Lottery EV question

hi @all

let's say a lottery participant has 10 tickets which costs him all in all $1 in a lottery game with 2100 tickets. There will be a draw with 2 winners who get $450 each.

What is the player's EV?

I am confused about the 2 winners of the draw

thank you as always in advance for your help

cheers

There are C(2100, 2) = 2,203,950 pairs of winning tickets.
There are three possible results as far as the player is concerned:

1. There are 45 pairs where the player has both tickets; the ER (expected return) of this is 45 / 2,203,950 x 900

2. There are 10 x 2090 = 20,900 pairs where the player has one ticket; the ER is 20,900 / 2,203,950 x 450.

3. There are C(2090, 2) pairs where the player wins nothing.

The total ER is 9,445,000 / 2,203,950 = 30/7.
Since the player paid 10 for the tickets, the EV is (30/7 - 10) = -40/7.

Quote: ThatDonGuy

There are C(2100, 2) = 2,203,950 pairs of winning tickets.
There are three possible results as far as the player is concerned:

1. There are 45 pairs where the player has both tickets; the ER (expected return) of this is 45 / 2,203,950 x 900

2. There are 10 x 2090 = 20,900 pairs where the player has one ticket; the ER is 20,900 / 2,203,950 x 450.

3. There are C(2090, 2) pairs where the player wins nothing.

The total ER is 9,445,000 / 2,203,950 = 30/7.
Since the player paid 10 for the tickets, the EV is (30/7 - 10) = -40/7.

link to original post

thank you very much for the answer but I need to apologize that I did not explain correctly/understandable. I am really sorry

the player who has 10 tickets paid $1 for all the 10 tickets ( he did not pay $10 for the 10 tickets)

the 2 winner prizes will go to 2 separate players means if one player wins he can not win the 2nd prize

I hope I could clear it up now

cheers

edit
maybe I should explain the 2100 tickets draw
there will be 2100 tickets in a hat and one ticket will be drawn for the 1st winner then there will be drawn another ticket for the 2nd winner. if the 1st winner will win again it will not count and another ticket will be drawn

Quote: seven

thank you very much for the answer but I need to apologize that I did not explain correctly/understandable. I am really sorry

the player who has 10 tickets paid $1 for all the 10 tickets ( he did not pay $10 for the 10 tickets)

the 2 winner prizes will go to 2 separate players means if one player wins he can not win the 2nd prize

I hope I could clear it up now

cheers

edit
maybe I should explain the 2100 tickets draw
there will be 2100 tickets in a hat and one ticket will be drawn for the 1st winner then there will be drawn another ticket for the 2nd winner. if the 1st winner will win again it will not count and another ticket will be drawn

link to original post

Okay, there are three possible combinations of events that can happen in this one. What we're going to do here is express them in the simplest terms:

1.) The first ticket drawn results in the guy winning:

10/2100 = 0.00476190476

2.) He does not win on the first ticket, but then he does win on the second ticket drawn:

(2090/2100) * (10/2099) = 0.00474148687

3.) He does not win on either ticket drawn:

(2090/2100) * (2089/2099) = 0.99049660836

Okay, so we know that he either wins $450 (which is really $440 profits because he paid for the tickets) or that he loses $10. With that, we have:

((0.00476190476+0.00474148687)*440) - (10 * 0.99049660836) = -5.7234737664

This is only slightly worse than what ThatDonGuy ended up with and the reason why is simply that the guy cannot win twice. It also assumes that the original $10 for the tickets is sunk (pays on a For One basis) even if he wins.

Quote: Mission146

Quote: seven

thank you very much for the answer but I need to apologize that I did not explain correctly/understandable. I am really sorry

the player who has 10 tickets paid $1 for all the 10 tickets ( he did not pay $10 for the 10 tickets)

the 2 winner prizes will go to 2 separate players means if one player wins he can not win the 2nd prize

I hope I could clear it up now

cheers

edit
maybe I should explain the 2100 tickets draw
there will be 2100 tickets in a hat and one ticket will be drawn for the 1st winner then there will be drawn another ticket for the 2nd winner. if the 1st winner will win again it will not count and another ticket will be drawn

link to original post

Okay, there are three possible combinations of events that can happen in this one. What we're going to do here is express them in the simplest terms:

1.) The first ticket drawn results in the guy winning:

10/2100 = 0.00476190476

2.) He does not win on the first ticket, but then he does win on the second ticket drawn:

(2090/2100) * (10/2099) = 0.00474148687

3.) He does not win on either ticket drawn:

(2090/2100) * (2089/2099) = 0.99049660836

Okay, so we know that he either wins $450 (which is really $440 profits because he paid for the tickets) or that he loses $10. With that, we have:

((0.00476190476+0.00474148687)*440) - (10 * 0.99049660836) = -5.7234737664

This is only slightly worse than what ThatDonGuy ended up with and the reason why is simply that the guy cannot win twice. It also assumes that the original $10 for the tickets is sunk (pays on a For One basis) even if he wins.

link to original post

I don't understand why 3 possible combinations

the player with 10 tickets in case one of his tickets is drawn out of the hat will win $450 minus $1 equals $449 profit. the winner ticket is out and does not go back into the hat.

in case the next ticket which is drawn is also from the player who owns 10 tickets the ticket is out of the hat and another ticket is drawn until a 2nd winner is drawn

what is the EV of the player with 10 tickets?

