Is this a good prop bet ?

[im a model]

i wholly disagree. well at least in no limit. if someone has an ace or two aces, they are going to raise it and you are less likely to see a flop because everyone else may fold. so the flop that is not happening because of the guy with aces raising, is the same flop that is more likely to have a deuce. so the deuces are being flopped less often because when the deuce has the advantage on the flop (because the guy has aces in the whole), that flop isnt being seen.

[jogsxyz]

No point in arguing this point. There must be many of you who have seen millions of flops stored in poker utilities. Just write a program to count the aces and deuces.

[TheChad]

bump jimmyfingers, good post

[btmagnetw]

[ QUOTE ]
bump jimmyfingers, good post

[/ QUOTE ]no it wasn't. the effect would be <0.1% if any at all. and there's reason for both sides. do you see flops more when aces are dealt because these hands enter pots or do you see less flops because these hands raise and end things preflop?

[jimmyfingers]

[ QUOTE ]
[ QUOTE ]
bump jimmyfingers, good post

[/ QUOTE ]no it wasn't. the effect would be <0.1% if any at all. and there's reason for both sides. do you see flops more when aces are dealt because these hands enter pots or do you see less flops because these hands raise and end things preflop?

[/ QUOTE ]

obviously it depends on the game. what i said was simply to keep it in mind because it makes a difference, in some games moreso than others.

for example, look at heads up limit hold 'em. if there is one ace dealt, we know at least that one player will want to see a flop. if two aces are out, there's a good chance both will want to (if the aces are split), or a decent chance if both aces are with one player (at least one player will want to). three or more aces and a flop is guaranteed.

on the other hand, look at deuces. if one deuce is out it's less likely that player will want to see a flop. if two deuces, one player will want to see the flop if he has both, but if they're split it's unlikely both players will want to see a flop. if three deuces are out, one guy has a pair and another guy will have 2x. most 2x's won't see a flop (23, 24, 25, 26...).

i think it's obvious in this case that the aces will be less likely to flop.

[JayTee]

Isn't it impossible to lose $75 doing this 40 times. I'm assuming that it wasn't exactly 40.

I have a question. How do you determine the likelihood of winning 70% of the time out of 40 trials being a 55% favorite?

[filsteal]

Maybe this Aces vs. Deuces flop thing is best explained this way:

If, hypothetically, we dealt the flop EVERY HAND, even if the hand ended with the pre-flop betting, then yes, aces and deuces would flop equally often.

What everybody's saying is that when one or more aces are waiting atop the deck to be flopped, it makes it less likely that there will actually be a flop.

[filsteal]

[ QUOTE ]
Isn't it impossible to lose $75 doing this 40 times. I'm assuming that it wasn't exactly 40.

I have a question. How do you determine the likelihood of winning 70% of the time out of 40 trials being a 55% favorite?

[/ QUOTE ]

You'll win 28 or more trials (70%+) 3.86% of the time in the situation you described. You can get this using the binomial distribution. The math is messy, but Excel can do it for you.

[PairTheBoard]

[ QUOTE ]
He loses if the flop contains no j,7, or 2. This will happen:

40/52 * 39/51 * 38/50= 44.7%

So he wins the other 55.3% of the time.

[/ QUOTE ]

You win 40^3/(52*51*50) = 48.3% of the time.

He wins 51.7% of the time.

That gives him about a 3.4% edge. For every $100 you bet you can expect to lose $3.40.

PairTheBoard

[jjshabado]

[ QUOTE ]
[ QUOTE ]
He loses if the flop contains no j,7, or 2. This will happen:

40/52 * 39/51 * 38/50= 44.7%

So he wins the other 55.3% of the time.

[/ QUOTE ]

You win 40^3/(52*51*50) = 48.3% of the time.

He wins 51.7% of the time.

That gives him about a 3.4% edge. For every $100 you bet you can expect to lose $3.40.

PairTheBoard

[/ QUOTE ]

Why is it 40^3 instead of (40*39*38)?

We're trying to find out the chance that no J72 hits the flop, right? On the first card there are 40 such cards. One of them comes on the first card, so the next card has a 39/51 chance of not being a J72. And so on...

Right? Or where am I messing up?