Red and Black Puzzle
Poll
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10 members have voted
January 29th, 2020 at 10:46:27 AM
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A casino game has the following rules:
1. A 52-card deck is used.
2. Cards are dealt one at a time, without replacement, until the player chooses to end the game or the end of the deck is reached.
3. Every time a black card is dealt, the player must pay $1.
4. Every time a red card is dealt, the player wins $1.
5. The player may quit and walk away with his winnings (if any) at any point.
The question is, what is a fair price to pay for this game?
Fair warning that I can't think of a pure mathematical solution. I did mine in Excel.
Usual rules apply. I'd like to stress the 24-hold hold for previous winners of a beer.
The question for the poll is what do you think of the puzzle?
1. A 52-card deck is used.
2. Cards are dealt one at a time, without replacement, until the player chooses to end the game or the end of the deck is reached.
3. Every time a black card is dealt, the player must pay $1.
4. Every time a red card is dealt, the player wins $1.
5. The player may quit and walk away with his winnings (if any) at any point.
The question is, what is a fair price to pay for this game?
Fair warning that I can't think of a pure mathematical solution. I did mine in Excel.
Usual rules apply. I'd like to stress the 24-hold hold for previous winners of a beer.
The question for the poll is what do you think of the puzzle?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 29th, 2020 at 11:07:08 AM
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This doesn't seem like a winner for the casino. If you play all the way to the end, you break even every time. Thus, you could just stop playing at any point you are ahead. If you fall behind, or never get ahead, just play all the way through to the end and try again on a new deck.
That's like a player advantage of what, 100%? Because you can't lose... only break even at worst....
Or am I totally misunderstanding the game?
That's like a player advantage of what, 100%? Because you can't lose... only break even at worst....
Or am I totally misunderstanding the game?
EDIT: Not sure if I was supposed to spoilerize that or not.
January 29th, 2020 at 11:11:24 AM
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Quote: TigerWuThis doesn't seem like a winner for the casino. If you play all the way to the end, you break even every time. Thus, you could just stop playing at any point you are ahead. If you fall behind, or never get ahead, just play all the way through to the end and try again on a new deck.
That's like a player advantage of what, 100%? Because you can't lose... only break even at worst....
Or am I totally misunderstanding the game?
I agree with this answer. The fair price to play the game under the initial set of conditions is $1 because the red cards and black cards are equally likely. I agree also with the point that the player literally cannot lose as long as he has $27 (could theoretically---almost impossible) lose the first 26 in a row and would need a dollar to place the first bet for the guaranteed cards of the opposite color.
If the player ever gets ahead, then the player should stop playing immediately or, in the alternative, not play until it is a completely new deck. Any further Even Money bets would be at a disadvantage until the deck evens out again.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
January 29th, 2020 at 12:26:47 PM
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Worst case scenario is you play the whole deck and break even. But sometimes you get to quit while you're ahead $x. Seems too complicated to work out the strategy without a computer. Even with a computer, I don't know how to work out the depth-dependent strategy, but for a simple flat quitting point you would quit at $4 for an EV of $2.18. If there was only 1 card left you would quit at $1, so the quitting point is lower towards the end of the shoe. I'd have to work backwards from there because the strategy at card N depends on the strategies at cards N+1 to 52.
the flat quitting point source code:
the flat quitting point source code:
from random import shuffle
reps = 1000000
deck = [-1]*26 + [1]*26
timesreached = dict(zip(range(27),[0]*27))
for rep in range(reps):
shuffle(deck)
runningtotal = 0
isreached = dict(zip(range(27),[0]*27))
for card in deck:
runningtotal += card
if runningtotal > 0:
isreached[runningtotal] = 1
for point, bit in isreached.iteritems():
timesreached[point] += bit
for point, times in timesreached.iteritems():
EV = point * (times/float(reps))
print("$%s quit = $%.3f ev" % (point, EV))
January 29th, 2020 at 12:42:43 PM
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Isn't this game similar to Faro? The game is easy to back count with a huge player EV if mid-deck entry is allowed.
For a house edge, what if the player picks a color that pays prior to starting, and the player loses $1.25 when a face card of the other color rolls, and wins $1.50 when the Ace of the selected color rolls, with all other cards of the selected color pay $1.
For a house edge, what if the player picks a color that pays prior to starting, and the player loses $1.25 when a face card of the other color rolls, and wins $1.50 when the Ace of the selected color rolls, with all other cards of the selected color pay $1.
Simplicity is the ultimate sophistication - Leonardo da Vinci
January 29th, 2020 at 4:27:10 PM
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Quote: WizardThe question is, what is a fair price to pay for this game?
There seems to be a misunderstanding about what I'm asking.
Yes, I agree if you could play this for free it would be a freeroll with the least bit of strategy. If you're negative, just wait until the end of the deck and you'll break even. It's easy to say the player should never settle for a loss (absent the fee to play).
However, I'm asking what would you pay to play for such a "freeroll." Think of literally a carnival game, like throwing a ball in a basket for a teddy bear. You pay a price and then it switches to another phase where you either win something or you don't.
In this case, the fair price is clearly more than $0, because the player could just wait to be up $1 and quit. If he's never up, then wait to the end and you push. However, it must also be less than $26, because there is less than a 100% chance of drawing all 26 red cards in the first 26 draws. The question, is where is that fair price between $0 and $26?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 29th, 2020 at 5:31:27 PM
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This isn't an answer, so I hope I'm not breaking the 24-hour rule, but I have a feeling there is no "simple solution" to this, any more than there is one to the Secretaries' (aka the Sultan's Dowry) Problem.
January 29th, 2020 at 6:56:18 PM
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My math skills peaked years ago. So here are my random thoughts...
The worst case scenario is you lose your 'entry fee'.
Without any proof, I feel the best strategy would be to wait until you approach the end of the deck. If you are up $4 with 6 cards to go, you know five of six cards are black, and maybe you stop there.
With one card left you have a 50% chance of winning $1 less fee, else lose fee
With two cards left you have a Y% chance of winning $2 less fee, where Y is the probability all 26 of the red cards are drawn in the first 50
With three cards left you have a Z% chance of winning $3 less fee, where Z is the probability all 26 of the red cards are drawn in the first 49
and so on
Another strategy to consider, to your advantage you have the knowledge as you go through the deck where you stand. Perhaps you hit a streak and happen to be up $8, maybe you stop there. What is the probability of being up $X at any time, where $1 <= X <= $26 At what value of X should you stop, knowing how many cards remain?
What is the fair price to play the game?
It should be more than 50 cents. But how much more?
I believe it should be less than $1. But how much less?
Good problem.
The worst case scenario is you lose your 'entry fee'.
Without any proof, I feel the best strategy would be to wait until you approach the end of the deck. If you are up $4 with 6 cards to go, you know five of six cards are black, and maybe you stop there.
With one card left you have a 50% chance of winning $1 less fee, else lose fee
With two cards left you have a Y% chance of winning $2 less fee, where Y is the probability all 26 of the red cards are drawn in the first 50
With three cards left you have a Z% chance of winning $3 less fee, where Z is the probability all 26 of the red cards are drawn in the first 49
and so on
Another strategy to consider, to your advantage you have the knowledge as you go through the deck where you stand. Perhaps you hit a streak and happen to be up $8, maybe you stop there. What is the probability of being up $X at any time, where $1 <= X <= $26 At what value of X should you stop, knowing how many cards remain?
What is the fair price to play the game?
It should be more than 50 cents. But how much more?
I believe it should be less than $1. But how much less?
Good problem.
January 29th, 2020 at 8:41:30 PM
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Let me add this, I think the optimal strategy depends on both how much money you have accumulated and total cards played. For example, if it's early in the game and you're $5 up, you have more of a chance to go higher than if you're up $5 late in the game.
