# Thread: Probability of new lottery

1. ## Probability of new lottery

Hi everyone,

I come from the Czech Republic and I received a business case at work. I am supposed to come up with a new lottery which would be a competitor to something similar like Lotto from UK National Lottery, which in the Czech Republic is called Loto.

I've been trying to calculate the probability of the old and of the new lottery for hours but I just cant get it right. Could some good soul from this forum please please help me?

With Loto here you have to pick 6 numbers from 1-49 and have to choose a color for the whole bet (black or white). You win most if you get all 6 numbers and the color. You win less if you guess 6 numbers and not the color. Then again 5 numbers and the color, and so on.
Could someone please explain the probability evaluation step by step for me please?

Now with the "new" lottery you will pick 5 numbers from 1-50 and you must pick a "dice" number 1-6. You win most when you get 5 numbers and the "dice" number, etc. etc. What is the probability evaluation in this case?

I really need a "for dummies" explanation. If some of you would be so kind to help me, you cant imagine how greatful I would be.

Thank you so very much

Tomas

2. ## Re: Probability of new lottery

Hello, Tomas!

I'll try to explain the "new" lottery.

Now with the "new" lottery you will pick 5 numbers from 1-50
and you must pick a "dice" number 1-6.
You win most when you get 5 numbers and the "dice" number, etc. etc.
What is the probability evaluation in this case?
Call the 5 chosen numbers "Winners". .The rest are called "Others".

There are: .$\displaystyle _{50}C_5 \:=\:{50\choose5} \:=\:\frac{50!}{5!\,45!} \:=\:2,\!118,\!760$ five-number choices,
. . and 6 outcomes for the die.

Hence, there are: .$\displaystyle 2,\!118,\!760 \times 6 \:=\:12,\!712,\!560$ possible outcomes.
. . This is the denominator $\displaystyle D$ for our probabilities.

Grand Prize: "5 numbers and die"
There is 1 way to choose the 5 Winners and 1 way to choose the die-number.
Hence, there is only $\displaystyle 1$ way to win the Grand Prize.

. . $\displaystyle P(\text{Grand Prize}) \:=\:\frac{1}{D}$

2nd Prize: "5 numbers and not-die"
There is 1 way to choose the 5 Winners and 5 way to choose a not-die number.
Hence, there are: .$\displaystyle 1 \times 5 \:=\:5$ ways to win 2nd Prize.

. . $\displaystyle P(\text{2nd prize}) \:=\:\frac{5}{D}$

3rd Prize: "4 numbers and die"
We want 4 winners and 1 Other. .There are: $\displaystyle {5\choose4}{45\choose1} \:=\:225$ ways.
And we want the die-number: $\displaystyle 1$ way.
Hence, there are: $\displaystyle 225$ ways to win 3rd Prize.

. . $\displaystyle P(\text{3rd Prize}) \:=\:\frac{225}{D}$

4h Prize: "4 numbers and not-die"
We want 4 winners and 1 Other: $\displaystyle 225$ ways.
And we want a not-die number: $\displaystyle 5$ ways.
Hence, there are: $\displaystyle 225 \times 5 \:=\:1125$ ways to win 4th Prize.

. . $\displaystyle P(\text{4th Prize}) \:=\:\frac{1125}{D}$

5th Prize: "3 numbers and die"
We want 3 Winners and 2 Others: $\displaystyle {5\choose3}{45\choose2} \,=\,9900$ ways.
And we want the die-number: $\displaystyle 1$ way.
Hence, there are: $\displaystyle 9900$ ways to win 5th Prize.

. . $\displaystyle P(\text{5th Prize}) \:=\:\frac{9900}{D}$

6th Prize: "3 numbers and not-die"
We want 3 Winners and 2 Others: $\displaystyle 9900$ ways.
And we want a non-die number: $\displaystyle 5$ ways.
Hence, there are: $\displaystyle 9900 \times 5 \:=\:49,\!500$ ways to win 6th Prize.

. . $\displaystyle P(\text{6th Prize}) \:=\:\frac{49,\!500}{D}$

As this point, I assume they are not awarding any further prizes.

If they are, I hope you have seen how the probabilities can be calculated.

3. ## Re: Probability of new lottery

Hi Soroban,

WOW! Thank you for the brilliant answer. I must say that explaining statistics in a way that even I can understand it truly a real talent! I modified the lottery a little bit so that now you pick from 5 numbers from 55 and have to guess a letter (instead of the dice number) in order to win the jackpot (in the Czech alphabet thats 32 letters). I made a excel chart so I hope it's all correct. I have one more question for you though, if you wouldn't mind... Could you try explaining in a similar way a situation, where there are two additional numbers that you have to guess (on top of the 5 normal numbers) in order to win the jackpot. The additional numbers would be ranging from 1 - 11. That means, in order to win the jackpot you need 5 numbers + 2 star numbers (1 - 11). This is something that the UK EuroMillions have and I would like to compare the probability of their lottery with the one I made up.

Thank you sooooooooooooooo much! I am just so greatful!