How the odds work...

[Cardcounter]

How much you win is just as important as how often you win. Lets say that I offered you a challenge where you paid me $1 to guess a number between 1 and 100 and if you get it right I pay you $1,000 even though I will win the bet 99% of the time I do not have the best of it. Why?
The 1% I lose the bet I will lose all my money and than some.
In order for me to have the best of it the times I lose can't wipe out all my wins.

 

[QFIT]

Incidentally, this is why banks don't lower credit card rates in a recession. If you pay the credit card bill (i.e. they win) they get 5% to 22% interest minus the cost of money and processing. If you default (common in a recession) they lose 100%, plus the interest the bank had to pay on the float.

 

[jack.jackson]

how about this..............

How much you win is just as important as how often you win. Lets say that I offered you a challenge where you paid me $1 to guess a number between 1 and 100 and if you get it right I pay you $1,000 even though I will win the bet 99% of the time I do not have the best of it. Why?
The 1% I lose the bet I will lose all my money and than some.
In order for me to have the best of it the times I lose can't wipe out all my wins.

If you paid me a $100 every time I got the # right, then that would give me a 1% edge right?

 

[Mimosine]

If you paid me a $100 every time I got the # right, then that would give me a 1% edge right?

i think so....

1 in 100 that pays 100 to 1 =

I think the EV equation would look like this:
(1/100)*$100 - (99/100)*-$1 = $0.01

right???

 

[stophon]

no because you only earn 99 dollars when you win because you paid the $1 to play

 

[UncrownedKing]

no because you only earn 99 dollars when you win because you paid the $1 to play

Exactly. If you win $100 dollars every 100 plays and you pay $1 per play, you would be under $100 dollars when you pay for the 100th play. So it would be break even. Either side would have 0% edge. This is exactly like the odds bet in craps. You would have a 1% edge if the payout was $101 every time you hit, because you would pay $100 to win $101. See how that works?

PS: It's amazing how y'all can calculate variance, EV of every rule variation, effects in EV after each card removed, but when it comes to the most simple probability question y'all fail. With the exception of stophon. :laugh:

PS: By the way, the chances of picking a number 1 - 100 is 1 in 101. So in this scenario, you would need a payout of 102 to 1 in order to have an advantage. 101 to 1 payout would be break even. In order to have an advantage, the payout must be greater than the odds. If the payout is the same as the odds, its break even. If it is anything less than the odds, then the house would have the advantage. Am I making any sense?

 

[EasyRhino]

PS: By the way, the chances of picking a number 1 - 100 is 1 in 101.

Uh, I think that would only be true if you were picking 0-100, inclusive.

Also, carefull if you're describing the bet as paying "100 to 1" or "100 for 1". The "for" language is from slot machines, and means your original bet is included in the payout figure.

 

[UncrownedKing]

Uh, I think that would only be true if you were picking 0-100, inclusive.

Also, carefull if you're describing the bet as paying "100 to 1" or "100 for 1". The "for" language is from slot machines, and means your original bet is included in the payout figure.

Picking 0-100 has the odds of 1 in 102. If you count 0 and 100 there are 102 numbers to pick from. Same for 1-100. If you count 1 and 100 there are 101 numbers, so the odds are 1 in 101.

 

[gordon5432]

Picking 0-100 has the odds of 1 in 102. If you count 0 and 100 there are 102 numbers to pick from. Same for 1-100. If you count 1 and 100 there are 101 numbers, so the odds are 1 in 101.

Check your arithmetic again. The # of inclusive numbers between x and y is (y-x)+1 so the odds of picking a number between 1 and 100 would be 1 in 100, not 101. In practice, a game which pays even money for predicting the flip of a coin is better than a game which pays 99 to 1 for picking a number between 1 and 100 given the player's limited bankroll because of the lower variance (similar to why a game of BJ is better than a game of video poker with the same HA.)

 

[Canceler]

Picking 0-100 has the odds of 1 in 102. If you count 0 and 100 there are 102 numbers to pick from. Same for 1-100. If you count 1 and 100 there are 101 numbers, so the odds are 1 in 101.

Let's simplify this down to 1-2. If I count all the numbers from 1 to 2 inclusive I should get 3? I just can't make that work out.

 

[stophon]

0 isn't a number between 1 and 100

 

[UncrownedKing]

Check your arithmetic again. The # of inclusive numbers between x and y is (y-x)+1 so the odds of picking a number between 1 and 100 would be 1 in 100, not 101. In practice, a game which pays even money for predicting the flip of a coin is better than a game which pays 99 to 1 for picking a number between 1 and 100 given the player's limited bankroll because of the lower variance (similar to why a game of BJ is better than a game of video poker with the same HA.)

You are right. I wasn't calculating (x-y) before adding 1. Sorry guys.

 

[Cardcounter]

In the top sceneraio I give the player a huge advantage because I said you pay me a $1 to guess a number between 1 and 100 to win a $1,000 giving the picker a 900% advantage. In the casino the payout would be like 50 to 1 all they way up to 90 to 1 for picking a number between 1 and 100. So the lose doesn't wipeout the win.

 

[stophon]

In the top sceneraio I give the player a huge advantage because I said you pay me a $1 to guess a number between 1 and 100 to win a $1,000 giving the picker a 900% advantage. In the casino the payout would be like 50 to 1 all they way up to 90 to 1 for picking a number between 1 and 100. So the lose doesn't wipeout the win.

the interesting thing is even though you have 900% adv. a kelly bet still has you betting slightly less than 1% of your bankroll

the fact that optimal betting can be simplified into a very simple algebraic formula really amazes me