Flushing Meadows: Part 3

Brian Alspach

Poker Digest Vol. 2, No. 6, Marc 12 - 25, 1999

In the preceding two articles, we saw that the probability of a flushing board occurring in hold'em - given no information about any of the players' hands - is 0.371. On the other hand, if a given player has two suited cards, the probability of a flushing board is 0.373, whereas if the player has two unsuited cards, the probability is 0.370. This prompted the following conversation with Bib Ladder.

``Now look here, professor, you've come up with different probabilities for what looks like the same thing to me. I remember the same kinda thing happened when we were calculating probabilities of low boards in Omaha. So what's goin' on here?''

``Well, Bib, let me see if I can think of an example which clearly illustrates the principle. Alright, suppose we are considering a freeway running through a city and the westbound traffic. Suppose the city has covered much of the freeway as it traverses the downtown area. There is one freeway stream entering the covered section, but coming out the other side of the covered section the freeway has branched into one freeway heading toward A and another freeway heading toward B. Now a traffic helicopter above the freeway notices that of the cars entering the covered section heading west, 7/20th emerge on the freeway heading toward A and the remaining 13/20th are on the freeway heading toward B. Thus, for an observer in the helicopter, the probability is 13/20 that a car entering the covered section will head toward B. This observer has no information about what is going on under the covered section.

Meanwhile, once the freeway is under cover, it first splits into two streams S and T to go around a diversion caused by the foundation of a large building. Each of the two streams S and T later splits to feed the freeways to both A and B. Suppose that one-fourth of the traffic takes stream S and three-fourths of the traffic takes stream T. Then suppose that of the traffic along stream S, one-fifth takes the freeway to A and four-fifths takes the freeway to B. Of the traffic along stream T, two-fifths heads toward A and three-fifths toward B.

Consider an observer who can see only the traffic stream along S and how it splits for the freeways to A and B. What she observes is that one-fifth of the traffic goes toward A and four-fifths goes toward B. So for her the probability is 1/5 that a car travelling along stream S will head toward A. Similarly, for an observer who can see only stream T, the probability is 2/5 that a car travelling along stream T will take the freeway toward A. Now let's see that the arithmetic makes sense. We have 1/4 of the traffic taking stream S follwed by 1/5 of that heading toward A and 4/5 toward B. This gives us 1/20 of the traffic heading toward A and 4/20 of the traffic heading toward B. Similarly, 3/4 of the traffic takes T and 2/5 of that then goes toward A giving 6/20 of the traffic heading toward A. Altogether this gives us 7/20 of the traffic heading toward A which is what observers in the helicopter see. The traffic heading toward B is 4/20 + (3/4)(3/5) = 13/20 of the total. Is this okay, Bib?''

``Well, it is dependent on where the observer is located.''

``That is precisely the point, Bib. It is analogous to our situation for flushes. The observer who knows nothing about the hands is essentially in the helicopter with much of the information occluded. On the other hand, a player upon looking at her cards is now seeing the stream of traffic corresponding to the two-card hands having been split into either two suited cards or two unsuited cards.

It is worth noting how the probabilities of the outcomes for the overall observer are calculated. We take the product of the various probabilities involved along the different paths leading to our outcome and then add them together. In the example above, we took (1/4)(4/5) + (3/4)(3/5) = 13/20 to obtain the probability of a car going to B for an observer in the helicopter.

In order to check this happens for the flushing boards we have been discussing, let's use the exact rational values in order to avoid errors arising from working with decimal approximations. The total number of two-card hands is C(52,2) = 1,326. Of these, $4\cdot C(13,2) = 312$ are suited and the remaining hands are unsuited. So the probability of two suited cards is 312/1326 and the probability of having two unsuited cards is 1014/1326. From our previous calculations (last article) we saw there are 2,118,760 boards, 790,251 flushing boards if the player has two suited cards, and 784,960 flushing boards if the player has two unsuited cards. So the probability of a flushing board is

\begin{displaymath}\frac{312\cdot 790,251 + 1,014\cdot 784,960}
2,118,760} = \frac{1,042,507,752}{2,809,475,760} = 0.371\end{displaymath}

which is the same number we calculated two articles ago. In fact, if one divides both the numerator and denominator of the preceding large fraction by 1081, one obtains


which is the same fraction we obtained earlier for the probability of a flushing board for an observer with no knowledge of any of the hands.

So hopefully, Bib, this resolves any lingering questions you may have had about these conditional probabilities we have encountered several times.''

``I want to think about it some, but there is a glimmer of understanding.''

``That's fine, Bib, but I am expecting that glimmer to turn into a blazing light of clarity.''

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