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Subject: Thought experiment on theory/probablity (picking cards) rss

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So I actually used to play a type of gambling game like this, but recently came across a game in 'mario party' where the four players each take a turn picking one of ten total items, three of the ten are 'losers', and if you pick the loser you are out. Last one left wins.

The card game we used to play was similar, deal all 52 cards out face down, and the aces are 'bad', and I pick 1, person b picks one, C picks 1, D picks 1, now I pick 2, B picks 2 until someone picks the ace and it resets back to 1.

My main question is, positioning certainly seems to matter, but I have no idea how to figure out the probability of where is 'the best' position. Especially in the mario party example, with only 10 total, and 3 are bad, I could make a strong argument why going first is the worst, but likewise could argue going first is the best...so what is the truth here?
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Perry Fergin
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Would it really make a difference when you go? Isn't the probability the same?
 
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Daniel Corban
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If there are only three bad cards, then being the fourth player is clearly the best position. Therefore, I suggest going last is always best.
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Aces reset back to 1, but do they lose? What loses?

Anyway, what's the argument for first being best? In Russian Roulette, you never want to go first.
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perrygf wrote:
Would it really make a difference when you go? Isn't the probability the same?

No, because you don't all draw at the same time. Imagine a deck that has only aces. Clearly, the first one who draws a card, loses.

It is true that in a deck, every position has an equal chance of holding a card that is an ace. So with each draw you have the same chance of drawing one, if everyone would keep drawing until the end. But since you don't, the fewer cards you have to draw, the less chance there is of drawing an ace. And the last player to draw in a turn mostly has to draw one card fewer than the first player, since the odds are that in a round where an ace is drawn, it won't be drawn before the last player's turn.
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Now imagine a variant. With four players, in the first round player 1 draws one card, then player 2 draws two cards, player 3three cards, and from then on everyone draws four cards, up to the end. In the last round, player 1 draws three cards, player 2 two cards, and player 3 the last one. Theoretically, that is, because at least one ace must have been drawn before player 2's last turn.

Who would have the best seat now?
 
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dragon0085 wrote:
deal all 52 cards out face down, and the aces are 'bad', and I pick 1, person b picks one, C picks 1, D picks 1, now I pick 2, B picks 2 until someone picks the ace and it resets back to 1.


Is 2 the max cards anyone would draw at a time, or does it increase by one each round of no-aces?
 
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This is weird and I probably have the maths wrong, but in your second example where picking an ace ends the game, all the positions seem to have the have the same odds. The probability of losing is the chance of picking the ace times the chance that no-one before you picked it. So:

1st position: 4/52 * 1 = 0.0769230769230769
2nd position: 4/51 * 51/52 = 0.0769230769230769
...
nth position: 4/(53-n) * (53-n)/52 = 0.0769230769230769
...
49th position: 4/4 * 4/52 = 0.0769230769230769

Bizarre!




 
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sbszine wrote:
2nd position: 4/51 * 48/52 = ...

I think this is how it would work.
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True dat.
 
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If there is only 1 item which is bad and that the number of items N is a multiple of the number of players P, then your seat number does not matter. Everyone has an equal chance of picking the bad item.

If there is only 1 item which is bad and that the number of items N is NOT a multiple of the number of players P, then the farther you are the better, since the first (N mod P) players will be drawing one item more than you.

If there are multiple items that are bad and that the game ends when one player is out, then the previous two theories mostly hold. You basically want to be in the later seats. (even if the number of items is a multiple of the players, you usually want to be last)

Now, if there are multiple items that are bad and that you play until a single player remains it becomes trickier. To not lose at the first bad item, the previous theory holds, so you basically want to be in the later seats. But then, everything resets, the player next in line becomes first player, and the previous player becomes last. I would guess that it's becoming so random that the chances should equalize pretty quickly as the number of items grows.

All in all, I'd say that you want to be in the later seats if possible, because it boosts the probability that you will not be the first player to go. And then, good luck !
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There's only actually 120 cases to consider with the Mario Party version of the question, so I wrote a program to check them all. Player 1 wins in 26 cases, player 2 wins 28, player 3 wins 31 and player 4 wins 35. It's better to go later.
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zenpunk wrote:
dragon0085 wrote:
deal all 52 cards out face down, and the aces are 'bad', and I pick 1, person b picks one, C picks 1, D picks 1, now I pick 2, B picks 2 until someone picks the ace and it resets back to 1.


Is 2 the max cards anyone would draw at a time, or does it increase by one each round of no-aces?


No, it keeps going up till someone hits it. Also, the game is NOT over justbecause one person is out, its last man standing wins.

grasa_total wrote:
There's only actually 120 cases to consider with the Mario Party version of the question, so I wrote a program to check them all. Player 1 wins in 26 cases, player 2 wins 28, player 3 wins 31 and player 4 wins 35. It's better to go later.


