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 01-14-2015, 12:26 AM #1 ilikexd ​   Join Date: Apr 2006 Location: Reno, Nevada Age: 25 Posts: 3,073 Probability Question In a standard 52-card deck, what are the odds that out of five cards drawn at random, exactly three will be hearts? exactly four? all five? I think I've gotten the answer but I need to be absolutely 100% certain about it.
 01-14-2015, 12:37 AM #2 EzExZeRo7497 ⠀     Join Date: Dec 2010 Location: Singapore, SG Age: 22 Posts: 6,839 Re: Probability Question There are 20 ways (5! / 3!) to have three hearts on the board, 5 ways (5! / 4!) to have 4 hearts on the board, 1 way to have 5 hearts on the board. P(3 hearts) = (39/52 * 38/51 * 13/50 * 12/49 * 11/48) * 20 = 16.3085% P(4 hearts) = (39/52 * 13/51 * 12/50 * 11/49 * 10/48) * 5 = 1.07293% P(5 hearts) = (13/52 * 12/51 * 11/50 * 10/49 * 9/48) = 0.049519807% Edit: Actually not sure about the three hearts one... Last edited by EzExZeRo7497; 01-14-2015 at 12:51 AM..
 01-14-2015, 12:54 AM #3 llyair Wiki Staff     Join Date: Jun 2014 Posts: 307 Re: Probability Question I'm getting 10 ways, (5 3) for 5 items in combinations of 3 = (5! / 3!2!) P(exactly 3 hearts) = 0.08154
 01-14-2015, 01:02 AM #4 Jtehanonymous Hunger Games Hunty     Join Date: Jan 2007 Location: La Florida Age: 27 Posts: 3,678 Re: Probability Question 10 ways is correct for 5 choose 3, yeah, so take eze's expression and multiply by 10 instead. The other two look good. __________________ I buy all FFR Trading Cards I don't own: See My FFR TCG Binder
 01-14-2015, 01:03 AM #5 EzExZeRo7497 ⠀     Join Date: Dec 2010 Location: Singapore, SG Age: 22 Posts: 6,839 Re: Probability Question Yeah, my mistake. Should be 5C3, 5C4 and 5C5.
 01-14-2015, 01:43 AM #6 ilikexd ​   Join Date: Apr 2006 Location: Reno, Nevada Age: 25 Posts: 3,073 Re: Probability Question Thanks everyone.
 01-14-2015, 05:23 AM #7 reuben_tate Kawaii Desu Ne?     Join Date: Dec 2007 Location: The Kawaiian Island~ Age: 26 Posts: 4,118 Re: Probability Question Although this problem is already solved, here is an alternative solution which might be helpful for solving future similar problems. So the probability of getting X hearts in a hand of 5 cards is equal to: The number of possible hands with X hearts ------------------------------------------ The number of possible hands overall The number of possible hands is simple, you have 52 cards and you have to pick 5, so there are C(52,5) such possible hands. Now consider the number of possible hands with X hearts. There are 13 hearts in the deck and 52-13 non-hearts in the deck. Thus, you must pick X hearts from 13 possible hearts and 5-X hearts from the 52-13 non-hearts, i.e., there are C(13,X) ways to pick out the hearts for the hand and C(52-13, 5-X) ways of picking the non-hearts for the hand. So together, there are C(13,X) * C(52-13, 5-x) ways of forming a 5 card hand with X hearts. So, using the formula above, the probability of getting a hand with X hearts is: C(13,X) * C(52-13, 5-X) ----------------------- C(52,5) __________________ AMA: http://ask.fm/benguino Happening now! Click to join!
01-14-2015, 07:17 AM   #8
blindreper1179
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Re: Probability Question

Fucking math nerds. And they said you wouldn't use math outside of school!
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 01-14-2015, 10:01 AM #9 Charles Claythorne FFR Player     Join Date: Apr 2013 Location: New York Age: 23 Posts: 135 Re: Probability Question you dont need math, you can just simulate 10,000 trials on excel like a pro __________________
 01-14-2015, 10:53 AM #10 grizz13114 FFR Veteran     Join Date: Aug 2013 Posts: 215 Re: Probability Question Goddamn Statistics are killing me
01-14-2015, 10:56 AM   #11
rushyrulz
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Re: Probability Question

Quote:
 Originally Posted by Charles Claythorne you dont need math, you can just simulate 10,000 trials on excel like a pro
Why stop at 10,000?