Keep an eye on the time of the OP. I think past winners will jump in after the 24 hours. Where is OD?
Keep an eye on the time of the OP. I think past winners will jump in after the 24 hours. Where is OD?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 29th, 2020 at 8:48:26 PM
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the complete depth-dependent solution
Let EV(n,k) be the EV of the state where there are n cards left and you are up k dollars. The question is asking for EV(52,0)
EV(0,0) = 0
EV(1,1) = 1
EV(1,-1) = 0
Let R and B be the number of red and black cards remaining.
R+B = n
R-B = k
R = (n+k)/2
B = (n-k)/2
pb = (n-k)/(2n)
pr = (n+k)/(2n)
EV(n,k) = max(k, pb*EV(n-1,k+1) + pr*EV(n-1,k-1))
So I see how wiz did it with a spreadsheet but I'll write a program instead.
I get an EV of 3.5128 with the following strategy:
quit at $7 with 43 or more cards remaining
quit at $6 with 31-42 cards remaining
quit at $5 with 21-30 cards remaining
quit at $4 with 13-20 cards remaining
quit at $3 with 7-12 cards remaining
quit at $2 with 2-6 cards remaining (the program says $3 with 3 cards left but the only possibilities there are $1 and $3 so a quit point of $2 is the same thing)
quit at $1 with 1 card remaining
source code at pastebin / gNuyXzA2
Let EV(n,k) be the EV of the state where there are n cards left and you are up k dollars. The question is asking for EV(52,0)
EV(0,0) = 0
EV(1,1) = 1
EV(1,-1) = 0
Let R and B be the number of red and black cards remaining.
R+B = n
R-B = k
R = (n+k)/2
B = (n-k)/2
pb = (n-k)/(2n)
pr = (n+k)/(2n)
EV(n,k) = max(k, pb*EV(n-1,k+1) + pr*EV(n-1,k-1))
So I see how wiz did it with a spreadsheet but I'll write a program instead.
I get an EV of 3.5128 with the following strategy:
quit at $7 with 43 or more cards remaining
quit at $6 with 31-42 cards remaining
quit at $5 with 21-30 cards remaining
quit at $4 with 13-20 cards remaining
quit at $3 with 7-12 cards remaining
quit at $2 with 2-6 cards remaining (the program says $3 with 3 cards left but the only possibilities there are $1 and $3 so a quit point of $2 is the same thing)
quit at $1 with 1 card remaining
source code at pastebin / gNuyXzA2
January 29th, 2020 at 9:58:17 PM
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I think I have finally come up with a solution to this, and to do that, I'm going to cheat. If the player wants a red card, I'm going to cheat and look at the bottom of the deck. If the bottom of the deck is a black card, then I know the player wins $1 at some point and quits.
Last Card Black: .5
We know that 50% of the time the last card will be red, which is the only condition under which our player does not win automatically. In order for the dealer side to break even with the 51 remaining cards, we need all 26 cards that are good for the dealer to come out before the other side ever leads by one. In other words, if all 26 of the player's cards come out before 25 dealer cards do, then the player wins.
nCr(26,26)*nCr(25,24)/nCr(51,50) = 0.4901960784313725 * .5 = 0.24509803921
We know that (26/52*25/51) =0.24509803921 of the time, the last two cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 24 cards that are bad for the player do.
nCr(26,26)*nCr(24,23)/nCr(50,49) = 0.48 * 0.24509803921 = 0.11764705882
We know that (26/52*25/51*24/50) = 0.11764705882 of the time, the last three cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 23 cards that are bad for the player do.
nCr(26,26)*nCr(23,22)/nCr(49,48) = 0.4693877551020408 * 0.11764705882 = 0.05522208883
We know that (26/52*25/51*24/50*23/49) = 0.05522208883 of the time, the last four cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 22 cards that are bad for the player do.
nCr(26,26)*nCr(22,21)/nCr(48,47) = 0.4583333333333333 * 0.05522208883 = 0.02531012404
We know that (26/52*25/51*24/50*23/49*22/48) = 0.02531012404 of the time, the last five cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 21 cards that are bad for the player do.
nCr(26,26)*nCr(21,20)/nCr(47,46) = 0.4468085106382979 * 0.02531012404 = 0.01130877882
We know that (26/52*25/51*24/50*23/49*22/48*21/47) = 0.01130877882 of the time, the last six cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 20 cards that are bad for the player do.
nCr(26,26)*nCr(20,19)/nCr(46,45) = 0.4347826086956522 * 0.01130877882 = 0.00491686035
Running Total: .5+0.24509803921+0.11764705882+0.05522208883+0.02531012404+0.01130877882+0.00491686035 = 0.95950295007
We know that (26/52*25/51*24/50*23/49*22/48*21/47*20/46) = 0.00491686035 of the time, the last seven cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 19 cards that are bad for the player do.
nCr(26,26)*nCr(19,18)/nCr(45,44) = 0.4222222222222222 * 0.00491686035 = 0.0020760077
0.0020760077 + 0.95950295007 = 0.96157895777
We know that (26/52*25/51*24/50*23/49*22/48*21/47*20/46*19/45) = 0.0020760077 of the time, the last eight cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 18 cards that are bad for the player do.
nCr(26,26)*nCr(18,17)/nCr(44,43) = 0.4090909090909091 * 0.0020760077 = 0.00084927587
0.00084927587+0.96157895777 = 0.96242823364
We know that (26/52*25/51*24/50*23/49*22/48*21/47*20/46*19/45*18/44) = 0.00084927587 of the time, the last nine cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 17 cards that are bad for the player do.
nCr(26,26)*nCr(17,16)/nCr(43,42) = 0.3953488372093023 * 0.00084927587 = 0.00033576022
0.00033576022 + 0.96242823364 = 0.96276399386
We know that (26/52*25/51*24/50*23/49*22/48*21/47*20/46*19/45*18/44*17/43) = 0.00033576023 of the time, the last ten cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 16 cards that are bad for the player do.
nCr(26,26)*nCr(16,15)/nCr(42,41) = 0.380952380952381*0.00033576022 = 0.00012790865
0.00012790865+0.96276399386 = 0.96289190251
We know that 0.00012790865 of the time, the last 11 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 15 cards that are bad for the player do.
nCr(26,26)*nCr(15,14)/nCr(41,40) = 0.3658536585365854* 0.00012790865 = 0.00004679584
0.96289190251+0.00004679584 = 0.96293869835
We know that 0.00004679584 of the time, the last 12 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 14 cards that are bad for the player do.
nCr(26,26)*nCr(14,13)/nCr(40,39) = 0.35*0.00004679584 = 0.00001637854
0.00001637854+0.96293869835= 0.96295507689
We know that 0.00001637854 of the time, the last 13 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 13 cards that are bad for the player do.
nCr(26,26)*nCr(13,12)/nCr(39,38) = 0.3333333333333333* 0.00001637854 = 0.00000545951
0.00000545951+0.96295507689=0.9629605364
We know that 0.00000545951 of the time, the last 14 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 12 cards that are bad for the player do.
nCr(26,26)*nCr(12,11)/nCr(38,37) = 0.3157894736842105*0.00000545951 = 0.00000172405
0.00000172405+0.9629605364 = 0.96296226045
We know that 0.00000172405 of the time, the last 15 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 11 cards that are bad for the player do.
nCr(26,26)*nCr(11,10)/nCr(37,36) = 0.2972972972972973*0.00000172405 = 0.000000512555405405405410065
0.000000512555405405405410065+0.96296226045=0.962962773005405405405410065
We know that 0.000000512555405405405410065 of the time, the last 16 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 10 cards that are bad for the player do.