Im curious to see the results, but 120 sounds way too low to me.

The first guy alone has 3/10 going down, 7/10 surviving, now, the 2nd guy already has tons of different options, if the 1st guy hit, its 2/9going down, 7/9 being safe, whereas if 1st didn't, it 3/9 down, 6/9 being safe.
 
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dragon0085 wrote:
grasa_total wrote:
There's only actually 120 cases to consider with the Mario Party version of the question, so I wrote a program to check them all. Player 1 wins in 26 cases, player 2 wins 28, player 3 wins 31 and player 4 wins 35. It's better to go later.


Im curious to see the results, but 120 sounds way too low to me.

The first guy alone has 3/10 going down, 7/10 surviving, now, the 2nd guy already has tons of different options, if the 1st guy hit, its 2/9going down, 7/9 being safe, whereas if 1st didn't, it 3/9 down, 6/9 being safe.
There's no replacement, is there?
I.e., once someone selects a losing card, cards already selected don't get shuffled back in, do they?

If there is no replacement, then for selecting 3 things among 10 spots which they can be chosen (i.e., first card selected, second card selected, ..., tenth card selected) the number of possible combinations in which that might happen is a fairly well known and proven mathematical formula. Typically this formula is denoted as C(n,k) or nCk (in the Mario party example n=10, k=3) and is fairly easy to calculate or just look up: http://lmgtfy.com/?q=10+choose+3 . Also note that it doesn't matter if we look for the spots where the 3 losing cards are selected or the spots where the 7 safe cards are selected as C(10,3) = C(10,7): http://lmgtfy.com/?q=10+choose+7 .
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dragon0085 wrote:
No, it keeps going up till someone hits it. Also, the game is NOT over justbecause one person is out, its last man standing wins.

Yeah, that's what I figured. And I think the math is tons more complex (for the 52 card deck version anyway) than the simple equations stated here.

For each player you have to consider their chance of drawing a bad card on a later round too, even if you survive the first pick, cus everyone else may too and it will get back to you, but now with more picks per player. Then it gets even more complicated when someone does draw an ace... then you start a new round, with a different start player and 1 less ace in the deck.

To figure out the actual probability, at the start, that each player will win ... that wrecks my brain. Intuition says later has to be better, but, I don't know if I'm capable of the actual maths... time for a Monte Carlo sim, methinks...
 
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zenpunk wrote:
dragon0085 wrote:
No, it keeps going up till someone hits it. Also, the game is NOT over justbecause one person is out, its last man standing wins.

Yeah, that's what I figured. And I think the math is tons more complex (for the 52 card deck version anyway) than the simple equations stated here.

For each player you have to consider their chance of drawing a bad card on a later round too, even if you survive the first pick, cus everyone else may too and it will get back to you, but now with more picks per player. Then it gets even more complicated when someone does draw an ace... then you start a new round, with a different start player and 1 less ace in the deck.

To figure out the actual probability, at the start, that each player will win ... that wrecks my brain. Intuition says later has to be better, but, I don't know if I'm capable of the actual maths... time for a Monte Carlo sim, methinks...



Curious how close the probabilities are, I wrote a quick MC simulation and used the Mario Party scenario to test it out.

Got 0.2166, 0.2331, 0.2588, and 0.2915 for the probabilities of players A through D winning respectively. These match to the first three decimals the exact probabilities that grasa_total supplied a few days ago.

Before running on the card problem I just wanted to verify that I understand procedure correctly. I'll use an example:

Player A draws one card, not an ace, players B - D do the same in order with none drawing an ace. 48 cards left in the deck. Player A draws one card, not an ace, draws a second card, again not an ace. Players B - D do the same in order, still no aces drawn. 40 cards left in deck. Player A draws three cards, one at a time, none are aces. Player B draws a card, not an ace, draws second card, it is an ace. Player B is eliminated. Player C now draws 1 card and the cycle repeats as it did at the beginning, but now the game state is the deck has 35 cards, three of which are aces, there are three players and order is C - D - A. Is this correct?
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ronjosen wrote:
I just wanted to verify that I understand procedure correctly. I'll use an example:

Player A draws one card, not an ace, players B - D do the same in order with none drawing an ace. 48 cards left in the deck. Player A draws one card, not an ace, draws a second card, again not an ace. Players B - D do the same in order, still no aces drawn. 40 cards left in deck. Player A draws three cards, one at a time, none are aces. Player B draws a card, not an ace, draws second card, it is an ace. Player B is eliminated. Player C now draws 1 card and the cycle repeats as it did at the beginning, but now the game state is the deck has 35 cards, three of which are aces, there are three players and order is C - D - A. Is this correct?


