Calculating Odds of Getting at Least One Pair in Pai Gow

[bland]

TL;DR Summary

This looks like it should be simple but it has me flummoxed. I have been unable to find anything online to give me a clue.

To be sure there are very complex odds calculators that give you useful information but nothing that addresses what I want to know, which is simply what are the odds of getting at least one pair in Pai Gow.

Simply put. There are 16 pairs of tiles. The 32 tiles are dealt out at 4 tiles per hand. And I'm wanting to know what are the odds of one of the hands receiving at least one pair.

I began by putting reducing this to simply picking out 4 tiles out of a stack of 32 and asking do I have a pair.

So the first tile goes down. Then there are 31 tiles left so the odds of drawing a pair with the first two tiles are 1 chance in 31. That seemed OK then I thought the odds then of the third tile matching the first is 1 chance in 30, but then there is the second tile that could have been match. It got confusing so I started a different tack...

I thought I'd do the two possible pairs separately. So I figured let's just match the first tile only, that means there are 31 tiles and 3 chances to match, so that about 1 chance in 10 that seems right. But now there was also 2 chances in 30 that could match the 2nd tile. So I have approx 1 chance in 10 plus 1 chance in 15. But now I don't know how to combine them. I'm sure this is very simple but I'm still confused.

 

[anuttarasammyak]

I have no knowledge of the game. If your question is probability of finding at least one pair in 4 tiles from 16 pairs tiles, I think it is
$$16(\frac{4}{32})^2 = \frac{1}{4}$$
Do I take it too simple ?

 

[bland]

Yes you have accurately grasped my problem. Thank you for that, no wonder I couldn't work it out, I shall attempt to understand your equation.

 

[PeroK]

Summary:: This looks like it should be simple but it has me flummoxed. I have been unable to find anything online to give me a clue.

To be sure there are very complex odds calculators that give you useful information but nothing that addresses what I want to know, which is simply what are the odds of getting at least one pair in Pai Gow.

Simply put. There are 16 pairs of tiles. The 32 tiles are dealt out at 4 tiles per hand. And I'm wanting to know what are the odds of one of the hands receiving at least one pair.

I began by putting reducing this to simply picking out 4 tiles out of a stack of 32 and asking do I have a pair.

So the first tile goes down. Then there are 31 tiles left so the odds of drawing a pair with the first two tiles are 1 chance in 31. That seemed OK then I thought the odds then of the third tile matching the first is 1 chance in 30, but then there is the second tile that could have been match. It got confusing so I started a different tack...

I thought I'd do the two possible pairs separately. So I figured let's just match the first tile only, that means there are 31 tiles and 3 chances to match, so that about 1 chance in 10 that seems right. But now there was also 2 chances in 30 that could match the 2nd tile. So I have approx 1 chance in 10 plus 1 chance in 15. But now I don't know how to combine them. I'm sure this is very simple but I'm still confused.

If I've understood correctly, there are 16 pairs of tiles and you want the probability that;

a) You have one pair and two different tiles.

Plus

b) You have two pairs of tiles.

There are a number of ways to do this. The simplest perhaps is to note that there is only one further possibility, which is to have four different tiles.

c) Four different tiles.

We can calculate the probability of that and note that $P(a) + P(b) = 1 - P(c)$.

To calculate $P(c)$ the first tile can be anything, the second tile mustn't be the same as the first (that's $30/31$); the third time must be different from both the first two (that's $28/30$); and the last one must be different from the first three (that's $26/29$). So:
$$P(c) = \frac{30}{31}\cdot \frac{28}{30} \cdot \frac{26}{29} = \frac{728}{899} \approx 0.810$$So that
$$P(a) + P(b) = \frac{171}{899}$$
We can also calculate $P(a)$ and $P(b)$ directly. This gives:
$$P(a) = \frac{168}{899}, \ \ P(b) = \frac{3}{899}$$

 

[PeroK]

Yes you have accurately grasped my problem. Thank you for that, no wonder I couldn't work it out, I shall attempt to understand your equation.

The answer in post #2 is not correct.

 

[bland]

If I've understood correctly, there are 16 pairs of tiles and you want the probability that;

a) You have one pair and two different tiles.

