Probability of winning with multiple ticketsPrev TopicNext Topic

• Stack47

  

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  Apr 1, 2026, 3:35 am

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  "If you hold two tickets for Pick 2, betting on 37 as well as 65, your odds of winning are 1 in 50. The calculation is 2 [number of distinct bets you have made] divided by 100 [number of possible outcomes]. 2 divided by 100 is 1/50. 2/100=1/50=0.02=2%"

  Don't know how else it could be explained better!

  But it looks like some here just don't understand why a 2% chance of winning a game with 100 outcomes is a 1 in 50 chance. Hopefully explaining how buying 10 chances gives an actual 1 in 10 chance won't short circuit anyone's mind. By using all the digits in either the first or second position there are only 9 ways to lose.

  "Some people have something to say while other people feel the need to say something"
• Stack47

  

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  In response to Tucker Black

  Quote: Originally posted by Tucker Black on Mar 30, 2026

  Each ticket is a 1 in 300M chance to win. It doesn't matter whether two tickets were bought by one person or two different people. Each ticket is a 1 in 300M chance.

  If you buy two tickets (different numbers) then you have a 1 in 150M chance of winning because you are holding TWO TICKETS. Each ticket alone is 300M but you have TWO of them. 300M divided by 2 is 150M.

  If you buy a third ticket, your chances of winning improve to 100M. Not 75M (half of 150M), it's 100M because you have three distinct tickets, each of which has 300M odds. 300M divided by 3 is 100M.

  "If you buy two tickets (different numbers) then you have a 1 in 150M chance of winning because you are holding TWO TICKETS. Each ticket alone is 300M but you have TWO of them. 300M divided by 2 is 150M."

  Thought this thread would had a great discussion on how odds are applied to various possible outcomes, but instead of using math, some only could offer "school yard bully's" lame insults. That being said, let's try to prove how buying two tickets actually reduces the expressed odds in half. 

  For starters what percentage of outcomes do two tickets have; divide 2 by 300,000 and that equals 0.000000006666666666666666666666666666667%. Because that is the percentage of 2 tickets, divide 1 by that percentage and get 150,000. Probably much easier calculating for a pick 3 game. Two tickets divided by 1000 outcomes is a percentage of  0.2% and 1 divided by 0.2% is 500. 

  Odds are sometimes expressed as "ways to win compared to ways to lose" so with ten pick 3 combos there are 10 combos that can win and 990 that can't, a 1.0101% chance with odds of 1 in 99. Using the total chances divided by the number of bought for expressing odds, 25 Powerball combos have odds of 1 in 11,688,054. If that's not correct, how come the odds of the 25 possible combos that match five numbers without the powerball have the same 11,688,054 to 1 odds?

   

  

  

  

  

  "Some people have something to say while other people feel the need to say something"
• Tucker Black

  

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  In response to Stack47

  Quote: Originally posted by Stack47 on Apr 10, 2026

  "If you buy two tickets (different numbers) then you have a 1 in 150M chance of winning because you are holding TWO TICKETS. Each ticket alone is 300M but you have TWO of them. 300M divided by 2 is 150M."

  Thought this thread would had a great discussion on how odds are applied to various possible outcomes, but instead of using math, some only could offer "school yard bully's" lame insults. That being said, let's try to prove how buying two tickets actually reduces the expressed odds in half. 

  For starters what percentage of outcomes do two tickets have; divide 2 by 300,000 and that equals 0.000000006666666666666666666666666666667%. Because that is the percentage of 2 tickets, divide 1 by that percentage and get 150,000. Probably much easier calculating for a pick 3 game. Two tickets divided by 1000 outcomes is a percentage of  0.2% and 1 divided by 0.2% is 500. 

  Odds are sometimes expressed as "ways to win compared to ways to lose" so with ten pick 3 combos there are 10 combos that can win and 990 that can't, a 1.0101% chance with odds of 1 in 99. Using the total chances divided by the number of bought for expressing odds, 25 Powerball combos have odds of 1 in 11,688,054. If that's not correct, how come the odds of the 25 possible combos that match five numbers without the powerball have the same 11,688,054 to 1 odds?

   

  

  

  

  

  Excellent observation.

  The probability of winning the jackpot with one ticket is equal to the probability of matching all five white balls times the probability of matching the Powerball. There are 26 Powerballs, so the probability of matching the Powerball is 1/26.

  Let X be equal to the probability of matching all five white balls. Thus, the probability of winning the jackpot is X*1/26.

  The probability of winning a million dollars (5+0) is the probability of matching all five white balls times the probability of not matching the Powerball. The probability of not matching the Powerball is 25/26. Therefore, the probability of winning one million dollars is X*25/26. This number (1 in about 11 million) is exactly equal to X*1/26 (jackpot) times 25. That's why the two numbers are the same.

