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Derren Brown's coin tossing game
The coin tossing game Derren Brown played in the How to win the lottery programme is known as Penney's game and it is an example of a non-transitive game.
In most games if player 1 is better than player 2, and player 2 is better than player 3, then it follows that player 1 is better than player 3. This is called the transitive property.
Not all games are like this. For example Rock-Paper-Scissors. Although rock beats scissors, and scissors beats paper, rock does not beat paper. So it's not transitive. If you know which of the three your opponent is going to go with, you can always beat him.
Penney's game is the same, albeit more complicated.
You'll remember than player 1 chooses a three coin sequence. Player 2 follows the algorithm Brown explained. The algorithm doesn't guarantee player 2 a win, it just makes it more likely that he will win than player 1 will win.
For example: if player 1 chooses HHH and player 2 chooses THH. Player 1 can only win if the sequence HHH comes up immediately in the first three tosses (with probability 1/8). If a T comes up at any point in the first three tosses then player 1 is doomed because player 2 now only needs HH whereas player 1 needs HHH - but this includes the HH player 1 needs. So player 2 will win 7 times out of 8 on average. (This scenario is the most convincing win for player 2 and was the one used in the programme. I wonder if Brown either fiddled it so that player 1 chose HHH, or kept filming it until he found a player who did.)
The same argument applies to TTT vs HTT.
For the other sequences, the idea is essentially the same. To win you choose a sequence that gets completed before your opponent's. So you need the last two items in your sequence to be the first two in your opponent's, so that once you've achieved your last two, he still has one more to go, but you're finished. (This happens when you put the switched letter in front of your opponent's sequence, making your last two his first two. And it explains why you ignore his last letter: you've already won by then, so it doesn't matter what his last one is.) Then you need your first item to "block" his sequence from being completed, so you make it the opposite of his second item. So if player 1 chooses TTH player 2 chooses HTT. If they start with a H then player 1 is doomed -- to complete his sequence he now needs TT (and then H) but the TT completes player 2's sequence first (and we don't care about player 1's final H). If they don't get the TT after the initial H, then they must have got an H straight away or TH. In this case they're back where they started -- with an H -- so player 2 still has the advantage. If they start with a T then it all depends on the second coin toss. If it's an H, then player 1 has been "blocked" and has to start again, but player 2 is now in the position just described -- he only needs TT, whereas player 2 needs TTH. So player 1 can only win when the initial two tosses are TT, which has probability 1/4. So player 2 wins three times out of four.
In most games if player 1 is better than player 2, and player 2 is better than player 3, then it follows that player 1 is better than player 3. This is called the transitive property.
Not all games are like this. For example Rock-Paper-Scissors. Although rock beats scissors, and scissors beats paper, rock does not beat paper. So it's not transitive. If you know which of the three your opponent is going to go with, you can always beat him.
Penney's game is the same, albeit more complicated.
You'll remember than player 1 chooses a three coin sequence. Player 2 follows the algorithm Brown explained. The algorithm doesn't guarantee player 2 a win, it just makes it more likely that he will win than player 1 will win.
For example: if player 1 chooses HHH and player 2 chooses THH. Player 1 can only win if the sequence HHH comes up immediately in the first three tosses (with probability 1/8). If a T comes up at any point in the first three tosses then player 1 is doomed because player 2 now only needs HH whereas player 1 needs HHH - but this includes the HH player 1 needs. So player 2 will win 7 times out of 8 on average. (This scenario is the most convincing win for player 2 and was the one used in the programme. I wonder if Brown either fiddled it so that player 1 chose HHH, or kept filming it until he found a player who did.)
The same argument applies to TTT vs HTT.
For the other sequences, the idea is essentially the same. To win you choose a sequence that gets completed before your opponent's. So you need the last two items in your sequence to be the first two in your opponent's, so that once you've achieved your last two, he still has one more to go, but you're finished. (This happens when you put the switched letter in front of your opponent's sequence, making your last two his first two. And it explains why you ignore his last letter: you've already won by then, so it doesn't matter what his last one is.) Then you need your first item to "block" his sequence from being completed, so you make it the opposite of his second item. So if player 1 chooses TTH player 2 chooses HTT. If they start with a H then player 1 is doomed -- to complete his sequence he now needs TT (and then H) but the TT completes player 2's sequence first (and we don't care about player 1's final H). If they don't get the TT after the initial H, then they must have got an H straight away or TH. In this case they're back where they started -- with an H -- so player 2 still has the advantage. If they start with a T then it all depends on the second coin toss. If it's an H, then player 1 has been "blocked" and has to start again, but player 2 is now in the position just described -- he only needs TT, whereas player 2 needs TTH. So player 1 can only win when the initial two tosses are TT, which has probability 1/4. So player 2 wins three times out of four.
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