nCr(26,26)*nCr(10,9)/nCr(36,35) = 0.2777777777777778* 0.000000512555405405405410065 = 0.000000142376501501501514185953453453453557
0.000000142376501501501514185953453453453557+0.962962773005405405405410065 = 0.962962915381906906906924250953453453453557
We know that 0.000000142376501501501514185953453453453557of the time, the last 17 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 9 cards that are bad for the player do.
nCr(26,26)*nCr(9,8)/nCr(35,34) = 0.2571428571428571* 0.000000142376501501501514185953453453453557 = 0.0000000366111003861
0.0000000366111003861+0.962962915381906906906924250953453453453557 = 0.962962951993007293006924250953453453453557
We know that 0.0000000366111003861 of the time, the last 18 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 8 cards that are bad for the player do.
nCr(26,26)*nCr(8,7)/nCr(34,33) = 0.2352941176470588* 0.0000000366111003861 = 0.000000008614376561
0.000000008614376561+0.962962951993007293006924250953453453453557 = 0.962962960607383854006924250953453453453557
We know that 0.000000008614376561 of the time, the last 19 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 7 cards that are bad for the player do.
nCr(26,26)*nCr(7,6)/nCr(33,32) = 0.2121212121212121*0.000000008614376561 = 0.000000001827292
0.000000001827291+0.962962960607383854006924250953453453453557 = 0.962962962434674854006924250953453453453557
We know that 0.000000001827292 of the time, the last 20 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 6 cards that are bad for the player do.
nCr(26,26)*nCr(6,5)/nCr(32,31) = 0.1875* 0.000000001827292 = 0.00000000034261725
0.00000000034261725+0.962962962434674854006924250953453453453557 = 0.962962962777292104006924250953453453453557
We know that 0.00000000034261725 of the time, the last 21 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 5 cards that are bad for the player do.
nCr(26,26)*nCr(5,4)/nCr(31,30) = 0.1612903225806452* 0.00000000034261725 = 0.0000000000552608467741935616497
0.0000000000552608467741935616497+0.962962962777292104006924250953453453453557 = 0.962962962832552950781117812603153453453557
We know that 0.0000000000552608467741935616497 of the time, the last 22 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 4 cards that are bad for the player do.
nCr(26,26)*nCr(4,3)/nCr(30,29) = 0.1333333333333333*0.0000000000552608467741935616497 = 0.00000000000736811290322580637793177419354794501
0.00000000000736811290322580637793177419354794501+0.962962962832552950781117812603153453453557 = 0.96296296283992106368434361898108522764710494501
We know that 0.00000000000736811290322580637793177419354794501 of the time, the last 23 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 3 cards that are bad for the player do.
nCr(26,26)*nCr(3,2)/nCr(29,28) = 0.103448275862069* 0.00000000000736811290322580637793177419354794501 = 0.00000000000076221857619577332765200778642931768833704115682569
0.00000000000076221857619577332765200778642931768833704115682569+0.96296296283992106368434361898108522764710494501
=0.96296296284068328226053939230873723543353426269833704115682569
We know that 0.00000000000076221857619577332765200778642931768833704115682569 of the time, the last 24 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 2 cards that are bad for the player do.
nCr(26,26)*nCr(2,1)/nCr(28,27) = 0.0714285714285714*0.00000000000076221857619577332765200778642931768833704115682569 = 0.0000000000000544441840139837873403269505799990448238518989356153
0.0000000000000544441840139837873403269505799990448238518989356153+0.96296296284068328226053939230873723543353426269833704115682569=0.9629629628407377264445533760960775623841142617431608930557613053
We know that 0.0000000000000544441840139837873403269505799990448238518989356153 of the time, the last 25 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 1 card that is bad for the player does.
nCr(26,26)*nCr(1,0)/nCr(27,26) = 0.037037037037037* 0.0000000000000544441840139837873403269505799990448238518989356153 = 0.0000000000000020164512597771752924497383924522853517370858865396
0.0000000000000020164512597771752924497383924522853517370858865396+0.9629629628407377264445533760960775623841142617431608930557613053=
0.9629629628407397428958131532713700121225067140285126301416478449
We know that not real G*****N often will all 26 red cards be on the bottom, but when they are, the 26 black cards are on top and the player automatically ties, but temporarily has to lose $26 and have a stroke first because something damn near impossible happened.
CONCLUSION
In conclusion, the probability of one side or the other leading at some point is approximately 96.29629628407397428958131532713700121225067140285126301416478449%, because of this, that same amount is also the fair price to pay for the +1/-1 game because that is the probability that the player will be ahead one card. It is always optimal for the player to quit when ahead one card because there is a negative expectation on the following card.
Provided the player has sufficient money to play out the deck, the roughly 3.70370371592602571041868467286299878774932859714873698583521551% probability of the wrong card being flipped first and never relinquishing the lead to the other side (but sometimes being tied) is irrelevant because the player does not ever lose $1 (or more) in this scenario.
The player either wins $1 or loses nothing in all cases. The probability of winning a dollar is roughly .9629629284074, the expected win is the same as the probability since our base unit is one dollar. Therefore, the fair value to play this game is a cost of 96.2962984074 cents.
Last Card Black: .5
We know that 50% of the time the last card will be red, which is the only condition under which our player does not win automatically. In order for the dealer side to break even with the 51 remaining cards, we need all 26 cards that are good for the dealer to come out before the other side ever leads by one. In other words, if all 26 of the player's cards come out before 25 dealer cards do, then the player wins.
nCr(26,26)*nCr(25,24)/nCr(51,50) = 0.4901960784313725 * .5 = 0.24509803921
We know that (26/52*25/51) =0.24509803921 of the time, the last two cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 24 cards that are bad for the player do.
nCr(26,26)*nCr(24,23)/nCr(50,49) = 0.48 * 0.24509803921 = 0.11764705882
We know that (26/52*25/51*24/50) = 0.11764705882 of the time, the last three cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 23 cards that are bad for the player do.
nCr(26,26)*nCr(23,22)/nCr(49,48) = 0.4693877551020408 * 0.11764705882 = 0.05522208883
We know that (26/52*25/51*24/50*23/49) = 0.05522208883 of the time, the last four cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 22 cards that are bad for the player do.
nCr(26,26)*nCr(22,21)/nCr(48,47) = 0.4583333333333333 * 0.05522208883 = 0.02531012404
We know that (26/52*25/51*24/50*23/49*22/48) = 0.02531012404 of the time, the last five cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 21 cards that are bad for the player do.
nCr(26,26)*nCr(21,20)/nCr(47,46) = 0.4468085106382979 * 0.02531012404 = 0.01130877882
We know that (26/52*25/51*24/50*23/49*22/48*21/47) = 0.01130877882 of the time, the last six cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 20 cards that are bad for the player do.
nCr(26,26)*nCr(20,19)/nCr(46,45) = 0.4347826086956522 * 0.01130877882 = 0.00491686035
Running Total: .5+0.24509803921+0.11764705882+0.05522208883+0.02531012404+0.01130877882+0.00491686035 = 0.95950295007
We know that (26/52*25/51*24/50*23/49*22/48*21/47*20/46) = 0.00491686035 of the time, the last seven cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 19 cards that are bad for the player do.
nCr(26,26)*nCr(19,18)/nCr(45,44) = 0.4222222222222222 * 0.00491686035 = 0.0020760077
0.0020760077 + 0.95950295007 = 0.96157895777
We know that (26/52*25/51*24/50*23/49*22/48*21/47*20/46*19/45) = 0.0020760077 of the time, the last eight cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 18 cards that are bad for the player do.
nCr(26,26)*nCr(18,17)/nCr(44,43) = 0.4090909090909091 * 0.0020760077 = 0.00084927587
0.00084927587+0.96157895777 = 0.96242823364
We know that (26/52*25/51*24/50*23/49*22/48*21/47*20/46*19/45*18/44) = 0.00084927587 of the time, the last nine cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 17 cards that are bad for the player do.
