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Probability

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joe1 | 20:56 Fri 18th Dec 2009 | Science
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I know that on every spin of a coin the chances are 50/50 heads or tails. But on say 100 spins there will be close to 50 /50 of each.
So if after say 20 spins and all heads - are the chances of a tail on next still 50/50?

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In theory yes, but in practice 20 spins all the same might well indicate that the coin has been cast with a defect.
Yes, but the chances of tails on the next spin is not the same as the whole exercise of spinning 21 times.

Even spinning a coin twice already reduces the chance of both being heads to 1 in 3.
wildwood ,
The chance of two heads after 2 tosses is 1 in 4 or 1:3. Not 1 in 3 as you state.
The 4 possibilities are:
HH
HT
TH
TT
As you see, only one in four of those is HH.

As far as the OP's question is concerned:
Yes 50/50, but I agree with Jumbuck re possibility of a biassed coin since chance of 20 heads in a row with a normal coin is 1 in 2^20 or 1 in 1,048,576
Ahhh yes, but to quote Terry Pratchett...

"Million-to-one chances crop up nine times out of ten"
The coin has no memory, it does not matter if you spin 1000 heads, the chance of a head is still 50% on the next spin. (ok pedants I'm ignoring the hole "edge" thing). The key here is the phrase "from the outset" if you said from the outset what is the chance of 20 heads followed by a tail then the answer is astronomical but after each spin you are back to 50/50.
My favourite roulette story (and if it isn’t true it could easily be) is of the chap who put $10 on red and let it ride. He got ten reds in a row and walked off with a profit of $10,240. Even so, that table made a fortune for the casino. Why? Because as his reds piled up all the other punters were sure that the next spin MUST produce a black and so made increasingly large bets on black. They were simply unable to understand that it was 50/50 every time no matter what had gone before.
There is a very well known probability theory, often used as a Roulette system known as the Martingale. In brief the theory, which is centred on the even money bets in Roulette (Red/Black, Odd/Even, High/Low) suggests that doubling one’s stake after a loss will eventually result in recovering all previous losses. In practice it does not work because in many cases, before the recovery is made, the punter either runs out of money or he reaches the house’s table limit and cannot bet any higher. A more sophisticated system known as the Labouchere employs a stakes plan to limit losses, but is still not a genuine system proven to work.

A book entitled “Thirteen Against the Bank” by Norman Leigh tells the exploits of a group of punters who took on the casinos successfully employing a reverse Labouchere staking system (too involved to detail here) covering all the even money bets on the roulette table. This was a successful strategy and depended on the fact that “long runs” occur within 50:50 probabilities and the system exploited those long runs whenever they occur.

These long runs are similar to the tossing of a coin you describe. Every toss is 50:50, but from time to time long runs of heads or tails will occur. The cumulative odds of those runs occurring (and the winnings if they are exploited) are high, but they do occur.

In theory the longer a series of tests (coin tosses) is, the closer to 50:50 will be the results. But long runs do occur, even though every toss is 50:50.
You're right, New Judge.

A Martingale works in theory, but not in practice.

Both the Martingale and the coin spinning work on the principle of BIG NUMBERS

The bigger number of "goes" in your sample, the more likely there will be a tendency towards a mean outcome.

If you flipped a coin 2 times, you would not be too shocked if the result was 100% heads, 0% tails.

If you flip a coin 1,000 times, you would expect the results to tend towards 50/50.

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