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06-13-2019 , 04:24 PM
Once you get picked for the game, they deal you 2 cards face up and you have the option of stopping or continuing until you get to 5th card.
2nd card you win 100
3rd card you win 200
4th card you win 400
5th card win you 20k.

Each card must lower in rank from Ace to 2. You lose if tie.

What are the odds of making it to the 5th card?

Extreme bonus points: EV of being picked for promotion?
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06-13-2019 , 04:45 PM
Lots of combos of card ordering possible, this would require an answer from a computer simulation.

At first, they deal you 2 cards in a sequence I guess, and you're losing if second rank is equal or higher than first? Also, for EV you would need to know your stake, is it 100?
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06-13-2019 , 05:12 PM
This is what I got:
1 : 52.9688 %
2 : 33.238099999999996 %
3 : 10.9706 %
4 : 2.3954 %
5 : 0.4271 %

1 means you fail at second card, in this case you get nothing.
2 means you fail at third card, and so on... finally, 5 means you got all 5 right.
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06-13-2019 , 05:13 PM
They deal 2 cards at 1st, 1 at time after if you continue. Yes you lose if next card ties or is higher.

There's no stake to know. Free casino promotional game
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06-13-2019 , 05:54 PM
So, if you lose at 4th card, do you win money for first 3 or nothing at all? If you lose all, that would complicate things, since then current sequence of cards would be very important, it would be a bad choice to draw one more in some cases.

Still, apriori probability of 0.43% to get all 5 right holds, but getting there from A7 is much harder than from AQ. Game would be much more dynamic if you can lose everything when you get a wrong rank, you would need EV calc on every street for each deck config as you'd be playing.
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06-13-2019 , 05:58 PM
After the 4th card win, do you have an option to stop?
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06-13-2019 , 06:04 PM
You draw for card #4 and lose, you lose previously "won" money. Option to stop after cards 2-4. But if you ever got to 4 you're suppose to always go for it because card 5 gets you a 50 times increase from card 4

I wasn't sure how possible it would be to get the EV of the game. Was pretty sure there was someone on here smarter than me who could get possibility of 5th card.
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06-13-2019 , 06:12 PM
Actually, this might even be easier to solve on paper than I thought since for each position it just matters what is the current rank. For example, KJ5 is equal to 765 in terms of completing the sequence.
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06-13-2019 , 06:44 PM
Quote:
Originally Posted by MMMed13
You draw for card #4 and lose, you lose previously "won" money. Option to stop after cards 2-4. But if you ever got to 4 you're suppose to always go for it because card 5 gets you a 50 times increase from card 4
.
But if card 4 was a 2, forcing you to draw card 5 means you lose all.

Suppose card 4 was a 3. Then there are 4 winning cards for 20K and 44 losing cards to lose 400. This has EV= 1300 but on a utility basis drawing may not appeal to everybody- that is the choice for a 100% chance to win 400 versus a 8.3% chance for 20K depends on one's utility function
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06-13-2019 , 07:36 PM
Echoing what others have said, it shouldn't be hard to come up with the optimal strategy for this game to maximize EV given your cards so far.
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06-13-2019 , 08:55 PM
Quote:
Originally Posted by statmanhal
But if card 4 was a 2, forcing you to draw card 5 means you lose all.

Suppose card 4 was a 3. Then there are 4 winning cards for 20K and 44 losing cards to lose 400. This has EV= 1300 but on a utility basis drawing may not appeal to everybody- that is the choice for a 100% chance to win 400 versus a 8.3% chance for 20K depends on one's utility function
I didn't think I needed to point out the 2 on the turn. They might not even let you draw.

Utility vs EV argument makes sense but was just talking in terms of EV
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06-14-2019 , 10:57 PM
We can define V(k,C) as the value obtained applying optimal strategy of being at round k (k=2,3,4) with C being the last card drawn. Next, we reason backward and start with k = 4. In order to find the optimal strategy we compare the expected value of the prize we obtain by drawing with the one we get if we don't draw and choose of course the maximum value.

As already stated, at round 4 we should draw with everything but a 2 (notice that we might have at most a Jack at round 4). So we have for instance that V(4,2) = 400 (we don't draw), while V(J,4) = 15000 (we draw and we have 3/4 of winning the big prize).

Next, we analyze step 3, and again we compare how much likely is to pass from (3,C) to (4,D) (D<C), where we know how much is the expected winning. Again, we compare the draw vs don't draw options. Eventually, we calculate V(3,C) for every C and, in turn, we also can calculate V(2,C). This is what I get (alert: I might have made some mistake along the path).

Code:
 card    value2    value3    value4
    A    0.0000    0.0000     0.000
    K 1853.9320    0.0000     0.000
    Q 1361.5238 6155.1020     0.000
    J  967.0748 4930.6122 15000.000
    T  659.7007 3842.1769 13333.333
    9  428.5170 2889.7959 11666.667
    8  262.6395 2073.4694 10000.000
    7  151.1837 1393.1973  8333.333
    6  100.0000  848.9796  6666.667
    5  100.0000  440.8163  5000.000
    4  100.0000  200.0000  3333.333
    3  100.0000  200.0000  1666.667
    2  100.0000  200.0000   400.000
When you see a value in the table above greater than the prize of that round, it means that you should draw. At round 4, you draw everything but a 2. At round 3, you draw 5 or higher and at round 2 you draw 7 or higher.
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