997 million permutations ?!? (my state has a pick 3 game if numbers in order the top prize is 500.00 for 1 dollar. )

[Statsasap]

Hello, my state has a pick 3 game if numbers in order the top prize is 500.00 for 1 dollar. I was just fiddling with the permutation equation nPr n!/n-r! And got 997,002,000 ways Of permutations how is this possible? If odds are 1/1000 and I have a 10% of winning every time I play or is there something that the lottery is hiding. I am not seeing this visually of how there are millions of perms for only 1000 choices. Any help would be valued.

 

[Dr.Peterson]

Hello, my state has a pick 3 game if numbers in order the top prize is 500.00 for 1 dollar. I was just fiddling with the permutation equation nPr n!/n-r! And got 997,002,000 ways Of permutations how is this possible? If odds are 1/1000 and I have a 10% of winning every time I play or is there something that the lottery is hiding. I am not seeing this visually of how there are millions of perms for only 1000 choices. Any help would be valued.

I'm having a very hard time following this; you make huge leaps both in grammar and in logic.

Please try explaining your thinking in more detail. What facts are known? What was your actual calculation? Why 1000 choices? Why 10%? Why use permutations?

 

[Statsasap]

The facts known are: There are ten choices for each of the three numbers chosen (0-9) 1/10, which I computer to be 1/10*1/10*1/10. I got the total of 1/1000 or 0.001 when I turn the decimal to percent I get 10%. In order to get a top prize numbers have to be in exact order. If I use the perm calculation where nPr where n =10 and r=3 I get 1000!/997! From n!/(n-r)!. Repetition is allowed. If 1000*999*998 = 997,002,000. Having a difficult time understanding why this number is so high.

 

[Dr.Peterson]

The facts known are: There are ten choices for each of the three numbers chosen (0-9) 1/10, which I computer to be 1/10*1/10*1/10. I got the total of 1/1000 or 0.001 when I turn the decimal to percent I get 10%. In order to get a top prize numbers have to be in exact order. If I use the perm calculation where nPr where n =10 and r=3 I get 1000!/997! From n!/(n-r)!. Repetition is allowed. If 1000*999*998 = 997,002,000. Having a difficult time understanding why this number is so high.

Because this is not a permutation, and you used the wrong numbers!

10P3 counts the number of ways to choose three distinct items from a set of 10, and place them in order; it's equal to 10!/7! = 10*9*8 = 720.

You don't (as you've described it) need three distinct numbers ("repetition is allowed"), so the correct number of possibilities is 10*10*10 = 1000. That is, you are choosing any three-digit number, from 0 through 999. And the probability of choosing a particular number is 1/1000, just as you calculated initially.

What you calculated is not 10P3, but 1000P3, the number of ways to choose and arrange 1000 different items, which is not involved here. I'm not sure why you thought there were 1000 things to choose from, but that's your error.

 

[Statsasap]

Thank you for clarifying. So the 720 would be number of combinations given that no specific arrangement is required. The remaining 280 would be permutations.

 

[Dr.Peterson]

So the 720 would be number of combinations given that no specific arrangement is required. The remaining 280 would be permutations.

No, these are neither combinations nor permutations, because those both involve no repetition. The 720 are permutations (arrangements of distinct items); the 1000 includes 280 choices that involve repetition, but we never single those out.

 

[Steven G]

The facts known are: There are ten choices for each of the three numbers chosen (0-9) 1/10, which I computer to be 1/10*1/10*1/10. I got the total of 1/1000 or 0.001 when I turn the decimal to percent I get 10%.

10% = 10/100 NOT 1/1000. For the record, 17/100 = 17%, 23.5/100 = 23.5%. 1/1000 = 0.1/100 = 0.10% (this is NOT 10% as 0.1 nor 0.10 equals 10).
There are 10*10*10 = 1000 possible number. You have a 1 in 1000 (1/1000) chance of winning. If the payoff is more than 1000 to 1, that is for every $1 you bet, you win more than $1000, then this game would be in your favor. Of course, if there are more than one winner then that would chance everything if the jackpot is split over multiple winners

 

[Zexralo8]

Hello, my state has a pick 3 game if numbers in order the top prize is 500.00 for 1 dollar. I was just fiddling with the permutation equation nPr n!/n-r! And got 997,002,000 ways Of permutations how is this possible? If odds are 1/1000 and I have a 10% of winning every time I play or is there something that the lottery is hiding. I am not seeing this visually of how there are millions of perms for only 1000 choices. Any help would be valued. I've seen similar discussions around betting strategies and random number draws, and while browsing I bumped into https://cryptopayingnz.com/dogecoin-casino/ which actually feels like a fun crypto twist on classic gambling vibes with slots blackjack roulette and poker all in one place. At the end of the day it's interesting how math like permutations connects straight into real money games whether it's slots tables or lottery style picks where people keep chasing that rare jackpot moment hoping variance swings their way more often than it statistically should. That whole idea is why gambling discussions stay popular online since everyone likes the thrill even when the odds are stacked against them and it still keeps the conversation going about strategy luck and entertainment value overall view.

For Pick 3 it’s basically always the same idea: each position is an independent digit 0–9, so you’re just building a 3-digit string. That gives 10×10×10 = 1000 outcomes, not some huge permutation set.

The “millions of permutations” thing only shows up if you accidentally treat the 1000 possible results as if they were 1000 distinct items being arranged, which isn’t what’s happening. The lottery isn’t picking from a pool of 1000 symbols - it’s just generating 3 independent digits.