How to measure "luckiness"?

[TomCollins]

I'm trying to measure luckiness between several independent events. I found a good way of determining luck on a series event that has a probability of p.
Suppose p(n,x,w) = The probability of having x wins in n trials with the odds of winning an individual trial as w. I define the "luck" of having z wins out of n trials is SUM(p(n,i,w),i=0 to x-1) + .5 * p(n,z,w).

So for flipping a fair coin twice,
Luck(0,2,.5) = .125
Luck(1,1,.5) = .5
Luck(2,0,.5) = .875

And lets suppose I have another event where i win if I roll a 5 or a 6 on a dice. For 2 trials:
Luck(0,2,.33) = .2224
Luck(1,1,.33) = .6660
Luck(2,0,.33) = .9436



Now suppose I roll a dice twice and win both times, and flip a coin and lose both times. What would be the best way to determine my "luck" of the 2 combined events?

[ QUOTE ]
SUM(p(n,i,w),i=0 to x-1) + .5 * p(n,z,w).

[/ QUOTE ]

do you mean:

SUM(p(n,i,w),i=0 to z-1) + .5 * p(n,z,w).

?

[TomCollins]

[ QUOTE ]
[ QUOTE ]
SUM(p(n,i,w),i=0 to x-1) + .5 * p(n,z,w).

[/ QUOTE ]

do you mean:

SUM(p(n,i,w),i=0 to z-1) + .5 * p(n,z,w).

?

[/ QUOTE ]

My bad, typo.

cool, just making sure I follow, I will think about it now

[skitzo444]

Wouldn't "luckiness" just be a un-probable out come for a period of time. Like flipping a coin and getting heads 25 times??

[TomCollins]

PRetty much. If you got 25 heads in a row, you would be very lucky- 100%. If you got 25 tails in a row, you would be very unlucky - 0%. If you got 12 heads, you would probably be around 45% on luck.

[SheetWise]

How can you have a serious discussion about luck without compensating for past events based upon the law of the conservation of luck?

[TomCollins]

After playing poker for the past few years, trust me, this law is bogus.

[SheetWise]

[ QUOTE ]
After playing poker for the past few years, trust me, this law is bogus.

[/ QUOTE ]
You luck account still has a balance [img]/images/graemlins/wink.gif[/img]?

[Siegmund]

In backgammon, a player's luck is calculated (when a match is analyzed after its completion by computer) by comparing the player's equity before he throws the dice with what his equity would be if he made the best move after throwing.

The game ends, of course, with one player reaching an equity of zero; there's a nice tidy theorem that says the final result of the game = (luck of player 1) - (luck of player 2) - (cost of mistakes by player 1) + (cost of mistakes by player 2).

The same definition could be applied to poker, after computers have some idea of how to estimate the equity of a position in the middle of a multiplayer hand, which is pretty far beyond Poki. If you are looking at a series of events with some payoff table associated with different sequences of events, you might find the same definition useful.

For further details on the backgammon definition of luck, see the article at http://www.math.columbia.edu/~zare/luckequity.html