View Full Version : MotionX Poker Odds Question

gowanis

Aug 15, 2008, 01:38 PM

OK here's a gameplay question.

Scenario: your 3rd roll is coming up. you have two pair, lets say KKQQ and the dealer also has two pair AAJJ. If you keep your two pair and roll the one die, you have a 2/6 (33.33%) chance of improving to a full house. So, is it smarter to throw away your two queens and roll 3 dice to find a 3rd king ?

I was good at this back in college, but it has been a long time. I think the way to calculate this is that there are 216 possible outcomes of rolling 3 dice, so we need to find out how many of those contain at least one King. If you were doing it with two dice, it would be 11/36, but how do you figure out how many of the 216 outcomes contain at least one king without writing it all out and counting?

Is my logic correct? Any math pros in here that can help me out?

Mal

Aug 15, 2008, 02:02 PM

I don't know what the sides of the dice are on MotionX poker, but assuming they're traditional 6 sided dice and there's one King on each, then you have a 1/6 chance of getting a King on each of those. Add those together, and you have a 3/6 (50%) chance of getting at least one King. That's pretty good odds, far better than in traditional card-based poker. In regular poker, your odds of getting a King in one out of three cards drawn randomly from a stack are far worse. There are two Kings remaining if you already have two, so you have a 2/41 chance (the dealer has 4 cards and you have 4, plus you've already seen and removed from the deck three others, so 52-11=41) of getting a King as the first card. The second one the chances get trickier, but assuming you didn't draw a King the first time, you now have a 2/40 chance. On the last one you have a 2/39 chance. Thus, your overall chances are right around 15% (don't ask for the exact calculations, I did it on the calculator, but it's 2/41 + 2/40 + 2/39). On MotionX Poker then, go for it, but in regular poker, there's a few more variables. Remember though, if someone else is playing (in MotionX it's just the dealer I take it) the chances of them having one of those Kings and thus your chances being lower is also fairly reasonable, so your estimated chances should be calculated as less than 10%, probably less than 7%.

jW

gowanis

Aug 15, 2008, 02:09 PM

thanks for the reply.

This game uses five 6-sided dice (9,T,J,Q,K,A) and each die is independent, so you don't use up Kings and the dealer's roll is also independent.

I wasn't sure if it was correct to add the 1/6 + 1/6 + 1/6 to get the 50%. If so, its clearly better to throw away one of your pairs to go for 3 of a kind with three dice. Of course the dealer gets the last roll and i'm not factoring that in, but I was just curious about this one basic calculation.

cjoshuav

Aug 15, 2008, 02:17 PM

Well, you have 3 shots all at 1/6 or one shot at 1/3 . I always go for the full house.

Joshua

gowanis

Aug 15, 2008, 02:27 PM

Well, you have 3 shots all at 1/6 or one shot at 1/3 . I always go for the full house.

Joshua

but that might be the wrong move. you have a 33% chance of improving to a full house or a 50% chance of improving to at least 3 of a kind. of course, the dealer will change his final move based on what you do, so i'm still not sure what the best choice is.

cjoshuav

Aug 15, 2008, 02:56 PM

From what I remember of probability, a 1/6 chance three times is not the same as a 1/2 chance...going to google probability now :D

gowanis

Aug 15, 2008, 03:08 PM

From what I remember of probability, a 1/6 chance three times is not the same as a 1/2 chance...going to google probability now :D

There are 216 possible outcomes of rolling three 6-sided dice. The number of those out of 216 which contain at least one of the value you are looking for is your answer. I just don't know how to count them up without writing out every combination.

diesel

Aug 15, 2008, 03:44 PM

I have no idea how poker dice works, but i have extensive knowledge of texas hold'em poker and poker odds and I will mention one point which may or may not be relavent as far as poker dice is concerned. If poker dice involves betting money in between rolls which would be the equivalent of betting rounds pf, flop, turn, and river as in texas hold'em then you shouldn't simply let probabilities dictate next action but the probability of improving to the best hand in relationship to money in the pot and the cost to see the next card, in other words how much you will need to potentially call in relationship to the pot that you can win in relationship to the probability that you can win the hand.

