Probability calculation

Dale · 20th December 2007, 12:50 PM

Ten new pennies are placed in a plastic cup, shaken and tossed on a smooth tabletop with no obstructions of any kind to influence the spread.
The distribution of heads and tails is counted.
What is the probability, in three runs of 20 tosses, that I will get two nine and one combinations in two of the runs and three nine and one combinations in a third?
I've done this toss many, many times because, apart from the obvious head and tail count, the distributions themselves are interesting; how many 7s and 3s, 6s and 4s, the occasional 8 and 2, etc. (also because I've got a lot of time on my hands.) but, I've never before gotten a nine and one.
Then just this morning I decided to introduce an anomaly; I substituted a Canadian penny for one of the US coins.
The Canadian coin is identical in diameter and thickness but, of course, has different images on front and back (obverse in coin person language.) It also has a pronounced ridge around the outside edges front and back.
I designated the image of the queen as heads and the maple leaf as tails.
I did the three runs of 20 throws and got the results noted above.
Now, I find it difficult to attribute such a profound difference in results to the simple substitution of one almost identical coin; especially since the procedure I'm using has not changed.
Could it be that the physical mixing of the coins has been altered due to the one oddball .
Are aliens watching over my shoulder and influencing the toss?
Is Shirley McLaine somehow involved?
My dear bride of many years has suggested that this may be my "lucky day" and I should run out and buy a lottery ticket.
Any comments? Cheers, D

GreedyAlgorithm · 20th December 2007, 01:08 PM

Well, let's make some assumptions. First, that each of the real pennies has approximately the same chance of landing on heads when tossed in the above manner, which we'll call p. This is probably not 0.5, since there is some chance the penny spins, and it's well known that a spinning coin is quite biased toward one side or the other. The pennies, we'll say, are independent of each other (this might be close enough to true). Tossing once, you expect to get at least 9 of one side with probability p^10+10*p^9*(1-p)^1+(1-p)^10+10*(1-p)^9*p^1. For reference, if p=0.5, that's about 2.15%, and it grows as p gets further from 0.5.

Even at p's lowest value, getting exactly three 9-and-1 combinations is 1140*0.0215^3*(1-0.0215)^17 = 0.78%, which while small is not altogether unreasonable. But on the basis of your first toss series you should definitely wonder whether the fake coin is influencing things somehow and do more tosses to confirm or rule out your guess.

boooeee · 20th December 2007, 01:13 PM

Does "nine and one" mean "9 heads and 1 tail", OR does it mean either "9 heads and 1 tail" or "9 tails and 1 head"?

Modified · 20th December 2007, 02:25 PM

That probability that in three runs of 20 tosses you will see something that looks unusual and unlikely is probably fairly high.

Dale · 20th December 2007, 02:29 PM

Modified, Until today that has not been my experience with this procedure. Try it yourself. D

Modified · 20th December 2007, 02:34 PM

Originally Posted by Dale

Modified, Until today that has not been my experience with this procedure. Try it yourself. D

Well, how many times have you done it?

Jeff Corey · 20th December 2007, 10:24 PM

This is how Dr. Jahn got his PEAR shaped distribution. Cherry-picking.

Southwind17 · 20th December 2007, 11:25 PM

Originally Posted by Dale

What is the probability, in three runs of 20 tosses, that I will get two nine and one combinations in two of the runs and three nine and one combinations in a third?

What's the relevance of the three runs? Why are you analyzing your results this way? Why are you not simply considering it as a series of 60 straight tosses?

The answers to these questions don't affect the fact that a '9/1' combination in 7 out of 60 tosses seems extraordinary, especially if these were your first 60 tosses with the rogue coin, as you indicate, but they might throw a little light on the efficacy of your experiment.

Have you repeated the experiment much since you made the OP? I would guess that you're 'back' generally to the original statistics.

Ivor the Engineer · 21st December 2007, 03:01 AM

This follows a Binomial distribution:

P(X=r) = nCr . p^r . (1-p)^(n-r)

Where:

n is the number of coins per trial

r is the number of heads (or tails) in each trial

p is the probability of obtaining a head or tail.

nCr = n!/(r!(n-r)!)

If both 9 heads and 9 tails are being considered, the probability is multiplied by two.