Plus

b) You have two pairs of tiles.

There are a number of ways to do this. The simplest perhaps is to note that there is only one further possibility, which is to have four different tiles.

c) Four different tiles.

We can calculate the probability of that and note that $P(a) + P(b) = 1 - P(c)$.

To calculate $P(c)$ the first tile can be anything, the second tile mustn't be the same as the first (that's $30/31$); the third time must be different from both the first two (that's $28/30$); and the last one must be different from the first three (that's $26/29$). So:
$$P(c) = \frac{30}{31}\cdot \frac{28}{30} \cdot \frac{26}{29} = \frac{728}{899} \approx 0.810$$So that
$$P(a) + P(b) = \frac{171}{899}$$
We can also calculate $P(a)$ and $P(b)$ directly. This gives:
$$P(a) = \frac{168}{899}, \ \ P(b) = \frac{3}{899}$$

I am actually after something in between a) and b) although although b) is also good to know.

But I can see what I was doing wrong now by looking at your working out.

Out of the four tiles I just want at least one pair, I don't care if the other two match or not. I don't need to exclude a second pair, if that makes sense. For example it's as if I bet someone I would draw a pair from the four tile, if I drew two pair, it doesn't matter.Strangely enough this talk of 'at least' reminded me of a very simple problem that I estimate 95% of people I ask get wrong that is...

There are six single black socks and six single white socks in a box in a completely dark room with you. You need to grab a matching pair in the dark. Q: What is the minimum quantity of single socks that you need to take to be absolutely certain that you have at least one matching pair. When asked almost everyone gets this wrong.

 

[PeroK]

I am actually after something in between a) and b) although although b) is also good to know.

But I can see what I was doing wrong now by looking at your working out.

Out of the four tiles I just want at least one pair, I don't care if the other two match or not. I don't need to exclude a second pair, if that makes sense. For example it's as if I bet someone I would draw a pair from the four tile, if I drew two pair, it doesn't matter.

That's what I've calculated. a) is exactly one pair and b) is two pairs. So, at least one pair is a + b.

Strangely enough this talk of 'at least' reminded me of a very simple problem that I estimate 95% of people I ask get wrong that is...

There are six single black socks and six single white socks in a box in a completely dark room with you. You need to grab a matching pair in the dark. Q: What is the minimum quantity of single socks that you need to take to be absolutely certain that you have at least one matching pair. When asked almost everyone gets this wrong.

If you take three socks, they can't all be different!

 

[PeroK]

There are six single black socks and six single white socks in a box in a completely dark room with you. You need to grab a matching pair in the dark. Q: What is the minimum quantity of single socks that you need to take to be absolutely certain that you have at least one matching pair. When asked almost everyone gets this wrong.

There is another solution: you could switch the light on!

 

[bland]

That's what I've calculated. a) is exactly one pair and b) is two pairs. So, at least one pair is a + b.If you take three socks, they can't all be different!

OK that's excellent thank you.

Yes pretty obviously 3 but ask someone and they'll probably say 7, although I can also get other answers. They'll say 7 and I'll say, no it's 3 and they'll say "but what if you get three black socks". And I'll say well you have a pair then don't you. It's crazy I know, try it.

 

[bland]

There is another solution: you could switch the light on!

Yes, this is why it's always better to add that it's not a trick. Try it and see! I even tell people before I ask it that almost no one ever gets this right, I explain it's just what it sounds like. Still they say 7, almost every time.

 

[PeroK]

We can also calculate $P(a)$ and $P(b)$ directly. This gives:
$$P(a) = \frac{168}{899}, \ \ P(b) = \frac{3}{899}$$

I'll show you how I calculated $P(b)$. In general, there are two approaches: counting and direct probabilities. I like to use direct probabilities unless it gets too difficult.