For example, and my apologies in advance if this has no application in the poke dice world, but suppose you're on the flop with a flush draw on a drawy board with a possible straight, for example JKT and you suspect that your opponent could have a made straight but you figure that if you hit your flush then you will win the hand. Suppose the pot has $1,000 and your opponent leads out with a $500 half pot bet. the pot odds are 1,500:500 or 3:1. the probability of hitting your flush with one card to come is about 4:1 (9 outs is about a 19%) or with two cards to come it's 2:1 (9 outs with 2 cards to come is in the range of 36% but for simplicity sake we round down to 33% which translates into 2:1 odds). If you are determined to ride the hand to the river and see both cards then you have the right odds to call the $500 bet because you are getting 3:1 on your money with only 2:1 odds to make your hand, thereby making this a profitable play and plus EV (expected value) in the long run. if you are only going to see one card, then you do not have the right odds to call, because you are only getting 3:1 on your money but the odds of hitting your hand at 4:1 are greater than the potential payoff at 3:1.

sorry if the above is confusing but a simple way of looking at this is supposed you have a 10 sided dice, and I tell you that i will roll the dice 10 times, and if i roll a 5, you have to pay me 9 dollars however if i roll anything other than a 5, i will pay you 1 dollar. statistically, with 10 rolls, i should hit the 5 10% of the time, or 9:1 odds. the payout will also be 9:1 or breakeven, meaning i will have paid you 9 dollars and you will have paid me 9 dollars after the conclusion of 10 rolls. now however, suppose i say, you pay me 10 dollars if i hit the 5, and i still only pay you 1 dollar for any other roll. now the payoff is 10:1 which is greater than the 9:1 odds of making my roll. at the end of 10 rolls, i should have $10 and you have $9. This makes the play profitable and plus EV in the long run.

of course the above only really works out in the long run, when you have put your money in with the best odds, you will be plus EV in the long run, however keep in mind, in the short term, suck outs will happen. with 10 rolls of the 10 sided dice, most likely you will not always get one of each value. however with 100,000 rolls of the dice, the distribution should fall more within the statistical norm.

i hope the above makes sense and should be relevant if poker dice has betting action, though the odds to hit the best hand could be different in poker dice vs traditional texas holdem, the same principles should hold true and that is to not only use probability to determine the next course of action, but the probability of winning the hand along with the pot odds to determine if you have the right odds to make a call or to see the next card given the betting action.

cjoshuav

Aug 15, 2008, 05:47 PM

OK, that's a better way of thinking it. The possible combinations are:

6x6x6=216

of those possible combinations, only these would meet your needs (let's say we're going for a king:

K _ _

K K _

K K K

_ K _

_ K K

_ _ K

K _ K

Which, to me, seems like 7/216. That seems very low. I'm not sure what I'm missing here, but the odds certainly shouldn't drop below 1/6...

Joshua

gowanis

Aug 15, 2008, 06:21 PM

for each blank you have provided there could be 6 different values, so its way more than 7/216.

Night Spring

Aug 15, 2008, 06:21 PM

OK, that's a better way of thinking it. The possible combinations are:

6x6x6=216

of those possible combinations, only these would meet your needs (let's say we're going for a king:

K _ _

K K _

K K K

_ K _

_ K K

_ _ K

K _ K

Which, to me, seems like 7/216. That seems very low. I'm not sure what I'm missing here, but the odds certainly shouldn't drop below 1/6...

Joshua

I think you are on the right track here, except:

K K K = only one possible way to get this

K K _

K _ K = 5 possible ways each. I think. (KK1, KK9, KKTen, KKJ, KKQ. Repeat for each position.)

_ K K

K _ _

_ K _ = Um... not sure how many ways there are, but many. 5x5, maybe?

_ _ K

So you get 1 + (3*5) + (3*5*5). Anyone got a calculator?

Solver

Aug 15, 2008, 08:54 PM

I think you are on the right track here, except:

K K K = only one possible way to get this

K K _

K _ K = 5 possible ways each. I think. (KK1, KK9, KKTen, KKJ, KKQ. Repeat for each position.)

