This problem begins with explanation, and the question is written on the end ---

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This problem begins with explanation, and the question is written on the end
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Asymetric information -- Previously, uncertainty was present because no one knew with certainty the outcome of some event. For example, when you and your opponent flip a coin in a game, neither one of you know the outcome.



However, there is another source of uncertainty, and that is when your opponent knows something that you do not, and vice versa.



This is nicely illustrated in card games. Keep in mind that in a standard deck there are 52 cards comprised of 13 “ranks” (from low to high: 2,3,4,5,6,7,8,9,10,J,K,Q,A), with each rank having 4 “suits” (clubs, diamonds, hearts, spades, or “c,d,h,s”)



Consider the game of Guts, a simplified version of Texas Hold’em. A pair (such as 99) beats all lower pairs, and also beats any unpaired hand (like AJ) . Thus 99 loses to KK, but beats 55 and AJ.



Starting pot: $100

Your starting hand: JcJd (pair of jacks, suits are irrelevant)



You bet $100. (pot now $200)

Your opponent raises for all his chips for $600 total (pot now $800)

You must call $500 to force him to turn over his cards to see who wins. (Pot would then be $1300)



Notice that your opponent knows exactly what his hand is, but you do not. Thus, calculating your probability of winning (needed to calculate EV) sounds impossible.



Except that it is possible, using “range theory.”



Suppose you have some additional information. Your opponent in this situation is Tight Tim. Suppose you know that Tight Tim only raises that big when he has AA, KK, QQ, or AK. These card combinations are his “range” in this situation.



If you call, you will only win if he turns over AK, and will lose to all other hands in his range.



What are the probabilities of him holding each of the different possible hands in his range?



Consider the combinations:

AK = AcKc, AcKd, AcKh,AcKs,

AdKc,AdKd, AdKh, AdKs

AhKc, AhKd, AhKh, AhKs

AsKc, AsKd, AsKh, AsKs

=16 combinations of AK (or any other 2 unpaired cards)



KK= KcKd,KcKh, KcKs,

KdKh, KdKs,

KhKs

= 6 combinations of KK (or any other pair)



Thus his range and combinations are:

AK 16

AA 6

KK 6

QQ 6

Total combos in his range = 34.



We can now convert this into probabilities of cards he may hold:



prob(AK) = 16/34

prob(AA) = 6/34

prob(KK) = 6/34

prob(QQ) = 6/34



Since you win if he holds AK, and you lose if he holds AA, KK, or QQ, we can simplify the probabilities to



prob(win)= 16/34

prob(lose)= (6+6+6)/34 = 18/34



Finally we can write the EV equation:



EV = -$Call + prob(win)($ final pot) + prob(lose)($0)

= -$500 + 16/34($1300) + 18/34($0)

= -500 + $611 = +$111



Thus, calling is a +EV move. (Folding is EV=0)



Notice how the presence of AK in his range made calling profitable for you.





Question. Modify the above example according to the following, and write out the EV equation, calculate, and state whether a risk neutral person would call.



i) Tight Tim’s range is only AA, KK, QQ. Be sure you construct the win/loss probilities correctly.



ii) Tight Tim’s range is AA, KK, QQ, JJ and 10-10. Be sure you construct the win/loss probilities correctly. This is tricky because, a) you are holding Jc and Jd, which will reduce the number of his combos, and b) if you both have JJ, you split the pot. So be sure you treat JJ as a separate term in the EV equation.



iii) The same scenario as the example, but this time you are playing Loose Larry. His range in this situation is AK, AQ, any pair (22-AA).



iv) The same scenario, except you are playing Polarized Paul. His range is AA,KK,QQ and 72.





Hint: Make sure you correctly count up the total combos for each question.

Explanation / Answer

Part one: The probability of a win is 0. The expected payoff for a loss is 0 0+0=0. The probability of win=0/18 of loss=18/18=1 Part two:six possible combos for AA,KK,QQ,1010. One possibility for a pair of JJ since you hold one. Thus, 25 possible outcomes. The probability (p) of a win is 6/25 for a loss=18/25 and a tie=1/25. So the equation is (6/26)1300+(1/25)650+(18/25)0=338 Part three: 16 combos of AK and AQ. 6 combos for each pair except jj which has one possibility. Ploss=18/105 Pwin=86/105 Ptie=1/105. Expected payoff=(18/105)0+(86/105)1300+650(1/105)=1071 Part four:pwin=16/34 ploss=18/34 expected payoff=611

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