Win the 3-Digit Lottery: Jackpots & Deuces!

[musicgold]

Homework Statement

This is not a homework problem. I am trying to understand an example given in a book.

Relevant Equations

It is about figuring out the chances of not getting a partial prize.

There is a 3-digit lottery that picks three numbers from the following: 1, 2, 3, 4, 5, 6, 7. The lottery has prizes for the jackpot and deuces. A ticket has a deuce when it has two of the three picked numbers. (see attached pictures if you want more context).

There are 35 possible combinations, and for every jackpot combination, there are 12 possible deuces (12/35). The book talks about a person buying 7 tickets of this lottery and claims that:
Chances of getting no deuces in the seven tickets = 5.3%
Chances of getting all 7 tickets being deuces = 0.1%

I tried to calculate the case of no deuces, but I couldn't get the correct answer. Here is what I did:

Chance of getting the first ticket with non-deuce values and the jackpot = 22/35
Chance of getting the second ticket ... = 21/34
and so on.

I got the following expression

$$ = \frac {22 } {35 } . \frac {21 } {34 } . \frac {20 } {33 } . \frac {19 } {32 } . \frac {18 } {31 } . \frac {17 } {30 } . \frac {16 } {29} $$
$$ = 0.025 $$

What am I missing?
Thanks

 

[PeroK]

This doesn't make a lot of sense to me. If the tickets are different, then the probability of being a deuce is not independent for each ticket. With seven tickets you can guarantee getting at least one deuce.

The calculations carried out would apply to, say, buying one ticket on seven consecutive weeks. In that case, the tickets being deuce or not are independent.

 

[PeroK]

PS I think he is assuming that you can pick the same ticket twice. That does make them independent.

 

[Office_Shredder]

I agree that the seven tickets can contain duplicates, so each ticket is independent (because I got 5.3% that way). Probably the mental model of the person writing the problem is you go into the gas station and they print you a ticket at random.

Also, note the first part only says no deuces, it doesn't say you can't have a jackpot.

 

[musicgold]

This doesn't make a lot of sense to me. If the tickets are different, then the probability of being a deuce is not independent for each ticket. With seven tickets you can guarantee getting at least one deuce.

The calculations carried out would apply to, say, buying one ticket on seven consecutive weeks. In that case, the tickets being deuce or not are independent.

The case is about buying multiple tickets of a draw. The book is discussing a case where some syndicates buy thousands of tickets of a draw. The example, is a simpler version of that. A person has bought 7 tickets of the 3-digit lottery and we are trying to see the chances of getting a deuce.

 

[PeroK]

The case is about buying multiple tickets of a draw. The book is discussing a case where some syndicates buy thousands of tickets of a draw. The example, is a simpler version of that. A person has bought 7 tickets of the 3-digit lottery and we are trying to see the chances of getting a deuce.

Okay, but one syndicate buying the same ticket twice doesn't make a lot of sense to me. Although, I suppose it depends on the rules of the lottery.

 

[musicgold]

Okay, but one syndicate buying the same ticket twice doesn't make a lot of sense to me. Although, I suppose it depends on the rules of the lottery.

When there are 10 million possibilities (7 number lottery) and a syndicate buys 300,000 tickets using the quick pick option, there will be some duplicate tickets.

Also, here is the actual case the book is talking about:
https://www.theatlantic.com/business/archive/2016/02/how-mit-students-gamed-the-lottery/470349/

Given this, is my approach to calculating the no deuce probability incorrect?

 

[PeroK]

Given this, is my approach to calculating the no deuce probability incorrect?

Yes, see post #4.

 

[musicgold]

Also, note the first part only says no deuces, it doesn't say you can't have a jackpot.

If I exclude the jackpot number, then there are 23, non-deuce combinations.

$$ \frac{23 } { 35} . \frac{22 } {34 } . \frac{21 } {33 } . \frac{20 } {32 } . \frac{19 } {31 } . \frac{18 } {30 } . \frac{17 } {29 }. $$

$$=0.036 $$

what am I doing wrong?

 

[PeroK]

If I exclude the jackpot number, then there are 23, non-deuce combinations.

$$ \frac{23 } { 35} . \frac{22 } {34 } . \frac{21 } {33 } . \frac{20 } {32 } . \frac{19 } {31 } . \frac{18 } {30 } . \frac{17 } {29 }. $$

$$=0.036 $$

what am I doing wrong?

You are assuming that each ticket you buy is different from the others. Although, in that case, whether one has a deuce or not is definitely not independent of the others.

 

[Office_Shredder]

I think the 5.3% assumes you do not attempt to coordinate the ticket picks. Each one is independently picked.

(23/35)^7 gives what you said is supposed to be the answer.

It doesn't seem like you're doing anything wrong other than the question being a bit confusing.

 

[musicgold]

I think the 5.3% assumes you do not attempt to coordinate the ticket picks. Each one is independently picked.

(23/35)^7 gives what you said is supposed to be the answer.

Ah! got it. I didn't realize that it is possible to pick the same ticket for 7 times, so the proportion of deuce tickets (12/35) doesn't change.

Thanks