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Old 12-07-2005, 07:58 PM
SrGuapo SrGuapo is offline
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Default Theoretical stats question re standard deviation

Say I offer you a coin flipping game, and you profit 1 if you flip heads, and 0 if you flip tails. Your expected win is .5, and the standard deviation is .5 (I think I am right so far).

Now, say I offer you a game that now involves two flips. If you flip heads twice, you profit 2. Otherwise, all other combos (TT, HT, TH) you profit 0. So, your expected profit is still .5, but what is your standard deviation?

I made a spread sheet where I ran this a thousand times, and the answer comes out to around .86 (plus or minus, not enough trials to get accurate). How does one arrive at this number via a calculation?

How this relates to poker is take a SnG, where a generally accepted standard deviation seems to be 1.7 buyins, and say you cash 40% of the time, and let's say you expect the average profit to be $2.50 (when you cash). Now, let's say you are playing some wacky double shootout step tournament where you have to cash in a second SnG to cash out, and you still expect to cash 40% of the time, but the expected profit when you cash is $6.25. So, in the double shootout tournament you expect to win .4*.4*6.25=1, which is the same as the single SnG (.4*2.5=1). However, if the standard deviation of the single SnG is 1.7, what is it for the double shootout step tourney?

I hope my question is clear. Thanks for your help.
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  #2  
Old 12-07-2005, 09:22 PM
BruceZ BruceZ is offline
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Default Re: Theoretical stats question re standard deviation

[ QUOTE ]
Say I offer you a coin flipping game, and you profit 1 if you flip heads, and 0 if you flip tails. Your expected win is .5, and the standard deviation is .5 (I think I am right so far).

[/ QUOTE ]

Right. Variance(x) = E(x^2) - [E(x)]^2 =

0.5*1^2 + 0.5*0^2 - 0.5^2 = 0.25

Standard deviation = sqrt[Variance(x)] = sqrt(0.25) = 0.5.


[ QUOTE ]
Now, say I offer you a game that now involves two flips. If you flip heads twice, you profit 2. Otherwise, all other combos (TT, HT, TH) you profit 0. So, your expected profit is still .5, but what is your standard deviation?

[/ QUOTE ]

sqrt(0.25*2^2 + 0.75*0^2 - 0.5^2) = sqrt(3/4) = sqrt(3)/2 =~ 0.87.


[ QUOTE ]
I made a spread sheet where I ran this a thousand times, and the answer comes out to around .86 (plus or minus, not enough trials to get accurate). How does one arrive at this number via a calculation?

[/ QUOTE ]

See above.


[ QUOTE ]
How this relates to poker is take a SnG, where a generally accepted standard deviation seems to be 1.7 buyins, and say you cash 40% of the time, and let's say you expect the average profit to be $2.50 (when you cash). Now, let's say you are playing some wacky double shootout step tournament where you have to cash in a second SnG to cash out, and you still expect to cash 40% of the time, but the expected profit when you cash is $6.25. So, in the double shootout tournament you expect to win .4*.4*6.25=1, which is the same as the single SnG (.4*2.5=1). However, if the standard deviation of the single SnG is 1.7, what is it for the double shootout step tourney?

[/ QUOTE ]

You could have made this easier by telling us how much the buy-in is, instead of us having to deduce it. I assume that one tourney has a buy-in B so that:

sqrt[0.4*2.5^2 + 0.6*(-B)^2 - (0.4*2.5 - 0.6*B)^2] = 1.7*B

2.65*B^2 - 1.2*B - 1.5 = 0

B = $1.01. I'll assume that the buy-in is exactly $1 since the 1.7 is approximate. The standard deviation for the double tourney is:

sqrt[0.16*6.25^2 + 0.84*(-1)^2 - (0.16*6.25 - 0.84*1)^2]

=~ $2.66 or 2.66 buy-ins.

Note that unlike the first 2 examples, since there is now a buy-in, the EV is less for the double tournament (0.16*6.25 - 0.84*1 = $0.16) than for a single tournament (0.4*2.5 - 0.6*1 = $0.40).
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  #3  
Old 12-08-2005, 12:05 PM
SrGuapo SrGuapo is offline
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Default Re: Theoretical stats question re standard deviation

Thanks for the response Bruce. Can you just explain the formula sqrt(0.25*2^2 + 0.75*0^2 - 0.5^2)? If I have a series where the outcome can only be 0 or 1, the standard deviation is going to be sqrt(prob of 0*0^2+prob of 1*1^2-mean^2)? Also, what does the formula look like if I have 3 outcomes?
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  #4  
Old 12-08-2005, 02:18 PM
SrGuapo SrGuapo is offline
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Default Re: Theoretical stats question re standard deviation

never mind, i figured it out, st dev is going to equal the square root of the sum of (x-mean)^2*(probabilities), which is what is represented in different form above
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  #5  
Old 12-08-2005, 02:19 PM
BruceZ BruceZ is offline
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Default Re: Theoretical stats question re standard deviation

[ QUOTE ]
Thanks for the response Bruce. Can you just explain the formula sqrt(0.25*2^2 + 0.75*0^2 - 0.5^2)?

[/ QUOTE ]

0.25*2^2 + 0.75*0^2 is E(x^2), the expected value of the square of the win. 0.5^2 is the square of the expected value of the win.


[ QUOTE ]
If I have a series where the outcome can only be 0 or 1, the standard deviation is going to be sqrt(prob of 0*0^2+prob of 1*1^2-mean^2)?

[/ QUOTE ]

Correct.


[ QUOTE ]
Also, what does the formula look like if I have 3 outcomes?

[/ QUOTE ]

For any number of outcomes, standard deviation is:

sqrt(prob1*value1^2 + prob2*value2^2 + prob3*value3^2 + ... - mean^2)

BTW, this is equivalent to:

sqrt[prob1*(value1-mean)^2 + prob2*(value2-mean)^2 + prob3*(value3-mean)^2 + ...]

The part inside the square root is the variance. This is the expected value of the squared difference between the values and the mean. That's why the standard deviation is called a ROOT MEAN SQUARE or rms average. It is the root of a mean of squares.

Note that this second form is sometimes preferable when working with real data, because the first form can be susceptible to roundoff errors. That is, you have to be careful that you don't subtract two very large numbers that are almost the same, because the difference might be so small that your calculator can't represent it, and you end up throwing away your whole result.
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