Quote: seven

I don't understand why 3 possible combinations

the player with 10 tickets in case one of his tickets is drawn out of the hat will win $450 minus $1 equals $449 profit. the winner ticket is out and does not go back into the hat.

in case the next ticket which is drawn is also from the player who owns 10 tickets the ticket is out of the hat and another ticket is drawn until a 2nd winner is drawn

what is the EV of the player with 10 tickets?

link to original post

Your original post stipulated that he has ten tickets. Is he going to get a refund on the nine that don't win? If you want to know his EV, then it includes the fact that he has purchased ten tickets.

Three combinations because:

His ticket could be the first drawn: 10/2100 is the probability of that event. 0.00476190476

His ticket could be the second drawn, which means some other ticket was drawn first: (2090/2100 * 10/2099) = 0.00474148687 (As you can see, this is slightly less likely than one of his tickets being the first drawn, mainly just because it includes another event happening)

Both tickets drawn are none of his ten.

Or, are you saying that the tickets cost a TOTAL of $1?

If that's the case, just replace "440" and "10":

((0.00476190476+0.00474148687)*440) - (10 * 0.99049660836) =

With "449" and "1", respectively.

Quote: Mission146

Quote: seven

I don't understand why 3 possible combinations

the player with 10 tickets in case one of his tickets is drawn out of the hat will win $450 minus $1 equals $449 profit. the winner ticket is out and does not go back into the hat.

in case the next ticket which is drawn is also from the player who owns 10 tickets the ticket is out of the hat and another ticket is drawn until a 2nd winner is drawn

what is the EV of the player with 10 tickets?

link to original post

Your original post stipulated that he has ten tickets. Is he going to get a refund on the nine that don't win? If you want to know his EV, then it includes the fact that he has purchased ten tickets.

link to original post

sorry for the confusion. he purchased 1 ticket for $1 and got another 9 for free so he has 10 tickets for $1 and he did not pay $10 for ten tickets. he does not get any refund.
as my confusion was because of the 2 winner option maybe we try the simple option with one prize winner of $450. in the hat are 2100 tickets and there will be drawn 1 ticket and the winner gets $450. what is the EV for a player with 10 tickets he paid $1 for all ten tickets ( he did not pay $10 for the ten tickets)?

Quote: seven

sorry for the confusion. he purchased 1 ticket for $1 and got another 9 for free so he has 10 tickets for $1 and he did not pay $10 for ten tickets. he does not get any refund.
as my confusion was because of the 2 winner option maybe we try the simple option with one prize winner of $450. in the hat are 2100 tickets and there will be drawn 1 ticket and the winner gets $450. what is the EV for a player with 10 tickets he paid $1 for all ten tickets ( he did not pay $10 for the ten tickets)?

link to original post

(449 * (10/2100)) - (1 * (2090/2100) = 1.14285714286 (This assumes only one winner of $450)

((0.00476190476+0.00474148687)*449) - (1 * 0.99049660836) = 3.27652623351 (This is ten tickets with two winners of $450, but the same person cannot win twice)

Quote: Mission146

Quote: seven

sorry for the confusion. he purchased 1 ticket for $1 and got another 9 for free so he has 10 tickets for $1 and he did not pay $10 for ten tickets. he does not get any refund.
as my confusion was because of the 2 winner option maybe we try the simple option with one prize winner of $450. in the hat are 2100 tickets and there will be drawn 1 ticket and the winner gets $450. what is the EV for a player with 10 tickets he paid $1 for all ten tickets ( he did not pay $10 for the ten tickets)?

link to original post

(449 * (10/2100)) - (1 * (2090/2100) = 1.14285714286 (This assumes only one winner of $450)

((0.00476190476+0.00474148687)*449) - (1 * 0.99049660836) = 3.27652623351 (This is ten tickets with two winners of $450, but the same person cannot win twice)

link to original post

thank you very much for the formula as it looks like it answers my question but please be so kind if you can give it to me ELI5
for example I know that a single zero roulette has a - 0.027 EV for the player means 2.7% house advantage
what is the EV for the Player in % of your result 1.14285714286 (This assumes only one winner of $450)
and 3.27652623351

cheers

Quote: seven

thank you very much for the formula as it looks like it answers my question but please be so kind if you can give it to me ELI5
for example I know that a single zero roulette has a - 0.027 EV for the player means 2.7% house advantage
what is the EV for the Player in % of your result 1.14285714286 (This assumes only one winner of $450)
and 3.27652623351

cheers

link to original post

You're welcome!

