Card Game

[jasonHoldEm]

Given: <ul type="square"> One standard deck of 52 cards One new car* 51 other people You [/list]

Problem #1

Everyone lines up (the first person in line is player 1, second is player 2, etc). Player 1 selects a card from the deck of 52, if it's the ace of spades he wins the car. If player 1 is unsuccessful his card is discarded and player 2 will now select a card (from a deck of 51), this process will continue until someone pulls the ace of spades and wins the car.

Where do you want to stand in line to maximize your chance of winning the car?

Problem #2

Everyone lines up again. Player one goes to the deck and selects one card, if it is the ace of spades he wins the car, if not his card is replaced and the deck is reshuffled. Player two will now go to the deck and select TWO cards if either of them are the ace of spades he wins the car, if not his cards are replaced and the deck is reshuffled. This process continues where each player selects cards equal to their place in line (third person selects 3, fourth selects 4, etc).

Where do you want to stand in line to maximize your chance of winning the car?

(I'm not sure of the answer to the second problem. /images/graemlins/crazy.gif )

Have fun, post your answers before you peek at anyone else's.

Peace,
J

* You won't actually win a car, this is just for fun.

[junkmail3]

1. Last
2. Last

Both chances you have a 1/1 chance of winning. It's the best probability.

Now I don't know if I would really want to, but you know. From a statistical standpoint, you have the best chance here.

[junkmail3]

Oh, and what kind of car. It might change my answer.

[GrannyMae]

just a guess:

1. all have an equal chance if the deck is shuffled properly, so i would guess the last position is best.

2. same as #1 for same reason

[BradleyT]

1. First position is the only one that is guaranteed to always have a chance to win the car.

2. Again first position is the only one guaranteed to have a chance however the ability to redraw in a later position probably increases your odds. The 52nd person would never win because the odds of the 51st, 50th, 49th, 48th, etc not picking the ace are astronomical.

No clue how to solve the 2nd but I'd assume you'd say something like:
Player 1 has 1/52 chance
Player 2 has 2/52 chance
If I'm player 3 I have 3/52 chance on my picks but to get my chance players one and 2 have to fail so my odds are X%.

As a guess I'd say position 4.

[TheRake]

#1 = 27th
#2 = Not sure, but seems like it should be 27 also.

Truthfully I am not sure of either of the guesses /images/graemlins/smile.gif

TheRake

[jasonHoldEm]

Answers (?)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
In the first problem everyone has an equal chance of pulling the ace of spades (odds will be 51:1 for everyone because the cards are not replaced). I'm pretty sure it doesn't matter where you stand, although the argument to go first seems logical.

The second problem requires more math than I can handle, but here is my general logic. Odds for the first player are 51:1, the second player will have the combined odds of 51:1 and 50:1 but he will only get to pick 51 times in 52 (because the first player will have won the other time).

I'm sure there is a way to figure out your max EV by comparing your combined odds of pulling the ace of spades with the chances that someone in front of you will win first. I'm almost positive this problem will have a specific answer for a position in line, but we might need bruceZ to give us the answer. /images/graemlins/blush.gif

J

[jek187]

1) All spots are equal:
1st person: 1/52
2nd person: 1/51*51/52
3rd person: 1/50*51/52*50/51
etc
These all come out to 1.92%

2nd:
7th
You have an 8.8% chance of winning in this spot, which is the best you can ask for. I just used a brute force method of computing all the chances w/a spreadsheet and am too lazy to type it out here.

[B00T]

I am no math major nor do I like typing out fractions but whatever...

For the first + second one, I want to be standing right before the chance = 50% of all the previous pullers chances combined.

First person has 1/52 chance, 2nd person has 1/51 chance, I wanna be standing before the person who first has the 50% chance in that plan. What spot it is I have no idea, but I want to go before the 50% threshold is assumed in all the previous people picking. I'm guessing I would want to be like the 23rd person on line.

For part 2, the chances of each person getting it increases after the first one. So once the total pulls = 26 taken,, so I'd say I want to be the 8th person in line or so.

Someone can do the math, but I think thats the right answer.

[daryn]

i disagree that all spots are equal in the first one. assuming there is only 1 car to give away, not all spots have an equal chance to win.

the first guy is 51:1 against winning, but the second guy is worse. of course if he gets to pick he is 51:1 also, but he won't even get to pick 1 out of 51 times.

you want to be first.