So for n=10, r=9 and p=0.5:

P(X=9) = 10C9 . 0.5^9 . (1-0.5)^(10-9) = 10 . 0.001953125 . 0.5 = 0.009765625

multiplying by 2 (because 9 heads and 1 tail is considered the same as 1 head and 9 tails) gives:

P(X=9) = 0.01953125

Over 20 tosses this is increased to 0.390625. I.e. a 39% chance of obtaining 9 heads and 1 tail, or 1 head and 9 tails.

Using the binomial distribution again for the n (20) trials and answering the question in the OP:

n = 20; r = 2; p = 0.01953125

P(X=2) = 0.050818

and:

n = 20; r = 3; p = 0.01953125

P(X=3) = 0.006074

There are 3 ways to have these results, so:

P(X=2 in 2 trials and X=3 in 1 trail) = 3 . P(X=2) . P(X=2) . P(X=3) = 4.70579E-05

Or about once in every 21250 runs.

(Ivor now waits for some probability guru to find the mistake(s!) in his calculations.)

Southwind17 · 21st December 2007, 05:23 AM

Originally Posted by Ivor the Engineer

This follows a Binomial distribution:

P(X=r) = nCr . p^r . (1-p)^(n-r)

Where:

n is the number of coins per trial

r is the number of heads (or tails) in each trial

p is the probability of obtaining a head or tail.

nCr = n!/(r!(n-r)!)

If both 9 heads and 9 tails are being considered, the probability is multiplied by two.

So for n=10, r=9 and p=0.5:

P(X=9) = 10C9 . 0.5^9 . (1-0.5)^(10-9) = 10 . 0.001953125 . 0.5 = 0.009765625

multiplying by 2 (because 9 heads and 1 tail is considered the same as 1 head and 9 tails) gives:

P(X=9) = 0.01953125

Over 20 tosses this is increased to 0.390625. I.e. a 39% chance of obtaining 9 heads and 1 tail, or 1 head and 9 tails.

Using the binomial distribution again for the n (20) trials and answering the question in the OP:

n = 20; r = 2; p = 0.01953125

P(X=2) = 0.050818

and:

n = 20; r = 3; p = 0.01953125

P(X=3) = 0.006074

There are 3 ways to have these results, so:

P(X=2 in 2 trials and X=3 in 1 trail) = 3 . P(X=2) . P(X=2) . P(X=3) = 4.70579E-05

Or about once in every 21250 runs.

(Ivor now waits for some probability guru to find the mistake(s!) in his calculations.)

What about Ivor, while he's waiting, commenting on whether Dale's description of this most extraordinary feat seems accurate?!

Beth · 21st December 2007, 07:33 AM

Originally Posted by Ivor the Engineer

(Ivor now waits for some probability guru to find the mistake(s!) in his calculations.)

Your calculations all look correct to me.

Dale · 21st December 2007, 10:38 AM

Question: Is there some restriction on length with this "Quick Reply"? I'm having trouble posting anything but one-liners.

Southwind17 · 21st December 2007, 11:05 AM

Originally Posted by Dale

Question: Is there some restriction on length with this "Quick Reply"? I'm having trouble posting anything but one-liners.

Well given what you've just posted runs into two lines in the Quick Reply box, I'd say not! Seriously, you should be able to type as much as you like, so far as I know.

blobru · 21st December 2007, 12:16 PM

Originally Posted by Dale

Ten new pennies are placed in a plastic cup, shaken and tossed on a smooth tabletop with no obstructions of any kind to influence the spread.
The distribution of heads and tails is counted.
What is the probability, in three runs of 20 tosses, that I will get two nine and one combinations in two of the runs and three nine and one combinations in a third?
I've done this toss many, many times because, apart from the obvious head and tail count, the distributions themselves are interesting; how many 7s and 3s, 6s and 4s, the occasional 8 and 2, etc. (also because I've got a lot of time on my hands.) but, I've never before gotten a nine and one.

As shown above, your latest result isn't too far out. But what I find odd is that you'd never gotten a single 9 and 1 before? Ten pennies, two sides, ten places for the 1, two ways to get 9 and 1:

2¹⁰ / 10 * 2 = 62

so you should expect on average to see one 9 and 1 every 62 tosses, or very nearly one per 3 runs of 20 tosses. But you hadn't seen any before?