That sounds correct to me, as per the OP's description.

Please share the MC code when you're done, if you don't mind. I'm just learning how to do it, and would probably learn a lot.
 
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zenpunk wrote:
ronjosen wrote:
I just wanted to verify that I understand procedure correctly. I'll use an example:

Player A draws one card, not an ace, players B - D do the same in order with none drawing an ace. 48 cards left in the deck. Player A draws one card, not an ace, draws a second card, again not an ace. Players B - D do the same in order, still no aces drawn. 40 cards left in deck. Player A draws three cards, one at a time, none are aces. Player B draws a card, not an ace, draws second card, it is an ace. Player B is eliminated. Player C now draws 1 card and the cycle repeats as it did at the beginning, but now the game state is the deck has 35 cards, three of which are aces, there are three players and order is C - D - A. Is this correct?


That sounds correct to me, as per the OP's description.

Please share the MC code when you're done, if you don't mind. I'm just learning how to do it, and would probably learn a lot.


While converting the simulation to the card problem I realized it was no more work to just compute the probabilities exactly by generating all possible game outcomes. Doing so I obtained 0.2277, 0.2409, 0.2561, and 0.2753 for the probabilities of players A - D winning respectively. (The exact numbers are 61644, 65218, 69335, and 74528 out of 270725). While I did some spot checks and the answers seem reasonable, I didn't want to spend a ton of time on this so I can't guarantee there's no error somewhere.

More than happy to share any code for the Mario Party simulation or the above calculation. Just so you know both were written in R and done pretty slapdash. If it will help you out as a learning experience I'm happy to convert the sim to C++ if that's what you code in. Should be a pretty quiet week and I need something to do between playing games.

Merry Christmas!





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Confirming the results of others. A quick Monte Carlo simulation of the Mario Party game gives about 22, 23, 26, 29 percent win chance for players 1-4 respectively. To put it another way, player 4 will win 32% more often than player 1. Not a very fair game it seems.
 
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dragon0085 wrote:
zenpunk wrote:
dragon0085 wrote:
deal all 52 cards out face down, and the aces are 'bad', and I pick 1, person b picks one, C picks 1, D picks 1, now I pick 2, B picks 2 until someone picks the ace and it resets back to 1.


Is 2 the max cards anyone would draw at a time, or does it increase by one each round of no-aces?


No, it keeps going up till someone hits it. Also, the game is NOT over justbecause one person is out, its last man standing wins.

grasa_total wrote:
There's only actually 120 cases to consider with the Mario Party version of the question, so I wrote a program to check them all. Player 1 wins in 26 cases, player 2 wins 28, player 3 wins 31 and player 4 wins 35. It's better to go later.


Im curious to see the results, but 120 sounds way too low to me.

The first guy alone has 3/10 going down, 7/10 surviving, now, the 2nd guy already has tons of different options, if the 1st guy hit, its 2/9going down, 7/9 being safe, whereas if 1st didn't, it 3/9 down, 6/9 being safe.


There are 10 cards in a deck, 3 are losers 7 are winners. There are only 120 possible ways to shuffle those cards. It is the same as the number of strings of three zeroes and seven ones, which is '10 choose 3'.

For 52 cards with 4 aces, there are '52 choose 4' which is 270,725 patterns.
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zenpunk wrote:
dragon0085 wrote:
No, it keeps going up till someone hits it. Also, the game is NOT over justbecause one person is out, its last man standing wins.

Yeah, that's what I figured. And I think the math is tons more complex (for the 52 card deck version anyway) than the simple equations stated here.

For each player you have to consider their chance of drawing a bad card on a later round too, even if you survive the first pick, cus everyone else may too and it will get back to you, but now with more picks per player. Then it gets even more complicated when someone does draw an ace... then you start a new round, with a different start player and 1 less ace in the deck.

To figure out the actual probability, at the start, that each player will win ... that wrecks my brain. Intuition says later has to be better, but, I don't know if I'm capable of the actual maths... time for a Monte Carlo sim, methinks...


Right. You get the problem. It is MUCH more complicated than it seems. Imagine you are the first guy, its easy to say you have a 3/10 chance of losing. Easy. But then say you don't lose. The probability tree grows HUGE as to whether no one else gets hit, 1 person gets hit, 2 people get hit, all of which make YOUR situation different.
 
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ronjosen wrote:
zenpunk wrote:
dragon0085 wrote:
No, it keeps going up till someone hits it. Also, the game is NOT over justbecause one person is out, its last man standing wins.

Yeah, that's what I figured. And I think the math is tons more complex (for the 52 card deck version anyway) than the simple equations stated here.