  If you buy 25 tickets, your chances of winning a million dollars can be much better than 1 in 11 million. That's the odds for just one ticket. It depends on what numbers you pick. If you make them as different as possible, then the odds to win a million dollars will be about 1 in 467,000. But if you make all the white numbers the same and pick 25 different Powerball numbers, then the odds to win the 5+0 prize are the same as one ticket (1 in 11 million) but you'll actually win $25 million (a terrible plan in California because you'll share $1 million with yourself 25 times).
• BobP

  

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  Way too far back to remember accurately, recall one of the states that offered two Pick-6 plays for a dollar showed half the normal odds for the dollar played. 
  
  BobP
  

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• Elizabeth03

  

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  The Atlantic 49 in Canada. I play this on Wednesday and Saturday. And the tickets cost $1.00!
• Stack47

  

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  In response to BobP

  Quote: Originally posted by BobP on Apr 11, 2026

  Way too far back to remember accurately, recall one of the states that offered two Pick-6 plays for a dollar showed half the normal odds for the dollar played. 
  
  BobP
  

  Learning to love yourself, is the greatest love all

  Whitney Houston

  When the Ohio Lottery started a 6/40 lotto game tickets were two for a $1. The first drawing was on April 9, 1986 and the secondary prizes were pari-mutual.

  "Some people have something to say while other people feel the need to say something"
• rcbbuckeye

  

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  In response to Stack47

  Quote: Originally posted by Stack47 on Apr 11, 2026

  When the Ohio Lottery started a 6/40 lotto game tickets were two for a $1. The first drawing was on April 9, 1986 and the secondary prizes were pari-mutual.

  I won about $900 on that Lotto game in the late 1980's. Got 5/6 correct.

  If I recall, jackpot was $250,000.

  JUST LOOK AT THE ODDS OF ANY JACKPOT GAME, THAT WILL TELL YOU EVERYTHING YOU NEED TO KNOW
• Stack47

  

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  "It depends on what numbers you pick."

  Someone gave an example of a pick 2 game (and most pick 3 games with front pair etc. bets) with 100 possible outcomes. Betting any 10 combos gets 10 to 1 odds, but are the odds the same for "any 10 combos" and 10 combos with 10 different digits in the first position?

  RCB reminded me of when I too hit 5 out of 6 playing the Ohio Lotto when it was a 6/40 game. Wrote down all 40 numbers on playing cards, mixed them up and drew six at a time and placed them on 7 lines. The strategy was I'd match 6 numbers regardless which numbers were drawn and hopefully all 6 were on the same line.

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• Stack47

  

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  In response to rcbbuckeye

  Quote: Originally posted by rcbbuckeye on Apr 11, 2026

  I won about $900 on that Lotto game in the late 1980's. Got 5/6 correct.

  If I recall, jackpot was $250,000.

  My 5 out of 6 was barely over $600 and had to pay taxes on it.

  "Some people have something to say while other people feel the need to say something"
• Tucker Black

  

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  In response to Stack47

  Quote: Originally posted by Stack47 on Apr 12, 2026

  "It depends on what numbers you pick."

  Someone gave an example of a pick 2 game (and most pick 3 games with front pair etc. bets) with 100 possible outcomes. Betting any 10 combos gets 10 to 1 odds, but are the odds the same for "any 10 combos" and 10 combos with 10 different digits in the first position?

  RCB reminded me of when I too hit 5 out of 6 playing the Ohio Lotto when it was a 6/40 game. Wrote down all 40 numbers on playing cards, mixed them up and drew six at a time and placed them on 7 lines. The strategy was I'd match 6 numbers regardless which numbers were drawn and hopefully all 6 were on the same line.

  For a Pick 2 game, as long as the numbers are different, 10 combos will have a 1/10 overall chance of winning (assuming the only prize is for hitting a straight). This is because the prize is for matching the numbers exactly.

  For secondary prizes where you match only some of the numbers, the probability of winning depends on what numbers you pick. Let's say you are playing Pick 3 and buy 2 combos and bet on a combination straight/box bet, and you make your tickets have three different digits (say you bet on 123 and 456). The probability of winning a straight bet is 1/500. The probability of winning a box bet is 1 in 83.333, or 1.2% (single ticket box bet odds divided by 2).

  But if you bet on 445 and 899, then the probability of winning a box bet is 1 in 166.666, or 0.6%. It's half as much, but if you win, you win twice as much money. The probability of winning the straight bet is the same, 1 in 500, because that prize is an exact match.
• Coin Toss

  

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  Tucker Black, Most players aren't as concerned about probabilities as you are or there'd be a lot less lotto games offered. There's a saying. "You bets your money and takes your chances" which regardless of the game the "house" is very well aware of.

