View Full Version : Help With Math Please
1p0kerb0y
09-20-2004, 09:46 PM
I was curious for a problem I am doing. I'm seeking either a mathematical equation or perhaps an online program that will help me to do a problem. Here it goes:
I have 13 cards, all hearts. The odds that I will draw the ace of hearts out is 1/13. If I shuffle and pick a card, and then repeat that process 13 times, the odds are that I will pick the ace out 1 time.
Is there a program or a calulation that will run random runs to show the variance associated with this promblem? For instance, maybe half of the time I pull the ace out once, 25% of the time I pull it out twice or not at all, etc.
Thanks to anyone who can help me.
Neil Stevens
09-20-2004, 10:05 PM
Probability of pulling an ace exactly once out of 13 tries:
(1/13)^1 * (12/13)^12 * 13 = 38%
(chance of getting the ace once) * (chance of getting non-ace 12 times) * (number of ways to pull once ace)
Probability of pulling an ace exactly twice:
(1/13)^2 * (12/13)^11 * 78 = 19%
Probability of pulling no aces:
(12/13)^13 = 35%
Hope this helps (and I hope I made no errors!)
BruceZ
09-20-2004, 10:22 PM
[ QUOTE ]
I was curious for a problem I am doing. I'm seeking either a mathematical equation or perhaps an online program that will help me to do a problem. Here it goes:
I have 13 cards, all hearts. The odds that I will draw the ace of hearts out is 1/13. If I shuffle and pick a card, and then repeat that process 13 times, the odds are that I will pick the ace out 1 time.
Is there a program or a calulation that will run random runs to show the variance associated with this promblem? For instance, maybe half of the time I pull the ace out once, 25% of the time I pull it out twice or not at all, etc.
Thanks to anyone who can help me.
[/ QUOTE ]
This is just the binomial distribution. The probability of pulling the ace exactly N times is:
P(N) = C(13,N)*(1/13)^N*(12/13)^(13-N)
Sum this to get the cumulative distribution P(>=N) or P(<=N).
In Excel, P(N) =BINOMDIST(N,13,1/13,FALSE)
Change FALSE to TRUE for the cumulative.
The actual variance of the distribution is 13*(1/13)*(12/13) = 12/13.
<font class="small">Code:</font><hr /><pre>
N P(N) P(<=N)
0 35.33% 35.33%
1 38.27% 73.60%
2 19.13% 92.73%
3 5.85% 98.58%
4 1.22% 99.80%
5 0.18% 99.98%
6 0.02% 100.00%
7 0.00% 100.00%
8 0.00% 100.00%
9 0.00% 100.00%
10 0.00% 100.00%
11 0.00% 100.00%
12 0.00% 100.00%
13 0.00% 100.00%
</pre><hr />
Neil Stevens
09-20-2004, 10:24 PM
Glad to see my numbers confirmed :-)
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