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# Odds? 2 of 3, pick 3

Topic closed. 26 replies. Last post 11 months ago by Stack47.

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New Member
Burlington, VT
United States
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February 25, 2016
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 Posted: February 27, 2016, 12:14 pm - IP Logged

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

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 5 7 9 28 5 8 7 29 5 9 7 30 6 5 7 31 6 7 5 32 7 0 5 33 7 1 5 34 7 2 5 35 7 3 5 36 7 4 5 37 7 5 0 38 7 5 1 39 7 5 2 40 7 5 3 41 7 5 4 42 7 5 5 43 7 5 6 44 7 5 7 45 7 5 8 46 7 5 9 47 7 6 5 48 7 7 5 49 7 8 5 50 7 9 5 51 8 5 7 52 8 7 5 53 9 5 7 54 9 7 5
United States
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February 11, 2016
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 Posted: February 27, 2016, 12:25 pm - IP Logged

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.

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?

New Member
Burlington, VT
United States
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February 25, 2016
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 Posted: February 27, 2016, 12:39 pm - IP Logged

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 you bet 50 cents each (so \$5 in total), if you win, you'll win either \$42 or \$83 depending on whether the 3 winning digits are distinct (\$42 win) or are there just two (e.g., 5-7-5 is \$83 win). 48 of the winning combos are \$42 wins and 6 are \$83 wins (i.e., 4.2% chance of winning \$42 and 0.6% chance of winning \$83).

Interesting! Thanks!

United States
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October 10, 2015
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 Posted: February 27, 2016, 5:59 pm - IP Logged

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
574 754
575* 755*
576 756
577* 757*
578 758
579 759

United States
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October 10, 2015
630 Posts
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 Posted: February 27, 2016, 6:04 pm - IP Logged

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 you bet 50 cents each (so \$5 in total), if you win, you'll win either \$42 or \$83 depending on whether the 3 winning digits are distinct (\$42 win) or are there just two (e.g., 5-7-5 is \$83 win). 48 of the winning combos are \$42 wins and 6 are \$83 wins (i.e., 4.2% chance of winning \$42 and 0.6% chance of winning \$83).

Interesting! Thanks!

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!

Kentucky
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February 14, 2006
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 Posted: February 27, 2016, 9:15 pm - IP Logged

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
574 754
575* 755*
576 756
577* 757*
578 758
579 759

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.

Lincoln, California
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June 27, 2015
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 Posted: February 27, 2016, 10:39 pm - IP Logged

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 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.

Findlay, Ohio
United States
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 Posted: February 28, 2016, 6:31 am - IP Logged

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 digits besides the 1 or 2, there are 8 digits x 6 ways = 48 combinations.

Of course X could be replaced with a 1 or a 2 to create a double digit combination. Each double can be drawn 3 different ways so 12X could be 122, 212, 221 OR 112, 121. 211. The 6 doubles combine with the 48 no-match to total the 54 possible straights that would have to be drawn to match your pair of 12X in any order or position.

~Probability=Odds in Motion~

Park City, UT
United States
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January 18, 2009
995 Posts
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 Posted: February 28, 2016, 1:29 pm - IP Logged

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 digits besides the 1 or 2, there are 8 digits x 6 ways = 48 combinations.

Of course X could be replaced with a 1 or a 2 to create a double digit combination. Each double can be drawn 3 different ways so 12X could be 122, 212, 221 OR 112, 121. 211. The 6 doubles combine with the 48 no-match to total the 54 possible straights that would have to be drawn to match your pair of 12X in any order or position.

It's great to see Thoth post again.

Jimmy

ORLANDO, FLORIDA
United States
Member #4924
June 3, 2004
5978 Posts
Online
 Posted: March 4, 2016, 4:20 am - IP Logged

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 digits besides the 1 or 2, there are 8 digits x 6 ways = 48 combinations.

Of course X could be replaced with a 1 or a 2 to create a double digit combination. Each double can be drawn 3 different ways so 12X could be 122, 212, 221 OR 112, 121. 211. The 6 doubles combine with the 48 no-match to total the 54 possible straights that would have to be drawn to match your pair of 12X in any order or position.

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

Texas
United States
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December 31, 2013
822 Posts
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 Posted: March 4, 2016, 7:38 am - IP Logged

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

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

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

~Robert A. Heinlein

Kentucky
United States
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February 14, 2006
7344 Posts
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 Posted: March 5, 2016, 3:37 pm - IP Logged

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

Me too!

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