Odds? 2 of 3, pick 3

Reply #26

Stack47 — Sat, 05 Mar 2016 20:37:17 GMT

Me too

Reply #25

Tialuvslotto — Fri, 04 Mar 2016 12:38:13 GMT

I'm a card-carrying member of the Thoth fan club

Reply #24

CARBOB — Fri, 04 Mar 2016 09:20:50 GMT

Welcome back! Where have you been? Hope you stick around.

Reply #23

jimjwright — Sun, 28 Feb 2016 18:29:20 GMT

It's great to see Thoth post again.

Jimmy

Reply #22

Thoth — Sun, 28 Feb 2016 11:31:37 GMT

Odds are 1 in 18.518 or 54 favorable against 946 unfavorable (reduces down to 1 : 17.518 as a ratio ) - however you want to look at it.

There are 45 boxed no-match pairs and 10 double digit pairs possible...

Suppose you play 12X boxed where X is unknown, there are 8 other digits besides 1 and 2 that could be drawn to give you a no-match combo. In this case we will say X = digit 3. All no matches can be drawn 6 different ways: 123, 132, 213, 231, 312, 321. Since X can be replaced by 8 other... [ More ]

Reply #21

AllenB — Sun, 28 Feb 2016 03:39:14 GMT

I think you are right. The Odds I posted were for each of the 3 possibilities in a Pick 3 game. Dividing that by 3 looks right; because, there are 3 chances of hitting. Not sure what the point is (here in California you get squat for 2 out of 3 in Pick 3 or 4.

Reply #20

Stack47 — Sun, 28 Feb 2016 02:15:34 GMT

Though my method of calculation the number of combos the 5-7 pair creates is different than yours, the results (54) are the same. I multiplied the other 8 digits times the 6 possible ways each creates and added the two 3 ways using the same digits creates. But what about the odds of another pair: 1-1?

I came up with nine 3 ways plus one triple for 28. The odds are obviously different on which 2 out of 3 digits are drawn.

Reply #19

ArizonaDream — Sat, 27 Feb 2016 23:04:44 GMT

Illinois also has a front pair and back pair option, it would only take 4 lines to play a pair front and back and either order. But you loose when the pair splits. I was thinking I had something worthwhile, but I'm not beating the odds on picking a pair by as much as I thought. Back to the drawing board. Wait a minute, I need to look at this again, may be something there after all.

Thanks all

Reply #18

ArizonaDream — Sat, 27 Feb 2016 22:59:26 GMT

I'm on board with those who said 54 in 1000 now. I had to map it out in a way I could see the matching pairs easily and then cross out the dups. 6 sets or 10, and then subtract the 6 duplicates. 1000/54 or 1 in 18.52 or 5.4%

057 075

157 175

257 275

357 375

457 475

557 575

657 675

757 775

857 875

957 975

507 705

517 715

527 725

537 735

547 745

557* 755

567 765

577 775*

587 785

597 795

570 750

571 751

572 752

573 753

574 75... [ More ]

Reply #17

Tucker Black — Sat, 27 Feb 2016 17:39:39 GMT

When I posted 10 results, I was getting ahead of myself. I was thinking of how you would actually play this given that a bet like that is not one of the games. But you can build a bet just like that by buying box tickets. So how many box tickets would you need to buy in order to cover all possible winning numbers that include the two you are betting on (5 and 7 in my example)? It is 10. In other words, 10 tickets purchased as a box will win if any one of the 54 are drawn (5.4%, or 1 in 18.5). If... [ More ]

Reply #16

Wisconsin3054 — Sat, 27 Feb 2016 17:25:28 GMT

These kinds of things are fun to think about. I now see what I have wrong here. This is the odds if you simply picked 2 distinct digits that they would both show up somewhere in the three picked. When you add in that you get to pick an additional digit and you are just looking for any of the two to hit tge odds should get better right?

Pretty sure only still.

Serge, can you explain your math on yours

Reply #15

Tucker Black — Sat, 27 Feb 2016 17:14:58 GMT

You're right, I messed up again! I don't know why I put criteria in my query that doesn't belong.

It should be 54:

count n1 n2 n3

1 0 5 7

2 0 7 5

3 1 5 7

4 1 7 5

5 2 5 7

6 2 7 5

7 3 5 7

8 3 7 5

9 4 5 7

10 4 7 5

11 5 0 7

12 5 1 7

13 5 2 7

14 5 3 7

15 5 4 7

16 5 5 7

17 5 6 7

18 5 7 0

19 5 7 1

20 5 7 2

21 5 7 3

22 5 7 4

23 5 7 5

24 5 7 6

25 5 7 7

26 5 7 8

27... [ More ]

Reply #14

SergeM — Sat, 27 Feb 2016 16:52:00 GMT

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

Reply #13

lakerben — Sat, 27 Feb 2016 15:43:09 GMT

Nm pays $5 for a pair.

Reply #12

Wisconsin3054 — Sat, 27 Feb 2016 14:32:14 GMT

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.

Reply #11

Tialuvslotto — Sat, 27 Feb 2016 13:49:01 GMT

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.

Reply #10

ArizonaDream — Sat, 27 Feb 2016 11:44:45 GMT

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.

Reply #9

jimjwright — Sat, 27 Feb 2016 10:54:47 GMT

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

Reply #8

ArizonaDream — Sat, 27 Feb 2016 06:53:56 GMT

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

Reply #7

Tucker Black — Sat, 27 Feb 2016 06:18:08 GMT

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

Reply #6

ArizonaDream — Sat, 27 Feb 2016 06:15:53 GMT

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.

Reply #5

Wisconsin3054 — Sat, 27 Feb 2016 05:16:51 GMT

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.

Reply #4

Tucker Black — Sat, 27 Feb 2016 04:28:41 GMT

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.

Reply #3

ArizonaDream — Sat, 27 Feb 2016 02:20:32 GMT

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.

Reply #2

AllenB — Sat, 27 Feb 2016 02:14:55 GMT

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

Reply #1

LOTTOKING2016 — Sat, 27 Feb 2016 02:04:59 GMT

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

Odds? 2 of 3, pick 3

ArizonaDream — Sat, 27 Feb 2016 02:00:38 GMT

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