This is a trader's statement

"Andrew Tobias told of people he knew who made a bundle buying a batch of options or "playing them long," then lost it all trying to repeat that success. If you have a 50-50 chance of winning and not losing once, you have only one chance in four of doing it twice in a row, and just one in eight of three consecutive wins. Thus the third try is Russian roulette with seven of eight chambers loaded. 10 doublings in a row increase the money a thousand-fold but you have only one chance in a thousand of doing it 10 times in a row."

with this reasoning if you play and lose your chance of losing twice in a row must be less?

Originally posted by luckystrike:

... if you play and lose your chance of losing twice in a row must be less?

No. Your chance of losing twice in a row IS less than losing once. However, your chance of losing twice in a row, given the fact that you have lost once already, is exactly the same chance as losing once.

It's called conditional probability and it differs to regular probability by assuming that certain events have already taken place.

For example, the probability of rolling a 6 followed by another 6, on a regular die, is 1/6 X 1/6 = 1/36.

The probability of rolling a 6 with 1 roll is 1/6.

But, the probability of rolling a 6 followed by another 6, given that your first roll was a 6, is 1/6 i.e the probability of rolling a 6 on your second roll.

Furthermore, the probability of rolling 100 6's in a row, given that your first 99 rolls were 6's, is still just 1/6.

Hope this helps

Geoff