[jek187]

[ QUOTE ]
the first guy is 51:1 against winning, but the second guy is worse. of course if he gets to pick he is 51:1 also, but he won't even get to pick 1 out of 51 times.

[/ QUOTE ]

The 2nd guy is 50:1, not 51:1 (the drawn cards are discarded.) All spots are equal due to this in problem #1.

[GrannyMae]

I'm pretty sure it doesn't matter where you stand

wtf is this? you told us to pick a spot!

boots answer sounds good tho.

just curious. is it Eugeenes car? if so, i don't want the stinkin' wreck.

u buy a new car yet with all your aruba money eugeene? i doubt it, but that's good since you will probably lose your roll this year.

http://users.telenet.be/eforum/emoticons4u/trans/fahr04.gif

[theghost]

1- first, so no one wins it before me.

2- guessing roughly 10th, to maximize cards I draw without too much risk that someone else wins first. (really wanted to peek before I put up this answer)

[jasonHoldEm]

[ QUOTE ]
[ QUOTE ]
I'm pretty sure it doesn't matter where you stand


[/ QUOTE ]
wtf is this? you told us to pick a spot!


[/ QUOTE ]

Sorry granny, if it makes you feel better everyone is right as long as they pick a spot. Unless of course first is the actually answer, note how I used the carefully placed "I'm pretty sure." /images/graemlins/smile.gif

[ QUOTE ]
is it Eugeenes car?

[/ QUOTE ]

Please refer to the original post, I clearly stated the prize was a (fake) new car. Eugeenes car wouldn't qualify...

http://www.wymondhamleics.free-online.co.uk/album/oldcar.jpg

[Cleveland Guy]

Problem #1 - Who cares. I have a 1 of 52 chance no matter where I am. They might as well all put a card in front of us and have us turn it up.

Problem #2 - 1+2+3+4+5+6+7+8+9+10 = 55

I want to be around 8th or 9th.

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
</font><blockquote><font class="small">In risposta di:</font><hr />
the first guy is 51:1 against winning, but the second guy is worse. of course if he gets to pick he is 51:1 also, but he won't even get to pick 1 out of 51 times.

[/ QUOTE ]

The 2nd guy is 50:1, not 51:1 (the drawn cards are discarded.) All spots are equal due to this in problem #1.

[/ QUOTE ]



oh alright, my mistake, you're right.

[Lori]

DARYN!!!! Too many beers?

#1 you have a lotto machine with 52 balls. Everyone pulls a number and doesn't look at it, the person who draws number 33 wins the car.
Everyone writes their name on the ball and hands it back in, the person who wrote on number 33 is revealed.
Do you think it matters when you pull?

#2 Can't be bothered to check Jek's math, but when guy #51 pulls he has a 51/52 chance of hitting, would you rather be in seat #52?

Also, guy #1 has a 1/52 chance of hitting, guy #2 has more than double that if he gets to pull.
Therefore he must be better than #1.
Repeat this for #2 and #4 and instinctively you can still see it's right.
However when you get to guy #8 you can see there have been 28 pulls already, which even instinctively is somewhere around 50%, so you're probably in trouble.

Lori

[pudley4]

[ QUOTE ]
1) All spots are equal:
1st person: 1/52
2nd person: 1/51*51/52
3rd person: 1/50*51/52*50/51
etc
These all come out to 1.92%

2nd:
7th
You have an 8.8% chance of winning in this spot, which is the best you can ask for. I just used a brute force method of computing all the chances w/a spreadsheet and am too lazy to type it out here.

[/ QUOTE ]

This is correct.

Game 1 should be self-explanatory

Game 2 is as follows:
Several step process.

1 - For each player, figure the chance of correctly picking the A, when choosing n/52 cards (where n=the order you are picking in). E.G 1/52 for the first player, 2/52 for the second, etc.

2 - For each player, figure out the chance of not picking the A (this is just 1 minus the number in step 1)

3 - Figure out the chance of none of the previous players winning (multiply all the losing probabilities for each previous player from step 2 together. E.G. For player 4, multiply the chance of player 1 not winning [51/52] times player 2 not winning [50/52] times player 3 not winning [49/52] - this is the probability that player 4 gets a chance to pick)

4 - For each player, multiply the chance of picking the correct card (step 1) by the chance of them actually getting to pick (step 3). Player 7 has the highest chance (~8.8%, as stated above)

[Lori]

Whew!

