Drawing for Dinner

[SamIAm]

At our Wednesday game, we realized that the local burrito-joint, for 1 night only, had $2 burritos. That's a good deal, but they won't deliver.

Andy, Bill, Carl, David, & Eric (Names changed to protect the notation.) all want food, but nobody wants to drive. We decide to draw for it, and Eric gets the 5[img]/images/graemlins/club.gif[/img]. He loses.

However, as soon as Eric grabs his keys, his girlfriend, Francine, says "Oh, get me some chips while you're out." This isn't fair, so we draw from the (now shuffled) deck for Francine. We agree that we'll break any ties with a coin-flip. She draws the J[img]/images/graemlins/spade.gif[/img], and Eric drives.

After Eric left, we asked the question: "Which method's better for Francine?" MethodA: We draw cards for ABCDEF. MethodB: We draw cards for ABCDE, shuffle, and draw for F, breaking ties with a coinflip.

I already know the answer, because we had fun talking about it for a while. [img]/images/graemlins/smile.gif[/img] I'll post our explanation soon.
-Sam

[JKDStudent]

What's better for Francine? She gets to draw from the full deck. If there are five people and low card loses, that means that four cards that beat the low card are gone. With a new deck, she has access to those cards.

End result, you're just adding outs, which is always +EV. [img]/images/graemlins/tongue.gif[/img]

[SamIAm]

[ QUOTE ]
End result, you're just adding outs, which is always +EV. [img]/images/graemlins/tongue.gif[/img]

[/ QUOTE ]Not true. You're adding cards that let her win, but you're also adding a card which lets her lose.

Keep thinking. [img]/images/graemlins/smile.gif[/img]
-Sam

[KenProspero]

ummmmmmmmm ..... I think ......

In your specific example, where F waited to see the 5 [img]/images/graemlins/club.gif[/img], F is better off with a 6 way draw.

Under all cases, in the original draw, she'd have a 1 in 6 chance of driving.

With the new draw, she has a 12/52 of losing outright (>1/6) PLUS either a 1/13 or 1/52 chance of a tie (in which she has a 50/50 shot of driving), in any case greater than 1/6.

Now, if we generalize, with 5 people drawing, you would expect, on average, that 1 card will be in each 5th of the deck. If suit rankings are used

Bottom fifth (11th card) -- 4 [img]/images/graemlins/heart.gif[/img] or lower
Fourth fifth (21st card) -- 7 [img]/images/graemlins/club.gif[/img] to 4 [img]/images/graemlins/spade.gif[/img]
Third fifth (32nd card) -- 9 [img]/images/graemlins/spade.gif[/img] to 5 [img]/images/graemlins/club.gif[/img]
Fourth fifth (42nd card) -- Q [img]/images/graemlins/heart.gif[/img] to 10 [img]/images/graemlins/club.gif[/img]
Top fifth (52nd card) -- A [img]/images/graemlins/spade.gif[/img] to Q [img]/images/graemlins/spade.gif[/img]

In this case, taking the midpoint of the band, for the bottom fifth, F expects to see the low person with the 6th card (i.e., 3 [img]/images/graemlins/diamond.gif[/img]), in which case, the expectation is that there is slightly greater than a 10% chance that she will drive, vs 16.7% in the original draw.

It's very late at night, so this whole analysis may not fly, but this is my impression.

[JKDStudent]

[ QUOTE ]
[ QUOTE ]
End result, you're just adding outs, which is always +EV. [img]/images/graemlins/tongue.gif[/img]

[/ QUOTE ]Not true. You're adding cards that let her win, but you're also adding a card which lets her lose.

Keep thinking. [img]/images/graemlins/smile.gif[/img]
-Sam

[/ QUOTE ]

Uh... no. You're adding four cards that let her win and one card that lets her tie, which then gets resolved by a coin flip. So, like I said, you're adding outs to the deck, which is +EV.

