

Say if someone offered you a deck of cards and it cost you $1 to draw if you hit an ace you get paid $5 and if you hit the ace of spades you get $20.
So it would be: 3/52 = (3x100)/52 = 6r12/100 = 0.06  0.012 = 0.057 1/52 = (1x100)/52 = 2r4/100 = 0.02  0.004 = 0.016 48/52 = (48x10)/52 = 9r12/10 = 0.90  0.012 = 0.88 0.057 x 5 = (0.057x1000) x 5 = 285/1000 = 0.28 0.016 x 20 = (0.016x1000) x 20 = 320/1000 = 0.32 0.88 x 1 = (0.88x100) x 1 = 88/100 =0.88 0.88 + 0.32 + 0.28 = 0.28 So your EV is 0.28. The percentages only add up to about 95% so i'm doing something wrong also is there an easier way of doing this in your head maybe using ratios or fractions? I don't know any other methods of dividing small numbers by big numbers other than multiplying and dividing which takes me a while without a calculator. Any advice is appreciated. 




Hi Guyguyson!
3 of 52 times you will win $4 ($5$1 entry fee) 1 of 52 times you will win $19 48 of 52 times I lose $1 .231+.365.923=.327 so I expect to lose 32.7 cents each time John (JWK24) SuperModerator
6 Time Bracelet Winner 




How did you get those numbers? I get 0.285 + 0.38 + 0.923 = 0.265 on a calculator for an expected value of 26.5 cents?





(3/52)*$4=.231
(1/52)*$19=.365 (48/52)*($1)=(.923) add those three up and you get .327 which is an expected loss of 32.7 cents. I think you're forgetting to take off the $1 entry when you win. if you get paid $5... really, it's a $4 gain, as it costs you $1 to try. Same with the $20 return (it's really $19 due to the entry fee of $1). John (JWK24) SuperModerator
6 Time Bracelet Winner 




In the book Poker math that matters he doesn't take off the entry fee.
If we do something simpler like the classic coinflip and say you bet $1 to win $2 then doing it the way the book says it's: 0.5*2 = 1 0.5*1= 0.5 = +50 cents In the theory of poker he gives the same example and the EV is +50 cents if I take off the entry fee it would be: 0.5*1=0.5 0.5*1=0.5 = 0 So I don't get what i'm missing? Do you only have to take off the entry fee if there's more than two outcomes? 




Hate bumping but I need to know if i'm doing EV calculations right. I'm not great at maths so sorry if JWK24 method should be easy to apply to the coinflip example. If someone could workout the coinflip example again it would help me see where i'm going wrong.





I'm not sure what book you're using... but the entry price absolutely must be taken into account.
a coin flip where it costs $1 each and you win $2 has an EV of ZERO. John (JWK24) SuperModerator
6 Time Bracelet Winner 




Say you flip the coin 100 times and it hits heads 50 times and tails 50 times you would lose $1 50 times and win 2$ 50 times for a profit of $50 how can that have an EV of 0?
Or did you think I meant you both put in $1 so the pool is 2$ Because that would be 0 EV you would just be trading back each others $1 stake. What i'm saying is you wager 1$ and if you win you get your stake which was one $1 plus another $1 if you lose you just lose your $1 stake. 




then you get $3 each time you win (if you want to end up with an EV of 50).... you have to take the entry price into account.
if each flip costs the same and the payout is 2X the buyin, a coin flip has an EV of ZERO. John (JWK24) SuperModerator
6 Time Bracelet Winner 




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You must ADD your LOSS to your PROFIT You are not doing this. 




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50 + 100 = 50. or 100 + 100= 0 Or lets do this with pot odds. Hero: 78s Villan: AKo Flop: Ah,4s,Js Pot: 400 villan bets 100 = 500. Pot: odds 5:1 Chance of improving: 4.2:1 Instant call. .18*500=90 .82*100= 82 = +8 If we do it your way: Hero: 78s Villan: AKo Flop: Ah,4s,Js Pot: 400 villan bets 100 Pot: odds 4:1 (100 entry fee) Chance of improving: 4.2:1 Instant fold. 0.18*400 (100 entry fee)= 72 0.82*100 = 82 = 10 




You are totally missing the point about my last reply:
You toss a coin 100 times...........OK 50 times are tails, you lose ......... = $50 + $0 = ($50) 50 times are heads, you win ......... = $50 + $100 = (+$50) On each line, first number is cost to you, second number is your win, third number is your profit. overall Profit = ($50) + (+$50) = $0 Hope this makes it clearer ? 




using the 78s scenario...
pot was 400, opp bet 100.. my pot equity is 100 (what I have to put in) / 600 (what pot will be once my chips are in it) = 16.7% I have 9 outs, so my hand equity is 9 (number of outs) * 2% per out = 18% so it's a call. to get the EV... 78s on the flop has 35.275% equity from pokerstove. .35275 * 600 = 211.65 chips on avg gained when I win the pot .64725 * 100 = 64.725 chips from the call when I lose the pot EV on the call is 211.65  64.275 = +147.375 chip EV from the call John (JWK24) SuperModerator
6 Time Bracelet Winner 




Guyguyson: you don't win $2 when you hit the coin you only win $1.
Look at it this way: you flip twice and invest $2 to do so. You win once, lose once, for a total payout of $2. In the end, you didn't win. If your math was right, and the flip would be +EV for 50 cts, then I would be winning money simply by not playing poker at all.... 




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Just realized where all this confusion might of stemmed from. What I meant was that you bet $1 on the flip of a coin if you win you get your $1 stake returned plus $2 for winning so for every win you have $3 returned to you. Basically you are getting 2/1 on your money for a 1:1 shot.
And in the Op the money you stake will also get returned to you along with the winnings. You guys were thinking if you win you only get the $2 for winning but don't have the $1 you wagered returned to you? 




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John (JWK24) SuperModerator
6 Time Bracelet Winner 







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If they do ? please point me in their direction ? 




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I would love to run in to someone who would lay those odds. 


