<p>If you could please help out with this math problem.</p>
<p>-You roll 2 dices
if you toss 2,3,12-you lose
if you toss 7 or 11 your first time-you win
if you toss 4,5,6,8,9,10 then you keep tossing
if you toss a 7 your second try-you lose</p>
<p>What is the probability of winning?-</p>
<p>I would also very apreciate if you could explain in this thread or show work so I could understand the problem in details.</p>
<p>isn’t this craps?</p>
<p>or something like that … it’s like a 49.7% chance of winning</p>
<p>isn’t this a homework problem?</p>
<p>I swear there’s a drinking game like this, I’m just too sober right now to think of it.</p>
<p>Well, your chances of winning on the first roll are (3/11) which is about 27%. Your chance of rolling again are (6/11), or about 55%. Considering the fact that you have the option of “rolling again” every time, the percent cannot be exact, because you can roll infinitley many times. But, you can approximate the %. Multiply (6/11)(3/11), which will give you about .15, or 15%. Continously square this number, because its a infinite sum. After squaring it about 3 times, you realize that the %s become too small, and therefore are unnecessary.
tl,dr: .27 + Σ(.55)(.27) to infinity, and then multiply by 100 for %</p>
<p>What if you roll an 11 the second time?</p>
<p>This is either a trick question or stated incorrectly. As stated, the player can only win on the very first try. That happens with a probability of 8/36 (I believe). Otherwise you keep rolling until you lose or produce a run-time error.</p>