
Incorrect solution for "Pennies"
Since the name of the game is to flip _until_ someone wins, the second scenario of
HHT
does not result in a tie, but in 'O' winning and 'X' losing. Thus, the odds are equal.
Simon Hall
Tuesday, November 19, 2002
The explanation is not clear, and you've misinterpreted it.
Each side is only concerned with THEIR OWN PENNIES.
So what the poster giving the reply is saying is that if 0 has three flips of his coin and the sequence comes up HHT he will have won once. If X has three flips of his own coin and the result comes up HHT he will have won once. Now it is possible that there is a tie; for example is O comes up HH and X comes up HT after two rounds the result is a tie. It is also possible that the game could be over before the three flips. So if O has flipped HH on his first two fliops and X has not flipped HT, or viceversa, then the game is over, and the third flips don't count.
It is when there is a tie (that is to say that both sides win) that the reason 0 has the odds in his favour become clear.
As we are only concerned with the last 2 flips it is clear that there is only one scenario where there is a tie.
0 flips HH with his coin, and X flips HT
Now they must make the third flip, for the tie breaker  and it is clear that 0 has a 50% chance of winning since if his coin shows up heads he has two heads on the run, but X has a 0% chance of winning on the third flip because as HIS last coin was a T it doesn't matter what he flips.
Now the odds are only skewed if after both X and 0 win the tie break is decided by flipping a third coin, and taking the second coin flipped into account. If the whole game starts again from zero, or if they are both considered to be winners, then the odds will be equal.
What we have in effect is a badly worded answer to an even worse worded question.
Stephen Jones
Tuesday, November 19, 2002
Even if a tie results in a restart the odds are not equal.
Each player has 1/4 chance of winning in exactly two tosses.
(Here by 'win' I mean get their sequence  if both players get their sequence it is a tie).
If we consider the chances of winning in exactly three tosses, 'you' have only one way to win (1/8th chance) with THH. 'I' have two ways to win: HHT and THT (2/8 chance).
The chances of winning in exactly four tosses are: 'you' have two chances (TTHH,HTHH) and 'I' have three chances (TTHT, HHHT, THHT).
As the original answer said, when the HT guy loses, he is set up to get a win next toss. The HT guy cannot win next toss.
David Clayworth
Thursday, December 12, 2002
Oops. Of course "the HH guy cannot win next toss".
David Clayworth
Thursday, December 12, 2002
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