nCr(26,26)*nCr(17,16)/nCr(43,42) = 0.3953488372093023 * 0.00084927587 = 0.00033576022
0.00033576022 + 0.96242823364 = 0.96276399386
We know that (26/52*25/51*24/50*23/49*22/48*21/47*20/46*19/45*18/44*17/43) = 0.00033576023 of the time, the last ten cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 16 cards that are bad for the player do.
nCr(26,26)*nCr(16,15)/nCr(42,41) = 0.380952380952381*0.00033576022 = 0.00012790865
0.00012790865+0.96276399386 = 0.96289190251
We know that 0.00012790865 of the time, the last 11 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 15 cards that are bad for the player do.
nCr(26,26)*nCr(15,14)/nCr(41,40) = 0.3658536585365854* 0.00012790865 = 0.00004679584
0.96289190251+0.00004679584 = 0.96293869835
We know that 0.00004679584 of the time, the last 12 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 14 cards that are bad for the player do.
nCr(26,26)*nCr(14,13)/nCr(40,39) = 0.35*0.00004679584 = 0.00001637854
0.00001637854+0.96293869835= 0.96295507689
We know that 0.00001637854 of the time, the last 13 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 13 cards that are bad for the player do.
nCr(26,26)*nCr(13,12)/nCr(39,38) = 0.3333333333333333* 0.00001637854 = 0.00000545951
0.00000545951+0.96295507689=0.9629605364
We know that 0.00000545951 of the time, the last 14 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 12 cards that are bad for the player do.
nCr(26,26)*nCr(12,11)/nCr(38,37) = 0.3157894736842105*0.00000545951 = 0.00000172405
0.00000172405+0.9629605364 = 0.96296226045
We know that 0.00000172405 of the time, the last 15 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 11 cards that are bad for the player do.
nCr(26,26)*nCr(11,10)/nCr(37,36) = 0.2972972972972973*0.00000172405 = 0.000000512555405405405410065
0.000000512555405405405410065+0.96296226045=0.962962773005405405405410065
We know that 0.000000512555405405405410065 of the time, the last 16 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 10 cards that are bad for the player do.
nCr(26,26)*nCr(10,9)/nCr(36,35) = 0.2777777777777778* 0.000000512555405405405410065 = 0.000000142376501501501514185953453453453557
0.000000142376501501501514185953453453453557+0.962962773005405405405410065 = 0.962962915381906906906924250953453453453557
We know that 0.000000142376501501501514185953453453453557of the time, the last 17 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 9 cards that are bad for the player do.
nCr(26,26)*nCr(9,8)/nCr(35,34) = 0.2571428571428571* 0.000000142376501501501514185953453453453557 = 0.0000000366111003861
0.0000000366111003861+0.962962915381906906906924250953453453453557 = 0.962962951993007293006924250953453453453557
We know that 0.0000000366111003861 of the time, the last 18 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 8 cards that are bad for the player do.
nCr(26,26)*nCr(8,7)/nCr(34,33) = 0.2352941176470588* 0.0000000366111003861 = 0.000000008614376561
0.000000008614376561+0.962962951993007293006924250953453453453557 = 0.962962960607383854006924250953453453453557
We know that 0.000000008614376561 of the time, the last 19 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 7 cards that are bad for the player do.
nCr(26,26)*nCr(7,6)/nCr(33,32) = 0.2121212121212121*0.000000008614376561 = 0.000000001827292
0.000000001827291+0.962962960607383854006924250953453453453557 = 0.962962962434674854006924250953453453453557
We know that 0.000000001827292 of the time, the last 20 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 6 cards that are bad for the player do.
nCr(26,26)*nCr(6,5)/nCr(32,31) = 0.1875* 0.000000001827292 = 0.00000000034261725
0.00000000034261725+0.962962962434674854006924250953453453453557 = 0.962962962777292104006924250953453453453557
We know that 0.00000000034261725 of the time, the last 21 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 5 cards that are bad for the player do.
nCr(26,26)*nCr(5,4)/nCr(31,30) = 0.1612903225806452* 0.00000000034261725 = 0.0000000000552608467741935616497
0.0000000000552608467741935616497+0.962962962777292104006924250953453453453557 = 0.962962962832552950781117812603153453453557
We know that 0.0000000000552608467741935616497 of the time, the last 22 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 4 cards that are bad for the player do.
nCr(26,26)*nCr(4,3)/nCr(30,29) = 0.1333333333333333*0.0000000000552608467741935616497 = 0.00000000000736811290322580637793177419354794501
0.00000000000736811290322580637793177419354794501+0.962962962832552950781117812603153453453557 = 0.96296296283992106368434361898108522764710494501
We know that 0.00000000000736811290322580637793177419354794501 of the time, the last 23 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 3 cards that are bad for the player do.
nCr(26,26)*nCr(3,2)/nCr(29,28) = 0.103448275862069* 0.00000000000736811290322580637793177419354794501 = 0.00000000000076221857619577332765200778642931768833704115682569
0.00000000000076221857619577332765200778642931768833704115682569+0.96296296283992106368434361898108522764710494501
=0.96296296284068328226053939230873723543353426269833704115682569
We know that 0.00000000000076221857619577332765200778642931768833704115682569 of the time, the last 24 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 2 cards that are bad for the player do.
nCr(26,26)*nCr(2,1)/nCr(28,27) = 0.0714285714285714*0.00000000000076221857619577332765200778642931768833704115682569 = 0.0000000000000544441840139837873403269505799990448238518989356153
0.0000000000000544441840139837873403269505799990448238518989356153+0.96296296284068328226053939230873723543353426269833704115682569=0.9629629628407377264445533760960775623841142617431608930557613053
We know that 0.0000000000000544441840139837873403269505799990448238518989356153 of the time, the last 25 cards will be red. In order for the player to win in this scenario, all 26 cards good for the player have to come out before the remaining 1 card that is bad for the player does.
nCr(26,26)*nCr(1,0)/nCr(27,26) = 0.037037037037037* 0.0000000000000544441840139837873403269505799990448238518989356153 = 0.0000000000000020164512597771752924497383924522853517370858865396
0.0000000000000020164512597771752924497383924522853517370858865396+0.9629629628407377264445533760960775623841142617431608930557613053=
0.9629629628407397428958131532713700121225067140285126301416478449
We know that not real G*****N often will all 26 red cards be on the bottom, but when they are, the 26 black cards are on top and the player automatically ties, but temporarily has to lose $26 and have a stroke first because something damn near impossible happened.
CONCLUSION
In conclusion, the probability of one side or the other leading at some point is approximately 96.29629628407397428958131532713700121225067140285126301416478449%, because of this, that same amount is also the fair price to pay for the +1/-1 game because that is the probability that the player will be ahead one card. It is always optimal for the player to quit when ahead one card because there is a negative expectation on the following card.
Provided the player has sufficient money to play out the deck, the roughly 3.70370371592602571041868467286299878774932859714873698583521551% probability of the wrong card being flipped first and never relinquishing the lead to the other side (but sometimes being tied) is irrelevant because the player does not ever lose $1 (or more) in this scenario.
The player either wins $1 or loses nothing in all cases. The probability of winning a dollar is roughly .9629629284074, the expected win is the same as the probability since our base unit is one dollar. Therefore, the fair value to play this game is a cost of 96.2962984074 cents.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
January 29th, 2020 at 10:13:41 PM
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Quote: WizardLet me add this, I think the optimal strategy depends on both how much money you have accumulated and total cards played. For example, if it's early in the game and you're $5 up, you have more of a chance to go higher than if you're up $5 late in the game.
