In Between

I just made a new page for a new game (as far as I know) called In Between. It could loosely be compared to Red Dog, in that the object is to have a middle card fall between cards to the left and right. However, there is no skill element and another two-card hand is throw in as a second way to win. Please click the link for the full rules and analysis.

You can play In Between for free at the OneTouch web site. Just scroll down to In Between.

I welcome all comments.

The question for the poll is would you play In Between? Multiple votes allowed.

There's nothing new here but it is a cute/fun little mixture of game mechanics. It's a pity the HE is so high. Sad for an online game.

Quote: Gialmere

article typos

In the last sentence of the first paragraph, replace the first "two" with "to".

The "Poker Bet Analysis" table is titled "Main Bet Analysis".

Thanks for the corrections and for putting them in a spoiler tag.

I tried a dozen hands on the demo and didn't get one "in between" winner. I did catch a pair of 9's on the bottom to extend the game a bit, but that's it.

If the Ace was high or low, would that wreck the edge?

Quote: Gialmere

There's nothing new here but it is a cute/fun little mixture of game mechanics. It's a pity the HE is so high. Sad for an online game.

yes, you are quite correct: High house is indeed a pity but at least we are being properly informed of that.

I'm on my phone so I didn't check everything, but your combinations for full house is wrong. 3744

For an online game it would have been so easy to set Imbetween or Pair as 2/1 and still have a house edge.

I am having a problem with the first number I tried to check in the Main Bet Analysis Table.

The Main Bet Analysis Table says that "In Between + Pocket Pair" has 1,262,976 combinations; I get that the number of combinations should be 1/3 of that.

Basically, I calculate that 1,262,976 is the number of combinations for three different ranks in the first three cards and a pair in the next two cards.

3 different ranks + pocket pair = c(13,3) * 4^3 * ( 10 * c(4,2) + 3 * c(3,2) )
= 18304 * 69 = 1,262,976

However, to qualify as an "In Between" the intermediate rank must be in the middle position of the first three cards That would occur with a frequency of 1/3. So, 1,262,976 / 3 = 420,992.

Did I screw up in my calculations?

Quote: miplet

I'm on my phone so I didn't check everything, but your combinations for full house is wrong. 3744

Good catch, thanks.

Quote: gordonm888

I am having a problem with the first number I tried to check in the Main Bet Analysis Table.

The Main Bet Analysis Table says that "In Between + Pocket Pair" has 1,262,976 combinations; I get that the number of combinations should be 1/3 of that.

Basically, I calculate that 1,262,976 is the number of combinations for three different ranks in the first three cards and a pair in the next two cards.

3 different ranks + pocket pair = c(13,3) * 4^3 * ( 10 * c(4,2) + 3 * c(3,2) )
= 18304 * 69 = 1,262,976

However, to qualify as an "In Between" the intermediate rank must be in the middle position of the first three cards That would occur with a frequency of 1/3. So, 1,262,976 / 3 = 420,992.

Did I screw up in my calculations?

Is it possible our total combinations are not the same? Mine is combin(52,3)*combin(49,2)*3 = 77968800.

Here is my formula for the probability: =(COMBIN(13,3)*4^3)*(10*COMBIN(4,2)+3*COMBIN(3,2))/( combin(52,3)*combin(49,2)*3). Do you disagree?

Quote: Wizard

Is it possible our total combinations are not the same? Mine is combin(52,3)*combin(49,2)*3 = 77968800.

Here is my formula for the probability: =(COMBIN(13,3)*4^3)*(10*COMBIN(4,2)+3*COMBIN(3,2))/( combin(52,3)*combin(49,2)*3). Do you disagree?

Yes. combin(13,3)*4^3 is the total number of combinations for which the first three cards will have different ranks. Three different ranks is necessary to win the In Between pay-out, but it is not sufficient. As I understand it, it is also required that the card with the middle rank be in the location between the other two cards. Given that you have three cards with three different ranks, that will happen with a frequency of 1/3.

Quote: gordonm888

Yes. combin(13,3)*4^3 is the total number of combinations for which the first three cards will have different ranks. Three different ranks is necessary to win the In Between pay-out, but it is not sufficient. As I understand it, it is also required that the card with the middle rank be in the location between the other two cards. Given that you have three cards with three different ranks, that will happen with a frequency of 1/3.

All your combinations will be 1/3 of Wizard’s.
There are 3 possible positions for the middle card. 2 on the ends and 1 in the middle. He multiplied the non in between combinations by 2. For combinations that didn’t involve 3 different ranks, he multiplied them by 3.

Quote: miplet

All your combinations will be 1/3 of Wizard’s.
There are 3 possible positions for the middle card. 2 on the ends and 1 in the middle. He multiplied the non in between combinations by 2. For combinations that didn’t involve 3 different ranks, he multiplied them by 3.

