Seeing the same flop

[MickeyHoldem]

During my last trip to my local casino the dealer flopped the same cards twice in a row. (they weren't in the same order) This led to an interesting discussion on the odds of this happening again. To make a long story short... we all agreed it was remote! (odds withheld to protect the ignorant) [img]/images/graemlins/grin.gif[/img]

A more interesting discussion insued about how long you would have to play before you had an even money (or better) chance of seeing 2 identical flops during the same session. (not necessarily back to back) Assume you see 30 flops an hour <insert chuckles here>, how long do you have to play? (assume order does NOT matter)

[gaming_mouse]

There are nCr(52,5)= 2598960 possible flops.

After n hands, we will have nCr(n,2) "attempts" at having two identical flops. Let nCr(n,2) = x.

Then, for a very close approximation to the solution, we need to solve:

1 - (2598959/2598960)^x = .5
(2598959/2598960)^x = .5
x*log(2598959/2598960) = log(.5)
x= log(.5)/log(2598959/2598960)
x= 1801461.44989

Returning to the definition of x, we solve for n:

(n^2 - n) = 3602922.89978
n^2 - n -3602922.89978 = 0

The positive solution is 949.56

At 30 hands/hr, that means you'll see the same flop about once every 31 hours.

gm

EDIT: woops. i made an arithemetic error calculating the logs the first time. Adjustments are above.

[MickeyHoldem]

[ QUOTE ]
There are nCr(52,5)= 2598960 possible flops.

[/ QUOTE ]

looking at flops... not boards!

[BruceZ]

This is the birthday problem with C(52,3) = 22100 days in the year. Solve

[ 22100*22099*22098*...*(22100-n+1) ] / 22100^n = 0.5

Your apporoximation seems to work well for the normal b-day problem.

[gaming_mouse]

Oh, got you.

NOTE: I think I still have arithmetic errors in the above solution anyway. x should be 1898

Anyway, I will redo the solution for flops.

nCr(52,3) = 22100, call this number "f", for number of flops.

Then, this time doing an exact rather than an approximate solution, we find the first x such that:

1 - ((f-1)/f)*((f-2)/f)*...*((f-x)/f) > .5

I wrote a simple Java program that gives us x = 175.

Shockingly, that gives us n between 19 and 20.

I find this quite hard to believe. Can anyone verify this?

gm

[BruceZ]

[ QUOTE ]
Shockingly, that gives us n between 19 and 20.

I find this quite hard to believe. Can anyone verify this?

gm

[/ QUOTE ]

I can verify that it's wrong. f=365 gives n=23 for the b-day problem, so it can't be 20 for f=22,100. I quickly found that n=70 gives 11%, and after that the intermediate values blow up in Excel, so you need a better program to get to 50%.

[gaming_mouse]

Bruce,

Thanks. I knew that was wrong. Silly, the way I wrote the program, I was actually finding n. So the answer in this case is 175. After 175 flops, there is a 50% chance that at least 2 will be identical.

gm

[BruceZ]

I got n = 185 tries gives 50% probability of duplication. I broke it into smaller pieces so Excel wouldn't choke.

EDIT: I now get n=175. See latter posts.

[gaming_mouse]

Here's the code, for anyone intersted:

<font class="small">Code:</font><hr /><pre>

public class Test {

public static void main(String[] args) {
double f = 22100;
double ans =2;
double lastTerm = (f-1D)/f;

while (true) {
lastTerm *= (f-ans)/f;
if ( (1-lastTerm) &gt; .5)
break;
ans++;
}

System.out.println(ans);
}
}
</pre><hr />

[gaming_mouse]

[ QUOTE ]
I got n = 185 tries gives 50% probability of duplication. I broke it into smaller pieces so Excel wouldn't choke.

[/ QUOTE ]

Sounds like you got rounding errors.