Uh oh, here we go: CBS claims Pats win coin flip at "impossible clip"

I don't really trust a CBS story, let alone one referencing a Boston Herald "article," so I took a look at it and I'm not really sure what data they were using, but it might be wrong.

First, it seems silly to point this out, but apparently CBS and the Boston Herald don't understand the basics, so I will. We're talking about extremely small sample sizes. 50/50 doesn't mean you win one, your odds to lose the next one go up. It's still 50/50 each time. Strange streaks happen. The Browns lost their first 11 coin flips in 2011.

Plus it's the Herald, so I am not sure I trust the accuracy of anything published on that toilet paper. So I took a look at Pro Football Reference.

2015 regular season: 5-2
2014 regular season: 12-4
2014 play-offs: 2-1
2013 regular season: 6-10
2013 play-offs: 0-2
2012 regular season: 8-8
2012 play-offs: 0-2

Regular Season: 33-25
Play-offs: 2-5

That comes out to 19-7 over 26 games if they included the play-offs. My guess is they included the play-offs and chopped off the first game of the 2014 season, where we lost the coin flip, because we had lost our previous 5 coin flips, and 19-12 isn't as sensational a claim. And 35-30 over the past 3 and a half years isn't really anything to get excited about. We didn't cry when we went 6-10, nor did we care when we went 12-4.

I don't know what the odds are of going 35-30 over 65 games, but I'm fairly confident that 99% of stuff in the Boston Herald is total ****ing trash.

Oswlek said:

If each toss remains 50/50 and none of those variables alter NE's ability to predict the outcome, then the odds remain unchanged.

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On individual flips you are right. On multiple flips where the Pats don't always call one side every time, it is no longer 50-50. It changes the odds you will get it right.

Statistics will always be affected when more variables are introduced into the equation. The 50-50 coin flip is one of the most simplified equation. It states that if you flip a coin ten times, the odds are that five will be heads and five will be tails. If you guess heads on all ten flips, you have 50% chance of being right.

But if you go on the assumption that 50% will be heads and 50% will be tails and you try to guess which one is going to be each on each flip and call heads five times and tails five times, the odds of you being right are far less than 50%.

Rob0729 said:

Huh?!? So are you arguing that if you flip a coin 50 times that the odds that it lands heads is 25 times AND if the Pats randomly call heads or tails on those tosses and the odds of the getting it right is 25 times? So so you could not be more wrong.

The 50-50 coin flip is simple variable equation based on minimal variables of one person flipping a coin over and over again and it coming down either heads or tails. Whenever you introduce new variables to an equation, the odds change. Some variables have very minimal effect and others have significant effect on the variables.

Everything in probability is effected by the variables that in the equation. That is why these football game predictor modules play out football games thousands of times because they need to calculate every variable.

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If a coin is flipped 50 times and the Patriots call it in the air each time, basic probability says they will get it right about 50% of the time (better yet, if the coin is flipped 500 times, they'll be even closer to 50% right). That's so straightforward that I must be misunderstanding what you're arguing about, but that's how I'm reading you. Think it about it: what if the Pats chose to call heads each of the 50 times vs. "randomly" deciding how to call it each time. Are you saying it would change their odds? Of course not, there's no guessing strategy to improve your chances in a series of fair coin tosses: calling out heads each time is just as good as calling out heads/tails based on how the spirit moves you. It's just like picking "random" numbers for your lottery ticket instead of 1-2-3-4-5-6. It truly doesn't matter.

Also, while football games are simulated, the coin toss problem is much simpler and probabilities can be calculated using the Bernoulli distribution; nobody's running simulations of coin tosses. Two factors makes the coin problem much easier than football games: 1) each toss is independent of every other toss, and 2) each toss has the same probability as every other toss, in other words you're repeating the same experiment. So the Bernoulli distribution can handle even an unfair coin, as long as the same unfair coin is used each time, but it can't handle different games between different teams having different probabilities.

I'm really not sure what else to tell you except that everything I'm saying is from a one semester basic probability and statistics class. You're using terms (like "variable") in a distinctly a non-probability way which makes me think you're trying to apply "general mathy/sciency knowledge" to the problem instead of a specific understanding of probability. If I'm wrong and I'm misunderstanding you, then I apologize for this whole post (which somehow turned into an essay).

FourierSeries said:

That's just not right.