Thank you.

I get the same numbers as the wizard using (52*51*50/6) * 3 for the first part and (49*48/2( for the second - 66300 * 1176.

	First Three	Pair	Non-Pair	Win+Pair	Win only	Pair only	Loser
	66 300			1 262 976	20 262 528	3 323 424	53 119 872
Trips	156	72	1 104			11 232	172 224
Pair	11 232	70	1 106			786 240	12 422 592
Singletons	54 912
Winners	18 304	69	1 107	1 262 976	20 262 528
Losers	36 608	69	1 107			2 525 952	40 525 056

Yes, of course. I agree, Charlie. Of course. But I never claimed that the probabilities in the WOO table were wrong. What I said was that, in the WOO table, the number of the combinations appeared to be too high by a factor of 3.

Using the terminology in your post, the total number of combinations for the first part should be (52*51*50/6) according to standard definitions and not (52*51*50/6) * 3. What is the justification for the ' * 3 ' multiplier? Of course, if you apply the ' * 3 ' multiplier when generating every combinations number in the combinations column, as has been done in this WOO table, then the numerical entries in the probability column will be unaffected. But that doesn't make the number of combinations in the published table correct.

I realize that, to the extent that readers want any information in this kind of WOO table, they want to know the probabilities. Almost no one, except math people, ever notice or look at the combinations column. So, I often feel that the WOO tables could be improved by omitting the combinations column entirely. But, in an article published on the internet, if one does choose to include a combinations column (perhaps as a guide for people trying to learn the combinations methodology) , then is it too much to ask that the combinations column be correct?

Quote: gordonm888

...the first part should be (52*51*50/6) according to standard definitions and not (52*51*50/6) * 3...

I agree with you as that's what I had, but then you needed to divide by 3 for the singletons and got fractions. In this case the order does matter. So AsKsQs and QsKsAs are winners whereas the other four (e.g. KsQsAs) are losers. Personally I might have put it back to 52*51*50, but this way the numbers agree with wizard's.

Quote: charliepatrick

I agree with you as that's what I had, but then you needed to divide by 3 for the singletons and got fractions. In this case the order does matter. So AsKsQs and QsKsAs are winners whereas the other four (e.g. KsQsAs) are losers. Personally I might have put it back to 52*51*50, but this way the numbers agree with wizard's.

Right, when I looked at this originally I thought: for the first three cards, this should be based on the number of combinations with permutations: 52*51*50 - which is exactly what you have said.

Personally, I think every number in the combinations column should either be divided by 3 - to give the number of combinations, as labeled - OR multiplied by two, to give the number of combinations, including order-dependent permutations of the first three cards. ( No change in probabilities either way.)

Maybe I'm slightly autistic or something, but I dislike it when someone publishes a number and labels it as the "Number of combinations" for some kind of condition or state, when it is clearly not; instead its some kluged concept like (combinations *3) or (permutations/2). Don't we already have enough false news in the world without WOO tables contributing to the clutter?

Quote: gordonm888

I realize that, to the extent that readers want any information in this kind of WOO table, they want to know the probabilities. Almost no one, except math people, ever notice or look at the combinations column. So, I often feel that the WOO tables could be improved by omitting the combinations column entirely. But, in an article published on the internet, if one does choose to include a combinations column (perhaps as a guide for people trying to learn the combinations methodology) , then is it too much to ask that the combinations column be correct?

I will never remove the combinations column when I know them. That would be against my religion. However, in games where card order matters, like this one, perhaps I should use permutations, since combinations would seem to imply order doesn't matter.

In the case of In Between, I used something in between (pun intended) combinations and permutations, since order only mattered for where one card feel relative to two others.

Will it satisfy you, Gordon, if I convert the In Between combinations to permutations?

I went ahead and multiplied all my numbers by 4 and called them permutations. I hope this will satisfy all concerned.

Quote: gordonm888

Maybe I'm slightly autistic or something, but I dislike it when someone publishes a number and labels it as the "Number of combinations" for some kind of condition or state, when it is clearly not; instead its some kluged concept like (combinations *3) or (permutations/2). Don't we already have enough false news in the world without WOO tables contributing to the clutter?

Now if we could only get lock manufacturers to change the name of “combination lock” to “permutation lock”

Quote: Wizard

I went ahead and multiplied all my numbers by 4 and called them permutations. I hope this will satisfy all concerned.

That's perfect.

Quote: Ace2

Now if we could only get lock manufacturers to change the name of “combination lock” to “permutation lock”

You're absolutely right!

If it did pay inbetween or pair 2/1 what would the house edge?

Quote: kingcreights

If it did pay inbetween or pair 2/1 what would the house edge?

About 1.15% - I don't really understand why they didn't think of it as it makes the payouts so simple with a House Edge that's much more reasonable.