Simple Probability Question

[onequickquestion]

Hi all,

I just have one quick probability question.

Say you are going to draw 10 cards and each card will have a number 1-20. Assume these are independent events, so you can draw multiple of the same number to no extent. Each numbered card has an equal 5% chance to be drawn.

If you are going to draw 10 at once, what would be the probability that say, 5 of them are number 1?

I know that if we ask the more simple question: What are the odds of drawing the same number back to back 5 times? The answer would simply be 0.05^5, but that doesn't take into account the other 5 draws that could have been the number we wanted. In theory, the answer to the bolded question should be slightly larger than 0.05^5, correct?

I tried doing something such as 0.05^5*(10/5) to account for 10 possible tries and 5 landing correctly, but I do not know if this is gives the correct results.

Thanks to anyone who can help. By the way, this question is out of pure curiosity

 

[Dr.Peterson]

It may be a quick question, but requires a long answer -- possibly a chapter or two of a book. There are a lot of different issues to be resolved, just to make the question itself clear.

What does it mean to draw 10 "at once", when your statement that the draws are independent and can repeat implies that you are selecting with replacement, one after another? Your question is more like rolling dice than cards.

Your answer is not "odds" but "probability". These are different.

And you're calculating the probability of drawing a specified card ("the number we wanted") 5 times out of 5, not of drawing the same card (which could be anything).

But the main thing you need to know (for the bold question) is the binomial distribution, which applies to this situation (assuming you want the probability of exactly 5 of a specified card, not "at least"). The multiplier you need is not 10/5, but [MATH]\binom{10}{5}[/MATH], also called [MATH]_{10}C_{5}[/MATH]. Are you familiar with combinations?

 

[onequickquestion]

It may be a quick question, but requires a long answer -- possibly a chapter or two of a book. There are a lot of different issues to be resolved, just to make the question itself clear.

What does it mean to draw 10 "at once", when your statement that the draws are independent and can repeat implies that you are selecting with replacement, one after another? Your question is more like rolling dice than cards.

Your answer is not "odds" but "probability". These are different.

And you're calculating the probability of drawing a specified card ("the number we wanted") 5 times out of 5, not of drawing the same card (which could be anything).

But the main thing you need to know (for the bold question) is the binomial distribution, which applies to this situation (assuming you want the probability of exactly 5 of a specified card, not "at least"). The multiplier you need is not 10/5, but [MATH]\binom{10}{5}[/MATH], also called [MATH]_{10}C_{5}[/MATH]. Are you familiar with combinations?

Hi, Thank you for your response!

Thanks for clarifying there is a technical difference between odds and probability. Also yes, an example with dice would have sufficed and probably been easier.

I'm not familiar with binomial distribution or combinations in a mathematical sense. I looked it up though, and feel I have a decent understanding of it now. I used this link: https://www.mathsisfun.com/combinatorics/combinations-permutations.html

Though, on this website, it differentiates combinations with and without repetition. In my case, the problem would include repetition. But for their example of combinations with repetition they did not use the formula [MATH]_{10}C_{5}[/MATH] (though they did with their combination without repetition). So I'm a bit confused on this .. you say to solve the problem I should use this, but this website explains it otherwise.

Could you elaborate? I understand you say this problem requires a chapter of two worth of learning, but I'm the type of person that usually when I'm shown the "how to" and explained a little bit, I understand it. Also, your keywords helped me a lot with just knowing what to research and learn about, so thanks for that!

 

[Dr.Peterson]

The fact that your problem allows repetitions does not mean that some part of the work to solve it won't involve combinations (without repetitions)! The _{10}C_5 is part of the formula for the binomial distribution, and relates to choosing (without repetition) 5 of the ten trials that will be "successes". You are not directly applying combinations to your dice rolls.

You didn't explicitly mention looking up the binomial distribution. Did you look here? https://www.mathsisfun.com/data/binomial-distribution.html You'll see there why combinations come into your problem.

 

[onequickquestion]

The fact that your problem allows repetitions does not mean that some part of the work to solve it won't involve combinations (without repetitions)! The _{10}C_5 is part of the formula for the binomial distribution, and relates to choosing (without repetition) 5 of the ten trials that will be "successes". You are not directly applying combinations to your dice rolls.

You didn't explicitly mention looking up the binomial distribution. Did you look here? https://www.mathsisfun.com/data/binomial-distribution.html You'll see there why combinations come into your problem.

Hi, I hadn't looked up binomial distribution. Though, I did watch a Khan Academy video which explained what you were doing and I believe helped me understand and solve my problem. I will look at this link tomorrow when I have more time to read and learn, but do you think you could fact check me in the mean time?

The conclusion I came to was 0.00007875 or 0.007875%. I got this from doing (0.05^5) * [MATH]\binom{10}{5}[/MATH]. For the multiplier which you introduced me to, I got an answer of 252. So (0.05^5) * 252 = 0.007875%. Is this the correct answer?

 

[Cubist]

The conclusion I came to was 0.00007875 or 0.007875%. I got this from doing (0.05^5) * [MATH]\binom{10}{5}[/MATH]. For the multiplier which you introduced me to, I got an answer of 252. So (0.05^5) * 252 = 0.007875%. Is this the correct answer?

Not quite correct yet, look a bit further down on the link @Dr.Peterson provided in post#4, specifically under the heading "Bias" (because the chances of getting a "1" are different to the chances of not getting a "1").

NB: I just want to emphasize that you are calculating the probability that exactly 5 cards will be "1". The calculation would be a little different if you actually want the probability that at least 5 cards will be number 1.