Probability of getting 3 out of 4 numbers correct

[Phys12]

Problem statement: In a lottery, players win a large prize when they pick four digits that match, in the correct order,four digits selected by a random mechanical process. A smaller prize is won if only three digits are matched. What is the probability that a player wins the small prize?

Solution: The total number of choosing 4 digits are: 10 * 10 * 10 * 10 = 10,000. Ways of choosing three digits correctly out of 4 is the same choosing one incorrectly. Ways of choosing the first digit incorrectly are 9, same for second, third and fourth. You add the total number of ways to get 9 + 9 + 9 + 9 = 36 and then divide by the total number of ways of choosing 4 digits to get the probability: 36/10000

How would you solve this problem by not considering the different ways of getting the digit incorrect, but by considering the ways of getting three digits correct?

My attempt: If I think about getting the digits correct, then for getting the first three digits correctly, I have 9 ways of picking it since the 4th digit can be any digit except for the correct one, similarly for the other three digits and another and then one more time. Giving us again 9 + 9 + 9 + 9 = 36 ways. Is that a/the correct way of thinking about it?

 

[GwtBc]

Think of this as four different trials. The odds of being successful in each trial are 0.1 or 10%. Now you want the probability of being successful in 3 but unsuccessful in one. So for example if you are successful in the first three trials but then fail in the last you have such a condition, and the odds of this happening are $0.1\cdot 0.1 \cdot 0.1 \cdot 0.9 = \frac{9}{10000}$. But you could also avail in the first two, then fail the 3rd test and then again be successful in the last, and that would also meet your conditions, the odds of it happening would again be $0.1\cdot 0.1 \cdot 0.9 \cdot 0.1 = \frac{9}{10000}$. So how many ways are there of this happening? Well you have four trials and you want to pick 3 to succeed in (or equivalently, 1 to fail in) so that's 4 pick 3, which is the same as 4 pick 1, i.e. we want $\binom{4}{3} = \binom{4}{1}=4$.

So the total chances of the event that you are successful in 3 trials and unsuccessful in the other are $\frac{9}{10000} \binom{4}{3} = \frac{36}{10000}$ as you calculated. I don't know if this was the thought process you were describing but I believe it's a pretty general way to think about it.

edit: If you have not seen this "4 pick 1" function before but are interested, I'm more than happy to discuss it.

 

[Phys12]

edit: If you have not seen this "4 pick 1" function before but are interested, I'm more than happy to discuss it.

I know what it is, I don't understand its application in this question. The "4 pick 1" function, as I understand it, is equivalent to saying that there are 4 choices, in how many ways can I pick 1 of them, if I do not care about the order. In this question, out of the 4 digits, we pick three of them, and that's (4 3) and the odds of being successful in each pick is 9/10^4, so we multiply the two. Is that correct?

 

[PeroK]

I know what it is, I don't understand its application in this question. The "4 pick 1" function, as I understand it, is equivalent to saying that there are 4 choices, in how many ways can I pick 1 of them, if I do not care about the order. In this question, out of the 4 digits, we pick three of them, and that's (4 3) and the odds of being successful in each pick is 9/10^4, so we multiply the two. Is that correct?

Another description of $\binom{n}{k}$ is:

If you have $n$ things and $k$ have a certain property, then that's the number of ways you can do it.

In this case, you have 4 things and you want 1 to be wrong and 3 to be right, then that is $\binom41$, which is equal to $\binom43$.

 

[GwtBc]

I know what it is, I don't understand its application in this question. The "4 pick 1" function, as I understand it, is equivalent to saying that there are 4 choices, in how many ways can I pick 1 of them, if I do not care about the order. In this question, out of the 4 digits, we pick three of them, and that's (4 3) and the odds of being successful in each pick is 9/10^4, so we multiply the two. Is that correct?

Correct!

 

[Ray Vickson]

Problem statement: In a lottery, players win a large prize when they pick four digits that match, in the correct order,four digits selected by a random mechanical process. A smaller prize is won if only three digits are matched. What is the probability that a player wins the small prize?

Solution: The total number of choosing 4 digits are: 10 * 10 * 10 * 10 = 10,000. Ways of choosing three digits correctly out of 4 is the same choosing one incorrectly. Ways of choosing the first digit incorrectly are 9, same for second, third and fourth. You add the total number of ways to get 9 + 9 + 9 + 9 = 36 and then divide by the total number of ways of choosing 4 digits to get the probability: 36/10000

How would you solve this problem by not considering the different ways of getting the digit incorrect, but by considering the ways of getting three digits correct?

My attempt: If I think about getting the digits correct, then for getting the first three digits correctly, I have 9 ways of picking it since the 4th digit can be any digit except for the correct one, similarly for the other three digits and another and then one more time. Giving us again 9 + 9 + 9 + 9 = 36 ways. Is that a/the correct way of thinking about it?

I think your solution ignores the problem stipulation "...digits that match, in the correct order, ..." The problem seems to be saying that order is important, so we need to be looking at permutations, not combinations. Assuming that the ten digits 0,1,...,9 can each be chosen only once, the total number of possibilities is $N = 10 \times 9 \times 8 \times 7 = 5040.$

Presumably, if the lottery selects $abcd$ you will a smaller prize if you pick either $abcX$ with $X \neq d$ or $Xbcd$ with $X \neq a$. What is not clear (and may vary according to Lottery rules) is whether $aXcd$ with ($X \neq b$) or $abXd$ (with $X \neq c$) could also win a small prize; they have three of the correct letters with $a$ before $b$ before $c$ before $d,$ but they do not occur in uninterrupted strings. Maybe that is not good enough---or maybe it is. If unsure, why not solve the problem both ways and explain the assumptions underlying each way?