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(This article was first published on <strong><a href="http://www.decisionsciencenews.com/2017/05/24/counterintuitive-probability-problem-day/"> R – Decision Science News</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p>TWO SHOT RUSSIAN ROULETTE WITH 2 BULLETS OR AN EQUAL CHANCE OF 1 or 3 BULLETS</p>
<p><a rel="nofollow" href="https://i0.wp.com/www.decisionsciencenews.com/wp-content/uploads/2017/05/plive2.png" target="_blank"><img sizes="(max-width: 381px) 100vw, 381px" src="https://i0.wp.com/www.decisionsciencenews.com/wp-content/uploads/2017/05/plive2.png?resize=381%2C421" alt="" class="wp-image-6177" data-recalc-dims="1"></a><br>
With <strong><em>p</em></strong> as the probability of dying on one shot, this figure shows how to get the probability of living through the game.</p>
<p>Peter Ayton is giving a talk today at the <a rel="nofollow" href="http://ljdm.info/" target="_blank">London Judgement and Decision Making Seminar</a></p>
<blockquote><p><strong>Imagine being obliged to play Russian roulette – twice (if you are lucky enough to survive the first game). Each time you must spin the chambers of a six-chambered revolver before pulling the trigger. However you do have one choice: You can choose to either (a) use a revolver which contains only 2 bullets or (b) blindly pick one of two other revolvers: one revolver contains 3 bullets; the other just 1 bullet. Whichever particular gun you pick you must use every time you play. Surprisingly, option (b) offers a better chance of survival.</strong> We discuss a general theorem implying, with some specified caveats, that a system’s probability of surviving repeated ‘demands’ improves as uncertainty concerning the probability of surviving one demand increases. Nonetheless our behavioural experiments confirm the counterintuitive nature of the Russian roulette and other kindred problems: most subjects prefer option (a). We discuss how uncertain probabilities reduce risks for repeated exposure, why people intuitively eschew them and some policy implications for safety regulation.</p>
</blockquote>
<p>We can see how many people would think the choice between (a) and (b) doesn’t matter. Naively one might think that 2 bullets leads to the same probability as choosing blindly between 1 and 3 bullets. But this is not true. See the graphic above and plug in different ps.</p>
<p>Or be lazy and let us do it for you. First, approve that this R function is correct:</p>
<p><code><br>
prob_live_game = function(bullets) {<br>
prob_die = bullets / 6<br>
prob_live = 1 - prob_die<br>
prob_die * 0 + prob_live * (prob_die * 0 + prob_live * 1)<br>
}<br>
</code></p>
<p>Now see below that the probability of living with 2 bullets is .44 while the probability of living with an equal chance of 1 or 3 bullets is .47</p>
<p><code><br>
> #probability living with 2 bullets<br>
> prob_live_game(2)<br>
[1] 0.4444444<br>
><br>
> #probability living with 1 bullet<br>
> prob_live_game(1)<br>
[1] 0.6944444<br>
><br>
> #probability living with 3 bullets<br>
> prob_live_game(3)<br>
[1] 0.25<br>
><br>
> #probability living with an equal chance of 1 or 3 bullets<br>
> .5 * (prob_live_game(1) + prob_live_game(3))<br>
[1] 0.4722222</code></p>
<p>The post <a rel="nofollow" href="http://www.decisionsciencenews.com/2017/05/24/counterintuitive-probability-problem-day/" target="_blank">Counterintuitive probability problem of the day</a> appeared first on <a rel="nofollow" href="http://www.decisionsciencenews.com/" target="_blank">Decision Science News</a>.</p>
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