For each player you have to consider their chance of drawing a bad card on a later round too, even if you survive the first pick, cus everyone else may too and it will get back to you, but now with more picks per player. Then it gets even more complicated when someone does draw an ace... then you start a new round, with a different start player and 1 less ace in the deck.

To figure out the actual probability, at the start, that each player will win ... that wrecks my brain. Intuition says later has to be better, but, I don't know if I'm capable of the actual maths... time for a Monte Carlo sim, methinks...



Curious how close the probabilities are, I wrote a quick MC simulation and used the Mario Party scenario to test it out.

Got 0.2166, 0.2331, 0.2588, and 0.2915 for the probabilities of players A through D winning respectively. These match to the first three decimals the exact probabilities that grasa_total supplied a few days ago.

Before running on the card problem I just wanted to verify that I understand procedure correctly. I'll use an example:

Player A draws one card, not an ace, players B - D do the same in order with none drawing an ace. 48 cards left in the deck. Player A draws one card, not an ace, draws a second card, again not an ace. Players B - D do the same in order, still no aces drawn. 40 cards left in deck. Player A draws three cards, one at a time, none are aces. Player B draws a card, not an ace, draws second card, it is an ace. Player B is eliminated. Player C now draws 1 card and the cycle repeats as it did at the beginning, but now the game state is the deck has 35 cards, three of which are aces, there are three players and order is C - D - A. Is this correct?


Yes, that is correct.

So those first probabilities are using the scenario for mario party, does that include a situation where it gets to '2nd round' meaning no one has gone out yet, and now 1st player is looking at a situation where he has to pick 3/6?

It'd be really interesting if somehow the distribution could be mapped out. Because in a situation where less people go down early 4th player gets progressively worse (eg, he is the first one to have a 100% loss if it gets to round 2 and no one gets out) whereas player 1 has 'best' odds for 1st/2nd round to survive, but also by virtue of picking first hurts his chances.
 
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dragon0085 wrote:


It'd be really interesting if somehow the distribution could be mapped out. Because in a situation where less people go down early 4th player gets progressively worse (eg, he is the first one to have a 100% loss if it gets to round 2 and no one gets out) whereas player 1 has 'best' odds for 1st/2nd round to survive, but also by virtue of picking first hurts his chances.


Grasa_total covered every case for Mario Party with her program. There really are only 120 cases. I am able to make a combinatorial argument for the same numbers, but it is pretty clunky.
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dragon0085 wrote:
zenpunk wrote:
dragon0085 wrote:
No, it keeps going up till someone hits it. Also, the game is NOT over justbecause one person is out, its last man standing wins.

Yeah, that's what I figured. And I think the math is tons more complex (for the 52 card deck version anyway) than the simple equations stated here.

For each player you have to consider their chance of drawing a bad card on a later round too, even if you survive the first pick, cus everyone else may too and it will get back to you, but now with more picks per player. Then it gets even more complicated when someone does draw an ace... then you start a new round, with a different start player and 1 less ace in the deck.

To figure out the actual probability, at the start, that each player will win ... that wrecks my brain. Intuition says later has to be better, but, I don't know if I'm capable of the actual maths... time for a Monte Carlo sim, methinks...


Right. You get the problem. It is MUCH more complicated than it seems.

No. No it is not. The answer has been presented to you several times over.

dragon0085 wrote:
ronjosen wrote:
zenpunk wrote:
dragon0085 wrote:
No, it keeps going up till someone hits it. Also, the game is NOT over justbecause one person is out, its last man standing wins.

Yeah, that's what I figured. And I think the math is tons more complex (for the 52 card deck version anyway) than the simple equations stated here.

For each player you have to consider their chance of drawing a bad card on a later round too, even if you survive the first pick, cus everyone else may too and it will get back to you, but now with more picks per player. Then it gets even more complicated when someone does draw an ace... then you start a new round, with a different start player and 1 less ace in the deck.

To figure out the actual probability, at the start, that each player will win ... that wrecks my brain. Intuition says later has to be better, but, I don't know if I'm capable of the actual maths... time for a Monte Carlo sim, methinks...



Curious how close the probabilities are, I wrote a quick MC simulation and used the Mario Party scenario to test it out.

Got 0.2166, 0.2331, 0.2588, and 0.2915 for the probabilities of players A through D winning respectively. These match to the first three decimals the exact probabilities that grasa_total supplied a few days ago.

Before running on the card problem I just wanted to verify that I understand procedure correctly. I'll use an example:

Player A draws one card, not an ace, players B - D do the same in order with none drawing an ace. 48 cards left in the deck. Player A draws one card, not an ace, draws a second card, again not an ace. Players B - D do the same in order, still no aces drawn. 40 cards left in deck. Player A draws three cards, one at a time, none are aces. Player B draws a card, not an ace, draws second card, it is an ace. Player B is eliminated. Player C now draws 1 card and the cycle repeats as it did at the beginning, but now the game state is the deck has 35 cards, three of which are aces, there are three players and order is C - D - A. Is this correct?


