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ArizonaDream

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Feb 26, 2016, 9:00 pm

What are the odds of matching 2 numbers, any position, any order playing a pick 3 game?
LOTTOKING2016

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Feb 26, 2016, 9:04 pm

Quote: Originally posted by ArizonaDream on Feb 26, 2016
What are the odds of matching 2 numbers, any position, any order playing a pick 3 game?

I have 7 combo from which always 2 digit matches so maybe 1:7
AllenB

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Feb 26, 2016, 9:14 pm

With 10 choices for 2 numbers wouldn't Your Odds Be 10 Squared or 1:100?
ArizonaDream

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Feb 26, 2016, 9:20 pm

Quote: Originally posted by AllenB on Feb 26, 2016
With 10 choices for 2 numbers wouldn't Your Odds Be 10 Squared or 1:100?

That would be the odds for a pick 2 game , but I think adding the 3rd number makes it easier.

My last prob/stat course was too long ago to remmber how to figure it out. It was my least favorite math course anyway.
Tucker Black

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Feb 26, 2016, 11:28 pm

Quote: Originally posted by ArizonaDream on Feb 26, 2016
What are the odds of matching 2 numbers, any position, any order playing a pick 3 game?

What state offers this game?

Nevertheless, let's assume a 50% house edge. The probability of winning this type of bet is 8/1,000. That means for a $1 bet, the state will award you 1,000/8/2=$62.50.
Wisconsin3054

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Feb 27, 2016, 12:16 am

Quote: Originally posted by Tucker Black on Feb 26, 2016
What state offers this game?

Nevertheless, let's assume a 50% house edge. The probability of winning this type of bet is 8/1,000. That means for a $1 bet, the state will award you 1,000/8/2=$62.50.

I don't think you're seeing this one right, Tucker. The odds should be much better. I believe it looks like this:

(10X10)/(3!).

1 in 16.7

Pretty sure anyway.
ArizonaDream

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Feb 27, 2016, 1:15 am

Not a real game. I'm just interested in knowing if something I'm watching is happening more often than one would expect based on probability.
Tucker Black

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Feb 27, 2016, 1:18 am

Quote: Originally posted by Wisconsin3054 on Feb 27, 2016
I don't think you're seeing this one right, Tucker. The odds should be much better. I believe it looks like this:

(10X10)/(3!).

1 in 16.7

Pretty sure anyway.

I messed up but not in that way. What I did was run a query on my table of 1,000 pick 3 numbers and found 8 that have a 5 and a 7 (my two chosen numbers) and have 3 different digits. The last requirement is not necessary (e.g., 5-5-7 should be a winner) and the revised query produces 10, which is 10/1,000 = 1 in 100 = 1%.

n1 n2 n3

0 5 7

1 5 7

2 5 7

3 5 7

4 5 7

5 5 7

5 6 7

5 7 7

5 7 8

5 7 9
ArizonaDream

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Feb 27, 2016, 1:53 am

Quote: Originally posted by Tucker Black on Feb 27, 2016
I messed up but not in that way. What I did was run a query on my table of 1,000 pick 3 numbers and found 8 that have a 5 and a 7 (my two chosen numbers) and have 3 different digits. The last requirement is not necessary (e.g., 5-5-7 should be a winner) and the revised query produces 10, which is 10/1,000 = 1 in 100 = 1%.

n1 n2 n3

0 5 7

1 5 7

2 5 7

3 5 7

4 5 7

5 5 7

5 6 7

5 7 7

5 7 8

5 7 9

There are matching numbers missing from your table. Any with the 7 before 5 are missing, and so are combos like 574, 573 etc.
jimjwright

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Feb 27, 2016, 5:54 am

Any pair made of 2 distinct digits equates to 54 straight combinations in Pick 3. It would be made up of 8 6-way combinations and 2 3-way combinations.

So I would say 1000/54 equates to 1 in 18.5

Jimmy
ArizonaDream

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Feb 27, 2016, 6:44 am

Quote: Originally posted by jimjwright on Feb 27, 2016
Any pair made of 2 distinct digits equates to 54 straight combinations in Pick 3. It would be made up of 8 6-way combinations and 2 3-way combinations.

So I would say 1000/54 equates to 1 in 18.5

Jimmy

The pairs need not be distinct digits. Even if they were distinct, I don't see how you'd have 54. There are 100 pairs, 00 through 99, 10 of which are doubles.
Tialuvslotto

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Feb 27, 2016, 8:49 am

There are 100 pairs in the P3 game: 90 pairs with 2 distinct digits and 10 same-digit pairs.

The 90 pairs with 2 different digits can be summarized as 45 "box" or "any-order" pairs.

Each winning combination has 3 pairs (if it is unmatched, 2 if it is a Double).

So you have 3 chances to match one of your 45 any-order pairs for unmatched (.72) or 1 chance for doubles (.27) or 2.43 chances overall.

45/2.43 = 18.5 like the J-man said.

"There is no such thing as luck; only adequate or inadequate preparation to cope with a statistical universe."

~Robert A. Heinlein
Wisconsin3054

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Feb 27, 2016, 9:32 am

Quote: Originally posted by Tialuvslotto on Feb 27, 2016
There are 100 pairs in the P3 game: 90 pairs with 2 distinct digits and 10 same-digit pairs.

The 90 pairs with 2 different digits can be summarized as 45 "box" or "any-order" pairs.

Each winning combination has 3 pairs (if it is unmatched, 2 if it is a Double).

So you have 3 chances to match one of your 45 any-order pairs for unmatched (.72) or 1 chance for doubles (.27) or 2.43 chances overall.

45/2.43 = 18.5 like the J-man said.

Although I think it would make more sense to look at odds seperately for distinct and pairs, as when you make your prediction you have do decide which you are choosing.
lakerben

50

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Feb 27, 2016, 10:43 am

Nm pays $5 for a pair.

I'm sitting here listening to Merle Haggard sipping some coffee!
SergeM

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Feb 27, 2016, 11:52 am

Quote: Originally posted by ArizonaDream on Feb 26, 2016
What are the odds of matching 2 numbers, any position, any order playing a pick 3 game?

14.4% is the best E(s=2), you can have for 2 digits correct for a pick 3 drawing.

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