Running It Twice and Insurance

[ev_slave]

Wildthought, MVD gives you a good example of why you are wrong, but here's the intution:

EV stands for "Expected Value," that is, how often you EXPECT to win before you see any more cards. In your example, your EV of the 2nd running has fallen after hero spikes his card, but that changed only AFTER you saw that he spiked his card. But, the EV of the second running only changes AFTER the first running, and so your example CANNOT disprove anything about the expected value BEFORE both runnings.

Further, your EV represents what your actual equity would converge to if you ran the possible board out an infinite number of times. Although the cards you deal on turn 1 will necessarily not constitute what appears on turn 2, both runs are possible if you run the hand only once. Thus, if you ran out 2 boards an infinite number of times, and the single board an infinite number of times, the various card combinations would occur with identical frequencies. As a result, EV is not affected.

[mvdgaag]

Yes... well explained ev_slave (with that name you should be able to [img]/images/graemlins/smile.gif[/img])

It's the same error in logic as this riddle induces... I'll post it here and give the answer with explanation later. Have fun.

You are in a quiz. There are three safes. One of them contains a million dollars. The other two are empty.

You get to pick one safe, A, B or C.

The quizmaster opens one of the other two safes and shows you it's empty.

You get the choice: opening the safe that you first chose or switch to the other one.

Should you stay with your safe, switch, or doesn't it matter?

[TomCowley]

Running it twice is bad if you have an edge in an uncapped game and aren't above your roll, unless the guy is likely to leave as soon as he beats you, but would play on after a split pot.

[WildThought]

I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

[DarkMagus]

[ QUOTE ]
Running it twice is bad if you have an edge in an uncapped game and aren't above your roll, unless the guy is likely to leave as soon as he beats you, but would play on after a split pot.

[/ QUOTE ]

Yeah, that's really the only reason that you would not want to run it twice. If you have the max buyin, and some bad players have more than the max buyin, you'd rather not run it twice so that you give yourself a bigger chance of doubling up and being able to play deep stacked against the bad players.

Once you've already built a stack up and have the bad players covered (or close to it), then you would definitely want to run it twice so you can have a better chance to keep playing deep stacked.

[DarkMagus]

[ QUOTE ]
I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

[/ QUOTE ]

Do an EV calc.

I'll simplify the math a little. Suppose there's $100 in the pot, one card to come on the river and you have some sort of draw with 4 outs, with 44 cards left in the deck (4 board cards and both players' cards gone) so you're 9.1% to hit.

If you run it once the EV is simply (0.091)*($100) = an EV of $9.10.

For running it twice:

The 9.1% you hit the first time, you win $50 of the pot but you're down to 3 outs out of 43 cards, which is 7.0%.
The 90.9% you don't hit the first time, you don't win the first pot but then you're up to 4 outs out of 43 cards, which is 9.3%.

So your EV for running it twice is:

(0.091)*[ $50 + (0.070)*($50) ] + (0.909)*[ $0 + (0.093)*($50) ] = $9.10

Exactly the same EV as if you'd run it once.

Moral: You're correct that hitting the first time reduces your odds of hitting the second time. But remember that missing the first time increases your odds of hitting the second time. These differences cancel each other out.

[mvdgaag]

[ QUOTE ]
I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

[/ QUOTE ]


U are not stupid... It's a very human flaw in reasoning. Look at my simplified example. The card IS left out after the first win/lose.

What you are missing is that you evaluate the situation AFTER you have more information. Before you do so you do not know what is going to happen the first time. Yes, your chances of winning are far lower, but so are your chances of losing.

You have to win BOTH the first AND the second time to win.
You have to lose BOTH the first AND the second time to lose.

After one lose or one win the first one you can calculate your odds (like you do), but that wouldn't be fair. When you make the decision to run it twice you are facing both deals.

[plexiq]

[ QUOTE ]
I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

[/ QUOTE ]

I think you are missing the following:

You are correct, that IF you hit one of your outs the first time, you will have a lower chance to hit one in the second run.

BUT: If you dont hit your outs the in first run, you will have a HIGHER chance to hit in the second run. This balances out with the first effect, such that the overall expectation remains neutral.

eg: You have 8 outs and run the river twice. After missing the first run, you will have 8/42 probability to hit in the second run - instead of 8/44.


I think the above might be more intuitive for you, but its basically overly complicated thinking. A better way to look at it imo:

If you are playing brick & mortar, the cards are already shuffled when you decide to run it twice. That is, the arrangement of cards will not change any more.

The probability of the first card being an out of yours is obv. 8/44. The probability of the second card being an out is also 8/44, obv (as long as we dont know the first card). So - at the moment you decide to run it twice - you are basically taking two smaller 8/44 gambles for half the pot each, instead of one big 8/44 gamble for the whole pot. It should be clear that this is EV-neutral and variance reducing.

Note to self: Read the last posts of a thread before replying. Yes, i basically just re-stated what is said in the posts above [img]/images/graemlins/wink.gif[/img] Meh, heh.

[WildThought]

Wow,

I thought I was smart about this and was readt to write an essay about the correct way of doing business. Thanks for the schooling, very instructive.

Andy

[mvdgaag]

[ QUOTE ]
Yes... well explained ev_slave (with that name you should be able to [img]/images/graemlins/smile.gif[/img])

It's the same error in logic as this riddle induces... I'll post it here and give the answer with explanation later. Have fun.

You are in a quiz. There are three safes. One of them contains a million dollars. The other two are empty.

You get to pick one safe, A, B or C.

The quizmaster opens one of the other two safes and shows you it's empty.

You get the choice: opening the safe that you first chose or switch to the other one.

Should you stay with your safe, switch, or doesn't it matter?


[/ QUOTE ]


No responses... I'll give the answer anyways:
You should switch.

When you first choose a safe you have 1/3 chance of winning and 2/3 chance that it's in one of the others.

If you had chosen the correct safe already the quizmaster could open one of both safes.

If you had chosen an incorrect safe the quizmaster could only have opened the one empty safe that remains. This information increases the chance that the other safe has the goods.

In other words:

When you don't switch: 1/3 of the time you chose the correct safe in the first place and win.

When you do switch: 1/3 of the time you chose the correct safe the first time and now lose. But 2/3 of the time you didn't and the only alternative safe left (the quizmuaster just took away one choice) must be the correct one.