Pick a Card...Any Card

first lets get one thing clear. The "optimum" time to bet, will be defined as the first moment that it pays off to bet. Else it would simply be when there is one card left


well this offers an equation system with 3 equations and 3 variables

let:

C = chance of winning the pot
P = The amount of bux in the pot
X = amount of cards picked

Chance is given as:

C = 1/(50-X)

Pot will be:

P = 10X + 10

We need to average 10 bux with each bet so the first time which it would pay off to bet would be the first time we win over 10 dollars on average

therefor:

P * S = 10

(winnings for P *S >10)

so we insert and get:
(10X+10)*1/(50-X)=10
(10X+10)/(50-X)=10
10X+10 = 10(50-X)
10X+10 = 500-10X
20X=490
X = 24.5

so we get condition: P*S = 10 for X=24.5

we needed a winning so: P*S > 10 for X>24.5

first time X>24.5 is when X=25

so answer would be when 25 cards have been picked

Well I would wait for the 49th card. If the card has been picked out before that, tough, but you don't lose 10 bux. If it's still there you have an even chance of picking it out of the two cards left, to win 245 bux, which is a great bet!

Originally posted by wishboneash
Well I would wait for the 49th card. If the card has been picked out before that, tough, but you don't lose 10 bux. If it's still there you have an even chance of picking it out of the two cards left, to win 245 bux, which is a great bet!

If you take card 49, average winnings are (500/2-10)/2 = 120

If you wait until card 50 your average winnings are (510/2)/2 = 127.50

Its always better to wait until the very last card if there is only £10 in the pot at first, and the losing cards are taken out.

when you pick a card do you keep it out of the hat .so that there are 49 cards left .until at the end there may be only one card left that could be the winner.

Originally posted by uzless
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.
....
To make the question more interesting perhaps we should say the pot starts with 100 bux in it.

If the pot starts with P money in it, and each player pays £10, getting half the pot back if he wins then:

If you wait until after go G

The chance of the pot lasting until after G is

49/50 * 48/49 * 47/48 *.... 49!/(49-G)! / (50!/(50-G)!) = (50-G)/50

If you go just after go G there is P + 10(G+1) in the pot, giving you potential winnings of
P/2 + 5(G+1)

If you go after G, there are 50-G cards in the pot, so your chance of winning is 1/(50-G) and you chance of losing your stake is (50-G-1)/(50-G)

Putting all this together, your expected winnings, if you wait until just after go g are:

(50-G)/50 * ((P/2 + 5(G+1))/(50-G) -10(50-G-1)/(50-G))

= (P/2 + 5(G+1) -500 +10G + 1)/50

= (P/2 + 15G -494) / 50

As G is positive in this expression, your winnings are maximized by waiting until the last go, whatever the money that is initially in the pot.

Originally posted by iamatiger
If you take card 49, average winnings are (500/2-10)/2 = 120

If you wait until card 50 your average winnings are (510/2)/2 = 127.50

Its always better to wait until the very last card if there is only £10 in the pot at first, and the losing cards are taken out.

If you wait till card 50 you are certain to win, otherwise there wouldn't be a card 50. But the person who takes card 49 still has an even money bet to win the money for a stake of 10bux, and he would win half of 49x10 bux or 245 bux or am I missing something?

ANSWER: Once 49 people have selected a card, pick a card.

It sounds to me like this is essentially a freeroll since you never actually have to commit your $10 until you want to pick. So just wait for a time when 49 people get it wrong, volunteer to pick next and you win all the money.

More specifically, I mean that as each person puts money in and loses, it becomes more an more +EV to play. It is neutral EV at the point in which the ratio of money in the pot is equivalent to the ratio of wins to losses in the hat.

On the first go, there is a 1/50 shot at winning $20 (if you paid) which is worth $0.40, but it costs you $10 to do so, so the total expectation is $0.40 - $10.00 = -$9.60. If you go second, the p(win) = 1/49 and the pot is $30 (if you paid) which is worth $0.61, but it costs you $10, so the total expectation is -$9.39.