Okay, you want to determine his EV whereas we have calculated his expected return. We know that he is betting a grand total of $1 paid for his ten tickets, so this will be a simple matter of dividing his expectation by the amount he is betting.

1.14285714286/1 = 1.14285714286 (or 114.2857%, rounded)

3.27652623351/1 = 3.27652623351 (or 327.6526%, rounded)

The fact that he only paid a dollar made this slightly easier. If he had paid a total of $2 for the tickets, here is what it would look like with one winner of $450:

(448 * (10/2100)) - (2 * (2090/2100) = 0.14285714285

Okay, so that solves for his expected return if he had paid $2 for the tickets, so with that:

0.14285714285/2 = 0.07142857142. (This means he has a 7.142857%, rounded, advantage)

Naturally, he would be at a disadvantage had he paid $10 for ten tickets, so let's see what that would look like:

(440 * (10/2100)) - (10 * (2090/2100) = -7.85714285714

-7.85714285714/10 = -0.78571428571 (This represents a 78.57143%, rounded, disadvantage, Yikes!)

Quote: Mission146

Quote: seven

thank you very much for the formula as it looks like it answers my question but please be so kind if you can give it to me ELI5
for example I know that a single zero roulette has a - 0.027 EV for the player means 2.7% house advantage
what is the EV for the Player in % of your result 1.14285714286 (This assumes only one winner of $450)
and 3.27652623351

cheers

link to original post

You're welcome!

Okay, you want to determine his EV whereas we have calculated his expected return. We know that he is betting a grand total of $1 paid for his ten tickets, so this will be a simple matter of dividing his expectation by the amount he is betting.

1.14285714286/1 = 1.14285714286 (or 114.2857%, rounded)

3.27652623351/1 = 3.27652623351 (or 327.6526%, rounded)

The fact that he only paid a dollar made this slightly easier. If he had paid a total of $2 for the tickets, here is what it would look like with one winner of $450:

(448 * (10/2100)) - (2 * (2090/2100) = 0.14285714285

Okay, so that solves for his expected return if he had paid $2 for the tickets, so with that:

0.14285714285/2 = 0.07142857142. (This means he has a 7.142857%, rounded, advantage)

Naturally, he would be at a disadvantage had he paid $10 for ten tickets, so let's see what that would look like:

(440 * (10/2100)) - (10 * (2090/2100) = -7.85714285714

-7.85714285714/10 = -0.78571428571 (This represents a 78.57143%, rounded, disadvantage, Yikes!)

link to original post

great! thanks again. that is exactly what I wanted to understand as it is an immense high +EV for the player. especially the 2 prizes version confused me and now I understand why as the +EV (player) is so high and I could not believe it.
now I can use your formula to get the player's EV if he has 9-8-7 .... etc tickets
stay safe

Thank you very much! You're welcome! Not only using the formula for that, but you'll also know how to do it in the future for different numbers of total tickets, individual tickets and/or cash spent.

Quote: Mission146

Thank you very much! You're welcome! Not only using the formula for that, but you'll also know how to do it in the future for different numbers of total tickets, individual tickets and/or cash spent.

link to original post

yes I did now the math for one prize and 9 tickets down to one ticket. this was very helpful, thanks again
but I didn't see or oversaw the formula for the 2 prizes
I saw this one you gave: ((0.00476190476+0.00474148687)*449) - (1 * 0.99049660836) = 3.27652623351 (This is ten tickets with two winners of $450, but the same person cannot win twice)
how can I do the math for 9-8-7 etc ?

cheers

Quote: seven

Quote: Mission146

Thank you very much! You're welcome! Not only using the formula for that, but you'll also know how to do it in the future for different numbers of total tickets, individual tickets and/or cash spent.

link to original post

yes I did now the math for one prize and 9 tickets down to one ticket. this was very helpful, thanks again
but I didn't see or oversaw the formula for the 2 prizes
I saw this one you gave: ((0.00476190476+0.00474148687)*449) - (1 * 0.99049660836) = 3.27652623351 (This is ten tickets with two winners of $450, but the same person cannot win twice)
how can I do the math for 9-8-7 etc ?

cheers

link to original post

For that one, you're going to have to go back to an earlier post and change the probabilities, because there are now fewer tickets.

