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Lots of questions on different subjects where you have to try to answer the latest question and if you are right then its your turn to ask the next question.

Moderators: playgirl, JUS, rodbox

#111034 by wildrover
Sun Nov 28, 2010 5:22 pm
Two gamblers are wagering on coin tosses - one selects heads and one tails. They are playing for £1,000 and first person to get to 10 wins the pot. One of the gamblers is called away with the score Heads 8 Tails 7 and they don't finish the game but they agree to split the pot based on their probability of winning when the game stopped. How should the pot be split?

#111308 by wildrover
Wed Dec 01, 2010 3:40 pm
As a clue, consider what is the maximum number of coin tosses that can occur before one side or the other must win.

#111314 by Stevie C
Wed Dec 01, 2010 4:21 pm
Is it £600 to the one on8 & £400 to the other?.

#111344 by wildrover
Wed Dec 01, 2010 10:22 pm
:( - not quite that simple.

#111346 by JUS
Wed Dec 01, 2010 11:05 pm
50:50 split as both have maximum of 4 throws to win :?

#111347 by wildrover
Thu Dec 02, 2010 12:49 am
no - the guy who is one ahead has a better chance of winning...

#111349 by pulse876
Thu Dec 02, 2010 1:22 am
Suppose H is the gambler who has selected Heads and T the gambler who has selected tails.

There must be at least two more coin tosses for anyone to reach 10 but at most four (ie when the score reaches 9-9 after three tosses).

H will win in the following circumstances.

(a) If the first two coin tosses are H, H (which is a 25% chance), or
(b) If the first two coin tosses are H, T in some order (50% chance) and then H wins the third toss (50% chance), ie 0.5*0.5 = 25% chance overall, or
(c) If the first two coin tosses are H, T in some order and then T wins the third toss (25% chance overall as in (b)) and then H wins the fourth toss (50% chance), ie 0.25*0.5=12.5% chance overall, or
(d) If the first two coin tosses are T, T (25% chance) and then H wins the next two tosses (25% chance), ie 0.25*0.25 = 6.25% chance overall.

So H would win (25% + 25% + 12.5% + 6.25%) = 68.75% of the time and hence the pot should be split accordingly.

#111385 by wildrover
Thu Dec 02, 2010 2:01 pm
..is the right answer (11/16 would also have been correct). After 4 rolls someone must win as if Heads hasn't got the 2 Heads he needs then Tails must have got the 3 Tail he needs. There are 16 possible combinations of 4 rolls and they are covered by pulse876's answer as follows:

a) HHHH HHHT HHTH HHTT - Heads wins every time
b) HTHT HTHH THHT THHH - Heads wins every time
c) HTTH THTH HTTT THTT - Heads wins 2 out of 4
d) TTHH TTHT TTTH TTTT - Heads wins 1 out of 4

so overall the probability is that Heads would win 11 times in every 16 and should get £687.50.

#119877 by Buzzie
Mon Apr 04, 2011 2:50 pm
On a standard 400m running track with semi-circular ends and lanes 1.25m wide, in a 400m race if all the competitors started on the start/finish line instead of being staggered, how much further would the competitor in lane 8 run than the competitor in lane 1?

#119880 by wildrover
Mon Apr 04, 2011 4:17 pm
I think it's 2 x pi x 7 x 1.25 = 54m 98cm?

#119881 by Buzzie
Mon Apr 04, 2011 4:20 pm
You would think correctly :D

#119882 by wildrover
Mon Apr 04, 2011 6:37 pm
Much further than I would have thought - but then again in a 400m race Lane 8 does start more than half way round the bend.

A man is given 4 saplings and told to plant them so that each sapling is the same distance from every other sapling. How did he do it?

#119884 by Bigal
Mon Apr 04, 2011 7:36 pm
Does the ground have to be flat ?

#119888 by JUS
Tue Apr 05, 2011 12:03 am
could you do things with a tetrahedron shaped layout with one planted above the rest?

#119890 by wildrover
Tue Apr 05, 2011 8:46 am
Yes - variant on an old favourite. Three are planted in an equilateral triangle around a central mound where the height of the mound is such that the distance from the central sapling to any tree is the same as one side of the triangle.
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