hypothetical wager question

[r3vbr]

consider an even money proposition bet in wich I have a 51% chance of winning. how many times do I need to bet, in order to be 99% sure to come of as a winner?

and what if the chance of winning were 52%?

[LetYouDown]

Haven't considered this type of question in a while, but for 51% I believe you need 6814 to be 99% confident of being ahead and 1715 at 52%.

[r3vbr]

[ QUOTE ]
Haven't considered this type of question in a while, but for 51% I believe you need 6814 to be 99% confident of being ahead and 1715 at 52%.

[/ QUOTE ]
thanks a lot for your reply

[BruceZ]

[ QUOTE ]
consider an even money proposition bet in wich I have a 51% chance of winning. how many times do I need to bet, in order to be 99% sure to come of as a winner?

[/ QUOTE ]

Make N bets so that your average winnings of 0.02*N will be greater than NORMSINV(99%) =~ 2.33 standard deviations above 0 by the normal approximation to the binomial distribution. <font color="red">The variance of the winnings for 1 bet is 0.51*(1)^2 + 0.49*(-1)^2 - 0.02^2 = 0.9996. The variance for N bets is 0.9996*N, and the standard deviation of the winnings for N bets is sqrt(0.9996*N).</font>

0.02*N &gt; NORMSINV(99%)*<font color="red">sqrt(0.9996*N)</font>

N &gt; [NORMSINV(99%)/0.02]^2 *0.9996

N = 13,525 bets.


[ QUOTE ]
and what if the chance of winning were 52%?

[/ QUOTE ]

<font color="red">The variance of the winnings for 1 bet is 0.52*(1)^2 + 0.48*(-1)^2 - 0.04^2 = 0.9984. The variance for N bets is 0.9984*N, and the standard deviation of the winnings for N bets is sqrt(0.9984*N).</font>


N &gt; [NORMSINV(99%)/0.04]^2 *<font color="red">0.9984</font>

N = 3378 bets.

[LetYouDown]

Hmm...I knew I didn't approach it in an efficient way, but why does:

=1 - BINOMDIST(3392,6784,0.51,TRUE) come out to 94.9%?

[BruceZ]

[ QUOTE ]
Hmm...I knew I didn't approach it in an efficient way, but why does:

=1 - BINOMDIST(3392,6784,0.51,TRUE) come out to 94.9%?

[/ QUOTE ]

Fine, but he asked for 99%, not 95%. BINOMDIST will blow up for that number.

[LetYouDown]

[ QUOTE ]
[ QUOTE ]
Hmm...I knew I didn't approach it in an efficient way, but why does:

=1 - BINOMDIST(3392,6784,0.51,TRUE) come out to 94.9%?

[/ QUOTE ]

Fine, but he asked for 99%, not 95%. BINOMDIST will blow up for that number.

[/ QUOTE ]
Right, but for me to get a number over 99%, I had to do:

=1 - BINOMDIST(6814,13628,0.51,TRUE) and
=1 - BINOMDIST(1715,3430,0.52,TRUE)

Are these wrong?

[BruceZ]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Hmm...I knew I didn't approach it in an efficient way, but why does:

=1 - BINOMDIST(3392,6784,0.51,TRUE) come out to 94.9%?

[/ QUOTE ]

Fine, but he asked for 99%, not 95%. BINOMDIST will blow up for that number.

[/ QUOTE ]
Right, but for me to get a number over 99%, I had to do:

=1 - BINOMDIST(6814,13628,0.51,TRUE) and
=1 - BINOMDIST(1715,3430,0.52,TRUE)

Are these wrong?

[/ QUOTE ]

I fixed my original post in red, and now I get results similar to these. I was using sqrt(N*p*q) as the standard deviation, but this is the standard deviation of the number of wins in N bets. We actually want the standard deviation of the winnings, which is very different.

I can't get Excel to evaluate these binomial expressions. How is it that you can evaluate them? This is what I tried first, and it should give the exact result.

[r3vbr]

i need to learn how to use the excel formulas... anyone know of a tutorial online for me to learn at least how to use the probability funcions?.. i tried it once and i didnt know what some variables meant.

[BruceZ]

[ QUOTE ]
i need to learn how to use the excel formulas... anyone know of a tutorial online for me to learn at least how to use the probability funcions?.. i tried it once and i didnt know what some variables meant.

[/ QUOTE ]

If you click on the fx next to the input window, it will take you to a list of functions, and you can click on these to bring up detailed help. If you type in = followed by the name of a function followed by open parentheses ( it will show you the input parameters, and if you then click the fx, it will open a window with a link to help on that function.

[alThor]

[ QUOTE ]
I can't get Excel to evaluate these binomial expressions. How is it that you can evaluate them? This is what I tried first, and it should give the exact result.

[/ QUOTE ]

What version? Excel 2003 SP1 works for me. I doubt it matters, but I have Analysis ToolPak Add-ins active, and am too lazy to see if it works without them.

My factorial and combinatorial functions have the usual limits, but they must have advanced the way binomdist works for big numbers.

alThor

[LetYouDown]

[ QUOTE ]
I can't get Excel to evaluate these binomial expressions. How is it that you can evaluate them? This is what I tried first, and it should give the exact result.

[/ QUOTE ]
Works for me in Excel (Office 2003). Do you have an older version of Excel?

[BruceZ]

[ QUOTE ]
[ QUOTE ]
I can't get Excel to evaluate these binomial expressions. How is it that you can evaluate them? This is what I tried first, and it should give the exact result.

[/ QUOTE ]

What version? Excel 2003 SP1 works for me. I doubt it matters, but I have Analysis ToolPak Add-ins active, and am too lazy to see if it works without them.

My factorial and combinatorial functions have the usual limits, but they must have advanced the way binomdist works for big numbers.

alThor

[/ QUOTE ]

I have 2002 SP3 with some add-ins, and I can't evaluate much above BINOMDIST(500,1000,51%,TRUE) without getting the #NUM! error.

[LetYouDown]

[ QUOTE ]
I have 2002 SP3 with some add-ins, and I can't evaluate much above BINOMDIST(500,1000,51%,TRUE) without getting the #NUM! error.

[/ QUOTE ]
Time to upgrade! Hehe...that has to be it. I have a barebones version of Office XP Pro and it works fine.