Log in

View Full Version : David Sklansky - Tournament Poker For Advanced Players


folding_the_nuts
05-22-2007, 09:01 AM
Hi!!
I've got some problems with one of david's calculations!
I don't know if I and my folks are just too stupid for this one or i miss something in the context because English isn't my native tongue.

My problem:

Chapter You're broke You're done:

two bets:
$200 to $100 on a coinflip
$120 to $100 on a coinflip

first bet is described in the book:
Expected value: 0,5 * (-100 $) + 0,5 * (200$) = 50

sounds pretty simple
According to this I Apply this formula on the second bet:

Expected value : 0,5* (-100$) + 0,5* (-120$) = 10

he now claims that if he had the money and would make both bets he would have an EV of 35

That's my problem. At first glance I see that the average of my bets should be 30

Using the old formula:
I would win one time bet 1
I would win one time bet 2
I would lose 2 times ( 1+2)

0,5* (-100$) + (0,25 * 120$) + ( 0,25* 200$) = 30

that's my reasoning

David makes it so without detailed explanations:

0,5* (-100$) + 0,25*( 20$) +0,25 * (320$) which gives us an EV of 35

Can someone help me with this one. I'm pretty pissed off that I don't get this calculation

Formal and informative replies are appreciated

Folding_the _nuts

PokrLikeItsProse
05-26-2007, 09:09 AM
I do not have the book with me, but I believe this is the section where you have to take the $120-for-$100 bet first before taking the second bet.

--50% of the time, you lose the first coin flip and don't get to make the second wager, so you lose $100.
--25% of the time, you win the first coin flip and lose the second, so you gain $120 from the first wager and lose $100 on the second wager for a net profit of $20.
--25% of the time, you win both, gaining $120 on the first bet and $200 on the second bet.

So the EV of taking a slightly positive wager before taking a massively positive wager that you can only take if you win the first wager is lower than if you skipped the slightly positive bet.