Help With Math Please

[1p0kerb0y]

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]

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]

[ 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(&lt;=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]

Glad to see my numbers confirmed :-)