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43 members have voted

charliepatrick
charliepatrick
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October 22nd, 2021 at 4:56:45 PM permalink
I like the idea that you think of the puzzle as an 18-bit number and are looking for how many 0's there are. If there were at least 18 people then one could consider that for every combination of 0's and 1's there is a complimentary number of 1's and 0's. In total those two combinations have 18 0's and 18 1's, so the average of those two is 9 or each. Thus the average for all combinations is 9.
In this example there are only sixteen people so any combination with 17 or 18 0's will have no people getting across rather than -1 or -2. Thus one needs to add in 18*1 and 1*2 = 20 before dividing by 218 giving the 7.000076 answer.
aceside
aceside
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October 23rd, 2021 at 1:51:58 PM permalink
Quote: charliepatrick

I like the idea that you think of the puzzle as an 18-bit number and are looking for how many 0's there are. If there were at least 18 people then one could consider that for every combination of 0's and 1's there is a complimentary number of 1's and 0's. In total those two combinations have 18 0's and 18 1's, so the average of those two is 9 or each. Thus the average for all combinations is 9.
In this example there are only sixteen people so any combination with 17 or 18 0's will have no people getting across rather than -1 or -2. Thus one needs to add in 18*1 and 1*2 = 20 before dividing by 218 giving the 7.000076 answer.

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Let me modify this puzzle a little bit. If all of the players cannot have footprint marks on the tempered glass, how many players can pass this bridge?
ThatDonGuy
ThatDonGuy
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October 23rd, 2021 at 2:30:20 PM permalink
Quote: aceside

Let me modify this puzzle a little bit. If all of the players cannot have footprint marks on the tempered glass, how many players can pass this bridge?
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Can you clarify this a little? Do you mean that all 16 cannot touch the same piece of tempered glass?
aceside
aceside
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October 23rd, 2021 at 2:37:42 PM permalink
Quote: ThatDonGuy

Quote: aceside

Let me modify this puzzle a little bit. If all of the players cannot have footprint marks on the tempered glass, how many players can pass this bridge?
link to original post


Can you clarify this a little? Do you mean that all 16 cannot touch the same piece of tempered glass?
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That is not what I meant. I mean the following players do not know which tempered glass panels have been stepped on, but they know the broken ordinary glass panels.
Wizard
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Wizard
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October 23rd, 2021 at 4:56:42 PM permalink
Quote: aceside

That is not what I meant. I mean the following players do not know which tempered glass panels have been stepped on, but they know the broken ordinary glass panels.
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I get it. This happened on the show, where an elderly player couldn't remember where the previous players stepped. They had to wear hospital slippers, which I think was to avoid leaving footprints.

Am I correct with at least the first five player's chances of survival?



Player Probability
1 0.00000381
2 0.00000763
3 0.00001526
4 0.00003051
5 0.00006094

Last edited by: Wizard on Oct 23, 2021
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
TwelveOr21
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October 23rd, 2021 at 5:06:05 PM permalink
Quote: Gialmere


You're playing a track and field board game with your family during the lockdown. If each of the runner pawns travels the indicated number of spaces every turn, at which numbered spot will all of the runners end up next to one another?
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Sorry, this is about a year old now, but.. what's the math surrounding the solution?
I've figured the answer is 19 for the spot they meet again, and the interval between those is always 30... and out of interest if any one of them is at any other spacing differential, they never meet within 15000 turns.

I figure the math behind working out the answer 'which square will they meet at' is related to the difference between the runner who moves the fastest, and the runner who moves the slowest.. but I'm not seeing a formula for that to verify.
Wizard
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Wizard
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October 23rd, 2021 at 5:56:41 PM permalink
Quote: Wizard

Am I correct with at least the first five player's chances of survival?

Player Probability
1 0.00000381
2 0.00000763
3 0.00001526
4 0.00003051
5 0.00006094


link to original post



I got impatient and finished the table. Am I right?



Player Probability
1 0.00000381
2 0.00000763
3 0.00001526
4 0.00003051
5 0.00006094
6 0.00011911
7 0.00023545
8 0.00046159
9 0.00089886
10 0.00175139
11 0.00345091
12 0.00693198
13 0.01418276
14 0.02923634
15 0.05993762
16 0.12152477
Total 0.23884892

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
aceside
aceside
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October 24th, 2021 at 12:27:37 AM permalink
Quote: Wizard

Quote: Wizard

Am I correct with at least the first five player's chances of survival?

