# Probability questions

This topic is 4988 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Two questions: 1. If nine races were ran, each with a field of nine: What's the odds on finishing in the top 3 twice? 2. If you have a deck of 52 cards and you're dealt 2 (pokect)cards and 3 more are turned up on the table(flopped), what's the chance of pairing either of your pocket cards with the 3 on the table?

##### Share on other sites
3. If someone asks completely arbitrary and unrelated questions in Maths and Physics, what's the probability that they're homework?

In answer to question 1, however, I would need the (presumably normal?) distribution of human running speeds and they're variation from race to race. It seems very likely that whoever wins the first race will win the second, since he's the faster runner . . . Indeed, as with most statistics questions, it's a really stupid question :)

Infact, being serious, let's assume I'm one of the 9 who are running every time, but that the other 8 aren't the same 8 each time, instead being selected at random from a large pool. Those two assumptions do not sit well together, but they are the only way to make the question answerable.

Then the odds of finishing in the top 3 twice (exactly twice, or atleast twice? It doesn't say...) is a fairly simple question . . .

At what point are you getting stuck? Show me some working and where you start to struggle and I will happily point you in the right direction . . . :)

##### Share on other sites
Heheh, no it's not homework, I don't have to worry about that anymore.

They are poker related questions. I used the race analagy at the start because I'm playing nine person tournaments. I want to know the answer to (1) and the workings behind it to see if I'm coming out above average.
For question one you can it take that all 9 players(runners :)) are of equal skill.

##### Share on other sites
For the first question, if all players are of equal skill and the game offer
euqal opportunities for everyone, then you finish first one out of 9 times.
Some goes for second, third place and so on. Hence you are expected
to finish in the top three 3/9 = 1/3 of the times. The probabillity for finish
in the top thrre if you play two tournaments are 1/3 * 1/3 = 1/9.

Now, I don't think this is of much relevance for your poker play, since there
are a lot of other factors. Most notably skill and luck. What you would like to
do is to play, perhaps fifty, tournaments and then estimate your winning chances
based on that. You'll also be able to get a rough estimate of how good your
expected winning chances are. (Known to statisticans as expected value and
standard deviation). Based on these estimates you will be able to find
the probability of finishing on the top three if you play N tournaments. You'll
also get your earnings per hour, per tournament and so on. Very valuable data
for any poker player.

(short answer, play lots of tournaments, look up the meaning of expected value,
standard deviation and normal distribution and then relate it to poker)

##### Share on other sites
Quote:
 Original post by MessageBox2. If you have a deck of 52 cards and you're dealt 2 (pokect)cards and 3 more are turned up on the table(flopped), what's the chance of pairing either of your pocket cards with the 3 on the table?

Two cases:

1) Pocket cards are not the same card
2) Pocket cards are the same card

Use the method of sampling without replacement for determining the probability of drawing x items in n trials, for a population of N things where m within the population are of the typ you want to draw. The formula for the probability is given by the hypergeometric distribution: (damn formatting is broken for some reason so I can't put in the neat form involving combination notation)
             m!             (N-m)!         -------- * --------------------         x!(m-x)!   (n-x)!((N-m)-(n-x))!p(x)   = -----------------------------------                       N!                    --------                    n!(N-n)!

There are two ways to solve case 1. a) work out the probability of matching just one of the cards and double it, or b) consider that there are 6 cards within the remaining 50 that will give you a match.

Case 1, method A: N=50, m=3, n=3, x=1 (substitute and solve, then double that number)
Case 1, method B: N=50, m=6, n=3, x=1 (should get that same answer as above)

Case 2: N=50, m=2, n=3, x=1

There are more interesting questions you could ask, like what is the probability that I will get 2 matches to either of the pocket cards. You should also give thought to the probability that two of the flop cards are identical and NOT in the initial 2 cards drawn. Then you can really spin yourself out by asking what is the probability of 2 flops cards being identical and greater than either of the pocket cards!

Have fun!

Timkin

##### Share on other sites
The AP's wrong. That's the probability of coming in the top 3 twice in 2 competitions.

The actual answer is (1/3)^2(2/3)^7*9C2 = 2^7*3^-9*36 = 0.234(3SF)

The expected value of number of times in the top 3 is ofcourse 3, so on that basis you're below average.

##### Share on other sites
Oops I've been an idiot. Sorry AP, didn't read all of your post, just the middle bit.

I should also add that if you want understanding of how to do these problems yourself, look into the Binomial distribution.

• ### What is your GameDev Story?

In 2019 we are celebrating 20 years of GameDev.net! Share your GameDev Story with us.

• 19
• 15
• 10
• 9
• 11
• ### Forum Statistics

• Total Topics
634097
• Total Posts
3015510
×