  Those who run the lotteries love it when players look for consistency in something that's designed not to have any. So many systems, so many theories, so few jackpot winners. 

  

  There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.
• Stack47

  

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  In response to Tucker Black

  Quote: Originally posted by Tucker Black on Apr 12, 2026

  For a Pick 2 game, as long as the numbers are different, 10 combos will have a 1/10 overall chance of winning (assuming the only prize is for hitting a straight). This is because the prize is for matching the numbers exactly.

  For secondary prizes where you match only some of the numbers, the probability of winning depends on what numbers you pick. Let's say you are playing Pick 3 and buy 2 combos and bet on a combination straight/box bet, and you make your tickets have three different digits (say you bet on 123 and 456). The probability of winning a straight bet is 1/500. The probability of winning a box bet is 1 in 83.333, or 1.2% (single ticket box bet odds divided by 2).

  But if you bet on 445 and 899, then the probability of winning a box bet is 1 in 166.666, or 0.6%. It's half as much, but if you win, you win twice as much money. The probability of winning the straight bet is the same, 1 in 500, because that prize is an exact match.

  "For a Pick 2 game, as long as the numbers are different,"

  What if the 10 combos played are only using 7 digits in the first position?  The math says `buying any 10 combos have the same 10% chance as any other 10 combos, but now only 70 of those possible 100 outcomes can be matched. It sure looks like those 10 combos have a zero chance of matching 30 possible outcomes. 

  Can't say if the 100 or so lottery players I know are as concerned about probability as you and especially not for "most players" that might number into the `100 millions. Still believe this thread could be a great discussion on how odds are applied to various possible outcomes. Wonder what "most players" are as concerned about frequency as a few here on LP?

  "Some people have something to say while other people feel the need to say something"
• Tucker Black

  

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  In response to Stack47

  Quote: Originally posted by Stack47 on Apr 13, 2026

  "For a Pick 2 game, as long as the numbers are different,"

  What if the 10 combos played are only using 7 digits in the first position?  The math says `buying any 10 combos have the same 10% chance as any other 10 combos, but now only 70 of those possible 100 outcomes can be matched. It sure looks like those 10 combos have a zero chance of matching 30 possible outcomes. 

  Can't say if the 100 or so lottery players I know are as concerned about probability as you and especially not for "most players" that might number into the `100 millions. Still believe this thread could be a great discussion on how odds are applied to various possible outcomes. Wonder what "most players" are as concerned about frequency as a few here on LP?

  Good question. It doesn't matter if the first digit is the same or not among the 10 combos. The game is to match both digits exactly. Betting on 00 to 09 also has a 10% chance of winning.

  Most lottery players don't care about this and don't visit LP at all. If even just half the lottery players did, this website would crash.

  LP members who don't care about this subject aren't here to make their displeasure known -- except for two of them, who are trolls and always have to make nasty or pointless comments.
• Stat$talker

  

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  "holding TWO TICKETS. Each ticket alone is 300M but you have TWO of them. 300M divided by 2 is 150M."

                                         

  Whut you're sayin izz...each one of thoze 2 ticketz would have to cover 150 Million combinationz ea.

  It only provez you can Divide...But YOU don't understand how Probability actually wErKz ..

  Wheaa did you git such reazonin, outta a Cracker -Jackz' Box?

   

  -Stat$talker 
• Stack47

  

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  In response to Tucker Black

  Quote: Originally posted by Tucker Black on Apr 14, 2026

  Good question. It doesn't matter if the first digit is the same or not among the 10 combos. The game is to match both digits exactly. Betting on 00 to 09 also has a 10% chance of winning.

  Most lottery players don't care about this and don't visit LP at all. If even just half the lottery players did, this website would crash.

  LP members who don't care about this subject aren't here to make their displeasure known -- except for two of them, who are trolls and always have to make nasty or pointless comments.

  Using the pick 2 example, buying ten lines and each tine has a different digit in the first position, there is a 100% probability of matching the digit drawn in the first digit position. Every line has a 1 in 10 chance of matching even if each line uses the same digit in the second position.  Every Powerball payout chart shows that 25 lines has odds of 1 in 11,688,054 and some will say as if 292,201,313 combos magically disappeared just can't comprehend why expressing 25 in 292,201,338 as 1 in 11,688,054 is the same. 

  Might be other examples proving how ten lines each with a unique two digit number has a 10% chance of winning, but haven't seen any examples proving that math is wrong.

  "Some people have something to say while other people feel the need to say something"