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# The spectacular coin toss

## Question

bonanova    76

There was a bet going on at Morty's last night,

about who could toss the most consecutive Tails

in 10 flips of a fair coin.

Davey went first and came up with H T T H T H T T T H -- 3 Tails in a row

Tom was next, and he flipped T T H T H H T T H H -- only 2 Tails in a row, but twice.

Next was Pete, who struck out completely with T H T H T H T H T H -- none.

Slim was next, starting out with 5 Tails; T T T T T H H T H H -- 5.

Then the unthinkable happened: Alex flipped 10 consecutive Tails.

Phil grabbed a pencil and tried to figure the odds, but he'd had too many beers.

So had Davey, but it didn't stop him from opining that none of them, in any of

their lifetimes, would see again what Alex had done.

They all agreed, and toasted Alex the rest of the night.

Slim felt slighted, saying his sequence was improbable enough for him to at

least have gotten a free beer. And Davey and Tom mumbled that it would be

tough to repeat what they'd done, too. Finally, Pete claimed that even his

result deserved a frosty one, "on the house."

But Matt the mathematician -- when he was sober at least -- would have none

of the whining. "Anybody could do what you blokes done t'night," he snarled,

"but it would take a bazillion years to do what Alex did."

Who was right?

Departing from my normal MO, here's my answer.

Alex's result was remarkable in that the "goal" was consecutive Tails.

But the others' claim that it would be difficult to repeat what they'd done was valid.

The probability of repeating any of the results is 1 in 1024.

## 5 answers to this question

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Guest

If you are only talking about the number of tails in a row, and not repeating anbody's particular sequence (except for the 10 in a row), then they are all wrong.

The best that any of them did - 5 in a row would be accomplished about 3 times out of 100. Tough, but not spectacular. Therefore they are all whiners and wrong about how tough their tosses were. Matt's wrong, because he overuses hyperbole and uses incorrect english. His toss isn't one in a bazillion years, but one in about a day's worth of consecutive tosses if each toss takes 10 seconds and the law of large numbers holds.

(for the math, I figured 1/1024 times the 10 in a row would hit. 10 seconds for each toss, 10 tosses each try, a full 1024 tries between "hits" = 102400 seconds. About 28 hours between hits.)

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bonanova    76

Great analysis!

But ...

Matt's wrong, because he overuses hyperbole and uses incorrect english.

Take it easy on Matt ... it took me 5 tries to develop his character.

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lol

let me just break this down for anyone still clueless:

every combination of ten binary flags is 2^10 or 1024.

Thus ANY combination of ten 0s or 1s (or tails/heads) has a 1 in 1024 chance of being realized, or flipped, in other words.

Thus 1111111111 isn't any more unlikely than 011010111. They have equal probabilities. Of course if you get the second one, you'll be like "okay" but if you get all 1's youll be like "whoa, crazy, what are the odds!" Just because it is a spectacular result. In this case, it was the goal which means it is desired to get.... and suddenly the odds of getting a non-all-heads are 1023/1024 and the odds of getting all heads are 1/1024

I have a question for you...

Jonny has flipped: 101101 (6 flips so far)

Albert has flipped: 0001000 (7 flips so far)

What is the probability that Albert will get MORE 1's IN A ROW than Jonny?

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bonanova    76
I have a question for you...

Jonny has flipped: 101101 (6 flips so far)

Albert has flipped: 0001000 (7 flips so far)

What is the probability that Albert will get MORE 1's IN A ROW than Jonny?

11/128

Here's why:

Johnny has two consecutive 1's, and, depending on his final 4 flips, he might end up with as many as 5.

Albert has 3 remaining flips and could end up with 1, 2, or 3 consecutive 1's.

For Albert to end up with MORE than Johnny,

[1] Albert would need 3 and

[2] Johnny would have to keep his total at 2.

Probability of Albert getting 3 1's is 1/8.

Johnny can flip 16 different outcomes.

Of those 16, 4 have the first two as 1's; of the other 12, 1 outcome has the last three as 1's.

Thus 5 of 16 possible outcomes give Johnny 3 or more consecutive 1's.

The other 11 outcomes leave Johnny's total at 2.

So the probability of [2] 11/16.

Since the events are independent, the joint probability is the product:

1/8 x 11/16 = 11/128.

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hooray!

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