Keep an eye on the time of the OP. I think past winners will jump in after the 24 hours. Where is OD?
I disagree and would suggest the optimal strategy is to always quit when you're a dollar ahead because the next card has a negative expectation. Let's say that the first card was good for you, so now you're ahead one dollar:
(25/51 * 2) = 0.98039215686
OR:
(25/51) - (26/51) = -0.01960784313
The expectation is a loss on the next card drawn and that never changes regardless of how much the player is ahead or how many cards are left in the deck. Similarly, the player's expectation on the next card is more than $1 if more cards are out that are bad for the player and the player is down money.
I'm also going to copy my conclusion from the previous post because that post is very long.
CONCLUSION
In conclusion, the probability of one particular side or the other leading at some point is approximately 96.29629628407397428958131532713700121225067140285126301416478449%, because of this, that same amount is also the fair price to pay for the +1/-1 game because that is the probability that the player will be ahead one card. It is always optimal for the player to quit when ahead one card because there is a negative expectation on the following card.
Provided the player has sufficient money to play out the deck, the roughly 3.70370371592602571041868467286299878774932859714873698583521551% probability of the wrong card being flipped first and never relinquishing the lead to the other side (but sometimes being tied) is irrelevant because the player does not ever lose $1 (or more) in this scenario.
The player either wins $1 or loses nothing in all cases. The probability of winning a dollar is roughly .9629629284074, the expected win is the same as the probability since our base unit is one dollar. Therefore, the fair value to play this game is a cost of 96.2962984074 cents.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
January 29th, 2020 at 10:34:52 PM
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great...Quote: Mission146I think I have finally come up with a solution to this, and to do that, I'm going to cheat.
more bucking cheaters
winsome johnny (not Win some johnny)
January 29th, 2020 at 10:47:32 PM
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Quote: Mission146I disagree and would suggest the optimal strategy is to always quit when you're a dollar ahead because the next card has a negative expectation. Let's say that the first card was good for you, so now you're ahead one dollar:
(25/51 * 2) = 0.98039215686
OR:
(25/51) - (26/51) = -0.01960784313
The expectation is a loss on the next card drawn and that never changes regardless of how much the player is ahead or how many cards are left in the deck. Similarly, the player's expectation on the next card is more than $1 if more cards are out that are bad for the player and the player is down money.
I'm also going to copy my conclusion from the previous post because that post is very long.
CONCLUSION
In conclusion, the probability of one particular side or the other leading at some point is approximately 96.29629628407397428958131532713700121225067140285126301416478449%, because of this, that same amount is also the fair price to pay for the +1/-1 game because that is the probability that the player will be ahead one card. It is always optimal for the player to quit when ahead one card because there is a negative expectation on the following card.
Provided the player has sufficient money to play out the deck, the roughly 3.70370371592602571041868467286299878774932859714873698583521551% probability of the wrong card being flipped first and never relinquishing the lead to the other side (but sometimes being tied) is irrelevant because the player does not ever lose $1 (or more) in this scenario.
The player either wins $1 or loses nothing in all cases. The probability of winning a dollar is roughly .9629629284074, the expected win is the same as the probability since our base unit is one dollar. Therefore, the fair value to play this game is a cost of 96.2962984074 cents.
Great problem Wiz!
I think you are mistaken Mission about quitting after being ahead. Assuming you are ahead and pocked that 1$ and forget about it. The remaining of the game is not necessariliy -EV. If you keep playing, you may get ahead more, or lose more in which case the worst that can happen is you finish - 1 $. Because of the variance, it is still potentially +EV.
Is it still +EV if you are 2$ ahead? 3$ ahead? that depends on the number of cards remaining. scrooge got it though!
January 30th, 2020 at 4:02:48 AM
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Quote: scroogethe complete depth-dependent solution
Thanks.
Last edited by: IndyJeffrey on Jan 30, 2020
January 30th, 2020 at 4:12:18 AM
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Quote: tyler498Great problem Wiz!
I think you are mistaken Mission about quitting after being ahead. Assuming you are ahead and pocked that 1$ and forget about it. The remaining of the game is not necessariliy -EV. If you keep playing, you may get ahead more, or lose more in which case the worst that can happen is you finish - 1 $. Because of the variance, it is still potentially +EV.
Is it still +EV if you are 2$ ahead? 3$ ahead? that depends on the number of cards remaining. scrooge got it though!
I'm pretty certain, if you are ahead at any point by definition of the rules, the remainder of the game is -EV from where you presently stand. If you are up $1 after one card, or up $1 after 27 cards, there would be more black than red cards left to draw.
January 30th, 2020 at 4:17:39 AM
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Quote: Mission146The player either wins $1 or loses nothing in all cases.
Yes, if you play the deck until one or fewer cards remain. What of the scenario where the first four cards are red, the player quits and takes the cash?
January 30th, 2020 at 4:27:22 AM
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Quote: scroogequit at $3 with 7-12 cards remaining
I submit you should quit with $2 with 8 cards remaining. In other words EV(8,2)=2. Note how I constructed my EV table. What is the lowest number of cards remaining you disagree with it?
| EV(0,0)=0 | EV(1,1)=1 | EV(2,2)=2 | EV(3,3)=3 | EV(4,4)=4 | EV(5,5)=5 |
| EV(1,-1)=0 | EV(2,0)=0.5 | EV(3,1)=1 | EV(4,2)=2 | EV(5,3)=3 | EV(6,4)=4 |
| EV(2,-2)=0 | EV(3,-1)=0.333333333333333 | EV(4,0)=0.666666666666667 | EV(5,1)=1.2 | EV(6,2)=2 | EV(7,3)=3 |
| EV(3,-3)=0 | EV(4,-2)=0.25 | EV(5,-1)=0.5 | EV(6,0)=0.85 | EV(7,1)=1.34285714285714 | EV(8,2)=2 |
I show drawing to (8,2) (5 blacks and 3 reds remaining) has an EV of 2 by standing and 1.964285714 by drawing.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 30th, 2020 at 4:34:25 AM
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Quote: IndyJeffreyI'm pretty certain, if you are ahead at any point by definition of the rules, the remainder of the game is -EV from where you presently stand. If you are up $1 after one card, or up $1 after 27 cards, there would be more black than red cards left to draw.
That would be true if you can draw only one more card and must quit. If you're up $1 after 27 cards, even if you drop to $0 on the next draw, there is a good chance to can rebuild back to $1 by drawing more.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 30th, 2020 at 4:56:53 AM
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Quote: IndyJeffreyIt should be more than 50 cents. But how much more?
I believe it should be less than $1. But how much less?
After some reflection the fair price must be > $1. If the fair price is < $1, you never lose. And that, by definition, is not fair.
January 30th, 2020 at 6:04:15 AM
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Quote: tyler498
Great problem Wiz!
I think you are mistaken Mission about quitting after being ahead. Assuming you are ahead and pocked that 1$ and forget about it. The remaining of the game is not necessariliy -EV. If you keep playing, you may get ahead more, or lose more in which case the worst that can happen is you finish - 1 $. Because of the variance, it is still potentially +EV.
Is it still +EV if you are 2$ ahead? 3$ ahead? that depends on the number of cards remaining. scrooge got it though!
Can someone please explain to me the concept of why the player's money is expected to improve if he continues playing when, regardless of the result of the next card, the player is expected to lose money any time the player is ahead?
Also, if this was a free game to play (no fee) is this assumed as something the player could only play once lifetime? Why could the player not simply opt to stop the game, put all of the cards back in the deck and then play it again? If the player can end the game and play it again anytime he wants to, then it would be the most expedient for the player to quit whenever ahead $1 and ask to start the game anew.
