# If we are drawing red and black cards out of an infinite deck, and we draw red with probability 4/5, what is E(num draws to draw 3 consecutive blacks)

This is the full question: John is drawing red and black cards out of an infinite deck. The probability of drawing a red card is 4/5. Calculate the expected number of draws to draw three black cards consecutively.

The solution says

"The correct answer is 125. The question revolves around geometric distribution – how many Bernoulli trials are required to get a “success” outcome. The “success” in this case is three consecutive draws of black cards. The expected value of a geometric distribution is easily calculated as E(x) = 1/p, where p is the probability of “success”.

In the given question: p = The probability of drawing 3 consecutive black cards = (1/5)3 = 1/125 Plugging into the expected value formula above we get 125. Therefore, the expected number of draws to draw 3 black cards consecutively is 125."

However, I feel like this is an oversimplification somehow (I would agree this is the case if we draw 3 cards at a time, but since there's a difference between getting to 3 Blacks if the first two draws are red vs. if they are black, it seems wrong to just simply state p = 1/125). I'm not sure where I'm going wrong here!

• The solution is incorrect, barring a very eccentric reading of the question. It's correct if you are drawing in groups of three and are asking how many groups of three you must choose to hit one that is all black. But that's really not how the question reads.
– lulu
Commented Jul 23 at 22:10
• If you are drawing one card at a time, Conway's algorithm would suggest an expectation of $5^3+5^2+5^1=155$ for three consecutive black cards because of the similar starting and ending of the string "BBB". The same algorithm would suggest an expectation of $5$ for one black card, $30$ for two consecutive black cards, and $780$ for four consecutive black cards. Commented Jul 24 at 23:42

The expectation is $$125$$ trials, with possible outcomes R, BR, BBR, and BBB.

R: $$80\%$$ of trials will fail on the first draw. This is $$100$$ trials, one draw each, for a total of $$100$$ draws.

BR: $$16\%$$ of trials will fail on the second draw. This is $$20$$ trials, two draws each, for a total of $$40$$ draws.

BBR: $$3.2\%$$ of trials will fail on the third draw. This is $$4$$ trials, three draws each, for a total of $$12$$ draws.

BBB: One trial in $$125$$ will succeed on third draw. Three draws.

The expectation is therefore $$100+40+12+3=155$$ draws.

The given answer of $$125$$ draws is correct only if three cards are drawn at a time, and each such draw has a success $$Pr$$ of $$(\frac15)^3$$

If the cards are drawn one at a time, I get an answer of $$155$$
If $$s$$ = start, $$a$$ = last card black, $$b$$ = last two cards black, $$c$$ = last three cards black, we get equations

$$s = 1 + (1/5)a + (4/5)s \tag1$$ $$a = 1 + (1/5)b + (4/5)s \tag2$$ $$b = 1+ (4/5)s \tag3$$

and solving the set of linear equations,
number of draws needed from start, $$s = 155$$

• Just to point out something that confused me for a moment: the number of cards drawn in the "draw 3 at a time" case should be higher than in the "draw 1 at a time" case, and it is, because that 125 is 125 draws of 3 cards each, meaning 375 cards total.
– Ian
Commented Jul 23 at 22:13
• @Ian: If we draw three at a time, the probability of "success" is $p= (1/5)^3 = 1/125$ and by the geometric distribution, expected trials to succed $=1/p = 125$ Oh, I see your confusion, we are to find the expected number of draws, not the exprcted # of cards drawn. Commented Jul 23 at 22:19
• My confusion was more simple, I was confused by the fact that $125<155$ but you would expect the inequality to go the other way. But that was because comparing 125 to 155 is apples to oranges. It is 375 vs 155 when you compare apples to apples.
– Ian
Commented Jul 23 at 22:31