Monty Hall problem: Difference between revisions
Content added Content deleted
(added Ursala) |
|||
| Line 1,206: | Line 1,206: | ||
Estimate: 4960/10000 wins for 'picking anew' strategy |
Estimate: 4960/10000 wins for 'picking anew' strategy |
||
Of course, this challenge could also be tackled by putting up a GUI and letting the user be the source of the randomness. But that's moving away from the letter of the challenge and takes a lot of effort anyway... |
Of course, this challenge could also be tackled by putting up a GUI and letting the user be the source of the randomness. But that's moving away from the letter of the challenge and takes a lot of effort anyway... |
||
=={{header|Ursala}}== |
|||
This is the same algorithm as the Perl solution. Generate two lists |
|||
of 10000 uniformly distributed samples from {1,2,3}, count each |
|||
match as a win for the staying strategy, and count each non-match as a win |
|||
for the switching strategy. |
|||
<lang Ursala> |
|||
#import std |
|||
#import nat |
|||
#import flo |
|||
rounds = 10000 |
|||
car_locations = arc{1,2,3}* iota rounds |
|||
initial_choices = arc{1,2,3}* iota rounds |
|||
staying_wins = length (filter ==) zip(car_locations,initial_choices) |
|||
switching_wins = length (filter ~=) zip(car_locations,initial_choices) |
|||
format = printf/'%0.2f'+ (times\100.+ div+ float~~)\rounds |
|||
#show+ |
|||
main = ~&plrTS/<'stay: ','switch: '> format* <staying_wins,switching_wins> |
|||
</lang> |
|||
Output will vary slightly for each run due to randomness. |
|||
<pre> |
|||
stay: 33.95 |
|||
switch: 66.05 |
|||
</pre> |
|||
=={{header|Vedit macro language}}== |
=={{header|Vedit macro language}}== |
||