_ K K

K _ _

_ K _ = Um... not sure how many ways there are, but many. 5x5, maybe?

_ _ K

So you get 1 + (3*5) + (3*5*5). Anyone got a calculator?

I think it goes something like this...

For the first dice the odds are 1 in 6.

That means that the odds are 5 in 6 that you will not make a K.

For two dice add the odds you will make it with the first dice with the odds you won't make it with second dice multiplied by the odds you will make it...

1/6+5/6*1/6 = 11/36

It starts getting tricky. For three dice add the odds you will make it with the first dice, the odds you will make it the second dice and the odds you won't make it with third dice multiplied by the odds you will make it...

1/6+5/36+(31/36*1/6) = 97/216

gowanis

Aug 15, 2008, 09:26 PM

It starts getting tricky. For three dice add the odds you will make it with the first dice, the odds you will make it the second dice and the odds you won't make it with third dice multiplied by the odds you will make it...

1/6+5/36+(31/36*1/6) = 97/216

i don't have the brainpower to fully understand that, but based on your ability to calculate the answer of 11/36 for two dice, i trust that your answer for three is correct.

therefore, the odds of improving your hand to at least 3 of a kind is 97/216 = 44.9% and is a much better move than keeping the two pair and going for the full house (33.33%).

the next step, of course would be to factor in what the dealer would do for a last move...

Night Spring

Aug 15, 2008, 09:35 PM

therefore, the odds of improving your hand to at least 3 of a kind is 97/216 = 44.9% and is a much better move than keeping the two pair and going for the full house (33.33%).

the next step, of course would be to factor in what the dealer would do for a last move...

Exactly. What is the possibility that the dealer will be able to beat my three of a kind? Does a full house guarantee a win?

gowanis

Aug 15, 2008, 09:46 PM

Exactly. What is the possibility that the dealer will be able to beat my three of a kind? Does a full house guarantee a win?

so essentially we have figured out the best move that will improve your hand enough to take the lead, but what we really want to know is which move will produce the best final outcome, knowing that the dealer will make the last move.

if you improve to 3 of a kind, i believe the dealer will hold 2 cards and roll three dice (being consistent with our findings).

if you improve to a full house (after keeping the 2 pairs and rolling the single die), the dealer will be forced to try for a straight, with much longer odds.

anyone care to run this through ?

bdsmyth

Aug 15, 2008, 11:00 PM

Wow, how interesting was that to read .... (not really)

Personally I've collected all dice sets, and all gems bar 2. The 14 day accumulator and the 10 streak win!. You've no idea how many times I've played just to get 9 in a row only to be beaten by the sodden dealer!!!!!!

So gonna finish this game in the next 10 days!

cjoshuav

Aug 16, 2008, 08:48 AM

This really is interesting stuff, and thanks for pointing out the error in my logic. Brainstorming is a good thing :)

-Joshua

DreamPod

Aug 16, 2008, 09:57 AM

if you improve to a full house (after keeping the 2 pairs and rolling the single die), the dealer will be forced to try for a straight, with much longer odds.

Actually, all the dealer would need would be a full house of his own to beat you (as he has a pair of Aces), or a four (or five) of a kind. He wouldn't need to go for the straight.

gowanis

Aug 16, 2008, 05:42 PM

Actually, all the dealer would need would be a full house of his own to beat you (as he has a pair of Aces), or a four (or five) of a kind. He wouldn't need to go for the straight.

That's true, but the dealer would specifically need the Ace (1/6 chance) because if you've made your full house, you have either kings or queens on top and if he gets a jack it won't help. I suppose a real computer simulation is the only way to really figure out the right move.

swiftaw

Aug 16, 2008, 05:51 PM

It's easier to do it the other way.

P(getting no kings) = 5/6 * 5/6 * 5/6 = 125/216

Thus P(at least 1 king) = 1 - P(no kings) = 1 - 125/216 = 91/216 = 0.42

gowanis

Aug 16, 2008, 07:14 PM

anyone wanna show their stats? i've never been able to get above 58%. I've been as low as 56%

http://i109.photobucket.com/albums/n63/gowanis/motionx-aug16.jpg