Burlington, VT United States Member #173303 February 25, 2016 17 Posts Offline

Posted: February 26, 2016, 3:29 pm - IP Logged

This is true for any game but here I will present results for pick 3. For one ticket and one drawing, it does not matter what numbers you pick. However, if you buy multiple tickets, you can adjust the probability of winning or losing based on whether you buy n tickets for one drawing or buy one ticket for each of the next n drawings. Let's say you bet $1 per ticket, bet on a permutation, and pick 3 digits that are different. The house edge is 50%, so the prize is $83.33.

Of the 1000 possible choices for pick 3, 720 of them have 3 different digits. You wish to buy all the possible combinations that have 3 different digits. You don't need to buy 720 tickets. You only need to buy 120 tickets, because 4-5-2 is the same win as 4-2-5, 5-4-2, 5-2-4, 2-4-5, and 2-5-4. 720 divided by 6 is 120.

Scenario 1: you buy 120 tickets with all the possible permutations of 3 distinct digits for one drawing. Hand the cashier $120. The probability of winning is 72%, and if you win, you win $83.33. You can't win less than that and you can't win more than once because all of your tickets are different. The probability of losing is 28%.

Scenario 2: buy 4-5-2 as a permutation for the next 120 drawings. Now you can win more than once. I ran a simulation (1 million trials) and produced the following:

win

win amt

probability

none

$0.00

48.55%

once

$83.33

35.21%

twice

$166.66

12.64%

3 times

$249.99

2.99%

4 times

$333.32

0.53%

5 times

$416.65

0.0684%

6 times

$499.98

0.0074%

7 times

$583.31

0.0003%

8 times

$666.64

0.0001%

As you can see, the probability of losing went way up, from 28% to 48.55%. The probability of winning $83.33 dropped significantly, from 72% to 35.21%. However, the probability of winning twice or more is 16.24% whereas in scenario 1 that is a zero.

In scenario 1, the expected value of winnings is 72% times $83.33 which is $60, half of your $120 (50% house edge, makes sense).

In scenario 2, the expected value of winnings is the sum of the product of win amt and probability for each row in the above table. That value is also $60. So you don't change the expected value; you can't. It's always 50%.

However, by buying 120 tickets over 120 drawings, you change the probability distribution of winnings, increasing the chance of losing and decreasing the chance of winning $83.33 in exchange for a 16% chance of winning $83.33 more than once. Even just twice will result in a profit of $46.66. You have a 3% chance of netting $130.

Scenario 1 is kinda silly because you are guaranteed to lose money. The only question is, do you lose all $120, or do you win $83.33 and only lose $36.67?

However, you can modify this and say "I'll buy just half of all the permutations for $60 and try to win $83.33". Similarly, if you buy one ticket for each of the next 60 drawings, your probability of losing will go up in exchange for getting a chance to win $83.33 more than once. You're simply increasing the stakes.

It takes about half an hour to run 1 million trials. I will post the results for buying one ticket for the next 60 drawings and see how that compares to buying 60 tickets for one drawing (i.e., if you win, you profit $83.33-$60=$23.33)

Burlington, VT United States Member #173303 February 25, 2016 17 Posts Offline

Posted: February 26, 2016, 4:06 pm - IP Logged

60 permutations in one drawing has a 36% chance of winning (half the 72% in the previous post) and a 64% chance of losing. Expected value is 0.36*83.33=$30, half of your $60 investment.

One permutation bet in the next 60 drawings:

win

win amt

probability

none

$0.00

69.70%

once

$83.33

25.24%

twice

$166.66

4.48%

3 times

$249.99

0.53%

4 times

$333.32

0.0451%

5 times

$416.65

0.0024%

6 times

$499.98

0.0002%

This one is much better. Spend $60 trying to win $83.33 in one drawing (profiting $23.33), or if you do 1 ticket for the next 60 drawings, the probability of losing doesn't go up that much, from 64% to 69.7%. The probability of winning once (which is still a profit albeit the smallest one) goes down from 36% to 25.24%. The probability of winning more than once is 5.06%. This is the one disadvantage compared to buying 1 ticket for the next 120 drawings, namely that the probability of winning $83.33 more than once drops from 16.24% to only 5.06%, just under 1/3 as much.

If you want me to run a simulation for you so you can compare two strategies, just let me know!

Madison, WI United States Member #172977 February 11, 2016 515 Posts Offline

Posted: February 27, 2016, 12:28 pm - IP Logged

I like this kinda stuff Tucker. I haven't had a chance to look real deeply at it yet but once I do I will post more about how I think we can use this kind of information.

Madison, WI United States Member #172977 February 11, 2016 515 Posts Offline

Posted: February 29, 2016, 2:07 pm - IP Logged

Ok, finally have a chance to sit down with this.

How I'm using this is looking at what it takes to get to a 50% probability to win based on the number of plays. I think this number can tell you how often you can expect to win over time based on how many plays you are entering each drawing. The fun for me is winning. If I want to have that feeling on average once a month, then I need to play enough numbers per drawing to reach that.

Why 50%? 50% is essentially your coin flip. Over time, I can expect to average winning every other coin flip. So you can take the number of plays to get a 50% probability to win and double that (which happens to also equal the number of plays to get a 75% probability to win if you want to shortcut and maybe more exact). If you play long enough, you should hit that often.

