#1
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Help With Math Please
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. |
#2
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Re: Help With Math Please
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!) |
#3
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Re: Help With Math Please
[ 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 /> |
#4
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Re: Help With Math Please
Glad to see my numbers confirmed :-)
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