Yes, that is correct.

So those first probabilities are using the scenario for mario party, does that include a situation where it gets to '2nd round' meaning no one has gone out yet, and now 1st player is looking at a situation where he has to pick 3/6?

It'd be really interesting if somehow the distribution could be mapped out. Because in a situation where less people go down early 4th player gets progressively worse (eg, he is the first one to have a 100% loss if it gets to round 2 and no one gets out) whereas player 1 has 'best' odds for 1st/2nd round to survive, but also by virtue of picking first hurts his chances.


For the Mario Party game?
Spend 5 minutes on it.

As you have been told multiple times, there are only 120 possibilities:
Winner
4 l1 l2 l3
3 l1 l2 s3 l4
4 l1 l2 s3 s4 l3
3 l1 l2 s3 s4 s3 l4
4 l1 l2 s3 s4 s3 s4 l3
3 l1 l2 s3 s4 s3 s4 s3 l4
4 l1 l2 s3 s4 s3 s4 s3 s4 l3
3 l1 l2 s3 s4 s3 s4 s3 s4 s3 l4
2 l1 s2 l3 l4
4 l1 s2 l3 s4 l2
2 l1 s2 l3 s4 s2 l4
4 l1 s2 l3 s4 s2 s4 l2
2 l1 s2 l3 s4 s2 s4 s2 l4
4 l1 s2 l3 s4 s2 s4 s2 s4 l2
2 l1 s2 l3 s4 s2 s4 s2 s4 s2 l4
3 l1 s2 s3 l4 l2
2 l1 s2 s3 l4 s2 l3
3 l1 s2 s3 l4 s2 s3 l2
2 l1 s2 s3 l4 s2 s3 s2 l3
3 l1 s2 s3 l4 s2 s3 s2 s3 l2
2 l1 s2 s3 l4 s2 s3 s2 s3 s2 l3
4 l1 s2 s3 s4 l2 l3
3 l1 s2 s3 s4 l2 s3 l4
4 l1 s2 s3 s4 l2 s3 s4 l3
3 l1 s2 s3 s4 l2 s3 s4 s3 l4
4 l1 s2 s3 s4 l2 s3 s4 s3 s4 l3
2 l1 s2 s3 s4 s2 l3 l4
4 l1 s2 s3 s4 s2 l3 s4 l2
2 l1 s2 s3 s4 s2 l3 s4 s2 l4
4 l1 s2 s3 s4 s2 l3 s4 s2 s4 l2
3 l1 s2 s3 s4 s2 s3 l4 l2
2 l1 s2 s3 s4 s2 s3 l4 s2 l3
3 l1 s2 s3 s4 s2 s3 l4 s2 s3 l2
4 l1 s2 s3 s4 s2 s3 s4 l2 l3
3 l1 s2 s3 s4 s2 s3 s4 l2 s3 l4
2 l1 s2 s3 s4 s2 s3 s4 s2 l3 l4
1 s1 l2 l3 l4
4 s1 l2 l3 s4 l1
1 s1 l2 l3 s4 s1 l4
4 s1 l2 l3 s4 s1 s4 l1
1 s1 l2 l3 s4 s1 s4 s1 l4
4 s1 l2 l3 s4 s1 s4 s1 s4 l1
1 s1 l2 l3 s4 s1 s4 s1 s4 s1 l4
3 s1 l2 s3 l4 l1
1 s1 l2 s3 l4 s1 l3
3 s1 l2 s3 l4 s1 s3 l1
1 s1 l2 s3 l4 s1 s3 s1 l3
3 s1 l2 s3 l4 s1 s3 s1 s3 l1
1 s1 l2 s3 l4 s1 s3 s1 s3 s1 l3
4 s1 l2 s3 s4 l1 l3
