placed to ensure a 17-string-winning one?Quote:steeldcodwheatley, let me try to re-phrase. I can hit 57%. I want a high probability of winning 17 games in a row. I'll accept your definition of "high probability".

How many selections would need to be made?

In the second case we need to know N.

In the former case, I doubt that the book has not put a deadline... So there probably IS a limit number of possible bets. Making the 2 million answer all but useless.

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February 3rd, 2014 at 7:22:24 AM
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kubikulann, as I stated, there is a contest with a cash award for having picked 17 consecutive winners. My question was how many picks would have to be made, of a winner in a game (1 per game), in order for a 57% success rate to hit a streak of 17 consecutive winners, with high probability?DO NOT blindly accept what has been spoken.
DO NOT blindly accept what has been written.
Think. Assess. Lead. DO NOT blindly follow.

February 3rd, 2014 at 7:27:47 AM
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"for having picked 17 consecutive winners" in what context? Is ther a limit on the number of picks you are allowed? Or a time limit when the contest is running? (Leading essentially to the same limit in # of picks)

Streaks are not a well-defined concept. I see two occurrences in statistics: "first appearance of a streak" or "existence of a streak in a predetermined # of trials". Streak by itself doesn't mean much...

Streaks are not a well-defined concept. I see two occurrences in statistics: "first appearance of a streak" or "existence of a streak in a predetermined # of trials". Streak by itself doesn't mean much...

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February 3rd, 2014 at 7:54:42 AM
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kubikulann, I'm sorry if I am not getting my question across clearly. I appreciate the patience.

I have a contest that I'm going to enter since it costs nothing to do so. The contest involves many selections provided each day amongst a variety of sports.

I, however, am just focused on NBA games and I would say that, on average, I will end up with 3 picks per day. So let's try this question in a different manner.

I'm going to make 90 picks within a month, hitting on 57% of them, what is the probability of hitting on a streak of 17 consecutive winners? Also, if I were to do this until I hit the grave (assume that I'm a vampire and will live forever), how many months would I need to make selections before I see a streak of 17 consecutive winners.

Thanks!DO NOT blindly accept what has been spoken.
DO NOT blindly accept what has been written.
Think. Assess. Lead. DO NOT blindly follow.

I have a contest that I'm going to enter since it costs nothing to do so. The contest involves many selections provided each day amongst a variety of sports.

I, however, am just focused on NBA games and I would say that, on average, I will end up with 3 picks per day. So let's try this question in a different manner.

I'm going to make 90 picks within a month, hitting on 57% of them, what is the probability of hitting on a streak of 17 consecutive winners? Also, if I were to do this until I hit the grave (assume that I'm a vampire and will live forever), how many months would I need to make selections before I see a streak of 17 consecutive winners.

Thanks!

February 3rd, 2014 at 8:53:48 AM
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The latest question is easier to answer. (Assuming total independence between the hits, that is, your result on one match is totally uncorrelated with your result on another). The model is called 'geometric', but with a twist.

Let us call p the probability of individual hit (here 57%) and q its opposite (here 43%).

Work in "double streaks" (DS): a sequence of hits followed by a sequence of misses. Your result history is a succession of such double streaks, and you are interested in the first appearance of one with 17+ hits. Call that a "success".

We will have to take into account the two cases: your series begins with a hit (prob p) or with a miss (q) in which case we must add a preliminary sequence of misses before the succession of DS begins. Mean length of the prelim sequence = 1/p.

Probability of a DS being a success : P = p^16 (the initial game of the streak is, by definition, a hit, so you need 16 more).

Expected number of trials (DS) before a success: N = 1/P - 1 = 8052.83

Mean length of a DS (as per the geometric law): 1/q + 1/p

But we are interested here in the mean length of a DS conditional on its being shorter than 17 hits (as these are the only ones to appear before the successful one). That is EC = ( 1/q + 16(1 - 1/(1-P)) ) + 1/p *--- to be checked out; done on a scrap of paper --*

Total expected length before the successful DS : N * EC if begin with a hit, N * EC + 1/p if begin with a miss.

Result: N*EC + q/p

= [1/P - 1] [(16 + 1/q - 16/(1-P) ) + 1/p] + q/p

= 32840 games (which gives 32857 at the end of the successful 17-streak) or 30 years and 5 months

This is an EXPECTATION. The question "how much time will I have to wait" has no deterministic answer, only a probability distribution. Be (very) lucky and it could be "no time"... Be (very) unlucky and it could be 1000 years.

------

Former model (90 games) in a next post.

Let us call p the probability of individual hit (here 57%) and q its opposite (here 43%).

Work in "double streaks" (DS): a sequence of hits followed by a sequence of misses. Your result history is a succession of such double streaks, and you are interested in the first appearance of one with 17+ hits. Call that a "success".

