# Thread: Probability of getting the highest card

1. ## Probability of getting the highest card

Hello,
I would appreciate if someone can give me some ideas about how to solve this probability problem:
50 Index Cards are marked with numbers from 1 upward (with none matching). They are randomly shuffled and turned over one at a time. What is the probability of choosing the highest number when it is turned?
Thank you very much

2. Originally Posted by August80
Hello,
I would appreciate if someone can give me some ideas about how to solve this probability problem:
50 Index Cards are marked with numbers from 1 upward (with none matching). They are randomly shuffled and turned over one at a time. What is the probability of choosing the highest number when it is turned?
Thank you very much
What do you mean? What is the probability that the first card turned is the highest card? Or, do you mean what is the likelihood that you will flip the highest card on it's value? Or...something else?

What I mean is that the 50 cards are going to be flipped one by one, but I have to determinate a strategy to guess which card is going to be the highest while the cards are turned one by one. And so determinate the probability that the card I choose is the highest. For example, if the first card that is flipped is marked as "10", the second is "20", and the third is "2", I can say that the next one will be the highest. But I need to determinate the probability that my guess is true.
Hope this can help.
Thank you,

4. hmm, I would do it like this. I am just a beginner so it is proparbly not correct.

Here goes.

Calculate the probability that the highest card is card nr. $n$ drawn.

There are total $50!$ different ways that the 50 cards could be shuffled.

EDIT: FIXED ERROR

There are $\frac{49!}{(49-n)!}$ different ways (or is it $\binom{49}{n-1}$ ways ?) to draw $n-1$ cards that are not the highest. Then there is $1$ way that you could draw the highest card.

So the probability that the highest card is the n-th card drawn is:

$\frac{\frac{49!}{(49-n)!}\cdot1}{50!}=\frac{1}{50}\cdot\frac{1}{(49-n)!}$

EDIT: You should probarbly disregard this picture.
Image attached, it looks like a bell curve

5. Thank you very much hjotur,

Best,
August80

6. Well, you should take what I said with a grain of salt.
First I wrote something completely different, then I discovered an error,
and then I wound up so confused I didnt know what I was doing.

So the general idea of how to do this is legit, but all the numbers I gave are probarbly way off.

7. I really appreciate the fact you are trying to help me.

I am definetely working it out to find the best answer.

Thanks,