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How to win Free Drinks with the 'Heads-Tails Probability Trick'......
........... or lots of money, or whatever............
If you flip a fair coin there is the same chance of getting Heads as Tails. So if you bet lots of people a drink based on one coin flip you would win as often as you lose and end up breaking even. If you flip a coin twice, there are equal choices of getting HH, HT, TH and TT. Still no advantage. Three flips still have exactly equal odds between HHH, HHT, HTH, THH, HTT, THT, TTH and TTT. What most people don’t realize is that this only applies when you flip the coin only three times and then start again if no one wins. If you just keep flipping until someone wins, the odds are very different. Let’s say you are betting someone a drink based on three flips and they predict HHH. You might choose HTH. Then the coin flips are: THHTTHHTTTHHH [nobody wins on the first three flips, so you keep flipping.......... they eventually win because HHH happens before HTH] But was this fair? If they choose HHH can you make your choice to give yourself an advantage? Yes you can! If you choose THH there is an 87.5% chance that you will win. After the first three flips, all eight outcomes are equally likely, so there is a 1/8 [= 12.5%] chance they will win straight away. After this, it is impossible for HHH to occur in a string of heads and tails before THH! Some examples from when I flipped a coin: TTHTHHH [your THH wins before their HHH occurs] THTHHH [ditto] In order to get HHH, there must already be a THH. So you have an 87.5% chance of winning. You can use this ‘cutting off at the pass’ technique for any prediction they make. You choose your guess so it ends in the same two that theirs starts with. Then pick your first one so that your guess is not a palindrome. There will always be a better chance of you winning! They choose You choose Odds you win HHH THH 87.50% HHT THH 75% HTH HHT 66.70% THH TTH 66.70% HTT HHT 66.70% THT TTH 66.70% TTH HTT 75% TTT HTT 87.50% So you always have a better than 50% chance of winning. Across all the options, you have a 74% chance of winning on average. This means that for every four times you bet someone, you will have to buy one drink but you will get three drinks for free. Therefore after one hundred bets you will be fifty free drinks ahead! It is an unfortunate corollary that after the first ten to twenty bets you will run out of friends to drink with. (C) Matt Parker 2010 |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Are you bored?
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Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
All that bloody figuring just for a free drink.
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Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Now I remember why I killed my statistics professor :D
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Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by The Dean
(Post 10898206)
What most people don’t realize is that this only applies when you flip the coin only three times and then start again if no one wins. If you just keep flipping until someone wins, the odds are very different. Am I missing something here? |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by weasel decentral
(Post 10899058)
This guy seems to be trying to tell us in very complicated language that if you flip a coin three times and heads comes up each time (or at least twice), then the statistical likelihood of tails on the next flip is very high. (or vice versa)
Am I missing something here? Each individual flip is 50:50 but the three-flips-sequence is very different. It works - the maths is correct. |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by The Dean
(Post 10899664)
Yes - you are missing the truth.
Each individual flip is 50:50 but the three-flips-sequence is very different. It works - the maths is correct. Anyhow i wasn't disagreeing with the premise, just that he made it needlessly complicated. Ask yourself this if you flipped a coin and got three heads in a row and had to bet on the outcome of the fourth, would you say heads or tails? It seems he has just proved what everyone would intuitively know. |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by weasel decentral
(Post 10899682)
Yeah its different in the sense it's an aggregate :rofl:
Anyhow i wasn't disagreeing with the premise, just that he made it needlessly complicated. Ask yourself this if you flipped a coin and got three heads in a row and had to bet on the outcome of the fourth, would you say heads or tails? It seems he has just proved what everyone would intuitively know. You're missing his point - read it again. If your opponent bets on HHH, and it doesn't come up in the first three flips, you then have an 87.5% chance of winning. |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by weasel decentral
(Post 10899058)
This guy seems to be trying to tell us in very complicated language that if you flip a coin three times and heads comes up each time (or at least twice), then the statistical likelihood of tails on the next flip is very high. (or vice versa)
