Running 'em three times

zepmetal · #1 03-15-2005, 02:18 PM

I understand that typically players choose to run them three times when it is a cointoss-like hand and they want to reduce their variance. It seems that regardsless of the situation (whether you're getting 1:1 odds to win, or 10:1) the long-term outcome will be the same running them three times or not. Is it faulty logic to say something like "but if he gets allin on the flop with a flush draw, if he hits it on the first run, hes now less likely to hit it on the 2nd run, so hes at a disadvantage". On the other hand, when he misses it on the 1st run, he's getting better odds to hit it on the 2nd run and 3rd run, thus compensating. My question is this: is there ever a situation where running them 3 times gives an advantage to one player and therefore a disadvantage to the other?

dkernler · #2 03-15-2005, 07:27 PM

I may be missing something here with the phrase 'run them three times', but we generally assume each hand (or trial, in probability terms) is independent of the others. Flipping a fair coin heads 5 times in a row does not make it more likely that the next flip will be tails.

I'm not sure if that's what you meant, so completely ignore this advice if I'm way off-track! [img]/images/graemlins/laugh.gif[/img]

thylacine · #3 03-15-2005, 08:43 PM

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My question is this: is there ever a situation where running them 3 times gives an advantage to one player and therefore a disadvantage to the other?

[/ QUOTE ]

No. EV is exactly the same as running once, even though the runs are NOT independent.

2ndGoat · #4 03-16-2005, 02:52 PM

Might be easier to see they're equal EV each time by thinking that they'll deal a river face down three times and then turn them all up at the same time. The 3nd river doesn't care about the 1st and 2rd, they may have well have been burn cards.

You could also see that if you're chasing a flush and it comes the first time, you do have a lower chance for the 2nd and 3rd times, but if it doesn't come you have a higher chance. You would have to do more arithmetic than I'm willing to do to prove it that way, but hopfully it makes sense conceptually.

2nd