It doesn't matter what the Pats or their opponents choose. The probability they win the toss is still 50%.

If they are at home, there are two possible scenarios:

Scenario 1 - opponent chooses heads. Both teams have a 50% chance of winning.
Scenario 2 - opponent chooses tails. Both teams have a 50% chance of winning.

If they are on the road:

Scenario 1 - Pats choose heads. Both teams have a 50% chance of winning.
Scenario 2 - Pats choose tails. Both teams have a 50% chance of winning.

Only one of those scenarios can occur each game.

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Wrong! It is a fifty percent chance that it lands heads or tails not that they guessed right. You are talking about two different things. I'm sorry, but I took stats in grad school.

If you call heads all the time, it is a 50% scenario over time because you aren't trying to predict which 50% of those coin flips will be heads. If you flip the coin 20 times, odds are 10 of them will be heads.

Now if you switch between calling heads and tails on those 20 flips, you are now trying to predict which of those 20 flips will be heads and which will be tails. It makes the odds actually much lower.

Pujo said:

If a coin is flipped 50 times and the Patriots call it in the air each time, basic probability says they will get it right about 50% of the time (better yet, if the coin is flipped 500 times, they'll be even closer to 50% right). That's so straightforward that I must be misunderstanding what you're arguing about, but that's how I'm reading you. Think it about it: what if the Pats chose to call heads each of the 50 times vs. "randomly" deciding how to call it each time. Are you saying it would change their odds? Of course not, there's no guessing strategy to improve your chances in a series of fair coin tosses: calling out heads each time is just as good as calling out heads/tails based on how the spirit moves you. It's just like picking "random" numbers for your lottery ticket instead of 1-2-3-4-5-6. It truly doesn't matter.

Also, while football games are simulated, the coin toss problem is much simpler and probabilities can be calculated using the Bernoulli distribution; nobody's running simulations of coin tosses. Two factors makes the coin problem much easier than football games: 1) each toss is independent of every other toss, and 2) each toss has the same probability as every other toss, in other words you're repeating the same experiment. So the Bernoulli distribution can handle even an unfair coin, as long as the same unfair coin is used each time, but it can't handle different games between different teams having different probabilities.

I'm really not sure what else to tell you except that everything I'm saying is from a one semester basic probability and statistics class. You're using terms (like "variable") in a distinctly a non-probability way which makes me think you're trying to apply "general mathy/sciency knowledge" to the problem instead of a specific understanding of probability. If I'm wrong and I'm misunderstanding you, then I apologize for this whole post (which somehow turned into an essay).

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Of course it changes the odds. Who is more likely to be right in the following scenario:

You flip a coin ten times:

Person A predicts heads on all ten flips

Person B predicts heads on only flip 1,4,5,7,&9

The odds are five of the flips will be heads. The odds are against that the five head flips will all be on flips 1,4,5,7&9. Person A is far more likely to get more right than Person B.

Rob0729 said:

On individual flips you are right. On multiple flips where the Pats don't always call one side every time, it is no longer 50-50. It changes the odds you will get it right.

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No, This is not just wrong. but spectacularly wrong. Calling heads each time vs. calling heads or tails depending on how you feel on each toss doesn't change the odds, just like calling red on the roulette wheel on each spin is no different from deciding red/black for each spin. You can see this by drawing a probability tree diagram (if you still disagree with me after this then whatever).

Let's say that you decide to call heads about half the time but at random (though it doesn't have to be half). Here is the probability tree diagram:


Because you deciding what to pick and the actual toss are independent events, the multiplication law of probability says you can multiply the two events' probabilities to get their intersection, formally: P(A∩B) = P(A)*P(B). So when you look at the possible combinations, the two in bold (call tails/lands tails and call heads/land heads) are the two where you win, and their combined probability is 25% + 25% = 50%.

I can refer you to Statistics 11th Edition by McClave and Sincich, which I have sitting on my shelf and which I used for my prob/stats classes, if you want an authoritative source, but whether you believe me or not, I'm calling it quits on debating it.

Rob0729 said:

Of course it changes the odds. Who is more likely to be right in the following scenario:

You flip a coin ten times:

Person A predicts heads on all ten flips

Person B predicts heads on only flip 1,4,5,7,&9

The odds are five of the flips will be heads. The odds are against that the five head flips will all be on flips 1,4,5,7&9. Person A is far more likely to get more right than Person B.

