Question about cycle length (aka hit frequency)

Hi all! Been a while since my last post. Just been enjoying the reading!

Can't recall the calculation for determining expected number of hands until I see (let's say) a straight AND a flush. Do I just add the individual frequencies, as I would with (say) 2 quads, or is there a more complex calculation for 2 different outcomes?

I'm trying to figure the expected cycle of a series of hands for FP Pick'Em (3 4oak, 3 FH, 3 FL, and 5 ST), and it seems like it would be a bit longer than the 3 quads, but not as long as all 14 separate hands. If you have a different example, I'm all ears, as long as you name the game and give enough detail that I can translate it to my game.

Thanks in advance!

I'd search for something like the cycle of hitting all 13 four of a kinds. It's much higher than one would intuitively expect, and it doesn't make any sense to me. I think Wizard or others may have written it out before.

Thanks, RS, for the reply! And thanks to all others for viewing!

I read through all of the posts regarding hitting one of each quad while searching for an answer to my question. While useful info, I don't believe it helps me here, as any quads, full houses, flushes, and straights will do.

Thanks again!

Quote: camapl

Thanks, RS, for the reply! And thanks to all others for viewing!

I read through all of the posts regarding hitting one of each quad while searching for an answer to my question. While useful info, I don't believe it helps me here, as any quads, full houses, flushes, and straights will do.

Thanks again!

If someone wrote out the math for hitting all quads, then you'd use the same type of formula for full houses, flushes, etc.

Quote: camapl

I'm trying to figure the expected cycle of a series of hands for FP Pick'Em (3 4oak, 3 FH, 3 FL, and 5 ST), and it seems like it would be a bit longer than the 3 quads, but not as long as all 14 separate hands.

I would like some clarification as to what you are trying to do.

Do you want the expected number of hands (using perfect strategy, presumably, as opposed to a strategy to minimize the number of hands for this result) before you get 3 or more quads, 3 or more fulls, 3 or more flushes, and 5 or more straights?
If so, then this looks like it can get rather complicated, and might probably be done better with a Monte Carlo approach (where you run the simulation millions of times to get an approximation). You can't just add the separate expected values together as they don't take into account the possibility of different hands occurring in different orders.
Even if it was just "one flush and one straight," you can't take the expected number of hands for a flush and add it to the expected number of hands for a straight as that does not take into account the possibility that you will get your first straight before your first flush.

Quote: ThatDonGuy

I would like some clarification as to what you are trying to do. Do you want the expected number of hands (using perfect strategy, presumably, as opposed to a strategy to minimize the number of hands for this result) before you get 3 or more quads, 3 or more fulls, 3 or more flushes, and 5 or more straights?

Yes, exactly - couldn't have put it better! And thank you, TDG, for your reply!

Quote:

If so, then this looks like it can get rather complicated, and might probably be done better with a Monte Carlo approach (where you run the simulation millions of times to get an approximation).

I was afraid you might say that! In fact, to complicate matters, at least one of the quads must be A, K, Q, or J's, although I'd be content if I could come up with an estimate based on your first quote above. I may just go with triple the quad cycle (or maybe 3.5?) as long as the probability of failure is relatively low on the rest. I am trying to get a handle on the duration on the play, as well as compare estimated values of playing quarter Pick'Em vs. dollar Bonus 8/5. Clearly Bonus would have a shorter expectation, but is it short enough to outweigh nearly a 1% deficit to return and the increased ROR? I hadn't brought up the series of Bonus hands, as I had convinced myself there was a formula for this that I just couldn't recall (or derive). I figured I could just plug'n'chug my way thru both sets of hands once I had the formula. Alas!

For those who may be interested, I finally figured out the expected cycle for the "simple" case of two unlike events, such as a straight and a flush. I used drawing from an urn with replacement to represent a series of independent hands. At first, what hung me up was trying to use only 2 colors. By using 3 colors (Red, Green, and Blue), I was able to cover all possibilities (Flush, Straight, and all others).

The expected number of draws until exactly 1 Red and at least 1 Green OR exactly 1 Green and at least 1 Red, or
E(R & G) = P(1st draw is Red) * [ 1 + E(# draws till 1 Green) ]
+ P(1st draw is Green) * [ 1 + E(# draws till 1 Red) ]
+ P(1st draw is Blue) * [ 1 + E(R & G) ]

So if the first draw is Red, then we have the cycle of Green plus 1. Likewise, if the first draw is Green, then we have the cycle of Red plus 1. However, if the first draw is Blue, then we have 1 plus that which we were trying to find the in the first place. Each of these is weighted by its respective probability of occurrence.

Isolating E(R & G) and simplifying gives:

E(R & G) = { 1 + P(Red) * E(Green cycle) + P(Green) * E(Red cycle) } / [ 1 - P(Blue) ]

For example, with 20 Reds, 30 Greens, and 50 Blues,
E(R & G) = 19/3, or 6.333

Back to Pick'Em: subbing Flush for Red, Straight for Green, and all other hands (incl. losers) for Blue gives an expected cycle of...

[edit to include:] 389.86, which is greater than the individual frequencies (Flush: 313.6, Straight: 197.4) and less than the sum (511.0).

(Sorry for having to edit. I'm on an iPhone and didn't want to risk losing the post when I jumped to my spreadsheet and back! lol)

Unfortunately, thatdonguy was right - a sim might be in order. The urn equation above compounds exponentially with the addition of each item and/or color. And I would need multiple colors and would need to treat each of the 14 items separately! 14! nested terms!?! No thanks! 3 or 4 wouldn't be too bad...

Brilliant...

Thank you for the kind post, lightningbolts! I knew if I kept at it, the light would eventually come on!

Speaking of which, I believe I owe RS some credit. I sincerely hope my response was not too dismissive!

After arriving at the equation below in a prior post, I was trying to come up with a simpler approach in order to tackle problems with a greater degree of difficulty...
E(R & G) = { 1 + P(Red) * E(Green cycle) + P(Green) * E(Red cycle) } / [ 1 - P(Blue) ]

What hit me was that I would come to an equivalent expression if I set it up as a series of two cycle lengths and that RS had inferred as much. Thanks again, RS. Sorry that I was too bullheaded to see it at first!

So, our first occurrence is the expected number of draws to get EITHER a red OR a green ball, given by
E(R or G) = 1 / [ P(Red) + P(Green) ]

Then our next occurrence is whatever is left given our first occurrence. If we got Red, then we add the cycle for Green and vice versa. So, the expected number of draws remaining is a weighted average, as follows:
E(remaining) = { P(Red) / [ P(Red) + P(Green) ] } * E(Green cycle)
+ { P(Green) / [ P(Red) + P(Green) ] } * E(Red cycle)

Adding E(R or G) and E(remaining) and simplifying:
{ 1 + P(Red) * E(Green cycle) + P(Green) * E(Red cycle) } / [ P(Red) + P(Green) ]
which is equivalent to E(R & G).*

*Note that [ P(Red) + P(Green) ] equals [ 1 - P(Blue) ].