Probability question
November 1st, 2012 at 6:01:02 PM
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Hey all, mathematically deficient human here looking for an answer to the following...
Assuming a player is dealt four cards from a continuously shuffled 4 deck shoe, what would his average total score/hand composition be assuming the following values:
Ace: 1 point
2 through 10: Pip value
JQK: 10 points
Thank you for your help/time!
Assuming a player is dealt four cards from a continuously shuffled 4 deck shoe, what would his average total score/hand composition be assuming the following values:
Ace: 1 point
2 through 10: Pip value
JQK: 10 points
Thank you for your help/time!
November 1st, 2012 at 6:05:39 PM
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To the nearest whole number, I get 26 as a mean value.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
November 1st, 2012 at 6:11:15 PM
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That was fast. Can you show me how you arrived at that?
November 1st, 2012 at 6:12:34 PM
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Me too. Even if one adjusts for the fact that you are less likely to get two of the same card it is still only slightly over 26.
November 1st, 2012 at 6:17:30 PM
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Quote: chrisjs87That was fast. Can you show me how you arrived at that?
The average card value is 6.53, and you're drawing 4 cards. I just took 6.53*4 and rounded.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
November 1st, 2012 at 6:19:58 PM
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add all values = 86
divide by 13 = 6.6
multiply by 4 = 26.4
divide by 13 = 6.6
multiply by 4 = 26.4
November 1st, 2012 at 6:26:50 PM
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Quote: Dynamiteadd all values = 86
divide by 13 = 6.6
multiply by 4 = 26.4
85, not 86. 85/13 =6.538 x 4 = 26.154 (rounded)
November 1st, 2012 at 7:16:42 PM
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The distribution from the polynomial
(x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10+x^10+x^10+x^10)^4
4d13
Like rolling 4, 13 sided dice (are there any?)
or just rolling one, 4 times
(x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10+x^10+x^10+x^10)^4
4d13
Like rolling 4, 13 sided dice (are there any?)
or just rolling one, 4 times
total ways prob cumulative
4 1 0.0000350 0.0000350
5 4 0.0001401 0.0001751
6 10 0.0003501 0.0005252
7 20 0.0007003 0.0012254
8 35 0.0012254 0.0024509
9 56 0.0019607 0.0044116
10 84 0.0029411 0.0073527
11 120 0.0042015 0.0115542
12 165 0.0057771 0.0173313
13 232 0.0081230 0.0254543
14 318 0.0111341 0.0365884
15 420 0.0147054 0.0512937
16 535 0.0187318 0.0700256
17 660 0.0231084 0.0931340
18 792 0.0277301 0.1208641
19 928 0.0324919 0.1533560
20 1065 0.0372886 0.1906446
21 1200 0.0420153 0.2326599
22 1384 0.0484577 0.2811176
23 1524 0.0533595 0.3344771
24 1623 0.0568257 0.3913028
25 1684 0.0589615 0.4502643
26 1710 0.0598719 0.5101362
27 1704 0.0596618 0.5697980
28 1669 0.0584363 0.6282343
29 1608 0.0563005 0.6845349
30 1524 0.0533595 0.7378943
31 1528 0.0534995 0.7913939
32 1299 0.0454816 0.8368755
33 1092 0.0382340 0.8751094
34 906 0.0317216 0.9068310
35 740 0.0259095 0.9327405
36 593 0.0207626 0.9535030
37 464 0.0162459 0.9697490
38 352 0.0123245 0.9820735
39 256 0.0089633 0.9910367
40 256 0.0089633 1.0000000
total 28561 1.0000000
winsome johnny (not Win some johnny)
November 1st, 2012 at 8:16:32 PM
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you are right, 85 instead of 86. I stand corrected. I added the ace twice
November 2nd, 2012 at 4:02:54 PM
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Reminds me of the old Blackjack Question "Whats the average 3 card hand" which by same determination is 19.6 ( a winning hand for the Player in general). But alas, its not very favorable to hit 16 or Double A5 vs. 6 and achieve 19 or 20, tho you might win with 17 or 14 respectively. ;o)
Some people need to reimagine their thinking.