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Because they have to specify on what flips (1, 4, 5, 7, and 9) they want heads, the odds are EXACTLY the same is if they'd called all heads (those odds being 2^10=1024 to 1). If, on the other hand, they got to say "5 heads and 5 tails" but not specify which tosses were which, then they'd have much better odds (10 choose 5 = 252, divided by 1024, which is 63/256, or approx 4.06 to 1) versus having to guess that all 10 flips would be heads.

This is why a team going 16-0 IS much less likely than a team going 8-8. There are more ways a team can go 8-8, but with picking coins, you have to decide WHEN each heads/tails comes, versus just saying how many there will be.

Pujo said:

No, This is not just wrong. but spectacularly wrong. Calling heads each time vs. calling heads or tails depending on how you feel on each toss doesn't change the odds, just like calling red on the roulette wheel on each spin is no different from deciding red/black for each spin. You can see this by drawing a probability tree diagram (if you still disagree with me after this then whatever).

Let's say that you decide to call heads about half the time but at random (though it doesn't have to be half). Here is the probability tree diagram:
View attachment 11153

Because you deciding what to pick and the actual toss are independent events, the multiplicative law of probability says you can multiply the two events' probability to get their intersection, formally: P(A∩B) = P(A)*P(B). So when you look at the possible combinations, the two in bold (call tails/lands tails and call heads/land heads) are the two where you win, and their combined probability is 25% + 25% = 50%.

I can refer you to Statistics 11th Edition by McClave and Sincich, which I have sitting on my shelf and which I used for my prob/stats classes, if you want an authoritative source, but whether you believe me or not, I'm calling it quits on debating it.

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Ok, if I am wrong. I am sorry. I will end the argument. Tired of arguing this either way.

Rob0729 said:

Of course it changes the odds. Who is more likely to be right in the following scenario:

You flip a coin ten times:

Person A predicts heads on all ten flips

Person B predicts heads on only flip 1,4,5,7,&9

The odds are five of the flips will be heads. The odds are against that the five head flips will all be on flips 1,4,5,7&9. Person A is far more likely to get more right than Person B.

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It's pretty simple - whether you pick heads or tails, you have a 50% chance of winning on each toss.

By the way, you can win by correctly selecting tails. You have ignorerd the fact that Person B wins if the coin comes up tails on tosses 2,3,6,8 and 10.

What if each individual ref always places the coin head's up or tail's up? Also, what if each individual ref has a controlled mimicking 'flip' action that results in the coin landing on a specific side more often? While most would never notice the tendency, what if someone was then to compile each refs flipping results and identified the tendencies? The 'heads - tails call' is now not completely random.

Yes, even this kind of information would not explain the Patriot coin win success rate. Given the variations of the ground/coin landing spot, it would make any tendency unlikely. I'm just sayin the Patriots could be doing better homework/smarter than everyone else .

Here's a little more analysis:
Using a confidence level. Giving the odds without knowing where it falls in the expected range with respect to sample size is kind of pointless. That's why we use a confidence interval.
For the coinflip:
P = h / h + t = outcome / tosses = 19/25 = 0.73
Now for a 99.7% confidence interval (or 3 Std Deviations), the standard, Z = 3
The maximum error would thus be E = Z / (2 x sqrt(n)), where n is the number of tosses. So, E = 3 / (2 x sqrt(25)) = 0.3
The interval that our probability (0.73) needs to be within (to be 99.7% likely to 'basically be due to chance') is given by: P - E < r < P + E, where r is the actual probability, 0.5 in this case. So... 0.73 - 0.3 < 0.5 < 0.73 + 0.3 which simplifies to 0.43 < 0.5 < 1.03.

What does that mean? It means that winning 19 out of 25 tosses is a reasonably outcome to confidence of 99.7%. You are looking at probability distribution, NOT having an even split (0.50%) because that isn't likely to happen on small sample sizes. The amount of deviation is dependent on sample size.

Here's something I whipped up quickly in rstudio:


This is a graph of 1000 outcomes of a coin tossed 25 times. I used a flat random generator to give a 0 or 1, picked 25 times, then summed the 1s to get the sample outcome for 25 tosses. I then simulated this 1,000 times (obviously with newly generated 'tosses), saved into a database that I then plotted.