Welcome back.

Lori

[BradleyT]

[ QUOTE ]
DARYN!!!! Too many beers?

#1 you have a lotto machine with 52 balls. Everyone pulls a number and doesn't look at it, the person who draws number 33 wins the car.
Everyone writes their name on the ball and hands it back in, the person who wrote on number 33 is revealed.
Do you think it matters when you pull?

[/ QUOTE ]

That's not how it works. Everyone doesn't pull their balls all at once. You pull a ball until #33 is found. Once it's found there's no need to keep pulling your balls.

[MrX]

haven't looked at any answers yet but here is my take

scenario 1: simple all 52 have exactly the same same chance of winning

scenario 2: complicated based on your likelihood of winning and also you likelihood of even getting a chance to draw cards. It will max out sooner than you think, probably around picks 7-10. I am gonna work this out for fun then post my answer and look at the other results.

MrX

[junkmail3]

[ QUOTE ]
[ QUOTE ]
1) All spots are equal:
1st person: 1/52
2nd person: 1/51*51/52
3rd person: 1/50*51/52*50/51
etc
These all come out to 1.92%

2nd:
7th
You have an 8.8% chance of winning in this spot, which is the best you can ask for. I just used a brute force method of computing all the chances w/a spreadsheet and am too lazy to type it out here.

[/ QUOTE ]

This is correct.

Game 1 should be self-explanatory


[/ QUOTE ]

Then I want to pick 18th. 18 is HOT!

[junkmail3]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
1) All spots are equal:
1st person: 1/52
2nd person: 1/51*51/52
3rd person: 1/50*51/52*50/51
etc
These all come out to 1.92%

2nd:
7th
You have an 8.8% chance of winning in this spot, which is the best you can ask for. I just used a brute force method of computing all the chances w/a spreadsheet and am too lazy to type it out here.

[/ QUOTE ]

This is correct.

Game 1 should be self-explanatory


[/ QUOTE ]

Then I want to pick 18th. 18 is HOT!

[/ QUOTE ]

(tonight)

[Lori]

That's not how it works. Everyone doesn't pull their balls all at once. You pull a ball until #33 is found. Once it's found there's no need to keep pulling your balls.


How about if these people work for party support and don't know what the ace of spades looks like.
They all pull a card and then run around trying to find out who has the ace of spades.

Eventually they find Lucy Jones and a winner is declared.

Which one has the best chance of having the A /images/graemlins/spade.gif?

Lori

[BradleyT]

[ QUOTE ]
That's not how it works. Everyone doesn't pull their balls all at once. You pull a ball until #33 is found. Once it's found there's no need to keep pulling your balls.


How about if these people work for party support and don't know what the ace of spades looks like.
They all pull a card and then run around trying to find out who has the ace of spades.

Eventually they find Lucy Jones and a winner is declared.

Which one has the best chance of having the A /images/graemlins/spade.gif?

Lori

[/ QUOTE ]

Easy - Lee Jones.

[MrX]

wrote a quick little spreadsheet to fig it out (scenario 2 that is), this shows a slight increased chance of player 7 having the best odds at about 8.8%. The bolded "1" at the bottom of the last column shows that all percentages added together equal 1 (100%). I think these are fun.