EDIT: I'm not trying to be argumentative, it's just that I'm apparently missing something in the way I'm looking at this.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
January 30th, 2020 at 6:48:58 AM
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Quote: Mission146Can someone please explain to me the concept of why the player's money is expected to improve if he continues playing when, regardless of the result of the next card, the player is expected to lose money any time the player is ahead?
Variance. All the course of the game the score will wander up and down. The key is to quit when you're at an up point, but don't settle for too little. I think I can say the player will be at +$2 at least 50% of the time at some point.
Quote:Also, if this was a free game to play (no fee) is this assumed as something the player could only play once lifetime? Why could the player not simply opt to stop the game, put all of the cards back in the deck and then play it again? If the player can end the game and play it again anytime he wants to, then it would be the most expedient for the player to quit whenever ahead $1 and ask to start the game anew.
Think of it as playing once in a lifetime.
Quote:EDIT: I'm not trying to be argumentative, it's just that I'm apparently missing something in the way I'm looking at this.
I obviously didn't phrase the puzzle well. I plan to make an Ask the Wizard question out of it. I am open to alternative wording from anyone who understands the puzzle in the first place.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 30th, 2020 at 8:00:16 AM
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Quote: WizardThink of it as playing once in a lifetime.
(Quote clipped, relevance)
Sounds good, thanks! I wouldn't even have attempted an answer had I thought of it that way. I figured that the player could quit the game and play it at a different time (as one could with a casino game) anytime that the player wants to.
Out of curiosity, can you tell if my answer is right assuming that the player can quit the game and start it anew anytime he wants to?
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
January 30th, 2020 at 9:21:31 AM
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Quote: Mission146Out of curiosity, can you tell if my answer is right assuming that the player can quit the game and start it anew anytime he wants to?
If the player could do that, the value would be $26, because the player would keep starting over until he got 26 reds in a row.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 30th, 2020 at 9:31:03 AM
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Quote: WizardIf the player could do that, the value would be $26, because the player would keep starting over until he got 26 reds in a row.
That’s not what I mean. I don’t mean that the player can start over and thereby negate any losses, just that he can quit playing and start a new game anytime he likes with the result of the current game final.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
January 30th, 2020 at 9:38:05 AM
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Quote: Mission146That’s not what I mean. I don’t mean that the player can start over and thereby negate any losses, just that he can quit playing and start a new game anytime he likes with the result of the current game final.
Then the value would be infinite, because the game has a positive expected value. Multiply that by an infinite number of games and you have infinite money.
Also, reminder that previous beer winners may post in just over an hour. Still hoping somebody new will get that beer. Perhaps Scrooge is right, in which case I'll award the beer retroactively. For now, I don't agree with his answer.
Last edited by: Wizard on Jan 30, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 30th, 2020 at 12:16:37 PM
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fixed bug
I found the bug in my program. [{}]*53 creates 53 references to the same dictionary, but I wanted separate dictionaries. So I changed the initialization of ev and then the program agreed exactly with every number in your table and said the game was worth 2.6245.
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28 cards left quit at $4
29 cards left quit at $5
30 cards left quit at $6
31 cards left quit at $5
32 cards left quit at $6
33 cards left quit at $5
34 cards left quit at $6
35 cards left quit at $5
36 cards left quit at $6
37 cards left quit at $5
38 cards left quit at $6
39 cards left quit at $5
40 cards left quit at $6
41 cards left quit at $5
42 cards left quit at $6
43 cards left quit at $7
44 cards left quit at $6
45 cards left quit at $7
46 cards left quit at $6
47 cards left quit at $None
48 cards left quit at $None
49 cards left quit at $None
50 cards left quit at $None
51 cards left quit at $None
52 cards left quit at $None
EV(52,0) = 2.6245
With an odd number of cards left you can only be up an odd number of dollars and with an even number of cards left you can only be up an even number of dollars. So for simplicity you could fudge any stopping point down by one and it would be functionally the same:
1-3 cards left quit at $1
4-9 cards left quit at $2
10-18 cards left quit at $3
19-29 cards left quit at $4
30-42 cards left quit at $5
43-46 cards left quit at $6
47+ cards left don't quit
I found the bug in my program. [{}]*53 creates 53 references to the same dictionary, but I wanted separate dictionaries. So I changed the initialization of ev and then the program agreed exactly with every number in your table and said the game was worth 2.6245.
0 cards left quit at $None
1 cards left quit at $1
2 cards left quit at $2
3 cards left quit at $1
4 cards left quit at $2
5 cards left quit at $3
6 cards left quit at $2
7 cards left quit at $3
8 cards left quit at $2
9 cards left quit at $3
10 cards left quit at $4
11 cards left quit at $3
12 cards left quit at $4
13 cards left quit at $3
14 cards left quit at $4
15 cards left quit at $3
16 cards left quit at $4
17 cards left quit at $3
18 cards left quit at $4
19 cards left quit at $5
20 cards left quit at $4
21 cards left quit at $5
22 cards left quit at $4
23 cards left quit at $5
24 cards left quit at $4
25 cards left quit at $5
26 cards left quit at $4
27 cards left quit at $5
28 cards left quit at $4
29 cards left quit at $5
30 cards left quit at $6
31 cards left quit at $5
32 cards left quit at $6
33 cards left quit at $5
34 cards left quit at $6
35 cards left quit at $5
36 cards left quit at $6
37 cards left quit at $5
38 cards left quit at $6
39 cards left quit at $5
40 cards left quit at $6
41 cards left quit at $5
42 cards left quit at $6
43 cards left quit at $7
44 cards left quit at $6
45 cards left quit at $7
46 cards left quit at $6
47 cards left quit at $None
48 cards left quit at $None
49 cards left quit at $None
50 cards left quit at $None
51 cards left quit at $None
52 cards left quit at $None
EV(52,0) = 2.6245
With an odd number of cards left you can only be up an odd number of dollars and with an even number of cards left you can only be up an even number of dollars. So for simplicity you could fudge any stopping point down by one and it would be functionally the same:
1-3 cards left quit at $1
4-9 cards left quit at $2
10-18 cards left quit at $3
19-29 cards left quit at $4
30-42 cards left quit at $5
43-46 cards left quit at $6
47+ cards left don't quit
January 30th, 2020 at 12:26:44 PM
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Quote: WizardThen the value would be infinite, because the game as a positive expected value. Multiply that by an infinite number of games and you have infinite money.
Also, reminder that previous beer winners may post in just over an hour. Still hoping somebody new will get that beer. Perhaps Scrooge is right, in which case I'll award the beer retroactively. For now, I don't agree with his answer.
My Excel sheet gives:
approximately 2.56348224265954
Edit: I just found a bug, too. Now, I get 2.62447554899392, which is Scrooge's answer
Edit: I just found a bug, too. Now, I get 2.62447554899392, which is Scrooge's answer
January 30th, 2020 at 4:33:21 PM
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Quote: ChesterDogMy Excel sheet gives:
approximately 2.56348224265954
Edit: I just found a bug, too. Now, I get 2.62447554899392, which is Scrooge's answer
I get that as well.
January 30th, 2020 at 4:49:28 PM
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Could one of you guys modify your calculation:
Same game but a much larger N, like 1000 instead of 52. Assuming it’s not difficult.
I’d be interested to see if the fair value converges.
Thanks in advance
Same game but a much larger N, like 1000 instead of 52. Assuming it’s not difficult.
I’d be interested to see if the fair value converges.
Thanks in advance
It’s all about making that GTA
January 30th, 2020 at 5:52:01 PM
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Quote: Ace2Could one of you guys modify your calculation:
Same game but a much larger N, like 1000 instead of 52. Assuming it’s not difficult.
I’d be interested to see if the fair value converges.
Thanks in advance
If it converges, it does so very slowly.