You may remember this:

Quote:

Okay, there are three possible combinations of events that can happen in this one. What we're going to do here is express them in the simplest terms:

1.) The first ticket drawn results in the guy winning:

10/2100 = 0.00476190476

2.) He does not win on the first ticket, but then he does win on the second ticket drawn:

(2090/2100) * (10/2099) = 0.00474148687

3.) He does not win on either ticket drawn:

(2090/2100) * (2089/2099) = 0.99049660836

Okay, so we know that he either wins $450 (which is really $440 profits because he paid for the tickets) or that he loses $10. With that, we have:

((0.00476190476+0.00474148687)*440) - (10 * 0.99049660836) = -5.7234737664

Okay, so now you want him to have nine tickets total, if I am understanding you correctly, so you have to change the probabilities:

1.) The first ticket drawn results in the guy winning:

9/2100 = 0.00428571428

2.) He does not win on the first ticket, but then he does win on the second ticket drawn:

(2091/2100) * (9/2099) = 0.00426937997

3.) He does not win on either ticket drawn:

(2091/2100) * (2090/2099) = 0.99144490573

Okay, so we will say that he spends one dollar on these nine tickets and will either win $449 (profits) or lose the $1:

((0.00428571428+0.00426937997) * 449) - (1 * 0.99144490573) = 2.84979241252

And, he's only spending a dollar, so for his Expected Return Percentage, you just move the decimal place two to the right and get 284.97924%, rounded.

CHANGE NUMBER OF TICKETS---Steps 1 Through 3 (You can also change the total number of tickets in the pool)

CHANGE EXPECTATION DUE TO CASH SPENT---Change the dollar amounts in the last step to reflect potential profit and cash spent.

Quote: Mission146

Quote: seven

Quote: Mission146

Thank you very much! You're welcome! Not only using the formula for that, but you'll also know how to do it in the future for different numbers of total tickets, individual tickets and/or cash spent.

link to original post

yes I did now the math for one prize and 9 tickets down to one ticket. this was very helpful, thanks again
but I didn't see or oversaw the formula for the 2 prizes
I saw this one you gave: ((0.00476190476+0.00474148687)*449) - (1 * 0.99049660836) = 3.27652623351 (This is ten tickets with two winners of $450, but the same person cannot win twice)
how can I do the math for 9-8-7 etc ?

cheers

link to original post

For that one, you're going to have to go back to an earlier post and change the probabilities, because there are now fewer tickets.

You may remember this:



Okay, so now you want him to have nine tickets total, if I am understanding you correctly, so you have to change the probabilities:

1.) The first ticket drawn results in the guy winning:

9/2100 = 0.00428571428

2.) He does not win on the first ticket, but then he does win on the second ticket drawn:

(2091/2100) * (9/2099) = 0.00426937997

3.) He does not win on either ticket drawn:

(2091/2100) * (2090/2099) = 0.99144490573

Okay, so we will say that he spends one dollar on these nine tickets and will either win $449 (profits) or lose the $1:

((0.00428571428+0.00426937997) * 449) - (1 * 0.99144490573) = 2.84979241252

And, he's only spending a dollar, so for his Expected Return Percentage, you just move the decimal place two to the right and get 284.97924%, rounded.

CHANGE NUMBER OF TICKETS---Steps 1 Through 3 (You can also change the total number of tickets in the pool)

CHANGE EXPECTATION DUE TO CASH SPENT---Change the dollar amounts in the last step to reflect potential profit and cash spent.

link to original post

thank you so much for taking the time to explain it again and ELI5, very very much appreciated!
Stay safe and God Bless You and Your Family

You're very welcome! Thanks for the questions and have a great day!

Quote: Mission146

You're very welcome! Thanks for the questions and have a great day!

link to original post

I am really sorry to bother you again. I did now the math but I am not sure if I am correct or I messed up! would you be so kind whenever you have the time to go over my results? would be very much appreciated.

1 prize EV
10 tickets EV is + 114.28 %
9 tickets EV is + 92.9 %
8 tickets EV is + 72 %
7 tickets EV is + 50 %
6 tickets EV is + 29 %
5 tickets EV is + 7.4 %
4 tickets EV is - 14 %
3 tickets EV is - 35.4 %
2 tickets EV is - 56.8 %
1 ticket EV is - 78.1 %

2 prizes EV
9 tickets EV is + 284.98 %
8 tickets EV is + 242.29 %
7 tickets EV is + 199.57 %
6 tickets EV is + 156.84 %
5 tickets EV is + 114.08 %
4 tickets EV is + 71.30 %
3 tickets EV is + 28.51 %
2 tickets EV is - 14.30 %
1 ticket EV is - 57.14 %

another question I am asking myself and @all if 2 Prizes of $450 or 1 Prize of $900 is more attractive for the participants?

cheers