Player Probability
1 0.00000381
2 0.00000763
3 0.00001526
4 0.00003051
5 0.00006094


link to original post



I got impatient and finished the table. Am I right?



Player Probability
1 0.00000381
2 0.00000763
3 0.00001526
4 0.00003051
5 0.00006094
6 0.00011911
7 0.00023545
8 0.00046159
9 0.00089886
10 0.00175139
11 0.00345091
12 0.00693198
13 0.01418276
14 0.02923634
15 0.05993762
16 0.12152477
Total 0.23884892


link to original post


Let me try this

Let x1=1st player's chances of survival;
Let x2=2nd player's chances of survival;
Let x3=3rd player's chance of survival;
... ...
Let x16=16th player's chance of survival, then we have
x1=0.5^18=3.8147E-06;
x2=(x1)^2+(1-x1)*0.5^17=7.62938E-06;
x3=(x1)^3+(x1)*(1-x2)*0.5^17+(1-x1)*(x2)*0.5^17+(1-x1)*(1-x2)*0.5^16=1.52587E-05;

I don't know how to calculate the exact value of x4, because it has 8 terms, but we can have a very good approximation because only the last term of each xn is non-negligible, so I calculated them and found the sum of all 16 terms:
x1 3.8147E-06
x2 7.62937E-06
x3 1.52586E-05
x4 3.05168E-05
x5 6.10317E-05
x6 0.000122056
x7 0.000244082
x8 0.000488045
x9 0.000975613
x10 0.001949323
x11 0.003891046
x12 0.007751811
x13 0.015383441
x14 0.030293581
x15 0.058751759
x16 0.11059998
Sum 0.230568986
Last edited by: aceside on Oct 24, 2021
gordonm888
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gordonm888
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October 24th, 2021 at 8:18:58 AM permalink
Quote: aceside

Quote: Wizard

Quote: Wizard

Am I correct with at least the first five player's chances of survival?

Player Probability
1 0.00000381
2 0.00000763
3 0.00001526
4 0.00003051
5 0.00006094


link to original post



I got impatient and finished the table. Am I right?



Player Probability
1 0.00000381
2 0.00000763
3 0.00001526
4 0.00003051
5 0.00006094
6 0.00011911
7 0.00023545
8 0.00046159
9 0.00089886
10 0.00175139
11 0.00345091
12 0.00693198
13 0.01418276
14 0.02923634
15 0.05993762
16 0.12152477
Total 0.23884892


link to original post


Let me try this

Let x1=1st player's chances of survival;
Let x2=2nd player's chances of survival;
Let x3=3rd player's chance of survival;
... ...
Let x16=16th player's chance of survival, then we have
x1=0.5^18=3.8147E-06;
x2=(x1)^2+(1-x1)*0.5^17=7.62938E-06;
x3=(x1)^3+(x1)*(1-x2)*0.5^17+(1-x1)*(x2)*0.5^17+(1-x1)*(1-x2)*0.5^16=1.52587E-05;

I don't know how to calculate the exact value of x4, because it has 8 terms, but we can have a very good approximation because only the last term of each xn is non-negligible, so I calculated them and found the sum of all 16 terms:
x1 3.8147E-06
x2 7.62937E-06
x3 1.52586E-05
x4 3.05168E-05
x5 6.10317E-05
x6 0.000122056
x7 0.000244082
x8 0.000488045
x9 0.000975613
x10 0.001949323
x11 0.003891046
x12 0.007751811
x13 0.015383441
x14 0.030293581
x15 0.058751759
x16 0.11059998
Sum 0.230568986

link to original post



Several of your terms are larger than Wizard's. And that's inconsistent because you are ignoring positive terms. So there seems to be a non-negligible discrepancy between the two answers.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
ThatDonGuy
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October 24th, 2021 at 9:46:22 AM permalink
Quote: aceside

Quote: ThatDonGuy

Quote: aceside

Let me modify this puzzle a little bit. If all of the players cannot have footprint marks on the tempered glass, how many players can pass this bridge?
link to original post


Can you clarify this a little? Do you mean that all 16 cannot touch the same piece of tempered glass?
link to original post


That is not what I meant. I mean the following players do not know which tempered glass panels have been stepped on, but they know the broken ordinary glass panels.
link to original post


This assumes that the 16 people do not discuss strategy in advance. If they can, then the strategy is, "When you see both panels at the same step, always use the one on the left"; the problem is now the same as the first one, and the expected number is slightly more than 7.


