Drawing for Dinner

[SamIAm]

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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

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Ok. Let's make sure you understand all the terms I used (made-up).

By "repeated", I don't mean she necessarily tied with Eric. I mean she got one of the cards she couldn't have gotten in Method A. I mean she got one of the cards held originally by another player.

Now, given that she gets a repeated card, she's equally likely to get any of the repeated cards. Since there are X of those, the chance she gets the LOWEST repeated card (Eric's) is 1/X. After she gets the repeat of Eric's, she loses the coin-flip with probability 1/2. Hence 1/2X.

The probability that she loses with MethodB is therefore:
[(52-X)/52 * 1/(X+1)] + [X/52 * 1/2X]

If that didn't make sense, let me know. Typing math without LaTex is hard. [img]/images/graemlins/smile.gif[/img]
-Sam

P.S. It should be pointed out that MCS (above) plays at this weekly game. That's why he's not making any "road-head" jokes. Hel[/i]l, I'm probably gonna get in trouble when he tattles on me for naming her "Francine".

[bkfizz02]

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By "repeated", I don't mean she necessarily tied with Eric. I mean she got one of the cards she couldn't have gotten in Method A. I mean she got one of the cards held originally by another player.

[/ QUOTE ]

Great. I assumed this also.
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Now, given that she gets a repeated card, she's equally likely to get any of the repeated cards. Since there are X of those, the chance she gets the LOWEST repeated card (Eric's) is 1/X. After she gets the repeat of Eric's, she loses the coin-flip with probability 1/2. Hence 1/2X.

[/ QUOTE ]
Ok, so the probability she loses given that she drew a repeated card is 1/2X. Now we just have to figure out the probability that she loses given she has drawn a non-repeated card. You must have this as the first term in your expression :

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The probability that she loses with MethodB is therefore:
[(52-X)/52 * 1/(X+1)] + [X/52 * 1/2X]


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Examining this term, I see you assign a branching ratio of (52-X)/52 to this event, which makes sense. Then the probability this unique card is lowest is simply 1/(X+1), as though X+1 played the game originally. This makes sense too.

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Typing math without LaTex is hard. [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]
I agree [img]/images/graemlins/wink.gif[/img]


Thanks for detailing your work. Now, can you help me find the flaw in my logic in my previous post? I think it stems from the fact that you assume a somewhat small number of players.

Specifically, in the case of redrawing, what if she draws a card that 2 (or 3 or 4) of her opponents have already drawn? Then she gets to face off with the loser (based on the way you described the action that evening). This is bad for her because now she is heads up with a coin, rather than 3 or 4 (or 5!) ways if there are multiple ties. I'm pretty sure that if she draws a card that has already been drawn multiple times, she loses her advantage. When you wrote 1/2X for cards that had been replaced, it assumed that only 1 of that rank was chosen originally. If 2 of that rank was chosen originally, the probability would have been 2/X * P(losing the coin flip) = 2/X * 1/2 = 1/X.

I am running a large simulation, and will plot the result shortly.
-bk

[bkfizz02]

For my simulation, I varied the number of opponents from 1 to 51. Then, for each scenario, I drew from the deck, noted if Francine lost and then reshuffled and repeated 40000 times.

I observe that in general Francine benefits slightly until the number of opponents grows sufficiently large.

The second plot is the same as the first, just on a log scale.




I really think the key is that she loses her edge once multiple people frequently draw the same low card.

You can look at my simulation at http://www.npl.uiuc.edu/~kiburg/simpProj.cxx
if there are concerns

<Edited 12:51 AM central : I just noticed a potential bug, but it seems to push the simulation even further in favor of Method A (assuming she doesn't want to take drive)>

[SamIAm]

Man, a computer simulation is simply not necessary. This is what happens when computer scientists try to be mathematicians. (Just kidding. I happen to be both. [img]/images/graemlins/smile.gif[/img])

If there are 51 opponents, then, in method A, she has to draw the lowest card to lose. In method B, she pretty much still has to draw the lowest card to lose, but now she'll probably tie with somebody, and get a coin-flip to escape her loss.

Her advantage INCREASES as the numbers of players increases. To follow the notation from my earlier argument, 1/2X becomes even better than 1/(X+1), AND the chance of her repeating is more likely.

As X increases, we weight more and more toward the better and better side.
-Sam

[SamIAm]

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Specifically, in the case of redrawing, what if she draws a card that 2 (or 3 or 4) of her opponents have already drawn?

[/ QUOTE ]I just read this part. I think you don't understand something. How can she repeat TWO opponents? We deal ABCDE, shuffle, then deal F. A!=B.

You don't think I just mean the number on the card, do you? The cards are ordered by suit.
-Sam

[bkfizz02]

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You don't think I just mean the number on the card, do you?

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Yes, I thought that 2 [img]/images/graemlins/diamond.gif[/img] vs 2 [img]/images/graemlins/heart.gif[/img] was a tie and resulted in a coin flip.

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The cards are ordered by suit.


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This makes the question much simpler. Your formula a few posts back is right, given that the 52 cards are ranked. This is a critical part of your question, and since you mentioned ties in other parts of the question I never once thought that suits mattered. I figured your Wednesday game was poker, not bridge [img]/images/graemlins/smile.gif[/img]

For comparison, I'll simply show the result given that all 52 cards are ordered. She chooses Method B:
,

By the way, I'm pretty sure I'm right if the cards of the same rank lead to coin flips.

<edited: I just reread the entire thread and found where you said "the cards are all distinct" which I read to mean "draw without replacement". >

[SamIAm]

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By the way, I'm pretty sure I'm right if the cards of the same rank lead to coin flips.

[/ QUOTE ]What does this mean? If cards of the same rank tie, then the method isn't even definied when 3 people get the same rank.

Whatever. I'm glad we cleared that up. You should know, if they draw for the button in any poker room, and you get the K[img]/images/graemlins/club.gif[/img] and I get the K[img]/images/graemlins/spade.gif[/img], there isn't going to be any coin flipping. I win. [img]/images/graemlins/smile.gif[/img]
-Sam

[MCS]

[ QUOTE ]
P.S. It should be pointed out that MCS (above) plays at this weekly game. That's why he's not making any "road-head" jokes. Hel[/i]l, I'm probably gonna get in trouble when he tattles on me for naming her "Francine".

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To be honest, MCS kinda hopes that they're not actually jokes. We'll ask "Francine" what F-name she'd prefer. I'm sure "Eric" has plenty of inappropriate F-names for women that he's stolen from porn.

It was kinda weird since they're not just abstract people to me and thus I actually got the roadhead/SIIHP visuals in my mind. [img]/images/graemlins/smile.gif[/img]

I havn't sat and thought about it but at first glance I wouldn't even consider the card values.

I'm thinking in the first examine F has a 1:6 chance of winning (6 players).

In the second example she has a 1:2 chance of winning (2 players).

then add in the fact that he has a 6 so she has 8 card types that can beat him anyhow.

I could be completely off but without thinking about it I think she obviously has a MUCH greater advantage in the second scenario.

[darthtuttle]

why not just pull out overcards to represent the ones that were drawn, shuffle up and deal.