The first thing to note is that if we have two pairs, then we have one of the following:
$$XXYY, XYXY, XYYX$$The second point is that each of these is equally likely. You might have to think about this. So, we could calculate the probability of getting $XXYY$ and multiply by three. I.e.
$$P(b) = 3\times P(XXYY)$$Finally, we can use the direct probabilities to calculate $P(XXYY)$:

The first tile can be anything, but the second must match the first. Then, the third tile can be anything, but the fourth must match the third. This gives:
$$P(XXYY) = \frac{1}{31}\cdot\frac{1}{29} = \frac{1}{899}$$Note that we can get $P(a)$ as:
$$P(a) = 1 - P(b) - P(c) = 1 - \frac{3}{899} - \frac{728}{899} = \frac{168}{899}$$A final exercise is to double check this by calculating $P(a)$ independently. Hint:
$$P(a) = 6 \times P(XXYZ)$$

 

[bland]

The reason that this came up is that I was using this online Pai Gow practice tool, and I've seen hundreds of Pai Gow games played live in a casino and I am assuming that this site is using a random generator to select the tiles, and yet the hands it deals both to itself and to the player look completely wrong, if you saw it in a casino you'd think something was up. So now I know the proper probabilities, I can check the actual odds against what I see coming up.

 

[anuttarasammyak]

The answer in post #2 is not correct.

Yea. I agree with you that
$$P(c)=\frac{\ _{16}C_4\ 2^4}{\ _{32}C_4}$$
$$P(b)=\frac{\ _{16}C_2}{\ _{32}C_4}$$
$$P(a)=\frac{\ _{16}C_1 (\ _{30}C_2 - \ _{15}C_1)}{\ _{32}C_4}$$
Thanks.

 

[PeroK]

Yea. I agree with you that
$$P(c)=\frac{\ _{16}C_4\ 2^4}{\ _{32}C_4}$$
Thanks.

The number of hands is $\binom{32}{4}$ and the number of two-pair hands is $\binom{16}{2}$. So:$$P(b)= \frac{ \binom{16}{2}}{\binom{32}{4}} = 0.003337 = \frac{3}{899}$$

 

[FactChecker]

Any time you have a problem that says "at least one pair", you should expect the answer to be one of two forms:
1) Sum of P(1 pair) + P(2 pair) + P(3 pair) + ...
or
2) 1 - P(0 pair)

 

[fresh_42]

Please stay on topic. I had to remove a couple of posts. If you want to discuss QM, start a thread in our QM forum. This is about mathematics, and as far as I can tell very classical mathematics.

 

[Moes]

So the first tile goes down. Then there are 31 tiles left so the odds of drawing a pair with the first two tiles are 1 chance in 31. That seemed OK then I thought the odds then of the third tile matching the first is 1 chance in 30, but then there is the second tile that could have been match. It got confusing so I started a different tack...

I also thought this method should work and I like figuring out what was wrong with the way I was thinking, so I went through it. I’m not sure why you got confused, but you were actually correct. The third tile would have a 1 in 30 chance of matching the first tile but also a 1 in 30 chance of matching the second, so it would have a 2 in 30 chance of matching either. Then the fourth tile would have a 3 in 29 chance of matching one of the other 3 tiles. The question is just how to combine these probabilities. That’s why it’s easier to do it the way PeroK did it. But this is the way I combined the probabilities and got the same answer.

1/31 + (2/30 × (1 - 1/31)) + (3/29 × (1 - ((1/31) + (2/30 × (1 - 1/31))))) = 171/899Never mind how I got that, but the point is even though there is an easier way to do it, this way also worked.

 

[Moes]

Yes pretty obviously 3 but ask someone and they'll probably say 7, although I can also get other answers. They'll say 7 and I'll say, no it's 3 and they'll say "but what if you get three black socks". And I'll say well you have a pair then don't you. It's crazy I know, try it.

For some reason people think of a pair as a kind of set where you need one of each instead of two of the same items. It’s interesting. But I actually don’t think their mistake is with the math.

 

[bland]

... The question is just how to combine these probabilities.

Yes, it was when I tried to combine the probabilities that I got confused. I'll work through your example

r some reason people think of a pair as a kind of set where you need one of each instead of two of the same items. It’s interesting. But I actually don’t think their mistake is with the math.

Yes people do answer as if the question was 'at least one non matching pair', but you can ask the question in such a way as to eliminate all possibility of this error, yet you'll still get 7 very often. But when no one has ever said that they thought I meant a non matching pair, so they do understand it must be a matching pair. It's just one of those strange things.