Doubt in expected value calculation of Bill Chen book

I am studing The mathematics of poker of Bill Chen, and I am at the chapter of pot odds and implied odds. I have this doubt: Everyone who have a little background on probability knows that the expected value is mathematical defined as

<x>=P1 x1 +P2 x2

where P1+P2=1, that means x1+x2=U where U is the whole universe of possible cases. When xi are dollars we are calculcating the expected value of our possible gain. In the book of Chen I found many examples where, in the calculation of expected value of the possible dollars gain, the sum of the probability of the whole outcomes does not equate 1. From the mathematical point of view this is an error and can lead to a wrong answer. So my question is, which is quibble that I did not understand on the calculation of the expected value in a poker hand? Namely, is there some approximation that I did not understand? I guess that is an approx in order to perform quick calculations, but I am not sure.

Example: There are two players A and B. Player A hold A♠,A⋄, player B hold 9♥,8♥. On the board we have K♥,7♠,♥,2♥. The game is 30\$-60\$ and the pot is 400\$. Imagine that the player knows the cards that the opponents owns (is a conceptual exercise). 35/44 is the probability favorable to the player A and 9/44 is the probability favorable to the player B.

If the player A checks, we have (this is taken from the book)

<\$(A)>=35/44 400\$ ~318\$

If the player A bet 60\$

<\$(A)>=35/44 (400\$+60\$+60\$)- 60\$ ~ 353\$

We can see that in the second case the probability does not sum up to 1.

• Can you please provide an example of one of the calculations in the book. I haven't read the book, but I have a mathematical background and may be able to provide some insight. Dec 18 '12 at 14:47
• Here is a possible explanation. Let's say three events can occur, with probabilities of 20%, 50% and 30%. The first event gives you \$1, the second gives you \$2 and the third gives you \$0. Then the expectation value is simply = 0.2*1 + 0.5*2. The probabilities don't add to 1, because the third term (0.3*0) is 0, and thus omitted from the equation. Dec 18 '12 at 14:52
• Yes this is one case, and I understand this. The problem is in the other cases. I guess that is an approx in order to perform quick calculation. Dec 18 '12 at 15:06

The calculations given in your example make sense. Except you write 53/44 when you should have written 35/44 I believe.

In the case that player A checks, the expectation value of profit will be:

<(A)>=(35/44)*400 + (9/44)*0 = (35/44)*400 = 318

In the case that player B bets \$60, and player A calls, the pot increases to 400 + 60 + 60. The expected total income from player A is therefore:

Expected Income = (35/44)*(400 + 60 + 60) + (9/44)0 = (35/44)(400 + 60 + 60).

However player A has to wager an additional \$60 in this case, regardless of the outcome. Therefore the expected PROFITS is thus,

<(A)> = Expected Income - bet = (35/44)*(400 + 60 + 60) - 60 = 353.

Alternative approach:

Expected Profits in case of win = 400 + 60 + 60 - 60 = 460

Expected Profits in case of loss = - 60

Therefore expectation of profits = (35/44)*460 - (9/44)*60 = 353

Explanation as to why approaches give the same answer:

The first approach gives the formula,

<(A)> = (35/44)*(400 + 60 + 60) - 60

= (35/44)*(400 + 60 + 60) - (35/44)*60 - (9/44)*60

= (35/44)*(400 + 60 + 60 - 60) - (9/44)*60

= (35/44)*(400 + 60) - (9/44)*60

Which is the formula used in the second approach.

• thanx, I made a doube typo. I fixed ;) Dec 18 '12 at 15:28
• Anyway your calculation is not correct. If B bet and A call the expected value of A is Expected Income = (35/44)*(400 + 60) - (9/44)60 Dec 18 '12 at 15:32
• No that's not correct. The first term in your expression represents total income, but your second term represents profits. You can't mix and match like that. Dec 18 '12 at 15:36
• Yes you right, indeed i fixed the comment. But in my opinion your calculation it is still incorrect. Dec 18 '12 at 15:39
• thanx, you right :) Dec 18 '12 at 15:53