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View Full Version : Bruce... can you verify this math for rolled-up heads-up?

cavalier
03-11-2003, 11:41 PM

What are the odds that your last 4 cards will give you a full house or quads?

There are 48 cards left in the deck.

Those 48 cards can yield (48,4) = 194,580 four-card hands.

In order for you to turn your trips into quads, one of the cards in the hand must to be the same rank as your starting trips. The other 3 can be anything. There are (47,3) = 16,215 possible combinations for the other three cards.

Therefore, the probability of getting the last one of your rolled-up trips is
16,215 / 194,580 = .0833 or about 1 in 12 hands.

Step 2 : Filling rolled-up trips with two or three of your opponents rank.
* Do not allow for a second pair in the other 2 cards.

There are 3 ways to make a pair out of the 3 remaining cards of his rank. Also, there are (45,2)=990 two-card combinations for the other 2 cards (45 since we leave out the one that makes your quads). This gives (3*990)=2970 possible hands. Since there are 11 other ranks that we don't want to pair, we have to remove the 6 pairs each rank can create. This leaves us 2970-(11*6) = 2904 hands.

2,904 / 194,580 = .0149 or 1 in 67.0

In order for your last four cards to include a pair that isn't the rank your opponent shows on 3rd, you would have 44 (4*11 ranks) cards to improve with ( we exclude his rank and your quad card ).

Those 44 cards can yield (44,4) = 135,751 four-card hands.

We have 11 ranks we can possibly pair. A pair can be made (4,2)=6 ways for each rank. The other two cards can be arranged (42,2)=861 ways. This gives us 11*6*861=56,826 hands in which we can improve by getting a pair, trips or quads of a rank other than the one we or our opponent started with.

56,826 / 135,751 = .4186 or 1 in 2.4

Finally, the chances of improving to a full house or quads when rolled-up in a heads up game with an opponent showing a rank other than your trips is

.0833 + .0149 + .4186 = .5168 or 51.7% of the time.

Note that his showing your quad card on 3rd takes away about 16% of your chances, leaving you a 43.35% chance of getting a full house, or 'other quads'.

If anything is wrong, or I overlooked something... please post!

BruceZ
03-12-2003, 07:47 AM
I think we can do the whole thing more simply like this:

P(full or quads) = 1 - P(no pair AND no quads) =
1 - [ P(no pair AND no quads AND no opponents card) +
P(no pair AND no quads AND 1 opponents card) ]

= 1 - [ (44/48)(41/47)(38/46)(35/45) +
(3/48)*4(44/47)(41/46)(38/45) ] = 31%

This is the whole answer, so we didn't have to separate out the probability of making quads which BTW is 8% as you said. You cannot add the 8% onto the probability of making a full house without quads as you did because this double counts the times you make both a full house AND quads.

Another problem I saw in your calculation is:

There are 3 ways to make a pair out of the 3 remaining cards of his rank. Also, there are (45,2)=990 two-card combinations for the other 2 cards (45 since we leave out the one that makes your quads). This gives (3*990)=2970 possible hands.

You cannot just multiply the 3*990 like this because you don't know which of the pairs is being chosen, and then the other cards can pick another card of this rank, so you count some combinations twice this way. This is also done again in step 3. I also didn't quite follow why you subtract off the cases where you get another pair, nothing wrong with that right? Although doing it this way you would have to make sure you don't double count the double pairs.

In the appendix to Super System, Mike Caro has the probability for improving to a full house as 32%. This is close to what we have here, but I don't think it is quite right because he says this is the final hand value, so it wouldn't include quads, so I think it should be lower. If he isn't taking the opponents up card into account the answer would be a little lower still. Perhaps he is taking some up cards into acount, but he doesn't say this. I'll think about it some more and then write to Mike.

irchans
03-12-2003, 10:00 AM
These kinds of probability problems are very tricky. It is very easy to miss a single number or case. I made several mistakes before finding an answer that I was comfortable with.

Bruce's method is great. Very simple. I think his numbers are off by one in a few places. When I fix those values I get:

Probability of improvement to quads or full house is
1 - ( 44 40 36 32 / (48 47 46 45) + 4 3 44 40 36/(48 47 46 45)) = 0.403022.