MrX



order cards chance to chance product
drawn win if you you get to
do draw draw
1 1 0.019230769 1 0.019230769
2 2 0.038461538 0.980769231 0.037721893
3 3 0.057692308 0.943047337 0.054406577
4 4 0.076923077 0.88864076 0.068356982
5 5 0.096153846 0.820283779 0.07887344
6 6 0.115384615 0.741410338 0.085547347
7 7 0.134615385 0.655862992 0.088289249
8 8 0.153846154 0.567573743 0.087319037
9 9 0.173076923 0.480254705 0.083121007
10 10 0.192307692 0.397133699 0.076371865
11 11 0.211538462 0.320761834 0.067853465
12 12 0.230769231 0.252908369 0.05836347
13 13 0.25 0.194544899 0.048636225
14 14 0.269230769 0.145908674 0.039283105
15 15 0.288461538 0.10662557 0.030757376
16 16 0.307692308 0.075868194 0.02334406
17 17 0.326923077 0.052524134 0.017171352
18 18 0.346153846 0.035352783 0.012237502
19 19 0.365384615 0.023115281 0.008445968
20 20 0.384615385 0.014669313 0.005642043
21 21 0.403846154 0.009027269 0.003645628
22 22 0.423076923 0.005381641 0.002276848
23 23 0.442307692 0.003104793 0.001373274
24 24 0.461538462 0.001731519 0.000799163
25 25 0.480769231 0.000932357 0.000448248
26 26 0.5 0.000484108 0.000242054
27 27 0.519230769 0.000242054 0.000125682
28 28 0.538461538 0.000116372 6.26619E-05
29 29 0.557692308 5.37102E-05 2.99538E-05
30 30 0.576923077 2.37564E-05 1.37056E-05
31 31 0.596153846 1.00508E-05 5.99183E-06
32 32 0.615384615 4.05898E-06 2.49783E-06
33 33 0.634615385 1.56115E-06 9.90727E-07
34 34 0.653846154 5.70419E-07 3.72966E-07
35 35 0.673076923 1.97453E-07 1.32901E-07
36 36 0.692307692 6.45518E-08 4.46897E-08
37 37 0.711538462 1.98621E-08 1.41326E-08
38 38 0.730769231 5.72945E-09 4.18691E-09
39 39 0.75 1.54254E-09 1.15691E-09
40 40 0.769230769 3.85636E-10 2.96643E-10
41 41 0.788461538 8.89929E-11 7.01675E-11
42 42 0.807692308 1.88254E-11 1.52052E-11
43 43 0.826923077 3.62028E-12 2.99369E-12
44 44 0.846153846 6.26586E-13 5.30188E-13
45 45 0.865384615 9.63979E-14 8.34212E-14
46 46 0.884615385 1.29766E-14 1.14793E-14
47 47 0.903846154 1.4973E-15 1.35333E-15
48 48 0.923076923 1.43972E-16 1.32897E-16
49 49 0.942307692 1.10747E-17 1.04358E-17
50 50 0.961538462 6.38927E-19 6.14353E-19
51 51 0.980769231 2.45741E-20 2.41015E-20
52 52 1 4.72579E-22 4.72579E-22
1

[pudley4]

Here are the numbers for each position:

<font class="small">Code:</font><hr /><pre>
win cards total cards % winning pick %losing pick %no previous winner %win

1 52 0.019230769 0.980769231 1 0.019230769
2 52 0.038461538 0.961538462 0.980769231 0.037721893
3 52 0.057692308 0.942307692 0.943047337 0.054406577
4 52 0.076923077 0.923076923 0.88864076 0.068356982
