Chase the royal or keep the flush?

[tech58]

FP. We have them, ALL noted and memorized. PLEEEEZZZ believe, we've got it . All the veterans are
either fully indoctrinated,fully irritated,fully ignoring,or fully pissed short of violent thoughts.
Myself i have a very serious question. The "free drink" part of your last quote,at the Hard Rocks in FL.?

[Gronbog]

Gronbog. After reading your posts here, I have a new appreciation for your calculations. You clearly understand the math behind video poker strategy.

Thanks for that. I do have a university degree in math with my major being computer science and a minor in combinatorics. Combinatorics is essentially the mathematics of probability which is mainly what we're dealing with when analyzing games of chance.

[Eduardo]

Thanks for that. I do have a university degree in math with my major being computer science and a minor in combinatorics. Combinatorics is essentially the mathematics of probability which is mainly what we're dealing with when analyzing games of chance.

All that education and you still can't give us the lucky Keno numbers. What a waste.

[Gronbog]

As a representative of our math. challenged members, i have a couple of questions.
First, my intuitive thought about the probability of a royal in this 100 play example (probably wrong but possibly shared with others) is that for a single hand the prob. is 1/47=2.1%.

Absolutely correct. There is one card out of the 47 cards remaining which gives you a royal.

So each of the 100 hands,played out of their own deck,would have an equal chance of happening at 2.1%.My question is why does it appear to me that if prob. came to pass there would be 2.1 royals when the true # is .884?

You are correct that each of the 100 lines has exactly the same chance of hitting the royal (see, you're not doing bad so far!). You are also correct that you can therefore expect an average of one royal every 47 lines which is an average of (1/47)*100=2.1 royals per 100 lines. That is the expected result, but the actual result could end up being anywhere between hitting all 100 to hitting none.

What was asked for was the probability of hitting a royal. More specifically, since it was in the context of a contest, I calculated the probability of hitting one or more royals among the 100 lines. The probability of hitting exactly one royal can also be calculated, but it didn't seem relevant given the context.

Second, my error may be due to a lack of the ability to understand your formula resulting in 11.6%.
The little inverted v symbol may have something to do with sq. root. Not sure.
Could you possibly either write out longhand or write it out in first year algebra and post pic. ?
Hope springs eternal, ignorance CAN be cured.

Sure, I can break it down.

The probability of an event is calculated by counting the number of ways the event can happen and dividing it by the number of ways all possibilities can happen. We saw above that, since there was one card among 47 that would produce the royal on a given line, the probability was then 1/47 for each line.

Now we want to calculate the probability of at least one royal among the 100 lines. Well, there are quite a few ways that can happen. If only one royal occurs, then there 100 ways that could happen. If two royals occur, then there 100 ways for the first one to happen and then 99 ways for the second one to happen. So there are 100 x 99 = 9900 ways for that to happen. We then need to go on and calculate the number of ways 3 royals, 4 royals, and so on up 100 royals hitting can happen. That's a lot of very error prone calculations.

Instead, I used two principles of probability theory to calculate it an easier way:
1) The probability of more then one independent event happening is calculated by multiplying the individual probabilities together.

2) If you know the probability of something happening (call it P), then the probability of it not happening is 1 - P.

So we want to know that probability of one or more royals hitting among the 100 lines. Well, the opposite of that is for exactly zero royals to hit, which means that we miss the royal 100 times in a row.

We know that the probability of a royal on each line is 1/47, so by rule 2 above, the probability of not hitting a royal is 1 - 1/47 = 46/47.

Now we need to know the probability of missing 100 royals in a row. By rule 1 above we multiply the probability of missing one line by itself 100 times. This is (46/47)^100 which means 46/47 to the power of 100. This equals 0.1164 or 11.64%. This is the probability of hitting no royals among the 100 lines.

We then go back to rule 2 to get the probability of one or more royals as 1 - 0.1164 = 0.8836 or 88.36%

I hope this helps. If not, I'm happy to answer questions.

[tech58]

Damn Gronbog you are one fine resource for this site. Keeping in mind that my math thru calc. was 50 years ago i followed every shred of your extremely clear post up to 46/47 which would be .9787234 to the power of 100. My math instructor in high school would have thrown an eraser at me if i asked this dumb of a question (legal back in the day)but how do you take a number to a power? Have to rely on you here, at this crucial juncture, since my math. & stat. texts disappeared two moves ago. Else i will have to google it.
Counting on our math. guru one more time.
Stay safe, we need you.

