256bits said:
If you bet $4 you are betting that the head will come up on the second toss.
If head is non first toss you lose.
If tail is on first toss and tail is on second toss, you lose.
I head is on second toss you win
Wouldn't that be 1 out of 3 to win?
For an $8 bet.
You lose if
H
TH
TTT
You win if TTH
chance of winning is 1/4
etc.
You are doing probability incorrectly and are interpreting the game incorrectly.
Here's how the game works. You say something like "I'll pay $10 to play that game". Suppose the person running the game accepts. You flip a coin. Get heads on the first toss and you get $2 back, for a net loss of $8. Heads on the second toss represents a net loss of $6, on the third, $2. Live past the third toss and you are making a net profit. Live past the 20th toss and you will have won over a million, minimum.
Suppose the person running the game says "Not enough" to your paltry offer. You offer more and more, but the answer always remains "Not enough." At what point should you claim "Too much" and walk away? (Answer is further down.)Here's how the probabilities work. First off, the probabilities for each of the distinct outcomes must sum to one. Look at what you have: 1/2+1/3+1/4+1/5+... That is the divergent geometric series. The sum is not one. So right off the bat something is wrong.
What you did wrong was to erroneously apply the principle of indifference. For example,
256bits said:
If head is non first toss you lose.
If tail is on first toss and tail is on second toss, you lose.
I head is on second toss you win
Wouldn't that be 1 out of 3 to win?
Here you are assuming that the three events (a) heads on the first toss, (b) tails on the first and second tosses, and (c) heads on the second toss are equiprobable events. Your 1 out of 3 would be correct if these events were equally probable. But they aren't. The probability of heads on the first toss is 1/2. For tails on the first and second tosses the probability is 1/4. This is also the probability for the case for tails on the first toss and heads on the second toss. The probability of winning the $4 payoff is 1/4.One way to determine how the amount one should pay to play a game is to compute the expected payoff from the game. For a discrete game such as this, the expected return is the sum of the payoffs weighted by their corresponding probabilities. For this game, the expected return is
\sum_{n=1}^{\infty} $1 \cdot 2^n\cdot\frac 1 {2^n} = \sum_{n=1}^{\infty} $1
You should be willing to pay any finite amount to play in this game!However, there is no gambling house in the world that could cover the potential payoff for this game. In the real world there would be some upper limit beyond which the game would end because you just won the house.
So, to make the game a bit more realistic, let's say that you win $2 if you first flip heads on the first toss, $4 for the second, $8 for the third, and so on. Flip 20 tails in a row and you are paid $1,048,576; game over. Now how much would you be willing to pay to play?