Help explaning something obvoius about probability

[kismet]

I recently got into an argument with a dealer of all people, and he claims that the odds of flopping a set change with the number of people in the hand. He says that with 10 people in the hand there is a better chance of your "outs" not even being in the deck. I agree with him and say "but if they ARE in the deck your chances of hitting them are just that much better". So according to him if there are 3 others in the hand there are 8 cards out for players plus one for the burn card. Making the first card have a 2/45 chance of hitting your set. I can't drive home how it doesn't matter what the burn card is, or what the players have since you haven't seen them. Anyone have any simple logic tricks to drive the point home, or know any good links about this topic? It is driving me nuts, plus there is $50 on the line. [img]/images/graemlins/smile.gif[/img] Maybe I should just let him keep thinking things like this and just play more poker heads up.

BTW the first poker book he picked up at the bookstore said the odds of flopping a set were 10:1. I have no idea where that comes from, but it really put some more doubt in my case.

[Benholio]

You could try using a smaller, more simple deck to explain it to him, and see if he still thinks that way. For instance, Take 5 red cards and 5 black cards, shuffle them up, and "deal" 8 cards to imaginary players face down. Ask him what the odds of the top card being red or black are. Do the same thing only "dealing" 4 cards to imaginary players.
If the red/black thing doesn't match the whole "flop a set" thing as easily, maybe just play with 4 jacks, 4 queens, and 4 kings, or something.
If none of this dumbing-down convinces him, then commence plan b: Play this guy as often as possible.

Ben

[aloiz]

Well, if you can't explain to him why you don't factor in other players or the burn card you might try the following...

Take four cards from a deck, say A,2,3,4. Deal out two one card hands. Ask him what are the odds that the top card in the remaining deck of two is an A? Ask him what are the odds that the bottom card in the remaing deck of two is an A? If he fails to come up with 1/4 then try this. Take two cards A,2. Ask him what the probability is that the top card is an A. Seperate the two cards. Ask him what the probability is that one of the cards is an A. If he can't get it at this point then tell him he's an idiot and go from there [img]/images/graemlins/grin.gif[/img]

Also since he insists on thinking about it from a conditional standpoint, use conditional probability to prove that the odds don't change.

First the odds that you flop a set or four of a kind are 11.76% or 7.5:1. 2*C(48,2)/C(50,3) + 48/C(50,3) Now if you can convince him that those numbers are correct, when only one player is being dealt in then you can do the following:

Given that you have a pair in the hole, and that you're playing against x number of players calcuate the odds that either the remain x players have none of your two remaing cards, 1 of your cards, or both of your cards amoung their hands. e.g. 3 other people dealt in besides you.
odds that both your cards remain in deck = C(48,6)/C(50,6) ~= .772
odds that one of your cards remain in deck = C(2,1) * C(48,5)/C(50,6) ~= .216
odds that none of your cards remain in deck = C(2,2) * C(48,1)/C(50,6) ~= .012

Now take those probabilites and multiply them by the odds of flopping a set or better given the individual cases.
Both your outs are in deck, and four people total have received cards = 2*C(42,2)/C(44,3) + 42/C(44,3) ~= .133
One of your outs remain in deck = C(43,2)/C(44,3) ~= .068
None of your outs remain in deck = 0

So take the probabilites of each case occuring, multiply it by the odds of flopping a set or better given that case, and add all those up and we get .7722*.1331 + .2155*.0681 + .012*0 = 11.75%
So it makes no difference whether there are 1,4 or 20 players in the hand on the odds that you flop a hand.

Also the burn card thing can be explained by the same method. Either the burn card happens to be an out, or not. Multiply the probability of it being an out by the odds of you hitting a set given that the burn was an out, and add that to the probability that the burn card wasn't an out times the odds of hitting given that the burn wasn't an out.

Good luck,
aloiz

[cardcounter0]

The correct answer is "You don't have to be real smart to be a dealer".

Tip the poor guy an extra buck the next pot you drag.

[kismet]

HAHAHAHA that is the winning answer by far!

[DPCondit]

[ QUOTE ]
The correct answer is "You don't have to be real smart to be a dealer".


[/ QUOTE ]

You beat me to the punch on that answer.

Geez, how often have you (kismet) played with dealers? Most of them aren't exactly the most cunning opponents that you'll ever run across (to put it mildly).

[kismet]

[ QUOTE ]

You beat me to the punch on that answer.

Geez, how often have you (kismet) played with dealers? Most of them aren't exactly the most cunning opponents that you'll ever run across (to put it mildly).



[/ QUOTE ]

Oh I play with em all the time. I am starting to think that odds don't work against people who understand them. I went all in with KK and a dealer flipped TT. Ten hit on the flop. I rebought and two hands later my AA lost to 99 when a 9 came on the river. So much for my bankroll.

[DPCondit]

[ QUOTE ]
I am starting to think that odds don't work against people who understand them.

[/ QUOTE ]

I assume you meant "don't" understand them. [img]/images/graemlins/smile.gif[/img]

Don