Brute force of 5,000,000 iterations gives the following:
percentage of shuffles that resulted in 3 hearts in the first 5 cards of the deck: 8.1381%
percentage of shuffles that resulted in 4 hearts in the first 5 cards of the deck: 1.0734%
percentage of shuffles that resulted in 5 hearts in the first 5 cards of the deck: 0.0486%
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Last edited by rushyrulz; 01-14-2015 at 11:32 AM..

 01-14-2015, 04:04 PM #12 Charles Claythorne FFR Player     Join Date: Apr 2013 Location: New York Age: 23 Posts: 135 Re: Probability Question because i only have 10,000 decks of cards duh I am also fairly surprised that five million trials is approaching the theoretical probabilities to only two significant digits. Statistics is weird!!! __________________
01-16-2015, 04:06 PM   #13
stargroup100
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Re: Probability Question

Quote:
 Originally Posted by Charles Claythorne I am also fairly surprised that five million trials is approaching the theoretical probabilities to only two significant digits. Statistics is weird!!!
Actually, we can calculate this as well.

Let n be 5000000. Let p be the probability of getting exactly 3 hearts for one draw, or 2717/33320 (as shown already in this thread).

Let z be the 1-a/2 percentile of a standard normal distribution, and a be the error percentile. The confidence interval is

p +/- z sqrt(p(1-p)/n)

For a 50% confidence level, z = 0.674. The confidence interval is between 8.146% and 8.163%.
For a 99% confidence level, z = 2.576. The confidence interval is between 8.123% and 8.186%.

In order for your confidence interval to shrink by a factor of 10 [due to sample size], or become closer to the theoretical value by one more significant digit, the second term needs to differ by a factor of 10. The variable n only appears once, and it is inside the root, so it's quite easy to see that the number of trials needs to be increased about 100-fold. However, it should be noted that this is less accurate for small values of n and probabilities close to 0 and 1. Going backwards we can estimate 50,000 trials is accurate to a few percent, and 500 trials is accurate to a few ten-percent intervals.

This is one of the rough approximations used for finding confidence intervals of binomial distributions. There are more accurate ones, but this one should give you a pretty good idea.

Wrote some javascript code in like 2 minutes for anyone who wants to try:
Code:
```<script>
// heartdraw.html
var start = Date.now();

var deck = 52;
var hearts = 13;
var errors = [];

// CHANGE THE FOLLOWING VARIABLES TO CHOOSE HOW MANY TRIALS AND DRAWS PER TRIAL
var trials = 5;
var totaldraws = 50000;

// Test trials
for (var n = 0; n < trials; n++) {
var success = 0;

for (var k = 0; k < totaldraws; k++) {
if (Math.random() < 2717/33320) success++;

// The below is commented out because it's brute force drawing and slower.
// However, I kept it in in case anyone wants to try other draws at the cost of not needing to calculate the theoretical probabilties.

/*var handhearts = 0;
var decksize = 52;
var heartsleft = 13;
for (var i = 0; i < 5; i++) {
if (Math.random() < heartsleft/decksize) {
heartsleft--;
handhearts++;
}
decksize--;
}
if (handhearts === 3) success++;*/
}

// Display trial results
if (trials <= 20) {				// change the numerical value for display option
var rate = success/totaldraws;
var theoretical = 2717/33320;
var error = (rate/theoretical-1)*100;
errors.push(error);

console.log('Out of ' + totaldraws + ' draws, ' + success + ' were successes.');
if (trials <= 5) {			// change the numerical value for display option
console.log('Percentage of successful draws: ' + rate);
console.log('Theoretical success rate: ' + theoretical);
console.log('Percent error: ' + error);
}
}
}

// Check trial cumulative stats
if (trials > 1) {
var maxerror = errors[0];
var minerror = errors[0];
var sum = errors[0];
for (var n = 1; n < errors.length; n++) {
if (errors[n] > maxerror) maxerror = errors[n];
else if (errors[n] < minerror) minerror = errors[n];
sum = sum + errors[n];
}
var mean = sum/errors.length;

console.log('------------ Trial overall results ------------');
console.log('Mean trial error (percent): ' + mean);
console.log('Range of trial errors (percent): ' + minerror + ' to ' + maxerror);
}

console.log('Time elapsed: ' + ((Date.now()-start)/1000) + ' seconds.');

</script>```
I also haven't done this stuff in a long ass time so if I made a mistake anywhere someone please correct me on it.
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Last edited by stargroup100; 01-16-2015 at 04:12 PM..

 01-16-2015, 04:28 PM #14 Xtreme2252 Typical Bronzie     Join Date: Mar 2007 Location: BC, Canada Age: 27 Posts: 181 Re: Probability Question I'd ballpark it in the range of like 0%-33% for all of the above. Good enough, right?

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