[New2vp]

"Taking something to a power" is shorthand for repeatedly multiplying a number by itself.

(46/47) x (46/47) = (46 x 46) / (47 x 47) = 2116/2209 = 0.9579 (approximately)

You can also write the above as (46/47)^2. The upside down "v" is called a caret (above the 6 on your keyboard) and is one way to show that you are taking the thing to the left "to a power." In the case that I showed, I am multiplying two numbers together with both of them being 46/47.

Similarly, (46/47) x (46/47) x (46/47) = (46 x 46 x 46) / (47 x 47 x 47) = 97,336/103,823 = 0.9375 (approximately). And, you guessed it, this can be written as (46/47)^3, because we are now multiplying 3 numbers together, with all 3 of them being 46/47.

In Gronbog's case he was multiplying 100 numbers with all of them being 46/47.

He could have written:
(46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x
(46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x
(46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x
(46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) x (46/47) = 0.1164 (approximately)

Since I'm sure you would not like to count these, it is a whole lot easier to read and understand if it is written as (46/47)^100.

You may have looked at the above and rather than counting the numbers, you might have counted them on one row and multiplied by the number of rows. And of course we all might remember when we learned to multiply, it was so that we didn't have to add the same number over and over again.

Exponentiation (another word for "taking something to a power") allows us not to have to multiply a number over and over again just like multiplication allows us a shortcut to addition.

On many calculators, the way you take something to a power is look for the key that has a small "x" with a smaller "y" above and to the right of the "x". If you do not have such a button, you might have to look for a different calculator. If the calculator is an app on your computer or phone, you can sometimes look for settings and switch it from "Standard" to "Scientific" and then you can find the button.

To use your calculator for this problem, you can enter 46 / 47 = ^ 100 =. Please note, I can't show the symbols in this forum that you have on your calculator. So for "/", you may have a division sign. And for "^" you may have the "x with the smaller y to the right and above the x."

[New2vp]

If only one royal occurs, then there 100 ways that could happen. If two royals occur, then there 100 ways for the first one to happen and then 99 ways for the second one to happen. So there are 100 x 99 = 9900 ways for that to happen. We then need to go on and calculate the number of ways 3 royals, 4 royals, and so on up 100 royals hitting can happen. That's a lot of very error prone calculations.

Gronbog, I enjoy your posts. I particularly like your example of walking a dog to explain the difference between what is expected to occur and what actually occurs. And I think your (2nd) explanation of how you calculated the probability of at least one royal was very helpful.

One small point, your combinatorics professor might not have been totally happy with the excerpt that I copied above. As I'm sure you know, the number of "ways" that you can get two royals out of 100 hands is 100 x 99 / 2 = 4950 rather than 9900. I know it is difficult to try and explain the difference between combinations and permutations when starting from first principles, especially when we are dealing with 100 play, but maybe it would be easier for others to use Triple Play instead.

With triple play, there would be 3 ways to get one royal, either on the first, second, or third hand.

And there are only 3 ways to get two royals, not 3 x 2 = 6, but 3 x 2 / 2 = 3. Either (1) both the first and second hands, (2) both the first and third hands, or (3) the second and third hands.

If we think of this as explained in the excerpt, like 3 ways to get the first royal and 2 to get the second, you can count them up like this:

(1) Both the first and second hands,
(2) Both the first and third hands,
(3) Both the second and first hands,
(4) Both the second and third hands,
(5) Both the third and first hands, or
(6) Both the third and second hands.

It is easier to see when this is laid out this way that (3) and (1) are the same, (5) and (2) are the same, and (6) and (3) are the same, but just in a different order. And that is the reason for the division by 2 in 3 x 2 / 2 = 3.

It is the same concept with 2 royals in 100 hands, so you get 100 x 99 / 2 = 4950. I'm sure by this time, no one wants me to list the 9900 different permutations and match up which ones are identical.

Gronbog, keep up the good work. I know it seems like a lot of people don't like to talk about the math part of video poker, but who knows? Maybe this is a welcome diversion from politics and epidemics for a couple! Even though the odds may be slightly tipped against that proposition.

[onemoretry]

I think I need a beer!

[Gronbog]

Thanks New2vp for explaining exponents (powers) and for your correction of my incorrect count of how many ways two royals can happen with 100 trials. Didn't I say that these calculations are error prone?

[olds442jetaway]

Make that 2 or even three! Peroni for me if you have any!