Can this Sum be reduced? It is calculating the expected value of a poker hand when you "Run it multiple times" for the last card.

Question

There is a scenario in poker (Texas Hold'em) where if two players go "all in", they have the option to deal the remaining cards multiple times and it results in multiple pots that they could possibly split.

So if you "run it twice", then you have 4 possible outcomes:

WW - Win both pots and you win the entire pot
WL - Win the first pot and the two players split the pot
LW - Win the second pot and the two players split the pot
LL - Lose both pots and your opponent wins the entire pot

If you "run it three times", you have 8 possible outcomes:

WWW - You win the entire pot
WWL - You win 2/3 of the pot
WLW - You win 2/3 of the pot
WLL - You win 1/3 of the pot
LWW - You win 2/3 of the pot
LWL - You win 1/3 of the pot
LLW - You win 1/3 of the pot
LLL - Your opponent wins the entire pot

And your ability to win a hand is determined by your "outs". These are undealt cards that will ensure your win. So if your opponent has a pair of Queens, and you have an Ace-three, you might have "three outs", since there are 3 undealt Aces that would make your hand better than your opponent.

So the formula for calculating your Expected Value with one more card to be dealt (the "river" card) can be written as:

$$ \sum_{i=1}^n { { i \over n } {n \choose i} } { { {}_z \mathrm{ P }_i \times {}_{(44-z)} \mathrm{ P }_{(n-i)} } \over {}_{44} \mathrm{ P }_n } $$

Where z is your number of outs, and n is the number of times you "run it" (where $ 1 \le n \le 44 $), and 44 is the number of undealt/unknown cards.

This is the case when two players hold two cards each and there are 4 community cards. 52 - 8 cards is 44 undealt cards.

And I know this reduces to $ z \over 44 $ for all values of n where $ 1 \le n \le 44 $

For example, here is the calculation for z = 5 and n = 33, which reduces to $ 5 \over 44 $

Is there a way to prove that the sum reduces to $ z \over 44 $?

Answer 1

\begin{align} \sum_{i=1}^n \frac{i}{n} \binom{n}{i} \frac{P(z,i) P(44-z,n-i)}{P(44,n)} &= \sum_{i=1}^n \frac{i}{n} \binom{n}{i} \frac{i!\binom{z}{i} (n-i)!\binom{44-z}{n-i}}{n!\binom{44}{n}} \\ &= \sum_{i=1}^n \frac{i}{n} \cdot \frac{\binom{z}{i} \binom{44-z}{n-i}}{\binom{44}{n}} \\ &= \sum_{i=1}^n \frac{i}{n} \cdot \frac{\frac{z}{i}\binom{z-1}{i-1} \binom{44-z}{n-i}}{\frac{44}{n}\binom{43}{n-1}} \\ &= \frac{z}{44}\cdot\frac{\sum_{i=1}^n \binom{z-1}{i-1} \binom{44-z}{n-i}}{\binom{43}{n-1}} \\ &= \frac{z}{44}\cdot\frac{\sum_{i=0}^{n-1} \binom{z-1}{i} \binom{44-z}{(n-1)-i}}{\binom{43}{n-1}} \\ &= \frac{z}{44}\cdot\frac{\binom{(z-1)+(44-z)}{n-1}}{\binom{43}{n-1}} \\ &= \frac{z}{44} \end{align} The next-to-last equality is an application of Vandermonde's identity.

Answer 2

Alternative approach:

Running the situation more than once alters the variance, but does not alter the expectation. This is why it is not uncommon for both players to favor running the situation more than once, even though Poker is a zero-sum-game (i.e. whenever somebody wins, somebody loses).

That is, it is plausible that both players may be risk-averse.

Anyway, the expectation of running a situation more than once can be proven to be equivalent to running the situation exactly once, without (much) Math.

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By Linearity of Expectation, the expectation of the sum is equal to the sum of the expectations, regardless of whether the events are dependent events.

So, if the expectation when running a situation once is $E$, suppose that instead, you run the situation $n$ times, where in each of the $n$ times, you are playing for

$$\frac{1}{n} \times ~\text{the size of the pot}.$$

Then, since the expectation of the sum is equal to the sum of the expectations, and since each individual running clearly has an expectation of

$$\frac{E}{n}$$

Then, you must have that the overall expectation is

$$\frac{E}{n} \times n.$$

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Edit

How do I know that each individual running has an expectation of $$\frac{E}{n} ~?$$

That is clearly the expectation of the first of the $n$ runs, because this run is identical to the run that would occur if the situation was only being run once.

Further, the expectation of (for example) the second run can not be Mathematically different than the expectation of the first run, because it can not be any more or less likely that the pertinent cards that (for example) the underdog needs show up in the 2nd running than the 1st running.

For example, if there is one card to come, and the underdog needs a spade to complete a flush, the probability that the 2nd card off of the deck is a spade can not be different than the probability that the 1st card off the deck is a spade.

This is despite the fact that if the 1st card off of the deck is a spade, then there is one less spade in the deck.

As a (perhaps) better example, if you have $52$ ping pong balls in a bucket, each with a distinct number, from $1$ through $52$, and you select the ping pong balls, one at a time, without replacement, the probability that the $n$-th ping pong ball selected is number-1 is exactly the same as the probability that the first ping pong ball selected is number-1, for any value of $n$ in $\{2,3,\cdots,52\}.$