3 s1 l2 s3 s4 l1 s3 l4
4 s1 l2 s3 s4 l1 s3 s4 l3
3 s1 l2 s3 s4 l1 s3 s4 s3 l4
4 s1 l2 s3 s4 l1 s3 s4 s3 s4 l3
1 s1 l2 s3 s4 s1 l3 l4
4 s1 l2 s3 s4 s1 l3 s4 l1
1 s1 l2 s3 s4 s1 l3 s4 s1 l4
4 s1 l2 s3 s4 s1 l3 s4 s1 s4 l1
3 s1 l2 s3 s4 s1 s3 l4 l1
1 s1 l2 s3 s4 s1 s3 l4 s1 l3
3 s1 l2 s3 s4 s1 s3 l4 s1 s3 l1
4 s1 l2 s3 s4 s1 s3 s4 l1 l3
3 s1 l2 s3 s4 s1 s3 s4 l1 s3 l4
1 s1 l2 s3 s4 s1 s3 s4 s1 l3 l4
2 s1 s2 l3 l4 l1
1 s1 s2 l3 l4 s1 l2
2 s1 s2 l3 l4 s1 s2 l1
1 s1 s2 l3 l4 s1 s2 s1 l2
2 s1 s2 l3 l4 s1 s2 s1 s2 l1
1 s1 s2 l3 l4 s1 s2 s1 s2 s1 l2
4 s1 s2 l3 s4 l1 l2
2 s1 s2 l3 s4 l1 s2 l4
4 s1 s2 l3 s4 l1 s2 s4 l2
2 s1 s2 l3 s4 l1 s2 s4 s2 l4
4 s1 s2 l3 s4 l1 s2 s4 s2 s4 l2
1 s1 s2 l3 s4 s1 l2 l4
4 s1 s2 l3 s4 s1 l2 s4 l1
1 s1 s2 l3 s4 s1 l2 s4 s1 l4
4 s1 s2 l3 s4 s1 l2 s4 s1 s4 l1
2 s1 s2 l3 s4 s1 s2 l4 l1
1 s1 s2 l3 s4 s1 s2 l4 s1 l2
2 s1 s2 l3 s4 s1 s2 l4 s1 s2 l1
4 s1 s2 l3 s4 s1 s2 s4 l1 l2
2 s1 s2 l3 s4 s1 s2 s4 l1 s2 l4
1 s1 s2 l3 s4 s1 s2 s4 s1 l2 l4
3 s1 s2 s3 l4 l1 l2
2 s1 s2 s3 l4 l1 s2 l3
3 s1 s2 s3 l4 l1 s2 s3 l2
2 s1 s2 s3 l4 l1 s2 s3 s2 l3
3 s1 s2 s3 l4 l1 s2 s3 s2 s3 l2
1 s1 s2 s3 l4 s1 l2 l3
3 s1 s2 s3 l4 s1 l2 s3 l1
1 s1 s2 s3 l4 s1 l2 s3 s1 l3
3 s1 s2 s3 l4 s1 l2 s3 s1 s3 l1
2 s1 s2 s3 l4 s1 s2 l3 l1
1 s1 s2 s3 l4 s1 s2 l3 s1 l2
2 s1 s2 s3 l4 s1 s2 l3 s1 s2 l1
3 s1 s2 s3 l4 s1 s2 s3 l1 l2
2 s1 s2 s3 l4 s1 s2 s3 l1 s2 l3
1 s1 s2 s3 l4 s1 s2 s3 s1 l2 l3
4 s1 s2 s3 s4 l1 l2 l3
3 s1 s2 s3 s4 l1 l2 s3 l4
4 s1 s2 s3 s4 l1 l2 s3 s4 l3
3 s1 s2 s3 s4 l1 l2 s3 s4 s3 l4
2 s1 s2 s3 s4 l1 s2 l3 l4
4 s1 s2 s3 s4 l1 s2 l3 s4 l2
2 s1 s2 s3 s4 l1 s2 l3 s4 s2 l4
3 s1 s2 s3 s4 l1 s2 s3 l4 l2
2 s1 s2 s3 s4 l1 s2 s3 l4 s2 l3
4 s1 s2 s3 s4 l1 s2 s3 s4 l2 l3
1 s1 s2 s3 s4 s1 l2 l3 l4
4 s1 s2 s3 s4 s1 l2 l3 s4 l1
1 s1 s2 s3 s4 s1 l2 l3 s4 s1 l4
3 s1 s2 s3 s4 s1 l2 s3 l4 l1
1 s1 s2 s3 s4 s1 l2 s3 l4 s1 l3
4 s1 s2 s3 s4 s1 l2 s3 s4 l1 l3
2 s1 s2 s3 s4 s1 s2 l3 l4 l1
1 s1 s2 s3 s4 s1 s2 l3 l4 s1 l2
4 s1 s2 s3 s4 s1 s2 l3 s4 l1 l2
3 s1 s2 s3 s4 s1 s2 s3 l4 l1 l2