This clearly shows how expected (vs actual) distribution works. The actual outcome is much less than what is thought of the expected outcome (even heads vs tails 50% of the time). An even split of heads vs tails only happened ~15% of the time and following a normal distribution outwards. Just as in the confidence interval calculations, having 19 heads come up on the graph is totally indistinguishable from pure chance.

And if that still didn't hammer that home:
If we look at the Patriots last 50 games, they won 29 of the toss. So: P = 29 / 50 = 0.58
E = 3 / (2 x sqrt (50)) = 0.212
0.368 < r < 0.792
Once again, r (0.5, expected odds of a coin toss) is within the 99.7% confidence interval.

SAMPLE SIZE!!!!

And to compare how the graph distribution changes with sample size:
Here's the same graph as above, but with 50 tosses rather than 25:

29 heads (or wins) is very much in the meat of the bell curve...

And for a sample size of 6:

only ~22% of the time will you get an even split.

Rob, you're looking at this the wrong way. It's a binomial distribution. You're seeming to obsess about there being 5 expected heads and 5 expected tails but that is just one of many outcomes clearly. In fact is more likely there is a 6-4 distribution than exactly 5 of each.

I guess I could be mixing up my statistics. Been too long since grad school I guess. I will just admit I am wrong and be done with it and not continue hijacking the thread.

I guess this is why I went into marketing. I can do an awesome online advertising campaign about the Pats being great coin flip predictors if it helps.

Pujo said:

No, while you're right that any game is 50/50, the chances are that over a large number of tosses, you should get close to an even number of heads and tails. With 25 flips there are 2^25=33,554,432 possible outcomes (e.g, HHHHHHHHHHHHHHHHHHHT, HHHHHHHHHHHHHHHHHHTT, etc). The chances of getting exactly 19 heads is 25 choose 19 = 177,100, the chances of getting 20 heads is 25 choose 20 = 53,130, etc. We add up all the possibilities from 19 to 25, which equals 245,506, then divide that the total number of possibilities (the 33 mil. number, which gives us 0.0073...). This isn't despite each flip being independent, it's precisely because of it (otherwise you'd have to calculate the odds using conditional probabilities). So yeah, the Pats are pretty lucky, but 25 coin flips isn't that large of a sample size. If they got 76/100 heads (same % of heads, but over more flips), that might be fishy.

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Especially when they aren't the ones calling half the flips.

Jackson 2 said:

As someone in the comments section pointed out, each weekly coin flip is an independent event. The odds are 50/50 for each weekly flip. The same person flipping the same coin 25 times in a row is different than 25 weekly coin flips by a different official each week. Moronic premise to a moronic article.

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I'm guessing that this is the same as when I'm dumb enough to stand at the roulette table and think that black is definitely going to come up just because red has hit the past 7 times. As you mention, it remains a 50/50 wager despite past results, although it is a bit difficult for me to always grasp that. I want to think that the odds have to change based on recent results, but I guess they don't.

supafly said:

I'm guessing that this is the same as when I'm dumb enough to stand at the roulette table and think that black is definitely going to come up just because red has hit the past 7 times. As you mention, it remains a 50/50 wager despite past results, although it is a bit difficult for me to always grasp that. I want to think that the odds have to change based on recent results, but I guess they don't.

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No no no no no. The question isn't "what are the odds of getting heads on the 26th flip", it's "What are the odds that out of 25 flips, at least 19 will land as heads/tails". Two very different questions. The later requires the previous states be accounted for since it is a sequence of tosses.

A (maybe) familiar from poker: Preflop, AA wins ~85% against 1 opponent (accounting all possible hands). You increase the opponents and the odds drop all the way down to less than 35% with 8 opponents. Now the more parallel situation is the likely hood of being all-in with AA and winning each one. In large tournament play, this is where you simply need to be lucky. As the tournament reaches the later stages, the blinds become so large that it's almost impossible to survive to the end without going all-in on many occasions. If you manage to get dealt AA a number of times and move all in preflop, you have to win EACH time or you are out of the tournament. The previous state is clearly accounted for, since you require to win a sequence. So make the odds a simple as possible: you end up all-in with AA against 1 opponent with a random hand. Here are the odds of winning x-times in a row:
1 : 85%
2 : 73%
3: 62%
4: 53%
5: 45%
... 10 : 20%
Of course you wouldn't likely be lucky enough to always have AA on your all-ins, so if you include a range of JJ, QQ, and KK as well, the odds drop dramatically still. This is (partly) is why serious luck is involved in winning a specific large tournament, but over the course of many many tournaments, the luck averages out and the better players winners win more than they lose.
etc etc. It's the same calculation as flipping x amount of heads/tails in a row. The previous state is necessary to the next state since the question isn't "what are the odds of the NEXT" rather "What are the odds this sequence could occur?"

