Calculating the Probability of Winning the Lottery Using Combinatorics

[Ciaran]

Hi there, my question is the following:
What are your chances of winning the following lottery? You choose a set of 10 distinct
numbers from the set S = {1, . . . , 50}, and the person running the lottery selects 6 (distinct) numbers at
random, also from the set S. If all six selected numbers are a subset of your chosen set of 10 numbers, then
you win.

Is the answer just 1/(49 choose 10)*(10 choose 6)? That sort of makes sense; it's the combined probability of choosing the right ten numbers from 49 and having 6 chosen from that ten.

 

[Jameson]

Hi there, my question is the following:
What are your chances of winning the following lottery? You choose a set of 10 distinct
numbers from the set S = {1, . . . , 50}, and the person running the lottery selects 6 (distinct) numbers at
random, also from the set S. If all six selected numbers are a subset of your chosen set of 10 numbers, then
you win.

Is the answer just 1/(49 choose 10)*(10 choose 6)? That sort of makes sense; it's the combined probability of choosing the right ten numbers from 49 and having 6 chosen from that ten.

Hi Ciarin,

Interesting problem! :)

First thing to note is that there are actually 50 possible numbers. 50-1 is equal to 49 but we want {1,2,...50} inclusively so there are 50 possible values.

Ok, whenever we have a probability we should divide the ways to construct a specific event divided by the total possible events. Let's start with the specific event. We need to get all 6 of the correct numbers, so that will be $\displaystyle {6 \choose 6 }$. Now we have to account for the 4 other numbers that we chose that weren't part of the winning 6. How many numbers are there to choose from now and how many non-winners are we choosing?

 

[Ciaran]

There will be 44 numbers to choose from as 50-6=44. And there are 4 non winning numbers we are choosing?

 

[Jameson]

There will be 44 numbers to choose from as 50-6=44. And there are 4 non winning numbers we are choosing?

Exactly, so I think that can be written as $\displaystyle {44 \choose 4}$. This now accounts for the winning numbers and losing numbers in the specific event we want.

The part I'm still thinking about now is how to construct the denominator. At first I thought the answer would simply be: \(\displaystyle \frac{{6 \choose 6}{44 \choose 4 }}{{50 \choose 6}}\) but now I see that this is an issue. Let me think for a second.

 

[Ciaran]

2 events are possible- winning and not winning!

 

[Jameson]

Ah ok I think I see it now. I believe we should divide by \(\displaystyle {50 \choose 10}\) because we are considering all the possible 10 number combos, both winning and losing, that are possible for us to pick.

Does that make sense?

 

[Ciaran]

Yeah that makes sense, because the denominator only considers picking the numbers, winning or not.

 

[Jameson]

Yeah that makes sense, because the denominator only considers picking the numbers, winning or not.

Correct. My apologies that I didn't guide you through my logic more than just saying it, but I was trying to figure out the solution as I was typing.

Anyway, I think the solution is \(\displaystyle \displaystyle \frac{{6 \choose 6}{44 \choose 4 }}{{50 \choose 10}}\) but I see that someone else is replying so maybe there is an error in my logic. We'll see. Something I was thinking about was the fact that if we chose ALL 50 numbers, then clearly we should always get the 6 winning numbers among them, right? I wanted to make sure that my method scaled up to this and it does. If we choose all 50 numbers then this becomes \(\displaystyle \displaystyle \frac{{6 \choose 6}{44 \choose 44 }}{{50 \choose 50}}=1\), which is intuitively correct. If we choose all the numbers, we'll have the 6 winners and the 44 losers, all the time and we'll always win so the probability is 1.

 

[Ciaran]

I would concur; I can follow every step of your logic clearly and it scales up so I can't see it being incorrect!

 

[Opalg]

Hi there, my question is the following:
What are your chances of winning the following lottery? You choose a set of 10 distinct
numbers from the set S = {1, . . . , 50}, and the person running the lottery selects 6 (distinct) numbers at
random, also from the set S. If all six selected numbers are a subset of your chosen set of 10 numbers, then
you win.

Is the answer just 1/(49 choose 10)*(10 choose 6)? That sort of makes sense; it's the combined probability of choosing the right ten numbers from 49 and having 6 chosen from that ten.

Suppose that you are present (or watching on television) when the winning numbers are drawn. When the first winning number is drawn, there are 50 possible choices , and only 10 of them are in your chosen set. So the chance of still having a hope of winning after the first draw is $\frac{10}{50}.$

Assume that you are still in the running after the first draw. Then the second number is drawn. There are 49 possible choices, of which 9 are on your list: probability of success is $\frac9{49}.$

Continuing in that way, the probability of being a winner after all six draws looks to me like \(\displaystyle \frac{10\cdot9 \cdot8 \cdot7 \cdot6 \cdot5\:} {50\!\cdot\!49\! \cdot\!48 \!\cdot\!47\! \cdot\!46\! \cdot\!45} = \frac{10\choose6}{50\choose6}.\)

 

[Jameson]

I just wrote a long post where I tried to test out both of our methods on a smaller set by brute force but found out after a typo that both our methods worked on this test. So then, just out of curiosity I plugged in both of our answers into Wolfram Alpha and wouldn't you know it...

\(\displaystyle \frac{10\choose6}{50\choose6}=\frac{1}{75670}\) Link.

\(\displaystyle \frac{{6 \choose 6}{44 \choose 4 }}{{50 \choose 10}}=\frac{1}{75670}\) Link.

:D