(61/62)^r*20, r = runs

gives the odds of not getting a 9 and 1 for however many runs.
How many runs do you estimate you did?

Quote:

Then just this morning I decided to introduce an anomaly; I substituted a Canadian penny for one of the US coins.
The Canadian coin is identical in diameter and thickness but, of course, has different images on front and back (obverse in coin person language.) It also has a pronounced ridge around the outside edges front and back.
I designated the image of the queen as heads and the maple leaf as tails.
I did the three runs of 20 throws and got the results noted above.
Now, I find it difficult to attribute such a profound difference in results to the simple substitution of one almost identical coin; especially since the procedure I'm using has not changed.
Could it be that the physical mixing of the coins has been altered due to the one oddball .
Are aliens watching over my shoulder and influencing the toss?
Is Shirley McLaine somehow involved?
My dear bride of many years has suggested that this may be my "lucky day" and I should run out and buy a lottery ticket.
Any comments? Cheers, D

Maybe the Canadian penny turned the Americans into socialists?

20th December 2007, 12:50 PM	#1
Dale Student Join Date: Feb 2006 Posts: 31	Probability calculation Ten new pennies are placed in a plastic cup, shaken and tossed on a smooth tabletop with no obstructions of any kind to influence the spread. The distribution of heads and tails is counted. What is the probability, in three runs of 20 tosses, that I will get two nine and one combinations in two of the runs and three nine and one combinations in a third? I've done this toss many, many times because, apart from the obvious head and tail count, the distributions themselves are interesting; how many 7s and 3s, 6s and 4s, the occasional 8 and 2, etc. (also because I've got a lot of time on my hands.) but, I've never before gotten a nine and one. Then just this morning I decided to introduce an anomaly; I substituted a Canadian penny for one of the US coins. The Canadian coin is identical in diameter and thickness but, of course, has different images on front and back (obverse in coin person language.) It also has a pronounced ridge around the outside edges front and back. I designated the image of the queen as heads and the maple leaf as tails. I did the three runs of 20 throws and got the results noted above. Now, I find it difficult to attribute such a profound difference in results to the simple substitution of one almost identical coin; especially since the procedure I'm using has not changed. Could it be that the physical mixing of the coins has been altered due to the one oddball . Are aliens watching over my shoulder and influencing the toss? Is Shirley McLaine somehow involved? My dear bride of many years has suggested that this may be my "lucky day" and I should run out and buy a lottery ticket. Any comments? Cheers, D
Dale Student Join Date: Feb 2006 Posts: 31

20th December 2007, 01:08 PM	#2
GreedyAlgorithm Muse Join Date: Aug 2005 Location: Redmond, WA Posts: 569	Well, let's make some assumptions. First, that each of the real pennies has approximately the same chance of landing on heads when tossed in the above manner, which we'll call p. This is probably not 0.5, since there is some chance the penny spins, and it's well known that a spinning coin is quite biased toward one side or the other. The pennies, we'll say, are independent of each other (this might be close enough to true). Tossing once, you expect to get at least 9 of one side with probability p^10+10p^9(1-p)^1+(1-p)^10+10(1-p)^9p^1. For reference, if p=0.5, that's about 2.15%, and it grows as p gets further from 0.5. Even at p's lowest value, getting exactly three 9-and-1 combinations is 11400.0215^3(1-0.0215)^17 = 0.78%, which while small is not altogether unreasonable. But on the basis of your first toss series you should definitely wonder whether the fake coin is influencing things somehow and do more tosses to confirm or rule out your guess.
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20th December 2007, 01:13 PM	#3
boooeee Dart Fener Join Date: Aug 2002 Location: The Lando System Posts: 2,396	Does "nine and one" mean "9 heads and 1 tail", OR does it mean either "9 heads and 1 tail" or "9 tails and 1 head"?
	__________________ my nerdy sports blog: betting market analytics

20th December 2007, 02:29 PM	#5
Dale Student Join Date: Feb 2006 Posts: 31	Modified, Until today that has not been my experience with this procedure. Try it yourself. D
Dale Student Join Date: Feb 2006 Posts: 31