Remember that 1 + 1/2 + 1/3 + 1/4 + ... also looks like it converges, but it diverges.
| Deck Size | Expected Value |
|---|---|
| 52 | 2.624476 |
| 54 | 2.675437 |
| 56 | 2.725591 |
| 58 | 2.774917 |
| 60 | 2.823387 |
| 80 | 3.267317 |
| 100 | 3.658141 |
| 120 | 4.011461 |
| 140 | 4.336463 |
| 160 | 4.638738 |
| 180 | 4.922541 |
| 200 | 5.191092 |
| 220 | 5.446318 |
| 240 | 5.690287 |
| 260 | 5.924171 |
| 280 | 6.149274 |
| 300 | 6.36639 |
| 320 | 6.576431 |
| 340 | 6.779929 |
| 360 | 6.977574 |
| 380 | 7.169741 |
| 400 | 7.356939 |
| 500 | 8.22942 |
| 600 | 9.018087 |
| 700 | 9.743271 |
| 800 | 10.418186 |
| 900 | 11.052045 |
| 1000 | 11.651523 |
| 1100 | 12.221679 |
| 1200 | 12.766435 |
| 1300 | 13.288908 |
| 1400 | 13.791628 |
| 1500 | 14.276678 |
| 1600 | 14.7458 |
| 1700 | 15.200467 |
| 1800 | 15.641938 |
| 1900 | 16.0713 |
| 2000 | 16.489496 |
| 2100 | 16.897355 |
| 2200 | 17.295608 |
| 2300 | 17.684903 |
| 2400 | 18.065818 |
| 2500 | 18.438872 |
| 2600 | 18.804533 |
| 2700 | 19.163224 |
| 2800 | 19.515328 |
| 2900 | 19.861196 |
| 3000 | 20.201147 |
| 3100 | 20.535476 |
| 3200 | 20.864451 |
| 3300 | 21.188323 |
| 3400 | 21.507322 |
| 3500 | 21.821661 |
| 3600 | 22.131539 |
| 3700 | 22.43714 |
| 3800 | 22.738637 |
| 3900 | 23.03619 |
| 4000 | 23.329951 |
| 4100 | 23.620062 |
| 4200 | 23.906653 |
| 4300 | 24.189852 |
| 4400 | 24.469775 |
| 4500 | 24.746534 |
| 4600 | 25.020233 |
| 4700 | 25.290972 |
| 4800 | 25.558844 |
| 4900 | 25.82394 |
| 5000 | 26.086343 |
| 10000 | 36.900218 |
| 12000 | 40.423958 |
| 14000 | 43.664319 |
| 16000 | 46.680337 |
| 18000 | 49.513015 |
| 20000 | 52.192206 |
| 22000 | 54.740449 |
| 24000 | 57.175249 |
| 26000 | 59.510532 |
| 28000 | 61.757584 |
| 30000 | 63.925711 |
| 32000 | 66.022685 |
| 34000 | 68.055085 |
| 36000 | 70.028531 |
| 38000 | 71.947875 |
| 40000 | 73.817335 |
| 42000 | 75.640611 |
| 44000 | 77.420965 |
| 46000 | 79.161293 |
| 48000 | 80.864178 |
| 50000 | 82.531938 |
| 51000 | 83.353307 |
| 52000 | 84.166661 |
| 53000 | 84.972231 |
| 54000 | 85.770236 |
| 55000 | 86.560885 |
| 56000 | 87.344378 |
| 57000 | 88.120906 |
| 58000 | 88.890651 |
| 59000 | 89.653788 |
| 60000 | 90.410484 |
| 61000 | 91.1609 |
| 62000 | 91.905189 |
| 63000 | 92.6435 |
| 64000 | 93.375973 |
| 65000 | 94.102746 |
| 66000 | 94.823949 |
| 67000 | 95.539708 |
| 68000 | 96.250145 |
| 69000 | 96.955377 |
| 70000 | 97.655517 |
| 71000 | 98.350672 |
| 72000 | 99.04095 |
| 73000 | 99.726449 |
| 74000 | 100.407269 |
| 75000 | 101.083504 |
| 76000 | 101.755246 |
| 77000 | 102.422582 |
| 78000 | 103.085599 |
| 79000 | 103.744378 |
| 80000 | 104.399001 |
| 81000 | 105.049545 |
| 82000 | 105.696085 |
| 83000 | 106.338695 |
| 84000 | 106.977445 |
| 85000 | 107.612403 |
| 86000 | 108.243638 |
| 87000 | 108.871213 |
| 88000 | 109.495191 |
| 89000 | 110.115633 |
| 90000 | 110.7326 |
| 91000 | 111.346148 |
| 92000 | 111.956334 |
| 93000 | 112.563212 |
| 94000 | 113.166836 |
| 95000 | 113.767258 |
| 96000 | 114.364528 |
| 97000 | 114.958695 |
| 98000 | 115.549806 |
| 99000 | 116.13791 |
| 100000 | 116.72305 |
| 101000 | 117.305272 |
| 102000 | 117.884619 |
| 103000 | 118.461132 |
| 104000 | 119.034853 |
| 105000 | 119.605823 |
| 106000 | 120.17408 |
| 107000 | 120.739663 |
| 108000 | 121.302608 |
| 109000 | 121.862954 |
| 110000 | 122.420734 |
| 111000 | 122.975985 |
| 112000 | 123.528741 |
| 113000 | 124.079034 |
| 114000 | 124.626897 |
| 115000 | 125.172362 |
| 116000 | 125.715461 |
| 117000 | 126.256224 |
| 118000 | 126.794681 |
| 119000 | 127.330861 |
| 120000 | 127.864793 |
EDIT: Now up to 120,000 cards
Last edited by: ThatDonGuy on Jan 30, 2020
January 30th, 2020 at 6:18:12 PM
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If you plot the log of Don's fair values vs the log of Don's deck sizes, the points lie on an almost straight line.
January 30th, 2020 at 6:26:36 PM
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I would like to congratulate Scrooge on getting the first correct answer, beating Chesterdog by 10 minutes. Scrooge--I owe you a beer. However, I won't say "no," if you pick up the second round as I think my response led you to seeing the light.
That is indeed very interesting. There is a whole branch of study on "random walk" problems. As I recall, there is a great deal of predictability of such walks.
Quote: ChesterDogIf you plot the log of Don's fair values vs the log of Don's deck sizes, the points lie on an almost straight line.
That is indeed very interesting. There is a whole branch of study on "random walk" problems. As I recall, there is a great deal of predictability of such walks.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 30th, 2020 at 7:40:33 PM
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I think I found a formula for this:
e^(Ln(x - 2h) / 2 - 1)
Where x is the number of cards in the deck and h is the Euler-Mascheroni constant of 0.577216
100 card deck : e^(Ln(100 - 1.15443) / 2 - 1) = 3.66
4000 card deck : e^(Ln(4000 - 1.15443) / 2 - 1) = 23.26
50,000 card deck : e^(Ln(100- 1.15443) / 2 - 1) = 82.26
Thanks, ThatDonGuy, for calculating and posting the additional numbers.
e^(Ln(x - 2h) / 2 - 1)
Where x is the number of cards in the deck and h is the Euler-Mascheroni constant of 0.577216
100 card deck : e^(Ln(100 - 1.15443) / 2 - 1) = 3.66
4000 card deck : e^(Ln(4000 - 1.15443) / 2 - 1) = 23.26
50,000 card deck : e^(Ln(100- 1.15443) / 2 - 1) = 82.26
Thanks, ThatDonGuy, for calculating and posting the additional numbers.