Satchboogie3 said:

No no no no no. The question isn't "what are the odds of getting heads on the 26th flip", it's "What are the odds that out of 25 flips, at least 19 will land as heads/tails". Two very different questions. The later requires the previous states be accounted for since it is a sequence of tosses.

A (maybe) familiar from poker: Preflop, AA wins ~85% against 1 opponent (accounting all possible hands). You increase the opponents and the odds drop all the way down to less than 35% with 8 opponents. Now the more parallel situation is the likely hood of being all-in with AA and winning each one. In large tournament play, this is where you simply need to be lucky. As the tournament reaches the later stages, the blinds become so large that it's almost impossible to survive to the end without going all-in on many occasions. If you manage to get dealt AA a number of times and move all in preflop, you have to win EACH time or you are out of the tournament. The previous state is clearly accounted for, since you require to win a sequence. So make the odds a simple as possible: you end up all-in with AA against 1 opponent with a random hand. Here are the odds of winning x-times in a row:
1 : 85%
2 : 73%
3: 62%
4: 53%
5: 45%
... 10 : 20%
Of course you wouldn't likely be lucky enough to always have AA on your all-ins, so if you include a range of JJ, QQ, and KK as well, the odds drop dramatically still. This is (partly) is why serious luck is involved in winning a specific large tournament, but over the course of many many tournaments, the luck averages out and the better players winners win more than they lose.
etc etc. It's the same calculation as flipping x amount of heads/tails in a row. The previous state is necessary to the next state since the question isn't "what are the odds of the NEXT" rather "What are the odds this sequence could occur?"

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Thanks for the clarification, but what's the answer to my roulette example?

I was under the impression that it's still a 50/50 prop, despite the recent dominance of one color continuing to come up.

Is this different than what @Jackson 2 was getting at?

First, the laws of physics don't apply for the Patriots and now apparently the laws of probability don't.

Since stupidity and ignorance reign supreme in today's NFL, I'm sure Goodell is somehow going to make this into a big deal.

supafly said:

Thanks for the clarification, but what's the answer to my roulette example?

I was under the impression that it's still a 50/50 prop, despite the recent dominance of one color continuing to come up.

Is this different than what @Jackson 2 was getting at?

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Yes, on a roulette table, the odds of the NEXT spin are independent of the prior (so 46ish% odds with the zero).

If you asked the question "what are the odds of 7 straight reds?", the answer is ~0.0044. So if you thought like the author of the article, you might think "those odds are 'impossible', is something going on?!", neglecting the fact that you expect to see a run like this once every 227 spins, which, if they spin the wheel once a minute, would happen more than 6 times a day on average.

The analysis in the article (0.0073 probability of winning at least 19 of 25 tosses) is correct. The issue is that no analysis on the expected distribution (or confidence interval) was given, which is what tells us what to expect given a certain sample size. As I showed above, 25 coin tosses is WAY too small a sample size to attribute 19 out of 25 to anything but chance. It's within the normal distribution. You expect a very wide range of results with small sample sizes. And of course if you look at the Patriots last 50 or 60 games, the win % comes down closer to 50% and draws further into the middle of the bell curve (normal distribution).

But since it's the Patriots, it can't just be blind luck, it has to be cheating...

supafly said:

I'm guessing that this is the same as when I'm dumb enough to stand at the roulette table and think that black is definitely going to come up just because red has hit the past 7 times.

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There are some things that just feel right to us even though intellectually we know they're wrong. That's one of them. Or if you buy a lottery ticket (which in and of itself you intellectually know is a waste) you're never going to pick 1-2-3-4-5-6, because, of course, those numbers will never come up! When obviously they have as much of a chance as any other set of numbers.

And just to go completely OT, have you ever heard of the Monty Hall problem? The solution to that just seems so wrong.

There's three curtains, behind one is a car, behind the other two are goats. You pick a curtain, say, #1. The host then reveals there's a goat behind #3, and gives you the option of changing your selection. Changing to #2 greatly improves your odds of winning the car. So strange.

Monty Hall problem - Wikipedia, the free encyclopedia