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Ask the Wizard #337

You are given:

  • There is an airplane eight miles directly above a surface to air missile, which is fired at that moment.
  • At all times, the airplane travels in a straight direction.
  • The airplane travels at 600 miles per hour.
  • The missile travels at 2000 miles per hour.
  • The missile always travels at an angle that directly faces the airplane.

Questions:

  • How far will the plane travel before struck by the missile?
  • How long will it take for the missile to strike the plane?
  • How long is the flight path of the missile?

"Anonymous" .

  • How far will the plane travel before struck by the missile? = 240/91 miles
  • How long will it take for the missile to strike the plane? = 2/455 hours
  • How long will the missile travel? = 800/91 miles

Here is my solution (PDF).

This question is asked and discussed in my forum at Wizard of Vegas.

I once saw 49 consecutive baccarat hands with 48 Player wins, not counting ties. What is the probability of that per shoe?

Mdawg from California

The average shoe has 80.884 total hands. The probability of a Tie is 0.095156, so if we take those out we can expect 73.18740 hands per shoe, not counting ties.

The probability of any 49 consecutive hands, not counting ties, having 48 Player wins is 1 in 21,922,409,835,345. However, there are 25.1874 possible starting points for these 49 hands, to make estimate. Thus, the probability of seeing the aforementioned event in a shoe is 1 in 870,371,922,467. This is not a hard and fast answer, but what I feel is a very good estimate.

Assume:

  • 90% of the public wears masks.
  • The probability of getting the coronavirus is 1% for wearers and 3% for non-mask wearers.

Somebody is chosen at random with the coronavirus. What is the probability he is a mask wearer?

"Anonymous" .

This is a classic Bayesian conditional probability question.

The answer is probability(somebody is a mask-wearer and has coronavirus)/probability(somebody has coronavirus) =

(0.9*0.01) / (0.9*0.01 + 0.1*0.03) = 75%.

A fair die is rolled ten times. What is the probability that the sequence of rolls is non-decreasing? That is, each roll is greater than or equal to the previous roll.

Gialmere

The answer is approximately 1 in 20135.

The answer is hard to explain, but can be solved by recursively finding the probability the last roll of any given roll is x, with every roll being equal or greater than the previous roll.

The following table shows that probability for rolls 1 to 10 and the last side rolled of 1 to 6.

Ten Non-Decreasing Rolls

Roll Number Side of 1 Side of 2 Side of 3 Side of 4 Side of 5 Side of 6 Total
1 0.166667 0.166667 0.166667 0.166667 0.166667 0.166667 1.000000
2 0.027778 0.055556 0.083333 0.111111 0.138889 0.166667 0.583333
3 0.004630 0.013889 0.027778 0.046296 0.069444 0.097222 0.259259
4 0.000772 0.003086 0.007716 0.015432 0.027006 0.043210 0.097222
5 0.000129 0.000643 0.001929 0.004501 0.009002 0.016204 0.032407
6 0.000021 0.000129 0.000450 0.001200 0.002701 0.005401 0.009902
7 0.000004 0.000025 0.000100 0.000300 0.000750 0.001650 0.002829
8 0.000001 0.000005 0.000021 0.000071 0.000196 0.000472 0.000766
9 0.000000 0.000001 0.000004 0.000016 0.000049 0.000128 0.000199
10 0.000000 0.000000 0.000001 0.000004 0.000012 0.000033 0.000050

The answer is in the lower right corner. To take it to more decimal places, the answer is 0.0000496641295788244, which equates to about 1 in 20,135.

This question is asked and discussed in my forum at Wizard of Vegas.