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David Sklansky - Tournament Poker For Advanced Players
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 |
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Re: David Sklansky - Tournament Poker For Advanced Players
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. |
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