5 52 0.096153846 0.903846154 0.820283779 0.07887344
6 52 0.115384615 0.884615385 0.741410338 0.085547347
7 52 0.134615385 0.865384615 0.655862992 **0.088289249**
8 52 0.153846154 0.846153846 0.567573743 0.087319037
9 52 0.173076923 0.826923077 0.480254705 0.083121007
10 52 0.192307692 0.807692308 0.397133699 0.076371865
11 52 0.211538462 0.788461538 0.320761834 0.067853465
12 52 0.230769231 0.769230769 0.252908369 0.05836347
13 52 0.25 0.75 0.194544899 0.048636225
14 52 0.269230769 0.730769231 0.145908674 0.039283105
15 52 0.288461538 0.711538462 0.10662557 0.030757376
16 52 0.307692308 0.692307692 0.075868194 0.02334406
17 52 0.326923077 0.673076923 0.052524134 0.017171352
18 52 0.346153846 0.653846154 0.035352783 0.012237502
19 52 0.365384615 0.634615385 0.023115281 0.008445968
20 52 0.384615385 0.615384615 0.014669313 0.005642043
21 52 0.403846154 0.596153846 0.009027269 0.003645628
22 52 0.423076923 0.576923077 0.005381641 0.002276848
23 52 0.442307692 0.557692308 0.003104793 0.001373274
24 52 0.461538462 0.538461538 0.001731519 0.000799163
25 52 0.480769231 0.519230769 0.000932357 0.000448248
26 52 0.5 0.5 0.000484108 0.000242054
27 52 0.519230769 0.480769231 0.000242054 0.000125682
28 52 0.538461538 0.461538462 0.000116372 6.26619E-05
29 52 0.557692308 0.442307692 5.37102E-05 2.99538E-05
30 52 0.576923077 0.423076923 2.37564E-05 1.37056E-05
31 52 0.596153846 0.403846154 1.00508E-05 5.99183E-06
32 52 0.615384615 0.384615385 4.05898E-06 2.49783E-06
33 52 0.634615385 0.365384615 1.56115E-06 9.90727E-07
34 52 0.653846154 0.346153846 5.70419E-07 3.72966E-07
35 52 0.673076923 0.326923077 1.97453E-07 1.32901E-07
36 52 0.692307692 0.307692308 6.45518E-08 4.46897E-08
37 52 0.711538462 0.288461538 1.98621E-08 1.41326E-08
38 52 0.730769231 0.269230769 5.72945E-09 4.18691E-09
39 52 0.75 0.25 1.54254E-09 1.15691E-09
40 52 0.769230769 0.230769231 3.85636E-10 2.96643E-10
41 52 0.788461538 0.211538462 8.89929E-11 7.01675E-11
42 52 0.807692308 0.192307692 1.88254E-11 1.52052E-11
43 52 0.826923077 0.173076923 3.62028E-12 2.99369E-12
44 52 0.846153846 0.153846154 6.26586E-13 5.30188E-13
45 52 0.865384615 0.134615385 9.63979E-14 8.34212E-14
46 52 0.884615385 0.115384615 1.29766E-14 1.14793E-14
47 52 0.903846154 0.096153846 1.4973E-15 1.35333E-15
48 52 0.923076923 0.076923077 1.43972E-16 1.32897E-16
49 52 0.942307692 0.057692308 1.10747E-17 1.04358E-17
50 52 0.961538462 0.038461538 6.38927E-19 6.14353E-19
51 52 0.980769231 0.019230769 2.45741E-20 2.41015E-20
52 52 1 0 4.72579E-22 4.72579E-22
</pre><hr />

