Card Game - Page 3

BradleyT · #1 10-06-2004, 03:22 PM

[ 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.

Smackdab · #2 10-06-2004, 03:40 PM

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

[/ QUOTE ]

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

MrX · #3 10-06-2004, 03:22 PM

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

Lori · #4 10-06-2004, 03:25 PM

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 [img]/images/graemlins/spade.gif[/img]?

Lori

BradleyT · #5 10-06-2004, 03:27 PM

[ 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 [img]/images/graemlins/spade.gif[/img]?

Lori

[/ QUOTE ]

Easy - Lee Jones.

pudley4 · #6 10-06-2004, 03:09 PM

[ 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)

junkmail3 · #7 10-06-2004, 03:23 PM

[ 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 · #8 10-06-2004, 03:23 PM

[ 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)

MrX · #9 10-06-2004, 03:32 PM

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

theghost · #10 10-06-2004, 03:01 PM

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)