20th December 2007, 11:25 PM	#8
Southwind17 Illuminator Join Date: Sep 2007 Location: Location: Location Posts: 4,461	Originally Posted by Dale What is the probability, in three runs of 20 tosses, that I will get two nine and one combinations in two of the runs and three nine and one combinations in a third? What's the relevance of the three runs? Why are you analyzing your results this way? Why are you not simply considering it as a series of 60 straight tosses? The answers to these questions don't affect the fact that a '9/1' combination in 7 out of 60 tosses seems extraordinary, especially if these were your first 60 tosses with the rogue coin, as you indicate, but they might throw a little light on the efficacy of your experiment. Have you repeated the experiment much since you made the OP? I would guess that you're 'back' generally to the original statistics.
	__________________ The views expressed here do not necessarily represent the unanimous view of all parts of my mind "Always" and "never" are two words that you should always remember never to use. Facts do not cease to exist because they are ignored. Why is it that when I ask for a pair of hands a brain comes attached?

20th December 2007, 02:25 PM	#4
Modified Illuminator Join Date: Sep 2006 Location: SW Florida Posts: 4,062	That probability that in three runs of 20 tosses you will see something that looks unusual and unlikely is probably fairly high.

20th December 2007, 02:34 PM	#6
Modified Illuminator Join Date: Sep 2006 Location: SW Florida Posts: 4,062	Originally Posted by Dale Modified, Until today that has not been my experience with this procedure. Try it yourself. D Well, how many times have you done it?

20th December 2007, 10:24 PM	#7
Jeff Corey New York Skeptic Join Date: Aug 2001 Posts: 13,794	This is how Dr. Jahn got his PEAR shaped distribution. Cherry-picking.

21st December 2007, 03:01 AM	#9
Ivor the Engineer Philosopher Join Date: Feb 2006 Location: South Britain, near the middle Posts: 9,553	This follows a Binomial distribution: P(X=r) = nCr . p^r . (1-p)^(n-r) Where: n is the number of coins per trial r is the number of heads (or tails) in each trial p is the probability of obtaining a head or tail. nCr = n!/(r!(n-r)!) If both 9 heads and 9 tails are being considered, the probability is multiplied by two. So for n=10, r=9 and p=0.5: P(X=9) = 10C9 . 0.5^9 . (1-0.5)^(10-9) = 10 . 0.001953125 . 0.5 = 0.009765625 multiplying by 2 (because 9 heads and 1 tail is considered the same as 1 head and 9 tails) gives: P(X=9) = 0.01953125 Over 20 tosses this is increased to 0.390625. I.e. a 39% chance of obtaining 9 heads and 1 tail, or 1 head and 9 tails. Using the binomial distribution again for the n (20) trials and answering the question in the OP: n = 20; r = 2; p = 0.01953125 P(X=2) = 0.050818 and: n = 20; r = 3; p = 0.01953125 P(X=3) = 0.006074 There are 3 ways to have these results, so: P(X=2 in 2 trials and X=3 in 1 trail) = 3 . P(X=2) . P(X=2) . P(X=3) = 4.70579E-05 Or about once in every 21250 runs. (Ivor now waits for some probability guru to find the mistake(s!) in his calculations.)
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21st December 2007, 05:23 AM	#10
Southwind17 Illuminator Join Date: Sep 2007 Location: Location: Location Posts: 4,461	Originally Posted by Ivor the Engineer This follows a Binomial distribution: P(X=r) = nCr . p^r . (1-p)^(n-r) Where: n is the number of coins per trial r is the number of heads (or tails) in each trial p is the probability of obtaining a head or tail. nCr = n!/(r!(n-r)!) If both 9 heads and 9 tails are being considered, the probability is multiplied by two. So for n=10, r=9 and p=0.5: P(X=9) = 10C9 . 0.5^9 . (1-0.5)^(10-9) = 10 . 0.001953125 . 0.5 = 0.009765625 multiplying by 2 (because 9 heads and 1 tail is considered the same as 1 head and 9 tails) gives: P(X=9) = 0.01953125 Over 20 tosses this is increased to 0.390625. I.e. a 39% chance of obtaining 9 heads and 1 tail, or 1 head and 9 tails. Using the binomial distribution again for the n (20) trials and answering the question in the OP: n = 20; r = 2; p = 0.01953125 P(X=2) = 0.050818 and: n = 20; r = 3; p = 0.01953125 P(X=3) = 0.006074 There are 3 ways to have these results, so: P(X=2 in 2 trials and X=3 in 1 trail) = 3 . P(X=2) . P(X=2) . P(X=3) = 4.70579E-05 Or about once in every 21250 runs. (Ivor now waits for some probability guru to find the mistake(s!) in his calculations.) What about Ivor, while he's waiting, commenting on whether Dale's description of this most extraordinary feat seems accurate?!
	__________________ The views expressed here do not necessarily represent the unanimous view of all parts of my mind "Always" and "never" are two words that you should always remember never to use. Facts do not cease to exist because they are ignored. Why is it that when I ask for a pair of hands a brain comes attached?

