Prove the rule of two
 

Is there a proof or is it just coincidence?

Re: Prove the rule of two
 

When you have n outs on the flop there are 47 unseen cards left in the deck. So you will get there on the turn n/47 times.

n/47 = n/50 + 3n/2350 = 2n/100 + 6n/4700 ~ 2n*1% (since 6n/4700~0% for small n)

Re: Prove the rule of two
 

How did you know the (3n/2350) part?

Re: Prove the rule of two
 

[ QUOTE ]
How did you know the (3n/2350) part?

[/ QUOTE ]47 * 50 = 2350.

Thus n/47 - n/50 = 50n/2350 - 47n/2350 = 3n/2350.

Re: Prove the rule of two
 

I think I know what the this rule is, but please give me the rule. Thanks

Re: Prove the rule of two
 

The rule of two is used for outs after the turn.

After the turn, you have seen your two cards plus the four cards on the board so there are 46 cards you haven't seen.

If you have 1 out, then your chances of hitting that card on the river are 1/46, or 2.17%, and for any larger number of outs you just multiply the number of outs by 2.17. So, the "rule of two" is actually the rule of 2.17, but just using 2 works for most purposes. You have to get above 5 outs before you're off by more than 1 percent and above 11 outs before you're off by more than 2 percent.

Now, as for the rule of 4, after the flop you have seen 5 cards instead of 6, so there are 47 unseen cards. The number of possible turn-river combinations is C(47,2) = 1081.

If you have n outs, the number of combinations that will make you a winning hand is 1081 - C(47-n,2), and then the percentage is of course that result divided by 1081.

The exact percentages for 1-9 outs are:
1 04.8
2 08.4
3 12.5
4 16.5
5 20.4
6 24.1
7 27.8
8 31.5
9 35.0

So just using the "rule of four" gets you within one percent of the exact mathematical value.