 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Counting One's Marbles (Posted on 2008-02-18) There are two closed boxes, filled with red and white marbles. One of the boxes is 2/3 red and the other is 2/3 white.

You're allowed to sample 5 marbles from the first box, and 30 from the second. As a result, you get 4 red and 1 white marble from the first box, and 20 red and 10 white from the second. Which is more likely to be the one with majority red marbles?

Assume the two boxes contain the same number of marbles.

 Submitted by Charlie Rating: 1.0000 (1 votes) Solution: (Hide) The large sample size from the second box gives a greater probability that that one has the majority red marbles, and that the proportion from the first box was a statistical fluke. Certainly any number of marbles in each box that's less than 30 would disallow 20 red and 10 white from being drawn from either box, and any number less than 60 would preclude 20 reds from being drawn from the majority white box. For larger numbers of marbles, the following table holds: ``` n prob of both probability observations b has more reds given A has|given B has given the more reds | more reds observations 30 0.000000000 0.029472443 1.000000000 33 0.000000000 0.014246340 1.000000000 36 0.000000000 0.011346332 1.000000000 39 0.000000000 0.010032430 1.000000000 42 0.000000000 0.009270018 1.000000000 45 0.000000000 0.008768876 1.000000000 48 0.000000000 0.008413313 1.000000000 51 0.000000000 0.008147555 1.000000000 54 0.000000000 0.007941237 1.000000000 57 0.000000000 0.007776355 1.000000000 60 0.000000002 0.007641534 0.999999686 63 0.000000012 0.007529226 0.999998406 66 0.000000035 0.007434220 0.999995330 69 0.000000076 0.007352801 0.999989633 72 0.000000141 0.007282248 0.999980613 75 0.000000233 0.007220521 0.999967753 78 0.000000353 0.007166061 0.999950726 81 0.000000503 0.007117658 0.999929385 84 0.000000681 0.007074354 0.999903728 87 0.000000887 0.007035385 0.999873870 90 0.000001120 0.007000130 0.999840008 93 0.000001377 0.006968084 0.999802400 96 0.000001656 0.006938828 0.999761335 99 0.000001956 0.006912012 0.999717126 102 0.000002273 0.006887344 0.999670088 105 0.000002606 0.006864576 0.999620535 108 0.000002952 0.006843497 0.999568770 111 0.000003311 0.006823926 0.999515081 114 0.000003679 0.006805706 0.999459739 117 0.000004055 0.006788702 0.999402997 120 0.000004438 0.006772797 0.999345088 123 0.000004827 0.006757888 0.999286224 126 0.000005220 0.006743884 0.999226600 129 0.000005615 0.006730704 0.999166393 132 0.000006013 0.006718279 0.999105763 135 0.000006412 0.006706546 0.999044852 138 0.000006811 0.006695447 0.998983788 141 0.000007210 0.006684934 0.998922688 144 0.000007607 0.006674960 0.998861652 147 0.000008003 0.006665487 0.998800771 150 0.000008397 0.006656476 0.998740126 inf 0.000049195 0.006296924 0.992248062 ``` The case of an infinite number of marbles in each box is equivalent to sampling with replacement, and it's still over 99% probable that the 20/10 observed box is the one with more red marbles. Note the first two columns of probabilities are the combined probabilities of both events, so, for example in the case of 30 marbles, the probability of the occurrence given the majority of reds is in box 2 is 0.029472443, rather than the 1 that would be the probability that box 2 would result in 20 red and 10 white being observed, as 0.029472443 is the probability that the 10 red/20 white box would result in the 4 red/1 white observation, multiplied by 1. DECLARE FUNCTION p30y20p10# (r#, w#) DECLARE FUNCTION p5y4p1# (r#, w#) DEFDBL A-Z FOR n = 30 TO 150 STEP 3 n1r = n * 2 / 3: n1w = n / 3 n2r = n1w: n2w = n1r p1 = p5y4p1(n1r, n1w) * p30y20p10(n2r, n2w) p2 = p5y4p1(n2r, n2w) * p30y20p10(n1r, n1w) PRINT USING "### #.######### #.######### #.#########"; n; p1; p2; p2 / (p1 + p2) NEXT p1 = 5 * (n1r / (n1w + n1r)) ^ 4 * n1w / (n1w + n1r) * 30045015# * (n2r / (n2w + n2r)) ^ 20 * (n2w / (n2w + n2r)) ^ 10 p2 = 5 * (n2r / (n2w + n2r)) ^ 4 * n2w / (n2w + n2r) * 30045015# * (n1r / (n1w + n1r)) ^ 20 * (n1w / (n1w + n1r)) ^ 10 PRINT PRINT USING "& #.######### #.######### #.#########"; "inf"; p1; p2; p2 / (p1 + p2) FUNCTION p30y20p10 (r, w) n = r + w p = 1 FOR i = 0 TO 19 p = p * (r - i) / n: n = n - 1 NEXT FOR i = 0 TO 9 p = p * (w - i) / n: n = n - 1 NEXT p30y20p10 = p * 30045015# ' comb(30,10) END FUNCTION FUNCTION p5y4p1 (r, w) n = r + w p = 1 FOR i = 0 TO 3 p = p * (r - i) / n: n = n - 1 NEXT FOR i = 0 TO 0 p = p * (w - i) / n: n = n - 1 NEXT p5y4p1 = p * 5 ' comb(5,1) END FUNCTION Subject Author Date re: Solution ThoughtProvoker 2008-02-22 00:11:13 Solution Eric 2008-02-21 03:39:38 solution ThoughtProvoker 2008-02-19 08:02:02 My Thoughts Dej Mar 2008-02-19 08:00:04 My Thoughts Praneeth 2008-02-19 06:17:55 Please log in:

 Search: Search body:
Forums (5)