21st December 2007, 07:33 AM	#11
Beth Philosopher Join Date: Dec 2004 Location: Flatland Posts: 5,307	Originally Posted by Ivor the Engineer (Ivor now waits for some probability guru to find the mistake(s!) in his calculations.) Your calculations all look correct to me.
	__________________ Beth "You are not the stuff of which you are made." Richard Dawkins, July 2005, 10:45 http://www.ted.com/talks/richard_daw..._universe.html

21st December 2007, 10:38 AM	#12
Dale Student Join Date: Feb 2006 Posts: 31	Question: Is there some restriction on length with this "Quick Reply"? I'm having trouble posting anything but one-liners.
Dale Student Join Date: Feb 2006 Posts: 31

21st December 2007, 11:05 AM	#13
Southwind17 Illuminator Join Date: Sep 2007 Location: Location: Location Posts: 4,461	Originally Posted by Dale Question: Is there some restriction on length with this "Quick Reply"? I'm having trouble posting anything but one-liners. Well given what you've just posted runs into two lines in the Quick Reply box, I'd say not! Seriously, you should be able to type as much as you like, so far as I know.
	__________________ The views expressed here do not necessarily represent the unanimous view of all parts of my mind "Always" and "never" are two words that you should always remember never to use. Facts do not cease to exist because they are ignored. Why is it that when I ask for a pair of hands a brain comes attached?

21st December 2007, 12:16 PM	#14
blobru Philosopher Join Date: May 2007 Posts: 6,591	Originally Posted by Dale Ten new pennies are placed in a plastic cup, shaken and tossed on a smooth tabletop with no obstructions of any kind to influence the spread. The distribution of heads and tails is counted. What is the probability, in three runs of 20 tosses, that I will get two nine and one combinations in two of the runs and three nine and one combinations in a third? I've done this toss many, many times because, apart from the obvious head and tail count, the distributions themselves are interesting; how many 7s and 3s, 6s and 4s, the occasional 8 and 2, etc. (also because I've got a lot of time on my hands.) but, I've never before gotten a nine and one. As shown above, your latest result isn't too far out. But what I find odd is that you'd never gotten a single 9 and 1 before? Ten pennies, two sides, ten places for the 1, two ways to get 9 and 1: 2¹⁰ / 10 * 2 = 62 so you should expect on average to see one 9 and 1 every 62 tosses, or very nearly one per 3 runs of 20 tosses. But you hadn't seen any before? (61/62)^r*20, r = runs gives the odds of not getting a 9 and 1 for however many runs. How many runs do you estimate you did? Quote: Then just this morning I decided to introduce an anomaly; I substituted a Canadian penny for one of the US coins. The Canadian coin is identical in diameter and thickness but, of course, has different images on front and back (obverse in coin person language.) It also has a pronounced ridge around the outside edges front and back. I designated the image of the queen as heads and the maple leaf as tails. I did the three runs of 20 throws and got the results noted above. Now, I find it difficult to attribute such a profound difference in results to the simple substitution of one almost identical coin; especially since the procedure I'm using has not changed. Could it be that the physical mixing of the coins has been altered due to the one oddball . Are aliens watching over my shoulder and influencing the toss? Is Shirley McLaine somehow involved? My dear bride of many years has suggested that this may be my "lucky day" and I should run out and buy a lottery ticket. Any comments? Cheers, D Maybe the Canadian penny turned the Americans into socialists?
blobru Philosopher Join Date: May 2007 Posts: 6,591	__________________ "Say to them, 'I am Nobody!'" -- Ulysses to the Cyclops "Never mind. I can't read." -- Hokulele to the Easter Bunny Last edited by blobru; 21st December 2007 at 12:22 PM.