You can work this out for every instance as I have done below (rounded to two digits):


Trial Wins Losses p(win) cost pot EV
1 1.00 49.00 0.02 10.00 10.00 -9.60
2 1.00 48.00 0.02 10.00 20.00 -9.39
3 1.00 47.00 0.02 10.00 30.00 -9.17
4 1.00 46.00 0.02 10.00 40.00 -8.94
5 1.00 45.00 0.02 10.00 50.00 -8.70
6 1.00 44.00 0.02 10.00 60.00 -8.44
7 1.00 43.00 0.02 10.00 70.00 -8.18
8 1.00 42.00 0.02 10.00 80.00 -7.91
9 1.00 41.00 0.02 10.00 90.00 -7.62
10 1.00 40.00 0.02 10.00 100.00 -7.32
11 1.00 39.00 0.03 10.00 110.00 -7.00
12 1.00 38.00 0.03 10.00 120.00 -6.67
13 1.00 37.00 0.03 10.00 130.00 -6.32
14 1.00 36.00 0.03 10.00 140.00 -5.95
15 1.00 35.00 0.03 10.00 150.00 -5.56
16 1.00 34.00 0.03 10.00 160.00 -5.14
17 1.00 33.00 0.03 10.00 170.00 -4.71
18 1.00 32.00 0.03 10.00 180.00 -4.24
19 1.00 31.00 0.03 10.00 190.00 -3.75
20 1.00 30.00 0.03 10.00 200.00 -3.23
21 1.00 29.00 0.03 10.00 210.00 -2.67
22 1.00 28.00 0.03 10.00 220.00 -2.07
23 1.00 27.00 0.04 10.00 230.00 -1.43
24 1.00 26.00 0.04 10.00 240.00 -0.74
25 1.00 25.00 0.04 10.00 250.00 0.00
26 1.00 24.00 0.04 10.00 260.00 0.80
27 1.00 23.00 0.04 10.00 270.00 1.67
28 1.00 22.00 0.04 10.00 280.00 2.61
29 1.00 21.00 0.05 10.00 290.00 3.64
30 1.00 20.00 0.05 10.00 300.00 4.76
31 1.00 19.00 0.05 10.00 310.00 6.00
32 1.00 18.00 0.05 10.00 320.00 7.37
33 1.00 17.00 0.06 10.00 330.00 8.89
34 1.00 16.00 0.06 10.00 340.00 10.59
35 1.00 15.00 0.06 10.00 350.00 12.50
36 1.00 14.00 0.07 10.00 360.00 14.67
37 1.00 13.00 0.07 10.00 370.00 17.14
38 1.00 12.00 0.08 10.00 380.00 20.00
39 1.00 11.00 0.08 10.00 390.00 23.33
40 1.00 10.00 0.09 10.00 400.00 27.27
41 1.00 9.00 0.10 10.00 410.00 32.00
42 1.00 8.00 0.11 10.00 420.00 37.78
43 1.00 7.00 0.13 10.00 430.00 45.00
44 1.00 6.00 0.14 10.00 440.00 54.29
45 1.00 5.00 0.17 10.00 450.00 66.67
46 1.00 4.00 0.20 10.00 460.00 84.00
47 1.00 3.00 0.25 10.00 470.00 110.00
48 1.00 2.00 0.33 10.00 480.00 153.33
49 1.00 1.00 0.50 10.00 490.00 240.00
50 1.00 0.00 1.00 10.00 500.00 500.00

The game becomes exactly break even when there are 25 losers in the hat and 1 winner because the pot will contain $250 + $10 for you to play. So the ratio of risk to reward is $10😳250 and the ratio of wins to losses is 1:25.

Originally posted by wishboneash
If you wait till card 50 you are certain to win, otherwise there wouldn't be a card 50. But the person who takes card 49 still has an even money bet to win the money for a stake of 10bux, and he would win half of 49x10 bux or 245 bux or am I missing something?

Not really, just that he only wins 235 bux because 10 bux of his win was his stake, and a dead cert to win 245 on go 50 is better than a 50% chance of winning 235 bux on go 49.

I think the ev on go 50 is too high there (you only win half of the pot) in fact all of the evs look to have the same problem perhaps, this might change your break even point a bit.