A popular prop bet going on at one of the local poker rooms I play it is for one player to pick 3 cards (rank only) and if one of them comes out on the flop they win. If not, they lose. I have tried calculating the odds different ways and while I am sure the person picking the cards has the edge I am not certain of the exact odds.

Can someone reply with the calculation?

13 ranks 3 chances...3 divided into 13 = 23% of picking a winner

lol

(I would have thought it was 50/50, either it will happen or it won't. Math can be deceptive.)

How can the person selecting the cards "have the edge" before the odds are set ?

(Fwiw, this is not just simple math; the person selecting can skew his picks towards lower or middle ranked cards, as a plethora of higher ranked cards dealt preflop may increase the likelihood of there being any flop. As in, if everyone gets rags, there is less likely to be a flop, so any actual flop seen should have a higher incidence of lower or middle ranked cards versus any "3 of 52" . I expect someone has done that math somewhere.)

One time I was playing poker, and there was a pair on the flop.

True story.

True odds are (1-P(all three cards miss))

So if you pick before the down cards are dealt, and a flop is sure to happen, the math isn't hard (there is a probability sub-forum that loves this sort of thing).

If the odds are only determined for hands that make it to a flop, then some Aces (kings?) may be out of the possibility for the flop which can skew the odds some.

Were this bet available in BJ, I'd be a thousandaire - I always get 9 6...

The easiest way to figure the odds on this one are to calculate how often the picks won't hit. If 3 ranks are chosen then there are 12 cards that win if any one of them comes out on the flop. If none come out, it's a loss.

The chances of one of the 12 NOT being the first card are 40/52.
Given that the first card did not hit, the chances of one of the 12 NOT being the second card are 39/51.
Given that the first two cards did not hit, the chances of one of the 12 NOT being the third card are 38/50.

So the chances of none of the chosen 12 cards being on the flop are 40/52 * 39/51 * 38/50 = 44.7%

Which means that the chances one of the chosen 12 cards being on the flop are 55.3%

Since you signed up here 8 years ago, I’d assume you’re past high school math by now?

You basically do the same calculation you would to find out your chances of hitting a 12 outer on the turn or river. It’s just more unknown cards to start with (52 instead of 47) and one extra street. So it’s 1-((40/52)*(39/51)*(38/50)) = 55.3%

The math obviously looks different if you know your two hole cards. Assuming that you would pick 3 cards not in your hand, the odds are 1-((38/50)*(37/49)*(36/48)) = 57%.

This bet assumes
1) that a flop happens, and
2) that the cards are picked without excluding the cards the guy picking them has.

For clarification it is an even money bet and made before any cards are dealt or known.

Thanks to Fog of War and madlex for real math answers.

I was dividing 12/52 (.23) + 12/51 (23.5) + 12/50 (.24) = 70.5 and knew that was way to high. I guess that's why I didn't pass Algebra II back in 1970.

Geeking around like this is killing poker. Just f-n gamble.

Never offer a picker even money, conversely gladly take even money as a picker.

You posed a wagering query, not a math puzzle.

The real math answers assume away that a flop is conditional on the cards held by at least two players. A picker "knows" a flop is a condition of the prop bet running, which should influence his choice of likely flop cards away from Aces and Kings for starters.

It also matters how many players are dealt in pre-flop, as the more players, the more likely that two low cards will pair up, inducing one player without high cards to see a flop. OTOH, a heads-up or short-handed table likely sees more flops containing high cards than a full ring table, as more mid-range hands are flop-worthy by players involved.

Those "real math answers" accordingly do not go far enough in stating the degree to which a correct pick is likely. If we accept that a flop is required for the prop bet to be live action, then the picker should go with cards unlikely to induce a flop happening.

Can anyone do the math analysis ? Fair wagering odds may even be as high as 3-2 in favor of the picker.

Keeping in mind at least two players will need hands that would see a flop, thus removing 2 - 4 of such "flop-generating" cards from the pool yet to flop. For sake of discussion, let's remove Aces and Kings from the picker's selection range among the pool yet to flop.

That is probably a significant increase in the picker edge. The cards in the sub-range of 2 - Q are on average more likely than the "real math answer" to actually flop than are either an Ace or King.

Actually I do not see how any of your scenarios affects the math at all. The bet is made while the 52 card deck is in the hands of the dealer. In fact, it is inconsequential that we are even playing holdem. It seems to me the bet would work exactly the same by just picking the 3 ranks and turning over 3 cards. Am I mistaken?

Not THIS sort of thing. Also its not a SUB forum.

You described the bet in the context of a flop. That the bet is made before any cars are dealt is irrelevant to handicapping the appearance of a range of cards on the flop, if there is one.

When a flop is dealt the whole 52 card deck is no longer available from which to deal, and the 4 to 18 cards that will NOT appear on the flop have certain characteristics which relate to whether a flop ever dealt in that hand.

The circumstances of getting to a flop are different than those of just dealing three random cards off the top of a full deck. That the wager outcome is conditioned on a hand going to a flop is information of use to the picker in selecting his range of picks.

Look the odds are clearly better than 55% for the picker regardless, so getting more specific does little to inform you better regarding which side you want on that even money bet.

If you want to whittle the bet down to a "fair" even money prop, charge the picker a Winner commission for his advantage, kind of like they do in baccarat to bettors picking Bank.

Thanks for taking the time to clarify. That is quite a nuanced observation that I had not considered. The end result is that removing the hands that do not see a flop, slightly alters the outcome.

Regardless, I'll keep picking up a red chip each time one of my cards hits and long as someone wants to bet.

What if you are playing Omaha?
Is the bet just before the flop? so you see your cards and how many are still in seeing the flop?

Or is it made after the shuffle, before any cards are dealt?

These sort of bets are usually running

You wouldn’t look at your cards and then choose the cards you want - as that would give away your poker hand range

The other factor is betting ppl away from seeing a flop, either the bank or the player who has selected the three cards he’s bet on, may look at his preflight hand and decide a flop isn’t desirable (as he has 23s, two of his cards) and so he will bet preflop to try and stop the flop

So it changes the odds for the prop

And Changes the dynamic of the poker game

Surely the prop is 2-way? So you both pick a card and get paid based on whatever rules you set.

That way it's pretty much EV 0 unless someone has some actual insight into certain cards being more likely to appear than others but over one evening of cards I'd say unless you're playing masssssive it cant be worth your time