*EDIT* Gah. I misunderstood the original question. When you said draw ABCDEF, I took that to mean she just takes a card from the already drawn deck. As in re-create the previously drawn cards, and then she takes one. My bad.

*SECOND EDIT* I still remain unconvinced. Only one person can lose, and that's the lowest card. All the others have to do is beat that card. That means that you're adding winning cards back into the deck, which is adding outs. Even if you assume the fifth tie card ALWAYS results in a loss, that's still a net of four winnings cards that are available. I'll take an open-ended straight draw over a gutshot any day of the week.

[SamIAm]

You can't just say "You're adding outs." because you're also adding non-outs. I'd give you partial credit for getting the right answer, but your work isn't correct. [img]/images/graemlins/smile.gif[/img] Ken, you're definitely not allowed to say "Assume there's one card in each 5th of the deck."

Here's how I thought about it. In MethodA, the cards are all distinct. In MethodB, the cards can either be all distinct, or Francine can get a repeated card.<ul type="square">[*]In the distinct case, she loses 1/(X+1) of the time. (Where X = number of opponents.)[*]In the repeated case, she loses 1/2X of the time. 1/X to tie with the lowest, 1/2 to lose the coinflip.[/list]MethodB is a weighted average of 1/(X+1) and 1/2X, so she loses less often in MethodB.
-Sam

[MCS]

What if Francine will just get Eric to drive anyway, even if she loses? How does that impact the math here? [img]/images/graemlins/smile.gif[/img]

[bkfizz02]

[ QUOTE ]

In MethodA, the cards are all distinct. In MethodB, the cards can either be all distinct, or Francine can get a repeated card.
<ul type="square">[*]the distinct case, she loses 1/(X+1) of the time. (Where X = number of opponents.)[/list]
[/ QUOTE ]
I agree with this.
[ QUOTE ]
<ul type="square">[*]In the repeated case, she loses 1/2X of the time. 1/X to tie with the lowest, 1/2 to lose the coinflip.[/list]

[/ QUOTE ]
This doesn't make sense to me...how do you get P(tying for low) = 1/X ?

In the limiting case, this doesn't make any sense. If X=51, then the entire deck will be dealt out (save for 1 card) so either 3 or 4 people will get a deuce.

Method A:
She has a 1/52 chance of losing
Details:
1/13th of the time she has a deuce, and if they flip coins in a fair way she will lose 1/4th of the time. so 1/13 x 1/4 = 1/52.

Method B:
Every time, there will be either 3 or 4 players who draw a deuce, and they will break the tie by coinflip. Now she draws a card from the complete, shuffled deck. 1/13th of the time she will draw a deuce. Then she flips a coin with the loser from the original 51. So now she has a 1/26th chance of being the loser. The reason for this is because she effectively only had to lose 1 coin flip in this method instead of losing two coinflips in method A, when she was included. At least this is true if I read your initial post correctly (translation: Method B picks a loser and then Francine plays off against him/her.)

Even if Francine got to play against all the losers the 1/13th of the time she drew a two, the probability of losing would be 1/52, as in method A, not 1/(2*51) = 1/102

[ QUOTE ]
MethodB is a weighted average of 1/(X+1) and 1/2X, so she loses less often in MethodB.

[/ QUOTE ]

Please explain your reasoning for assigning P(tying for low) = 1/2X
and
please explain why method B is the weighted average of 1/X+1 and 1/2x

Thanks.

[WackityWhiz]

The correct answer is...

If method A is used, all players must REDRAW

If method B is used, since she gets an advantage, she goes along with him if she doesn't draw the lowest... and gives him road head.

If she draws the lowest card in method B, then she drives by herself. After the night is over, she gets it stuck in her pooper.

[bkfizz02]

[ QUOTE ]
The correct answer is...

If method A is used, all players must REDRAW


[/ QUOTE ]

If each player redraws and ties are broken by fair coin flips, then isn't this equivalent to just dealing out X cards to X players and breaking ties with coin flips (or more cards?) ? You just have a 1/X of being lowest,right?