Predicting randomness

If they are "all 50%" why are the payoffs different (roulette)?

The payoffs in Roulette vary from 35 to1 down to 1 to 1.

Why the difference if it is "all 50%"

 

Black versus red is 50% minus the house number zero

 

could you spoon feed me here. I get 25% for 2 flips showing 2 heads.

1/2 *1/2 = 1/4

 

Correct and back to the example..

If you ask me to bet on a series of events and you give me odds (which is what you are doing when you propose that out of a series of 5 flips, three "heads have already come up). If I have only to see two more such events, if the risk/reward is suitable I will take the bet...

 

So you got your own numbers wrong while trying to belittle everyone else... hysterical

And the odds at every stage of the streak are still 50% so there is no risk reward edge

 

LOL.

 

You are correct, my typo

 

I think you are confused as to what a typo is and what bad math is

 

i did not intend to change the course of this thread. I was just curious. i think steve46 is clear in his explaination

 

You know I am wrong occasionally. I do make mistakes. Certainly that was my typo

And yet one has to wonder how it is that you can't understand that runs occur, but they cannot occur unless the intervening steps are in place...A run of 5 can't occur unless there have been 4 previous heads....Somehow the logic doesn't click for you..

You can point to my mistake and confuse the issue, but it doesn't change the logic...by all means continue as you were.