We will have to take into account the two cases: your series begins with a hit (prob p) or with a miss (q) in which case we must add a preliminary sequence of misses before the succession of DS begins. Mean length of the prelim sequence = 1/p.

Probability of a DS being a success : P = p^16 (the initial game of the streak is, by definition, a hit, so you need 16 more).

Expected number of trials (DS) before a success: N = 1/P - 1 = 8052.83

Mean length of a DS (as per the geometric law): 1/q + 1/p

But we are interested here in the mean length of a DS conditional on its being shorter than 17 hits (as these are the only ones to appear before the successful one). That is EC = ( 1/q + 16(1 - 1/(1-P)) ) + 1/p *--- to be checked out; done on a scrap of paper --*

Total expected length before the successful DS : N * EC if begin with a hit, N * EC + 1/p if begin with a miss.

Result: N*EC + q/p

= [1/P - 1] [(16 + 1/q - 16/(1-P) ) + 1/p] + q/p

= 32840 games (which gives 32857 at the end of the successful 17-streak) or 30 years and 5 months

This is an EXPECTATION. The question "how much time will I have to wait" has no deterministic answer, only a probability distribution. Be (very) lucky and it could be "no time"... Be (very) unlucky and it could be 1000 years.

------

Former model (90 games) in a next post.

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February 3rd, 2014 at 11:50:34 AM
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I haven't had time to review in that I am in the middle of another task. However, I wanted to take a minute and thank you for your time on this. I'm going to review and try to comprehend sometime later today.

Thanks again!DO NOT blindly accept what has been spoken.
DO NOT blindly accept what has been written.
Think. Assess. Lead. DO NOT blindly follow.

Thanks again!

February 3rd, 2014 at 3:16:59 PM
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kubikulann, after staring at this a while. I find the following.....I'm just too dumb. I can only assume that your calculation is correct. It certainly seems closer to reality than 2.1 million (no offense dwheatedly). Thanks again!

Maybe I'll mull it over with a glass of wine later. It couldn't hurt. :-)DO NOT blindly accept what has been spoken.
DO NOT blindly accept what has been written.
Think. Assess. Lead. DO NOT blindly follow.

Maybe I'll mull it over with a glass of wine later. It couldn't hurt. :-)

February 4th, 2014 at 2:39:03 AM
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Answer to the first part: what is the probability of 17 hits in a streak within a sequence of 90 guesses?

I will use another approach.

Consider that you know the number n of hits, and you ponder the probability that they are ordered such that there is a 17 (or more) streak.

The idea is to take the streak away. You are left with 90-17 guesses among which n-17 hits. These can be ordered in (

Each of these orders has a probability of p

Now:

That was for ONE streak, but we didn't watch the cases where there are more than one, or the cases of a 34+ streak. We counted some occurrences twice.

We might disregard those as negligible, but let's do it correctly.

By the formula of total probabilities, the Pr(some streak) is equal to Pr(any one) - Pr(any two) + Pr(any three) - etc. [Here we stop at 5 because 6 is impossible in 90 games.]

Pr(any two if n hits) = (for n>33) (

and generally,

Pr(any k streaks if n hits) = (for n>=17k) (

There you have it. The general answer is

SUM{k=1 to 5} (-1)

with, in your example, p=57% and q=43%.

This can be computed in an Excel spreadsheet, for example. The Pr will be very low indeed. :-)

I will use another approach.

Consider that you know the number n of hits, and you ponder the probability that they are ordered such that there is a 17 (or more) streak.

The idea is to take the streak away. You are left with 90-17 guesses among which n-17 hits. These can be ordered in (

^{90-17}_{n-17}) ways. In between these, you have to insert your streak. This happens before a miss or at the end, making 90-n+1 possibilities.Each of these orders has a probability of p

^{n}q^{90-n}(Binomial formula).Now:

That was for ONE streak, but we didn't watch the cases where there are more than one, or the cases of a 34+ streak. We counted some occurrences twice.

We might disregard those as negligible, but let's do it correctly.

By the formula of total probabilities, the Pr(some streak) is equal to Pr(any one) - Pr(any two) + Pr(any three) - etc. [Here we stop at 5 because 6 is impossible in 90 games.]

Pr(any two if n hits) = (for n>33) (

^{90-34}_{n-34}) (^{91-n}_{2})and generally,

Pr(any k streaks if n hits) = (for n>=17k) (

^{90-17k}_{n-17k}) (^{91-n}_{k})There you have it. The general answer is

SUM{k=1 to 5} (-1)

^{k-1}SUM{n=17k to 90} (^{90-17k}_{n-17k}) (^{91-n}_{k}) p^{n}q^{90-n}with, in your example, p=57% and q=43%.

This can be computed in an Excel spreadsheet, for example. The Pr will be very low indeed. :-)

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