It’s all about making that GTA
January 30th, 2020 at 8:02:18 PM
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Quote: Ace2I think I found a formula for this:
e^(Ln(x - 2h) / 2 - 1)
Where x is the number of cards in the deck and h is the Euler-Mascheroni constant of 0.577216
100 card deck : e^(Ln(100 - 1.15443) / 2 - 1) = 3.66
4000 card deck : e^(Ln(4000 - 1.15443) / 2 - 1) = 23.26
50,000 card deck : e^(Ln(100- 1.15443) / 2 - 1) = 82.26
Thanks, ThatDonGuy, for calculating and posting the additional numbers.
That amounts to sqrt(x-2h)/e or sqrt(x-1.15443)/2.7.. It's very close at higher numbers, but there the Euler-Mascheroni constant effect is negligible. How did it come up in your formula?
January 30th, 2020 at 8:12:20 PM
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The constant does make a difference at the lower numbers. I came up with the formula just by looking at the relationships/trends...I had no doubt it would use e.Quote: tyler498That amounts to sqrt(x-2h)/e or sqrt(x-1.15443)/2.7.. It's very close at higher numbers, but there the Euler-Mascheroni constant effect is negligible. How did it come up in your formula?
It’s all about making that GTA
January 30th, 2020 at 8:37:08 PM
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There was quite a bit of confusion about what I was trying to ask in the original post. Can anyone suggest a better wording?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 30th, 2020 at 11:35:30 PM
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Quote: WizardThere was quite a bit of confusion about what I was trying to ask in the original post. Can anyone suggest a better wording?
The description of the rules seems fairly clear.
Instead of asking what is a fair price to pay, I would ask what is the EV in case that wasn't clear.
More specifically, what is the EV per Deck, knowing the players will play a variable number of cards in each deck before restarting with a new one.
January 31st, 2020 at 1:12:56 AM
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Another interesting variant I am thinking of is what would be the game worth if you get to decide on which color to bet:
If you get it, you win 1$, if you don't, you lose 1$: obviously you would bet on whichever color is more present among the remaining cards, and this would be +EV because of the variance. I've run a quick simulation for small number of cards:
Cards/deck EV $ EV/Card
2 1 0.5
4 1.666666667 0.4166666667
6 2.2 0.3666666667
8 2.657142857 0.3321428571
10 3.063492063 0.3063492063
12 3.432900433 0.2860750361
14 3.773892774 0.2695637696
16 4.092152292 0.2557595183
18 4.391690662 0.2439828146
20 4.675463855 0.2337731928
22 4.945724039 0.2248056381
24 5.204233779 0.2168430741
My code is terribly sub-optimal so it gets slow above these numbers.
EV/card will clearly get lower as the deck gets larger, since the extra EV you get from an unbalanced deck is smaller with more cards (card counters know that all too well), but I have no idea to what it converges!
If you get it, you win 1$, if you don't, you lose 1$: obviously you would bet on whichever color is more present among the remaining cards, and this would be +EV because of the variance. I've run a quick simulation for small number of cards:
Cards/deck EV $ EV/Card
2 1 0.5
4 1.666666667 0.4166666667
6 2.2 0.3666666667
8 2.657142857 0.3321428571
10 3.063492063 0.3063492063
12 3.432900433 0.2860750361
14 3.773892774 0.2695637696
16 4.092152292 0.2557595183
18 4.391690662 0.2439828146
20 4.675463855 0.2337731928
22 4.945724039 0.2248056381
24 5.204233779 0.2168430741
My code is terribly sub-optimal so it gets slow above these numbers.
EV/card will clearly get lower as the deck gets larger, since the extra EV you get from an unbalanced deck is smaller with more cards (card counters know that all too well), but I have no idea to what it converges!
January 31st, 2020 at 1:29:58 PM
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Quote: tyler498The description of the rules seems fairly clear.
Thanks, at least one person thought so. I'll take your suggestions under advisement.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 31st, 2020 at 4:20:18 PM
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Quote: tyler498The description of the rules seems fairly clear.
Instead of asking what is a fair price to pay, I would ask what is the EV in case that wasn't clear.
I agree with both sentences. Especially since I was the one who didn't equate EV with fair price. You'd think as much as I read this forum, I'd know that basic concept.
January 31st, 2020 at 4:41:33 PM
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How about this:
A casino game has the following rules:
1. You can only play this game once.
2. A 52-card deck is used.
3. Cards are dealt one at a time, without replacement, until the player chooses to end the game or the end of the deck is reached.
4. Every time a black card is dealt, the player must pay $1.
5. Every time a red card is dealt, the player wins $1.
6. The player may quit and walk away with his winnings (if any) at any point.
If a player is behind, they would continue to play until the end of the deck at which point they will break even. If a player is ahead, it means more losing cards than winning cards remain. The question is, what is the optimal amount to be ahead before quitting?
A casino game has the following rules:
1. You can only play this game once.
2. A 52-card deck is used.
3. Cards are dealt one at a time, without replacement, until the player chooses to end the game or the end of the deck is reached.
4. Every time a black card is dealt, the player must pay $1.
5. Every time a red card is dealt, the player wins $1.
6. The player may quit and walk away with his winnings (if any) at any point.
If a player is behind, they would continue to play until the end of the deck at which point they will break even. If a player is ahead, it means more losing cards than winning cards remain. The question is, what is the optimal amount to be ahead before quitting?
Simplicity is the ultimate sophistication - Leonardo da Vinci
January 31st, 2020 at 5:28:41 PM
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Quote: AyecarumbaThe question is, what is the optimal amount to be ahead before quitting?
Thanks, but that's a different question. It depends on how many cards are left, as shown in Scrooge's strategy.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 1st, 2020 at 1:00:16 AM
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Quote: WizardThere seems to be a misunderstanding about what I'm asking.
Yes, I agree if you could play this for free it would be a freeroll with the least bit of strategy. If you're negative, just wait until the end of the deck and you'll break even. It's easy to say the player should never settle for a loss (absent the fee to play).
However, I'm asking what would you pay to play for such a "freeroll." Think of literally a carnival game, like throwing a ball in a basket for a teddy bear. You pay a price and then it switches to another phase where you either win something or you don't.
In this case, the fair price is clearly more than $0, because the player could just wait to be up $1 and quit. If he's never up, then wait to the end and you push. However, it must also be less than $26, because there is less than a 100% chance of drawing all 26 red cards in the first 26 draws. The question, is where is that fair price between $0 and $26?
My confusion is the contradiction between “fair”, which apparently means “the price point where there is equal risk to both parties in the wager”; and “price” which implies a vig, Doesn’t Charging to play makes this “fair” zero sum game “unfair”?
Simplicity is the ultimate sophistication - Leonardo da Vinci
February 1st, 2020 at 3:09:02 AM
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Quote: AyecarumbaDoesn’t Charging to play makes this “fair” zero sum game “unfair”?
It's not a zero sum game. The worst that can happen is you break even and the best is a win of $26.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 1st, 2020 at 7:09:58 AM
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Quote: WizardIt's not a zero sum game. The worst that can happen is you break even and the best is a win of $26.
Since the game is not “fair” to one side, should the “price” attempt to equalize the hold on both sides, or include an edge so the casino can pay the dealer, keep the lights on, etc.?
Simplicity is the ultimate sophistication - Leonardo da Vinci
February 1st, 2020 at 8:42:47 AM
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Quote: AyecarumbaSince the game is not “fair” to one side, should the “price” attempt to equalize the hold on both sides, or include an edge so the casino can pay the dealer, keep the lights on, etc.?
To me “fair price” means the game is EV of $0. In other words, what is the highest price you should pay to play the game before it becomes rational to not play. So the indifference point.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
February 1st, 2020 at 10:57:41 AM
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Fair Price might be interpreted that either the casino or the player should have a profit margin, but it's a math question so it's obviously asking for the EV of the game. I guess you could just ask what's the EV for the player who plays this game optimally. Thanks wizard for the hint. I'll buy the second round.
