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#11
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Oh sorry i thought 0-50 was better than 51-100. And I'm way off with my guess anyway. But should he not bet more hands than 33%?
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#12
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Whatever hand player two bets with means that the lowest number in player 1's calling range must be able to beat one third of player 2's range .
ie 2 bets with [0.5,1] then the best strategy for player 1 is to check call with [0.666666,1]; 0.66666 beats one-third of the numbers from [0.5,0.66666]and loses to (0.66666,1] |
#13
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[ QUOTE ]
I finally spotted another error that we were all making . In the last example , if player 1 bets with any number , then player 2 must call with any number ! ie , Hero bets everything from 0-1 . We mistakenly suggested that villain should call with 1/3-1 . This is blatantly wrong !! If villain hadn't posted the ante , then this assumption is certainly valid . However , since he posted an ante , calling with any number is better than foregoing the loss of that ante . To make this question and the previous question more interesting , we should add the stipulation that only Hero posts the ante and that he acts first . I can't believe , nobody picked up on this . [/ QUOTE ] I think the reason you have such problems is you don't look at a fold as -EV. It's EV=0 for that decision, but that choice makes the overall EV negative. In a hypothetical game where we both ante $1, you are dealt in the range of [1,50] and I'm dealt in the range of [51,100] When I bet $2, you should always fold. You consider folding EV = $0. Your strategy has an EV of -$1 |
#14
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[ QUOTE ]
For this week's game theory problem we will take a look at another situation . There are two players who pick numbers from 1-100 without replacement . Each player posts a $1 ante but player one must always check even though he's first to act . Player two has the option of betting the pot or checking behind . Given this knowledge , what strategy must player two employ to maximize his EV ? We may make the assumption that player one and two are playing optimally aside from the stipulation placed on player one . [/ QUOTE ] Optimal is: P1 [1,56] Fold [57,100] Call P2 [1,11] Bet [12,78] Fold [79,100] Bet P1 EV = -1/9 P1 can't improve by making any changes against P2 P2 can't improve by making any changes against P1 |
#15
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Here is the EV set up using the cash game definition of EV .
Player two bets with [x,1] and player one check calls with [(2x+1)/3,1] . Notice that the answer to this will be the same as in the discrete case . EV(P2)= 1*(1-x)*(2x+1)/3 + 3*(1-x)*(2-2x)/3*1/3 -3*(1-x)*(2-2x)/3*2/3 - (1-x)/3 There are 4 different product terms : The first is your EV|player 1 folds . The second is your EV|player 1 calls and you win The third is your EV|player 1 calls and you lose The fourth is your EV|player 2 checks After simplifying of the EV formula you should get EV(P2)=(-4x^2+6x-2)/3 EV' = -8x/3 +2 So x=3/4 . This means that player two should bet with 75-100 and player one should call with 84-100 . This is better than my previous attempt . |
#16
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Ok here is what I just don't seem to understand.
You apply a formula, get an answer. But do you do anything else to verify it actually answers the question? Your P1 [1,83] fold [84,100] call Your P2 [1,74] Check [75,100] Bet My P1 [1,56] Fold [57,100] Call My P2 [1,11] Bet [12,78] Check [79,100] Bet My P1 vs Your P2 EV = -0.094545 Your P1 vs My P2 EV = -0.118182 My Strategy wins 0.011815 Ante's per hand from yours while rotating positions. Maximal against your P1 "P1MO" [1,65] Bet [66,92] Check [93,100] Bet Maximal against your P2 "P2MO" [1,83] Fold [84,100] Call Your P1 vs Your P2 EV = +0.023636 Your P1 vs P1MO EV = -0.447879 P2MO vs Your P2 EV = +0.023636 Maximal would win 0.2357575 Ante's per hand from your strategies while rotating positions |
#17
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[ QUOTE ]
Here is the EV set up using the cash game definition of EV . Player two bets with [x,1] and player one check calls with [(2x+1)/3,1] . Notice that the answer to this will be the same as in the discrete case . [/ QUOTE ] Why do you assume P2 bets with [x,1]? why not [0,x] and [1-2x,1]? For this problem you if you have P2 bet only with 1, 99, and 100: Player 1 would optimally fold 1, call with 99, 100 and be indifferent to calling or folding with 2 thru 98. Since calling and folding would both produce the same EV for P1, folding means losing the ante, then P2's EV when betting with a 1 is (-3*2+1*97)/99 = 0.919191 That is clearly better than if P2 checked with a 1. [ QUOTE ] EV(P2)= 1*(1-x)*(2x+1)/3 + 3*(1-x)*(2-2x)/3*1/3 -3*(1-x)*(2-2x)/3*2/3 -(1-x)/3 There are 4 different product terms : The first is your EV|player 1 folds . The second is your EV|player 1 calls and you win The third is your EV|player 1 calls and you lose The fourth is your EV|player 2 checks After simplifying of the EV formula you should get EV(P2)=(-4x^2+6x-2)/3 EV' = -8x/3 +2 So x=3/4 . This means that player two should bet with 75-100 and player one should call with 84-100 . This is better than my previous attempt . [/ QUOTE ] Again, you get an answer, but not the correct answer. It's correct for your formula, but your formula isn't correct for the question. |
#18
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mykey1961 could you explain where you get the numbers (-3 and 2+1 in your equation below. thank you
(-3*2+1*97)/99 = 0.919191 |
#19
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[ QUOTE ]
mykey1961 could you explain where you get the numbers (-3 and 2+1 in your equation below. thank you (-3*2+1*97)/99 = 0.919191 [/ QUOTE ] Assume P2 has a value of 1: If P1 has a value of 99, or 100, P1 calls and beats P2 for a 3 ante profit. That's a 3 ante loss for P2. 99 and 100 are 2 numbers If P1 has a value of 2 thru 98, P1 has the same EV if it calls or folds. If it folds, it's loss is 1 ante, and that's a 1 ante win for P2. 2 thru 98 represents 97 numbers. When P2 has a 1, there are 99 possible numbers for P1 P2's expected profit is (-3 antes * 2 numbers + 1 ante * 97 numbers) / 99 numbers = 0.919191 ante per number This calculation is just for the case when P2 has a 1. |
#20
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Why not avoid betting 1-11 and bet with 67+ instead ?
I'm not sure the significance in why you're betting with 1-11 here . |
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