# Pick a Card...Any Card

uzless
Posers and Puzzles 22 Oct '10 18:16
1. uzless
The So Fist
22 Oct '10 18:165 edits
The local chess club is having a draw to raise some cash for a new chess set.

There are 50 cards contained inside a hat. One of the cards has "You're Winner!" marked on it. If you pick the winning card, you get half the money put into the pot so far.

It costs 10 bux to pick a card out of the hat and the pot already has 10 bux in it to start with. You can choose to pick a card anytime you want and you don't have to pay your 10 bux until you actually pick the card out.

So, if someone goes first, pays their 10 bux and picks the winning card, then no one else is out money because obviously they haven't paid. However, obviously there is only a 1 in 50 chance that the person picks the card on the first try.

The question is:

Is there an optimum time that you should pay your 10 bux to pick out a card that will maximize your expected winnings? Ie should you pick after maybe the 5th card is picked, the 10, 25th??

Or does it make no difference at all?

You can only pick once.
2. 22 Oct '10 21:08
Hmm
First go, chance is 1 in 50 that you will win, you will only break even if you do, so average winnings are (0 -10*49)/50

Second go, chance is 1 in 50 that first guy won it, so zero takings, 1 in 49 that you win, if you do you are \$5 up, so winnings are
(5 - 48*10)/50

Third go, chance is 2 in 50 that first or second guy won it, 1 in 48 that you win, if you do you are \$10 up, so winnings are (10 - 47*10) / 50

winnings at go G are:

((G - 1)*5 - (50-G)*10)/50

= (5G - 5 -500 + 10G)/50

= (3G - 101)/10

So to max winnings you need to maximize G, i.e. wait until go 50
If you win, there is \$500 in the pot and you get \$250 if you win, \$240 profit so on average winnings are 240/50 = \$24/5 = \$4.90, which is as predicted by the above formula.
3. 22 Oct '10 21:27
i thinking along those lines
4. PBE6
Bananarama
26 Oct '10 14:011 edit
The expected winnings W for any given pot size P is:

W(P) = (1/50)*(P/2-10) <--it's (P/2-10) because you have to spend \$10 to play

The expected losses L for any given pot size P is:

L(P) = (49/50)*(10)

The expected value E for any given pot size P is just the expected winnings minus the expected losses:

E(P) = (1/50)*(P/2-10) - (49/50)*(10)

For some theoretical pot size P*, the player will be indifferent to playing the game since the expected winnings and the expected losses will balance out, resulting in an expected value of 0. Setting E(P*)=0 and solving for P*, we have:

0 = (1/50)*(P*/2-10) - (49/50)*(10)

490 = P*/2-10

P* = 1000

When the pot size reaches 1000, the expected winnings, losses and expected value are:

W(1000) = (1/50)*(1000/2-10) = 490/50 = \$9.80
L(1000) = (49/50)*(10) = 490/50 = \$9.80

E(1000) = W(1000) - L(1000) = \$9.80 - \$9.80 = 0

Since W(P) increases linearly with increasing P, pot sizes above \$1000 are increasingly attractive to the player. However in order to get to a pot size above \$1000, a minimum of 99 players must have gone first and lost (i.e. you, as the 100th player, will add the last \$10 needed to bring the pot size above \$1000 since the pot is originally seeded with \$10). The probability of this happening is:

P(no wins) = (49/50)^99 = 13.5%

The probability of having this happen and then having you win on your turn is:

P(no wins, then you win) = (49/50)^99 * (1/50) = 0.27%

This is pretty bleak. However, even worse is the fact that both the expected value and the expected winnings are never positive for the first player!

W(20) = (1/50)*(20/2-10) = 0

E(20) = (1/50)*(20/2-10) - (49/50)*(10) = -490/50 = -\$9.80

Based on this fact alone, we can conclusively state the location of this game. Where are we?
5. Palynka
Upward Spiral
26 Oct '10 14:14
Originally posted by PBE6
The expected winnings W for any given pot size P is:

W(P) = (1/50)*(P/2-10) <--it's (P/2-10) because you have to spend \$10 to play

The expected losses L for any given pot size P is:

L(P) = (49/50)*(10)

The expected value E for any given pot size P is just the expected winnings minus the expected losses:

E(P) = (1/50)*(P/2-10) - (49/50)*(10)