It is hard to duplicate this number using bigleftie's method because there are so many cases. When I break down the cases I get:

qu = 1/48*4 = 0.0833333
tr3nooth = c[4, 3] * 3/48 * 2/47* 1/46* 44/45 = 0.000226128
tr2nooth = c[4, 2] * 3/48 * 2/47* 44/46* 40/45 = 0.0135677
tr2oth = c[4, 2] * 3/48 * 2/47* 44/46* 3/45 = 0.00101758
tr1pair = 3*11*6*40/c[48, 4] = 0.0407031
tr1trip = 3*11*4/c[48, 4] = 0.000678384
tr0pair = c[4, 2]* 44/48*3/47*40/46*36/45 = 0.244218
tr0twopair = 3* 44/48*3/47*40/46*3/45 = 0.0101758
tr0trips = 11*4*40/c[48, 4] = 0.00904512
tr0quad = 11/c[48, 4] = 0.000056532

where

tr3nooth = probability of three his rank
tr2nooth = probability of two his rank &amp; no other pair
tr2oth = probability of two his rank &amp; other pair
tr1pair = probability of one his rank &amp; other pair
tr1trip = probability of one his rank &amp; trips
tr0pair = probability of none his rank &amp; other pair
tr0twopair=probability of none his rank &amp; 2 pair
tr0trips = probability of none his rank &amp; trips

adding up these probabilities gives 0.403022.

Using two different methods to get the same answer is a great way to reduce the chance of error. If we really wanted to make sure this answer was good, a simulation or a proof would be the next steps.

BruceZ
03-12-2003, 01:31 PM
Thank you. I subtracted 3 each time so I didn't get a pair, but I forgot to subtract 1 for the card I already dealt, so the numbers should change by 4 each time not 3. It was my bowling night, and I had too many pitchers /forums/images/icons/tongue.gif

This is in line with Super System now. In fact we can get about this same number even if we don't worry about the opponent's card:

1 - (48/49)(44/48)(40/47)(36/46) = 40.2%

Then if we subtract off the 8% for the quads, we get 32% which is the Super System number. Actually we should subtract off a little less than 8% because of the times we make both a full and quads, but that's incidental.

BruceZ
03-12-2003, 03:44 PM
Also, disregard my comments about getting quads and a full house. These probabilities are mutually exclusive, so the probabilites can be added and subtracted.

cavalier
03-12-2003, 07:23 PM
I think we can do the whole thing more simply like this:P(full or quads) = 1 - P(no pair AND no quads) = 1 - [ P(no pair AND no quads AND no opponents card) + P(no pair AND no quads AND 1 opponents card) ]= 1 - (44/48)(41/47)(38/46)(35/45) + (3/48)*4(44/47)(41/46)(38/45) = 31%

[/ QUOTE ]

OK, yes... much easier.

This is the whole answer, so we didn't have to separate out the probability of making quads which BTW is 8% as you said.

[/ QUOTE ]

At least I got something right... *grin*

You cannot add the 8% onto the probability of making a full house without quads as you did because this double counts the times you make both a full house AND quads.

[/ QUOTE ]

Ah, yes. Overlooked that.

Another problem I saw in your calculation is:There are 3 ways to make a pair out of the 3 remaining cards of his rank. Also, there are (45,2)=990 two-card combinations for the other 2 cards (45 since we leave out the one that makes your quads). This gives (3*990)=2970 possible hands. You cannot just multiply the 3*990 like this because you don't know which of the pairs is being chosen, and then the other cards I can pick another card of this rank, so you count some combinations twice this way. This is also done again in step 3.

[/ QUOTE ]

I don't follow you here. If we have a pair of his rank, we can have any other 2 cards to complete the hand. If you pick another card of his rank, that's fine isn't it? Depending on his rank and yours, you might have not filled your initial trips, but used 2 of those cards to fill HIS rank. Either way, you are full.

I also didn't quite follow why you subtract off the cases where you get another pair, nothing wrong with that right?

[/ QUOTE ]

Honestly, I kept toggling back and forth on this. I wasn't sure if I should count them or not since you only 'need' 1 pair to fill and I was afraid that including both pairs would inflate the odds. I see now that you are full with one pair or two, so it doens't inflate anything. It's a perfectly valid hand.

