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#2
12-07-2005, 09:22 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: Theoretical stats question re standard deviation

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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).

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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.

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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?

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

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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?

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See above.

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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?

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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]