[MrX]

we must have written almost the same formula Jeff.

I like your formatting much better. I think we can consider the book close on this one.

MrX

[Smackdab]

[ QUOTE ]
Once it's found there's no need to keep pulling your balls

[/ QUOTE ]

Once I found it ...... I couldn't stop. /images/graemlins/wink.gif /images/graemlins/cool.gif /images/graemlins/blush.gif

[B00T]

Yea 8th is the pick. That is the mathmatical computer saying why. I guess I tried to prove why you want that pick in a logical way. Everyone else picking before you has a less than 50/50 shot of getting the car. Anyone picking after #8 has a more than 50% shot of that 1/52 having already come up.

So what about the first case? At what point do you start approaching that 50% threshold? My arguement here coincides with Lori's Cmon math boys...

[jasonHoldEm]

Awesome, nothing like some hardcore math and data to get the zoo running in top form.

My work here is done, thanks to all for the help on problem 2. Anyone who thinks the zoo is dead just needs to see a thread like this occasionally to see we still got it.

J

[GrannyMae]

1 1 0.019230769 1 0.019230769
2 2 0.038461538 0.980769231 0.037721893
3 3 0.057692308 0.943047337 0.054406577
4 4 0.076923077 0.88864076 0.068356982
5 5 0.096153846 0.820283779 0.07887344
6 6 0.115384615 0.741410338 0.085547347
7 7 0.134615385 0.655862992 0.088289249
8 8 0.153846154 0.567573743 0.087319037
9 9 0.173076923 0.480254705 0.083121007
10 10 0.192307692 0.397133699 0.076371865
11 11 0.211538462 0.320761834 0.067853465
12 12 0.230769231 0.252908369 0.05836347
13 13 0.25 0.194544899 0.048636225
14 14 0.269230769 0.145908674 0.039283105
15 15 0.288461538 0.10662557 0.030757376
16 16 0.307692308 0.075868194 0.02334406
17 17 0.326923077 0.052524134 0.017171352
18 18 0.346153846 0.035352783 0.012237502
19 19 0.365384615 0.023115281 0.008445968
20 20 0.384615385 0.014669313 0.005642043
21 21 0.403846154 0.009027269 0.003645628
22 22 0.423076923 0.005381641 0.002276848
23 23 0.442307692 0.003104793 0.001373274
24 24 0.461538462 0.001731519 0.000799163
25 25 0.480769231 0.000932357 0.000448248
26 26 0.5 0.000484108 0.000242054
27 27 0.519230769 0.000242054 0.000125682
28 28 0.538461538 0.000116372 6.26619E-05
29 29 0.557692308 5.37102E-05 2.99538E-05
30 30 0.576923077 2.37564E-05 1.37056E-05
31 31 0.596153846 1.00508E-05 5.99183E-06
32 32 0.615384615 4.05898E-06 2.49783E-06
33 33 0.634615385 1.56115E-06 9.90727E-07
34 34 0.653846154 5.70419E-07 3.72966E-07
35 35 0.673076923 1.97453E-07 1.32901E-07
36 36 0.692307692 6.45518E-08 4.46897E-08
37 37 0.711538462 1.98621E-08 1.41326E-08
38 38 0.730769231 5.72945E-09 4.18691E-09
39 39 0.75 1.54254E-09 1.15691E-09
40 40 0.769230769 3.85636E-10 2.96643E-10
41 41 0.788461538 8.89929E-11 7.01675E-11
42 42 0.807692308 1.88254E-11 1.52052E-11
43 43 0.826923077 3.62028E-12 2.99369E-12
44 44 0.846153846 6.26586E-13 5.30188E-13
45 45 0.865384615 9.63979E-14 8.34212E-14
46 46 0.884615385 1.29766E-14 1.14793E-14
47 47 0.903846154 1.4973E-15 1.35333E-15
48 48 0.923076923 1.43972E-16 1.32897E-16
49 49 0.942307692 1.10747E-17 1.04358E-17
50 50 0.961538462 6.38927E-19 6.14353E-19
51 51 0.980769231 2.45741E-20 2.41015E-20
52 52 1 4.72579E-22 4.72579E-22
1



http://users.telenet.be/eforum/emoticons4u/violent/sterb207.gif

[GrannyMae]

http://www.wymondhamleics.free-online.co.uk/album/oldcar.jpg


is that punk's castle on the left? also, i would absolutely love to own that car. seriously.
i'll bet you could recoup anything you might pay for it just by salvaging those intact windows and reselling them as spare parts.
(unless that is saran wrap)

http://users.telenet.be/eforum/emoticons4u/trans/fahr29.gif

[Lori]

That's Eugene's car.

Lori

[34TheTruth34]

[ QUOTE ]
all have an equal chance if the deck is shuffled properly, so i would guess the last position is best.


[/ QUOTE ]

wha???

[Cazz]

I got the same answer.

Simplied formula

insert formula into D2
=(ROW(D2)-1)/52*(1-SUM($D$1:D1))
copy and paste d2-d53
You can put 1-52 in C2 thru C53 and replace the stuff
inside the first () w/ C2.

[Homer]

I sort of skimmed the thread and didn't see any simple math provided for Problem 1...

Probability of winning

Contestant #1 = 1/52
Contestant #2 = (51/52)*(1/51) = 1/52
Contestant #3 = (51/52)*(50/51)*(1/50) = 1/52

...and so on.

-- Homer

[SamJack]

Jek and whoever agreed with him has the right answer.

Since homer provided the formula for #1.
Here it is for #2.

For N, the probability you will be the winner is:

(Probability that you will have ace spade out of the N cards you are allowed to pick) * (probability that the people who went before you did not win)

N/52 * (1 - Sum(probofwinning(1) to probofwinning(n-1)))

Sorry I can't type the mathamatical notation, but you get the idea.

SamJack

[pzhon]

Problem 2: A simpler way to express the probability of winning in position N is

(51/52)(50/52)(49/52)...((52-N+1)/52) * (N/52).

The probability that the Nth person wins divided by the probability that the N+1st person wins is

(52/(52-N)) (N/N+1).

This should switch from greater than 1 to less than 1 at the maximum. The maximum should be close to where this ratio equals 1 for some real N.

(52/(52-N)) (N/N+1) = 1
...
(N + 1/2)^2 = 52.25

N = Sqrt(52.25)-.5 = 6.728.

For smaller values including N=6, the Nth person wins less frequently than the N+1st person wins. So, person 7 does better than person 6.

For larger values including N=7, the Nth person wins more than the N+1st person wins. So, person 7 does better than person 8.

With c cards in the deck, the probability of winning is greatest for the person in the position closest to sqrt(c+.25). For c=20, sqrt(c+.25) = 4.5, and positions 4 and 5 have equal chances.