Or if you prefer:
All 26 ways player 1 can win:
1 s1 l2 l3 l4
1 s1 l2 l3 s4 s1 l4
1 s1 l2 l3 s4 s1 s4 s1 l4
1 s1 l2 l3 s4 s1 s4 s1 s4 s1 l4
1 s1 l2 s3 l4 s1 l3
1 s1 l2 s3 l4 s1 s3 s1 l3
1 s1 l2 s3 l4 s1 s3 s1 s3 s1 l3
1 s1 l2 s3 s4 s1 l3 l4
1 s1 l2 s3 s4 s1 l3 s4 s1 l4
1 s1 l2 s3 s4 s1 s3 l4 s1 l3
1 s1 l2 s3 s4 s1 s3 s4 s1 l3 l4
1 s1 s2 l3 l4 s1 l2
1 s1 s2 l3 l4 s1 s2 s1 l2
1 s1 s2 l3 l4 s1 s2 s1 s2 s1 l2
1 s1 s2 l3 s4 s1 l2 l4
1 s1 s2 l3 s4 s1 l2 s4 s1 l4
1 s1 s2 l3 s4 s1 s2 l4 s1 l2
1 s1 s2 l3 s4 s1 s2 s4 s1 l2 l4
1 s1 s2 s3 l4 s1 l2 l3
1 s1 s2 s3 l4 s1 l2 s3 s1 l3
1 s1 s2 s3 l4 s1 s2 l3 s1 l2
1 s1 s2 s3 l4 s1 s2 s3 s1 l2 l3
1 s1 s2 s3 s4 s1 l2 l3 l4
1 s1 s2 s3 s4 s1 l2 l3 s4 s1 l4
1 s1 s2 s3 s4 s1 l2 s3 l4 s1 l3
1 s1 s2 s3 s4 s1 s2 l3 l4 s1 l2

All 28 ways player 2 can win:
2 l1 s2 l3 l4
2 l1 s2 l3 s4 s2 l4
2 l1 s2 l3 s4 s2 s4 s2 l4
2 l1 s2 l3 s4 s2 s4 s2 s4 s2 l4
2 l1 s2 s3 l4 s2 l3
2 l1 s2 s3 l4 s2 s3 s2 l3
2 l1 s2 s3 l4 s2 s3 s2 s3 s2 l3
2 l1 s2 s3 s4 s2 l3 l4
2 l1 s2 s3 s4 s2 l3 s4 s2 l4
2 l1 s2 s3 s4 s2 s3 l4 s2 l3
2 l1 s2 s3 s4 s2 s3 s4 s2 l3 l4
2 s1 s2 l3 l4 l1
2 s1 s2 l3 l4 s1 s2 l1
2 s1 s2 l3 l4 s1 s2 s1 s2 l1
2 s1 s2 l3 s4 l1 s2 l4
2 s1 s2 l3 s4 l1 s2 s4 s2 l4
2 s1 s2 l3 s4 s1 s2 l4 l1
2 s1 s2 l3 s4 s1 s2 l4 s1 s2 l1
2 s1 s2 l3 s4 s1 s2 s4 l1 s2 l4
2 s1 s2 s3 l4 l1 s2 l3
2 s1 s2 s3 l4 l1 s2 s3 s2 l3
2 s1 s2 s3 l4 s1 s2 l3 l1
2 s1 s2 s3 l4 s1 s2 l3 s1 s2 l1
2 s1 s2 s3 l4 s1 s2 s3 l1 s2 l3
2 s1 s2 s3 s4 l1 s2 l3 l4
2 s1 s2 s3 s4 l1 s2 l3 s4 s2 l4
2 s1 s2 s3 s4 l1 s2 s3 l4 s2 l3
2 s1 s2 s3 s4 s1 s2 l3 l4 l1

All 31 ways player 3 can win:
3 l1 l2 s3 l4
3 l1 l2 s3 s4 s3 l4
3 l1 l2 s3 s4 s3 s4 s3 l4
3 l1 l2 s3 s4 s3 s4 s3 s4 s3 l4
3 l1 s2 s3 l4 l2
3 l1 s2 s3 l4 s2 s3 l2
3 l1 s2 s3 l4 s2 s3 s2 s3 l2
3 l1 s2 s3 s4 l2 s3 l4
3 l1 s2 s3 s4 l2 s3 s4 s3 l4
3 l1 s2 s3 s4 s2 s3 l4 l2
3 l1 s2 s3 s4 s2 s3 l4 s2 s3 l2
3 l1 s2 s3 s4 s2 s3 s4 l2 s3 l4
3 s1 l2 s3 l4 l1
3 s1 l2 s3 l4 s1 s3 l1
3 s1 l2 s3 l4 s1 s3 s1 s3 l1
3 s1 l2 s3 s4 l1 s3 l4
3 s1 l2 s3 s4 l1 s3 s4 s3 l4
3 s1 l2 s3 s4 s1 s3 l4 l1
3 s1 l2 s3 s4 s1 s3 l4 s1 s3 l1
3 s1 l2 s3 s4 s1 s3 s4 l1 s3 l4
3 s1 s2 s3 l4 l1 l2
3 s1 s2 s3 l4 l1 s2 s3 l2
3 s1 s2 s3 l4 l1 s2 s3 s2 s3 l2
3 s1 s2 s3 l4 s1 l2 s3 l1
3 s1 s2 s3 l4 s1 l2 s3 s1 s3 l1
3 s1 s2 s3 l4 s1 s2 s3 l1 l2
3 s1 s2 s3 s4 l1 l2 s3 l4
3 s1 s2 s3 s4 l1 l2 s3 s4 s3 l4
3 s1 s2 s3 s4 l1 s2 s3 l4 l2
3 s1 s2 s3 s4 s1 l2 s3 l4 l1
3 s1 s2 s3 s4 s1 s2 s3 l4 l1 l2