Let P(A,B) be the probability of getting to the point with A persons and B steps remaining.
P(18, 16) = 1
At any particular step, if there are N people remaining, either all of them make it through (with probability 1 / 2^N), or somebody does not (with probability 1 - 1 / 2^N), in which case N-1 people made it past that step, since everybody that went before that person made it, and once that person missed, the remaining people know which of the two panels at that step is good, since the other one is now shattered.

Probabilities of N people getting through:

1: 2,697,517,124,160,727,979,794,298,334,928,874,789,683,725,800,350,340,625 /
12,259,964,326,927,110,866,866,776,217,202,473,468,949,912,977,468,817,408
2: 1,964,212,080,163,879,415,864,883,856,458,098,671,857,036,517,779,546,916,875 /
200,867,255,532,373,784,442,745,261,542,645,325,315,275,374,222,849,104,412,672
3: 148,293,486,218,086,804,009,743,742,950,935,624,613,059,121,063,512,982,208,125 /
1,645,504,557,321,206,042,154,969,182,557,350,504,982,735,865,633,579,863,348,609,024
4: 1,285,367,138,214,125,938,247,419,046,043,506,562,122,293,397,493,370,198,165,875 /
6,739,986,666,787,659,948,666,753,771,754,907,668,409,286,105,635,143,120,275,902,562,304
5: 1,336,951,595,373,849,559,848,407,669,176,570,833,601,928,235,391,249,926,718,125 /
13,803,492,693,581,127,574,869,511,724,554,050,904,902,217,944,340,773,110,325,048,447,598,592
6: 170,354,180,873,343,918,394,627,482,029,532,555,019,181,269,707,182,608,153,875 /
14,134,776,518,227,074,636,666,380,005,943,348,126,619,871,175,004,951,664,972,849,610,340,958,208
7: 2,685,367,999,035,869,585,923,908,872,770,880,068,179,152,178,227,773,366,125 /
7,237,005,577,332,262,213,973,186,563,042,994,240,829,374,041,602,535,252,466,099,000,494,570,602,496
8: 5,260,280,403,685,160,364,200,323,669,276,904,586,048,518,440,857,885,075 /
1,852,673,427,797,059,126,777,135,760,139,006,525,652,319,754,650,249,024,631,321,344,126,610,074,238,976
9: 1,282,361,092,544,113,425,375,574,714,049,208,253,252,454,025,526,125 /
237,142,198,758,023,568,227,473,377,297,792,835,283,496,928,595,231,875,152,809,132,048,206,089,502,588,928
10: 38,876,263,223,777,991,428,570,948,450,414,447,717,888,243,475 /
15,177,100,720,513,508,366,558,296,147,058,741,458,143,803,430,094,840,009,779,784,451,085,189,728,165,691,392
11: 146,073,116,406,017,224,213,999,324,527,976,045,188,525 /
485,667,223,056,432,267,729,865,476,705,879,726,660,601,709,763,034,880,312,953,102,434,726,071,301,302,124,544
12: 67,497,028,115,768,980,735,398,222,216,368,115 /
7,770,675,568,902,916,283,677,847,627,294,075,626,569,627,356,208,558,085,007,249,638,955,617,140,820,833,992,704
13: 3,772,265,948,271,415,978,513,796,925 /
62,165,404,551,223,330,269,422,781,018,352,605,012,557,018,849,668,464,680,057,997,111,644,937,126,566,671,941,632
14: 24,594,251,415,116,327,595 /
248,661,618,204,893,321,077,691,124,073,410,420,050,228,075,398,673,858,720,231,988,446,579,748,506,266,687,766,528
15: 17,179,541,505 /
497,323,236,409,786,642,155,382,248,146,820,840,100,456,150,797,347,717,440,463,976,893,159,497,012,533,375,533,056
16: 1 /
497,323,236,409,786,642,155,382,248,146,820,840,100,456,150,797,347,717,440,463,976,893,159,497,012,533,375,533,056
The expected number is
119,285,438,051,491,600,965,426,158,899,637,755,285,530,818,964,278,692,622,932,395,089,512,340,802,876,506,402,971
/
497,323,236,409,786,642,155,382,248,146,820,840,100,456,150,797,347,717,440,463,976,893,159,497,012,533,375,533,056
= 0.23985946


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