Although doing it this way you would have to make sure you don't double count the double pairs.

[/ QUOTE ]

By double pairs, do you mean like 2 pairs of the same rank; which essentially means your last 4 cards were running quads? I think I can see how this would be a problem. There'd be 66 (11 ranks times 6 ways to make double pairs) that would be double counted, right?

In the appendix to Super System, Mike Caro has the probability for improving to a full house as 32%. This is close to what we have here, but I don't think it is quite right because he says this is the final hand value, so it wouldn't include quads, so I think it should be lower.

[/ QUOTE ]

Right. Technically, I was only trying to find the odds of getting your quads or a pair to fill you. However, the whole problem was really an exercise in trying to get better at this sort of calculating so it's no big deal. I'm just hoping to understand the math when this is all done. Knowing the right answer is just a bonus. /forums/images/icons/wink.gif

Thanks

cavalier
03-12-2003, 07:37 PM
tr3nooth = c[4, 3] * 3/48 * 2/47* 1/46* 44/45 = 0.000226128

[/ QUOTE ]

I am not clear on something.

If I were trying to 'read' the above calculation I would say something like...

the 3/48 is for getting one of the 3 of his cards left from the 48 in the deck.

The 2/47 is for getting one of the 2 ( we just took one ) of his cards left from the 47 remaining.

Likewise, the 1/46 is for getting his last card out of the remaining 46.

Finally, we can get any card for the last card... except 1 .

Am I right when I assume we only have 44 choices for the last card because we don't want our quad card?

BruceZ
03-13-2003, 07:56 AM
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You cannot add the 8% onto the probability of making a full house without quads as you did because this double counts the times you make both a full house AND quads.
--------------------------------------------------------------------------------

Ah, yes. Overlooked that.

Actually it is OK to add these probabilities the way you have defined them. Getting a full house without quads is mutually exclusive of getting quads.

--------------------------------------------------------------------------------
Another problem I saw in your calculation is:There are 3 ways to make a pair out of the 3 remaining cards of his rank. Also, there are (45,2)=990 two-card combinations for the other 2 cards (45 since we leave out the one that makes your quads). This gives (3*990)=2970 possible hands. You cannot just multiply the 3*990 like this because you don't know which of the pairs is being chosen, and then the other cards I can pick another card of this rank, so you count some combinations twice this way. This is also done again in step 3.
--------------------------------------------------------------------------------

I don't follow you here. If we have a pair of his rank, we can have any other 2 cards to complete the hand. If you pick another card of his rank, that's fine isn't it? Depending on his rank and yours, you might have not filled your initial trips, but used 2 of those cards to fill HIS rank. Either way, you are full.

You will count some of the same card combinations twice this way. First you pick a pair say Ks Kc, then pick the rest of the cards including another one of the same rank say the Kh . Then you pick another pair of this rank say Ks Kh, and then you pick the rest the cards same as before except that now you pick the Kc. This combination is counted twice. This is a common error, and it can be avoided by making sure that whenever you multiply combinations, that the combinations don't have any cards in common, thus avoiding overlaps like this.

I also didn't quite follow why you subtract off the cases where you get another pair, nothing wrong with that right?
--------------------------------------------------------------------------------

Honestly, I kept toggling back and forth on this. I wasn't sure if I should count them or not since you only 'need' 1 pair to fill and I was afraid that including both pairs would inflate the odds. I see now that you are full with one pair or two, so it doens't inflate anything. It's a perfectly valid hand.

I was actually thinking of two pairs of different ranks. This is a similar problem to the last one. When you pick a pair and then the remaining cards and get another pair, this could generate the same two pairs and other cards that you already counted.

irchans
03-13-2003, 08:53 AM
bigleftie wrote:
&gt;I am not clear on something...

tr3nooth = c[4, 3] * 3/48 * 2/47* 1/46* 44/45 = 0.000226128

Everything you wrote about the equation above is perfectly correct.

cavalier
03-13-2003, 07:04 PM
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cavalier
03-13-2003, 07:10 PM
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