All 35 ways player 4 can win:
4 l1 l2 l3
4 l1 l2 s3 s4 l3
4 l1 l2 s3 s4 s3 s4 l3
4 l1 l2 s3 s4 s3 s4 s3 s4 l3
4 l1 s2 l3 s4 l2
4 l1 s2 l3 s4 s2 s4 l2
4 l1 s2 l3 s4 s2 s4 s2 s4 l2
4 l1 s2 s3 s4 l2 l3
4 l1 s2 s3 s4 l2 s3 s4 l3
4 l1 s2 s3 s4 l2 s3 s4 s3 s4 l3
4 l1 s2 s3 s4 s2 l3 s4 l2
4 l1 s2 s3 s4 s2 l3 s4 s2 s4 l2
4 l1 s2 s3 s4 s2 s3 s4 l2 l3
4 s1 l2 l3 s4 l1
4 s1 l2 l3 s4 s1 s4 l1
4 s1 l2 l3 s4 s1 s4 s1 s4 l1
4 s1 l2 s3 s4 l1 l3
4 s1 l2 s3 s4 l1 s3 s4 l3
4 s1 l2 s3 s4 l1 s3 s4 s3 s4 l3
4 s1 l2 s3 s4 s1 l3 s4 l1
4 s1 l2 s3 s4 s1 l3 s4 s1 s4 l1
4 s1 l2 s3 s4 s1 s3 s4 l1 l3
4 s1 s2 l3 s4 l1 l2
4 s1 s2 l3 s4 l1 s2 s4 l2
4 s1 s2 l3 s4 l1 s2 s4 s2 s4 l2
4 s1 s2 l3 s4 s1 l2 s4 l1
4 s1 s2 l3 s4 s1 l2 s4 s1 s4 l1
4 s1 s2 l3 s4 s1 s2 s4 l1 l2
4 s1 s2 s3 s4 l1 l2 l3
4 s1 s2 s3 s4 l1 l2 s3 s4 l3
4 s1 s2 s3 s4 l1 s2 l3 s4 l2
4 s1 s2 s3 s4 l1 s2 s3 s4 l2 l3
4 s1 s2 s3 s4 s1 l2 l3 s4 l1
4 s1 s2 s3 s4 s1 l2 s3 s4 l1 l3
4 s1 s2 s3 s4 s1 s2 l3 s4 l1 l2


Your 52 cards with 4 aces is only minimally more complex, and again with the lack of replacement the exact number of possible card positions can easily be calculated (and has repeatedly in this thread - it's 270725). Note that that is not the number of distinct games, because you have more stoppers (Aces) than players to be eliminated, you can have a lot of duplicate games with distinct shuffles (there are only 51C3 or 20825 distinct games).
While that's too many to advise you to work out every possible game by hand, it's not too computationally burdensome, that, given sufficient error checking and oversight. A proper answer should be easily within grasp.
Edit:
I figured out the mistake I made
I now confirm ronjosen's results:
61644, 65218, 69335, 74528 (0.2277, 0.2409, 0.2561, 0.2753)
I have:
51309, 60944, 72196, 86276 (0.1895, 0.2251, 0.2666, 0.3188)


So I was off slightly, but able to easily correct my mistake.
My own work also indicates that drawing additional cards makes it significantly less fair to earlier players, as without the extra card draw I have:
65052, 66736, 68520, 70417 (0.2403, 0.2465, 0.2531, 0.2601)

My work also indicates that removing the extraneous ace also makes it slightly more fair to earlier players.
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Larry L
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tuckerotl wrote:

Your 52 cards with 4 aces is only minimally more complex, and again with the lack of replacement the exact number of possible card positions can easily be calculated (and has repeatedly in this thread - it's 270725). Note that that is not the number of distinct games, because you have more stoppers (Aces) than players to be eliminated, you can have a lot of duplicate games with distinct shuffles (there are only 51C3 or 20825 distinct games).


How do you make that adjustment (from 52C4 to 51C3)? If I try to follow that step, in order for the probabilities to hold up, it looks like there are 13 equivalent deals for each game, but if you added in the jokers would you have 54C4 deals but 53C3 games? Then there would be 13.5 deals per game which is why I'm having a hard time following.
 
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