F ...[text shortened]... on this fact alone, we can conclusively state the location of this game. [b]Where are we?
[/b]
Are you doing this with replacement?
6. PBE6
Bananarama
26 Oct '10 17:381 edit
Originally posted by Palynka
Are you doing this with replacement?
Yes. If the number of people exceeds the number of cards then doing it with replacement is the only way to go, but I suppose it would be more practical to do it without replacement for smaller groups.
7. forkedknight
Defend the Universe
26 Oct '10 18:12
Originally posted by PBE6
Yes. If the number of people exceeds the number of cards then doing it with replacement is the only way to go, but I suppose it would be more practical to do it without replacement for smaller groups.
I read the problem to mean there was no replacement, but your analysis looks correct if you assume there is.
8. PBE6
Bananarama
26 Oct '10 18:25
Originally posted by forkedknight
I read the problem to mean there was no replacement, but your analysis looks correct if you assume there is.
Yeah, even though it's not specific I think the "no replacement" interpretation is more believable. Having said that, in either case it's going to be difficult to get anyone start playing as the first player will break even at best and lose \$10 otherwise.

If that's the case, I contend that we are not at a chess club but at a _____. Anyone care to hazard a guess?
9. 26 Oct '10 22:141 edit
Assuming the losing cards are replaced in the hat:

If there is \$1450 in the pot and you play:
average amount won = -490/50 + 1460/100 = \$4.80

If you wait one turn (and you get to play) then the average amount you will win is:
49/50*(-490/50 + 1470/100) = \$4.802

if you wait two turns then the average amount you win is"
(49^2)/(50^2)*(-490/50 + 1480/100) = \$4.802

if you wait three turns, then the amount you win is
(49^3)/(50^3)*(-490/50 + 1490/100) = \$4.8000792

So you should go when there is \$1460 or \$1470 in the pot, i.e. on turn 146 or 147. If it actually gets to that turn and you bet, then on go 146 you would make \$4.90 on average and on go 147 you would make \$5.00 on average, but after that the increase in expected wnnings does not offset the chance of the next person taking the pot.
10. PBE6
Bananarama
27 Oct '10 15:17
No one wants to make a guess as to where we are? OK, here's the answer:

A. We're at a stag-and-doe!! That's the only place where gamblers spend money knowing full well that there is no hope of a positive return.
11. uzless
The So Fist
28 Oct '10 02:402 edits
Originally posted by PBE6
Yeah, even though it's not specific I think the "no replacement" interpretation is more believable. Having said that, in either case it's going to be difficult to get anyone start playing as the first player will break even at best and lose \$10 otherwise.

If that's the case, I contend that we are not at a chess club but at a _____. Anyone care to hazard a guess?
The question should be read as once a card is drawn it is not replaced...ie now there are only 49 cards left and your chances of picking the card increase.

Also, the first person that draws a card is obviously a maple leaf fan because even though there is 10 bux in the pot to start with, and even though we add his additional 10 bux to the pot to make 20, he only gets half of 20, meaning 10 bux, meaning he only gets his money back.

Why would he do this? Because he's a leaf fan obviously.

But, let's keep in mind this is a fundraiser so the odds of course are not expected to be in your favour.

To make the question more interesting perhaps we should say the pot starts with 100 bux in it.
12. PBE6
Bananarama
28 Oct '10 14:16
Originally posted by uzless
Also, the first person that draws a card is obviously a maple leaf fan because even though there is 10 bux in the pot to start with, and even though we add his additional 10 bux to the pot to make 20, he only gets half of 20, meaning 10 bux, meaning he only gets his money back.
This must mean something in French.
13. 28 Oct '10 16:32
Originally posted by PBE6
This must mean something in French.
That's what I thought of Shirov's game in Hoogeveen this afternoon. Alas for him, it was not French, it was a Grünfeld, and he lost it.

Richard
14. uzless
The So Fist
28 Oct '10 21:26
Originally posted by PBE6
This must mean something in French.
The pot has 10 bux in it.

The first person wants to draw a card so he pays his 10 bux and then takes a card.

The pot now has 20 bux in it. (10+10= 20)

He discovers he's drawn the winning card and is entitled to half the pot.

Half of 20 is 10.

He "wins" 10 bux. The same amount he paid.

Obviously a maple leaf fan.

Vous comprendez maintenant?
15. forkedknight
Defend the Universe
28 Oct '10 21:49
Originally posted by uzless
The pot has 10 bux in it.

The first person wants to draw a card so he pays his 10 bux and then takes a card.

The pot now has 20 bux in it. (10+10= 20)

He discovers he's drawn the winning card and is entitled to half the pot.

Half of 20 is 10.

He "wins" 10 bux. The same amount he paid.

Obviously a maple leaf fan.

Vous comprendez maintenant?
How do things change if you only pay the pot after you lose? i.e. The pot initially has \$10 in it. If you win, you get \$5, if you lose, you pay \$10.

For person #2, the pot has \$20 in it. If you win, you get \$10, if you lose, you pay \$10.

etc.