General Betting
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If a set of criteria returns 8% winners, can anyone tell me what the expected highest run of consecutive losers would be in 4000 trials.
To make it a bit more specific, what would be the highest number of consecutive losers in 4000 trials with at least a 50% chance of occurring?
If anyone can tell me how they made the calculation, that would be great too. I've thought about it but found that my probability statistics isn't what it used to be.
Thanks in advance.
To make it a bit more specific, what would be the highest number of consecutive losers in 4000 trials with at least a 50% chance of occurring?
If anyone can tell me how they made the calculation, that would be great too. I've thought about it but found that my probability statistics isn't what it used to be.
Thanks in advance.
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The expected is about 76.
The 50% mark is about 73, you have a 50% chance it will be 73 or less and 50% chance that it will be 73 or more.
The 50% mark is about 73, you have a 50% chance it will be 73 or less and 50% chance that it will be 73 or more.
Via a spreadsheet......
Label cell A1 Trials
Label cell B1 StrikeRate
Label cell C1 LLR
Enter 4000 in cell A2
Enter 8% in cell B2
Enter formula =LOG(A2)/(LOG(1-B2)*-1) in cell C2
Label cell A1 Trials
Label cell B1 StrikeRate
Label cell C1 LLR
Enter 4000 in cell A2
Enter 8% in cell B2
Enter formula =LOG(A2)/(LOG(1-B2)*-1) in cell C2
Statistics mean nothing
arecent survey found out 42% of people dont believe in them
arecent survey found out 42% of people dont believe in them
Tpo2rated,
I plugged in the labels, formula, and parameters but came up with 99.471 rather than farra's answer of 76, which is intuitively (and hopefully) closer to the mark.
Sorry to bug you with this, but possible typo in the formula?
I plugged in the labels, formula, and parameters but came up with 99.471 rather than farra's answer of 76, which is intuitively (and hopefully) closer to the mark.
Sorry to bug you with this, but possible typo in the formula?
That formula estimates an outer limit to the maximum LS which is different to the mathematical expected value (76) of the distribution of all possible maximum LS's.
I make a maximum LS of 99 a 0.6% chance of occurring and 99 or less 93%.
I make a maximum LS of 99 a 0.6% chance of occurring and 99 or less 93%.
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