Re: 100 vs 6500
 

I'm pretty sure I'd want Raymer on my 100 list before Tuan Le.

Re: 100 vs 6500
 

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I'm pretty sure I'd want Raymer on my 100 list before Tuan Le.

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Raymer was a pretty close alternate.

I made a conscious decision to try and mentally compensate for the mainstream bias I'm going to have. A former WSOP ME winner is immediately going to seem more attractive, particularly being a 2+2er, etc. It's definitely conceivable that he suffered unfairly because of that, but there's no way I'm going to get this exactly right based on little RGP scraps from Phillips/Negreanu/Seidel et al and a few dozens hands on TV. Tuan Le won two WPT events, he goes on the list.

Re: 100 vs 6500
 

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Guys, you don't even know what x, y, and z are. How could you possibly tell whether the bet is good or bad?


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Surely you read in my response where I not only said that I might be misunderstanding what the OP is saying <because values are not given> but also that I assume it is an even money bet when I say he's taking the worst of it.

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I think you may have the worst of it here.

I might be misunderstanding the bet though. Is it even money that one of these players will make the final table or not?



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Re: 100 vs 6500
 

It's not even money. He's laying what works out to 2:1 that none of the pros will make the final table.

Re: 100 vs 6500
 

Dont need to know x,y, and z. Unless x is huge relative to z and y you got way the worst of it.
Assumption. The hundred pros dont play any better than the 6500 randoms. We know this is not true. The chance of none of the 100 making the final 9 is equal to 1 minus the chance of the nine seats are all filled by randoms. This number is about 0.05 or one in twenty. For this to be a good bet for you, that 'x' number better be 40 or 50 times bigger than the 'y' number.

Re: 100 vs 6500
 

[ QUOTE ]
Dont need to know x,y, and z. Unless x is huge relative to z and y you got way the worst of it.
Assumption. The hundred pros dont play any better than the 6500 randoms. We know this is not true. The chance of none of the 100 making the final 9 is equal to 1 minus the chance of the nine seats are all filled by randoms. This number is about 0.05 or one in twenty. For this to be a good bet for you, that 'x' number better be 40 or 50 times bigger than the 'y' number.

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We figured that the average pro from my 100 is 3x more likely to make the final table than the average player from the remainder of the field (assuming a 6600-person field). So the chance of one of my names filling Seat 1 at the final table is equivalent to 300/6800, right? And the chance of this not being the case is 6500/6800?

So if that's not wrong then the chances of there not being a single name from my list at the final table would come to [(6500/6800)(6499/6799)(6498/6798)(6497/6797)(6496/6796)(6495/6795)(6494/6794)(6493/6793)(6492/6792)], or [66.6%].

Am I wrong?

Re: 100 vs 6500
 

RETRACTION.

Okay, I'm retracting my calculations. Did it in excel, but didn't check my work.
Not clear how to treat 3 times better. But your method is a good approximation.

Re: 100 vs 6500
 

You are assuming that there will be 6,600 in the field. From what we've heard so far about the size of the field to date, it doesnt sound like it will be full, which would give you a better chance to win than your buddy. Nice bet.

Re: 100 vs 6500
 

do u happen to know how many have presently entered?

Re: 100 vs 6500
 

But David Williams over Raymer? If your concern is a track record, that one doesn't make much sense to me.