#21
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Re: odds of getting a pocket pair 3x in a row
[ QUOTE ]
Now, using your formula, the chances of Ichiro having a 55 game hitting streak in a 162 game season (not missing any games) would then be 0.001106131, or, a little less than 1 in 900. Is that correct? [/ QUOTE ] No, I get 0.005815 or about 1 in 172. P(1) to P(54) = 0 P(55) = (0.866636)^55 P(56) = P(55) + (1 - 0.866636)*(0.866636)^55 P(n) = P(n-1) + [1 - P(n-56)]*(1 - 0.866636)*(0.866636)^55, for n > 56 ... P(162) = 0.5815% |
#22
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Re: odds of getting a pocket pair 3x in a row
Here's one that'll blow your mind. Earlier today I got KK 3 times in a row on bodog. And after the first hand I got KK on Pacific as well. I don't feel like doing the math again, but it was something like 0.000093% chance of getting KK 3 times in a row.
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#23
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Re: odds of getting a pocket pair 3x in a row
[ QUOTE ]
[ QUOTE ] Now, using your formula, the chances of Ichiro having a 55 game hitting streak in a 162 game season (not missing any games) would then be 0.001106131, or, a little less than 1 in 900. Is that correct? [/ QUOTE ] No, I get 0.005815 or about 1 in 172. P(1) to P(54) = 0 P(55) = (0.866636)^55 P(56) = P(55) + (1 - 0.866636)*(0.866636)^55 P(n) = P(n-1) + [1 - P(n-56)]*(1 - 0.866636)*(0.866636)^55, for n > 56 ... P(162) = 0.5815% [/ QUOTE ] Okay, I had an error in my formula. For some reason I was multiplying by (1-.86636)^2. In other words, my formula was: P(n) = P(n-1) + [1 - P(n-56)]*(1 - 0.866636)^2*(0.866636)^55, for n > 56 I get the same answer now. |
#24
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Re: odds of getting a pocket pair 3x in a row
Here's another question for BruceZ.
Given the same information about Ichiro, what is the median longest hitting streak during a 162 game season? I think I could figure this out using the following method: For each n, 0 to 162, find the probability that Ichiro will have a streak of this length using your formula. The probability that this is the longest streak of the season is P(n)-P(n+1). The median longest streak would be the first value where the sum from 0 to n of P(n)-P(n+1) >= .5. |
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