So, taking for example the pick 3 six way box play. My state offers only single drawing a day. $0.50 play wins $40

For 1 play per drawing, you hit 75% probability of a win with 233 plays.

for 2 plays per drawing, you hit 75% probability of a win with 116 plays.

and so on:

3 plays/draw = 77 plays

4 plays/draw = 58 plays

5 plays/draw = 46 plays

6 plays/draw = 39 plays

7 plays/draw = 33 plays

8 plays/draw = 29 plays

So, If I want to expect a win every month on average, I need to be playing 8 sets of numbers per day. If I play each of those at $.50 that amounts to $4 per day. Average month (30 days) i will spend $120, and hit once for $40.

What isn't making sense to me is that the Pick 3 brochure claims they pay out on average 48.2% of sales. If that is right then my math is not adding up, because you should be expecting to hit with about every $81 or so spent.

Maybe Tucker or someone else can steer me in the right direction here on what I'm doing wrong or if the lottery's numbers are off?

Either way, it seems that the only way to come out ahead over time would be to use a progressive betting system, so that when you do hit, there are times where you are playing maybe 4X on each number and win $160 instead of $40. But you would need a big bankroll to keep up with that because even coin flips can go one way for a while!

Madison, WI United States Member #172977 February 11, 2016 515 Posts Offline

Posted: March 10, 2016, 6:14 pm - IP Logged

Ok so this question has been nagging at my mind and maybe somebody can help make sense of this for me.

I'm trying to figure out how often you can expect to hit over time playing one combo 6-way boxed. There's two ways I see it in my mind and each leads to a different result.

First, thinking out 1000 draws, since I have 6 of 1000 possible combos, I say I would expect to hit 6 times every 1000 draws. This amounts to 1 time every 166 and 2/3 draws. This number equals the odds per day, 1 in 166 and 2/3.

Second, thinking daily, with 6 of 1000 possible combos, each day I have a 0.6% chance of hitting. If I use probability math, I would a expect a 50% chance of hitting in about 116 draws (94% chance of not hitting each day X 116 draws = about 50%). As I noted above, 50% gets you to the coin flip probability, and you expect to hit once every two of those over time. Double the 116 to 232 for how often you can expect to hit.

Ok, so funny thing. Typing it out like this actually led me to solve my own problem. In the second way of thinking, I am calculating the percentage chance of hitting at least once. So over 116 draws I would have a 50% chance of hitting at least once, sometimes more than once, which would lead me to believe my first way of thinking about it more accurately tells us how often we should expect to hit over time.

Anybody agree or disagree? I'm going to consider how this way of looking at it impacts the playing of multiple combos per draw and come back to this.

United States Member #116344 September 8, 2011 3926 Posts Offline

Posted: March 31, 2016, 6:43 pm - IP Logged

Quote: Originally posted by Wisconsin3054 on February 29, 2016

Ok, finally have a chance to sit down with this.

How I'm using this is looking at what it takes to get to a 50% probability to win based on the number of plays. I think this number can tell you how often you can expect to win over time based on how many plays you are entering each drawing. The fun for me is winning. If I want to have that feeling on average once a month, then I need to play enough numbers per drawing to reach that.

Why 50%? 50% is essentially your coin flip. Over time, I can expect to average winning every other coin flip. So you can take the number of plays to get a 50% probability to win and double that (which happens to also equal the number of plays to get a 75% probability to win if you want to shortcut and maybe more exact). If you play long enough, you should hit that often.

So, taking for example the pick 3 six way box play. My state offers only single drawing a day. $0.50 play wins $40

For 1 play per drawing, you hit 75% probability of a win with 233 plays.

for 2 plays per drawing, you hit 75% probability of a win with 116 plays.

and so on:

3 plays/draw = 77 plays

4 plays/draw = 58 plays

5 plays/draw = 46 plays

6 plays/draw = 39 plays

7 plays/draw = 33 plays

8 plays/draw = 29 plays

So, If I want to expect a win every month on average, I need to be playing 8 sets of numbers per day. If I play each of those at $.50 that amounts to $4 per day. Average month (30 days) i will spend $120, and hit once for $40.

What isn't making sense to me is that the Pick 3 brochure claims they pay out on average 48.2% of sales. If that is right then my math is not adding up, because you should be expecting to hit with about every $81 or so spent.

Maybe Tucker or someone else can steer me in the right direction here on what I'm doing wrong or if the lottery's numbers are off?

Either way, it seems that the only way to come out ahead over time would be to use a progressive betting system, so that when you do hit, there are times where you are playing maybe 4X on each number and win $160 instead of $40. But you would need a big bankroll to keep up with that because even coin flips can go one way for a while!

Stats are facts, but the projected percentiles has lot of nuances. Focusing on percentiles may not be a good waging strategy, rather focus on 'bet option' , straight or box. a straight with a lead digit is 100 lines, with Front pair is 10 lines.

georgia United States Member #65527 September 28, 2008 417 Posts Offline

Posted: March 31, 2016, 7:54 pm - IP Logged

that's a lot of odds, makes me dizzy but if it works for you, I once used a 36 number box group not hard to figure out, there 36 boxed combinations using one number , not counting doubles. that way I pick one number to hit , play 36 or 18 bucks win 80 or 40 if number falls. I will do 0's for ya, so you understand.

012,013,014,015,016,017,018,019,

023,024,025,026,027,028,029,034

035,036,037,038,039,045,046,047

048,049,056,057,058,059,067,068

069,078,079,089,

Play those an 0 falls that is not a double ya win, If you are real lucky can cut 10 or 15 numbers that don't hit that often. but takes lot of study.