Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**araneae****Member**- Registered: 2013-03-18
- Posts: 4

Probably an easy one for many of you out there...

Just a quick question on conditional probability... (I really should know this, I have a science PhD and stats is a big part of it!).

To use a deck example, lets say I wanted the full complement of kings in a deck with 4 tries, each time replacing the kings.

First try it`s 4/52 as it doesn`t matter which I take... then after replacing it`s now down to 3/52... then 2/52... then 1/52.

So to acquire all the kings (hearts, diamonds, clubs and spades) = (4/52)*(3/52)*(2/52)*(1/52)...(right?)

1. But lets say I had 40 tries to get all kings? - how would the stats change? - obviously my likelihood of being able to collect all kings is a lot higher, but how would I calculate it?

2. Likewise, what about in an infinite pool? - Like those Pokemon style collecting games where you have 100 cards to get, each with their own probability... lets say you bought 100 cards... 1000 cards... (I presume the answer is identical to the above question as it's still about % chance).

3. Again, using the Kings example, lets say I wanted all the kings, how would I calculate the likelihood of getting them all to for example a 50%, 90%, 95% and 99.9% confidence level? - how many cards would I have to select to be x% chance sure of getting all of them?

Thanks in advance

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,485

Hi;

Let's look at one question at a time.

But lets say I had 40 tries to get all kings? - how would the stats change? - obviously my likelihood of being able to collect all kings is a lot higher, but how would I calculate it?

Welcome to the forum. Do you understand the hypergeometric distribution?

*Last edited by bobbym (2013-03-18 22:01:28)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline

**araneae****Member**- Registered: 2013-03-18
- Posts: 4

Hey, thanks for replying

I was not familiar with it but have looked it up now. I'm not saying I understand it necessarily but the principles look similar to Mendelian genetics. Could you walk me through it please?

Thanks

Offline

**araneae****Member**- Registered: 2013-03-18
- Posts: 4

Ok, I had a more in-depth look on the Wikipedia page and I 'believe' I understand the principles (they gave a good one using Texas Hold'em as an example)... but maybe I'm missing something, that relies on a draw without replacement?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,485

The hypergeometric distribution is for sampling without replacement.

winners = 4 ( there are 4 kings )

population = 52 ( 52 cards in the deck )

trials = 40 (40 draws )

k = 4 (you want all 4 kings )

For 2) I will need more information.

*Last edited by bobbym (2013-03-19 00:02:12)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline

**araneae****Member**- Registered: 2013-03-18
- Posts: 4

So what about replacing them in the deck?

Say only 12 cards, each with an equal chance of getting chosen.

To get all of them once would be 1*(11/12)*10/12)*(9/12)*(8/12)*(7/12)*(6/12)*(5/12)*(4/12)*(3/12)*(2/12)*(1/12) = 0.00005372321 or roughly 1 in 18,000

...but that's if I had 12 chances to take a single card... what if I had 40...? How would my 1 in 18,000 chances improve?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,485

Hi;

If it is with replacement you would use the binomial distribution.

To get all of them once would be

I am sorry, I am not following you. Please be a bit more specific. An example would be nice.

*Last edited by bobbym (2013-03-19 07:38:18)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline

Pages: **1**