Am I missing something here? You choose your guess so it ends in the same two that theirs starts with. Then pick your first one so that your guess is not a palindrome. There will always be a better chance of you winning! Anywho, this is one I haven't got my head round yet... http://www.bbc.co.uk/news/magazine-24045598 Imagine Deal or No Deal with only three sealed red boxes. The three cash prizes, one randomly inserted into each box, are 50p, £1 and £10,000. You pick a box, let's say box two, and the dreaded telephone rings. The Banker tempts you with an offer but this one is unusual. Box three is opened in front of you revealing the £1 prize, and he offers you the chance to change your mind and choose box one. Does switching improve your chances of winning the £10,000? |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by Bahtatboy
(Post 10899701)
Yes. It's not so much statistics, as pure logic:
You choose your guess so it ends in the same two that theirs starts with. Then pick your first one so that your guess is not a palindrome. There will always be a better chance of you winning! Anywho, this is one I haven't got my head round yet... http://www.bbc.co.uk/news/magazine-24045598 Imagine Deal or No Deal with only three sealed red boxes. The three cash prizes, one randomly inserted into each box, are 50p, £1 and £10,000. You pick a box, let's say box two, and the dreaded telephone rings. The Banker tempts you with an offer but this one is unusual. Box three is opened in front of you revealing the £1 prize, and he offers you the chance to change your mind and choose box one. Does switching improve your chances of winning the £10,000? In that case, SWITCH - you are improving your chances from one-in-three to one-in-two. |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by The Dean
(Post 10899771)
In that case, SWITCH - you are improving your chances from one-in-three to one-in-two.
Your odds if you switch go from 1/3 to 2/3. Your first choice was a 1/3 shot, and if you assume you picked the wrong box (the most likely outcome) you are now switching to the right box (the most likely outcome) which means a 2/3 shot. You can only lose if your first guess was right (that's the 1/3 shot) The error is to assume that it's a stand alone guess between 2 doors, that the initial assumptions have no bearing on the outcome. Similar to the coin toss example. Its called variable change in statistics. |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by weasel decentral
(Post 10900166)
It must look like I am hounding you but here I am again :p
Your odds if you switch go from 1/3 to 2/3. Your first choice was a 1/3 shot, and if you assume you picked the wrong box (the most likely outcome) you are now switching to the right box (the most likely outcome) which means a 2/3 shot. You can only lose if your first guess was right (that's the 1/3 shot) The error is to assume that it's a stand alone guess between 2 doors, that the initial assumptions have no bearing on the outcome. Similar to the coin toss example. Its called variable change in statistics. I still see it as improving your original 1/3 to a better 1/2. |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by The Dean
(Post 10900176)
I agree you should switch, as I said, but I don't see the 2/3. There aren't TWO boxes with good prizes - and as I said the host must know which box to open (so he can always show you a box with a poor prize, whichever one you have chosen).
I still see it as improving your original 1/3 to a better 1/2. Its something that I guess you either get or don't. Anyhow it's a fairly well known and proved premise in statistics. There is even a film called '21' with Kevin Spacey about card counters in Vegas, based on a real life story about students who applied this method to beat the odds in blackjack. |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
From your earlier post:
"....you are now switching to the right box (the most likely outcome) which means a 2/3 shot...." Try explaining that to me again - I can be quite intelligent when really challenged. It's the 2/3 bit I don't get. Once the host opens a box, he isn't simply giving you the chance to switch - he is inviting you to play a completely new game. |
Re: How to win Free Drinks with the 'Heads-Tails Probability Trick'......
Originally Posted by The Dean
(Post 10900183)
From your earlier post:
"....you are now switching to the right box (the most likely outcome) which means a 2/3 shot...." Try explaining that to me again - I can be quite intelligent when really challenged. It's the 2/3 bit I don't get. Once the host opens a box, he isn't simply giving you the chance to switch - he is inviting you to play a completely new game. The possible scenarios of each box are (W for win, L for lose), imagine them laid out in a row. 1: WLL 2: LWL 3: LLW Picking scenario 1 to 3 and choosing box 2 (for example) in each case shows that you will win in one case (no 2) and lose in 2(no 1 and no 3). Now imagine he has opened a losing box in each case and asks you to switch or stick. There is only one scenario where you can lose, that's in no. 2 because you had already chosen the right box, in the other two you are switching from the losing position to the winning position. Hence why it is now a 2 out of 3 shot to win by switching. As there were 3 scenarios in play not a simple choice between 2 boxes. |
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